Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts)

Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 2 Banking (Recurring Deposit Accounts)

Banking (Recurring Deposit Accounts) Exercise 2A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Manish opens a Recurring Deposit Account with the Bank of Rajasthan and deposits ₹ 600 per month for 20 months. Calculate the maturity value of this account, if the bank pays interest at the rate of 10% per annum.
Solution:
Installment per month(P) = ₹ 600
Number of months(n) = 20
Rate of interest(r) = 10% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 1
The amount that Manish will get at the time of maturity
= ₹ (600×20) + ₹ 1,050
= ₹ 12,000 + ₹ 1,050
= ₹ 13,050

Question 2.
Mrs. Mathew opened a Recurring Deposit Account in a certain bank and deposited ₹ 640 per month for 4 ½ years. Find the maturity value of this account, if the bank pays interest at the rate of 12% per year.
Solution:
Installment per month(P) = ₹ 640
Number of months(n) = 54
Rate of interest(r)= 12% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 2
The amount that Manish will get at the time of maturity
= ₹ (640×54)+ ₹ 9,504
= ₹ 34,560 + ₹ 9,504
= ₹ 44,064

Question 3.
Each of A and B both opened recurring deposit accounts in a bank. If A deposited ₹ 1,200 per month for 3 years and B deposited ₹ 1,500 per month for 2 ½ years; find, on maturity, who will get more amount and by how much? The rate of interest paid by the bank is 10% per annum.
Solution:
For A
Installment per month(P) = ₹ 1,200
Number of months(n) = 36
Rate of interest(r) = 10% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 3
The amount that A will get at the time of maturity
= ₹ (1,200×36) + ₹ 6,660
= ₹ 43,200 + ₹ 6,660
= ₹ 49,860
For B
Instalment per month(P) = ₹ 1,500
Number of months(n) = 30
Rate of interest(r) = 10% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 4
The amount that B will get at the time of maturity
= ₹ (1,500×30) + ₹ 5,812.50
= ₹ 45,000 + ₹ 5,812.50
= ₹ 50,812.50
Difference between both amounts = ₹ 50,812.50 – ₹ 49,860
= ₹ 952.50
Then B will get more money than A by ₹ 952.50.

Question 4.
Ashish deposits a certain sum of money every month is a Recurring Deposit Account for a period of 12 months. If the bank pays interest at the rate of 11% p.a. and Ashish gets ₹ 12,715 as the maturity value of this account, what sum of money did money did he pay every month?
Solution:
Let Installment per month(P) = ₹ y
Number of months(n) = 12
Rate of interest(r) = 11% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 5
Maturity value = ₹ (y × 12) + ₹ 0.715y = ₹ 12.715y
Given maturity value = ₹ 12,715
Then ₹ 12.715y = ₹ 12,715
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 6

Question 5.
A man has a Recurring Deposit Account in a bank for 3 ½ years. If the rate of interest is 12% per annum and the man gets ₹ 10,206 on maturity, find the value of monthly instalments.
Solution:
Let Installment per month(P) = ₹ y
Number of months(n) = 42
Rate of interest(r) = 12% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 7
Maturity value= ₹ (y × 42) + ₹ 9.03y= ₹ 51.03y
Given maturity value = ₹ 10,206
Then ₹ 51.03y = ₹ 10206
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 8

Question 6.
(i) Puneet has a Recurring Deposit Account in the Bank of Baroda and deposits ₹ 140 per month for 4 years. If he gets ₹ 8,092 on maturity, find the rate of interest given by the bank.
(ii) David opened a Recurring Deposit Account in a bank and deposited ₹ 300 per month for two years. If he received ₹ 7,725 at the time of maturity, find the rate of interest per annum.
Solution:
(a)
Installment per month(P) = ₹ 140
Number of months(n) = 48
Let rate of interest(r) = r% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 9
Maturity value= ₹ (140 × 48) + ₹ (137.20)r
Given maturity value = ₹ 8,092
Then ₹ (140 × 48) + ₹ (137.20)r = ₹ 8,092
⇒ 137.20r = ₹ 8,092 – ₹ 6,720
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 10
(b)
Instalment per month(P) = ₹ 300
Number of months(n) = 24
Let rate of interest(r)= r% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 11
Maturity value = ₹ (300 × 24) + ₹ (75)r
Given maturity value = ₹ 7,725
Then ₹ (300 × 24) + ₹ (75)r = ₹ 7,725
⇒ 75 r = ₹ 7,725 – ₹ 7,200
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 12

Question 7.
Amit deposited ₹ 150 per month in a bank for 8 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and interest is calculated at the end of every month?
Solution:
Installment per month(P) = ₹ 150
Number of months(n) = 8
Rate of interest(r) = 8% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 13
The amount that Manish will get at the time of maturity
= ₹ (150 × 8) + ₹ 36
= ₹ 1,200 + ₹ 36
= ₹ 1,236

Question 8.
Mrs. Geeta deposited ₹ 350 per month in a bank for 1 year and 3 months under the Recurring Deposit Scheme. If the maturity value of her deposits is ₹ 5,565; find the rate of interest per annum.
Solution:
Installment per month(P) = ₹ 350
Number of months(n) = 15
Let rate of interest(r)= r% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 14
Maturity value= ₹ (350 × 15) + ₹ (35)r
Given maturity value = ₹ 5,565
Then ₹ (350 × 15) + ₹ (35)r = ₹ 5,565
⇒ 35r = ₹ 5,565 – ₹ 5,250
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 15

Question 9.
A recurring deposit account of ₹ 1,200 per month has a maturity value of ₹ 12,440. If the rate of interest is 8% and the interest is calculated at the end of every month; find the time (in months) of this Recurring Deposit Account.
Solution:
Installment per month(P) = ₹ 1,200
Number of months(n) = n
Let rate of interest(r) = 8% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 16
Maturity value = ₹ (1,200 × n) + ₹ 4n(n+1) = ₹ (1200n+4n2+4n)
Given maturity value= ₹ 12,440
Then 1200n+4n2+4n = 12,440
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 17
Then number of months = 10

Question 10.
Mr. Gulati has a Recurring Deposit Account of ₹ 300 per month. If the rate of interest is 12% and the maturity value of this account is ₹ 8,100; find the time (in years) of this Recurring Deposit Account.
Solution:
Installment per month(P) = ₹ 300
Number of months(n) = n
Let rate of interest(r)= 12% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 18
Maturity value= ₹ (300 × n)+ ₹ 1.5n(n+1)
= ₹ (300n+1.5n2+1.5n)
Given maturity value= ₹ 8,100
Then 300n+1.5n2+1.5n = 8,100
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 19
Then time = 2 years.

Question 11.
Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹ 2,500 per month for two years. At the time of maturity he got ₹ 67,500. Find:
(i) the total interest earned by Mr. Gupta
(ii) the rate of interest per annum.
Solution:
(i)
Maturity value = ₹ 67,500
Money deposited = ₹ 2,500 × 24= ₹ 60,000
Then total interest earned = ₹ 67,500 – ₹ 60,000 = ₹ 7,500 Ans.
(ii)
Installment per month(P) = ₹ 2,500
Number of months(n) = 24
Let rate of interest(r)= r% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 20

Banking (Recurring Deposit Accounts) Exercise 2B- Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Pramod deposits ₹ 600 per month in a Recurring Deposit Account for 4 years. If the rate of interest is 8% per year; calculate the maturity value of his account.
Solution:
Installment per month(P) = ₹ 600
Number of months(n) = 48
Rate of interest(r)= 8% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 21
The amount that Manish will get at the time of maturity
= ₹ (600 × 48) + ₹ 4,704
= ₹ 28,800 + ₹ 4,704
= ₹ 33,504

Question 2.
Ritu has a Recurring Deposit Account in a bank and deposits ₹ 80 per month for 18 months. Find the rate of interest paid by the bank if the maturity value of account is ₹ 1,554.
Solution:
Installment per month(P) = ₹ 80
Number of months(n) = 18
Let rate of interest(r) = r% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 22
Maturity value = ₹ (80 × 18) + ₹ (11.4r)
Given maturity value = ₹ 1,554
Then ₹ (80 × 18 ) + ₹ (11.4r) = ₹ 1,554
⇒ 11.4r  = ₹ 1,554 – ₹ 1,440
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 23

Question 3.
The maturity value of a R.D. Account is ₹ 16,176. If the monthly installment is ₹ 400 and the rate of interest is 8%; find the time (period) of this R.D Account.
Solution:
Installment per month(P) = ₹ 400
Number of months(n) = n
Let rate of interest(r)= 8% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 24
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 25
⇒ 1200n +4n2+4n= ₹ 48,528
⇒ 4n2+1204n = ₹ 48,528
⇒ n2+301n – 12132= 0
⇒ (n+337)(n-36)=0
⇒ n = -337 or n=36
Then number of months = 36 months = 3 years

Question 4.
Mr. Bajaj needs ₹ 30,000 after 2 years. What least money (in multiple of 5) must he deposit every month in a recurring deposit account to get required money after 2 years, the rate of interest being 8% p.a.?
Solution:
Let installment per month = ₹ P
Number of months(n) = 24
Rate of interest = 8% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 26
Maturity value = ₹ (P × 24)+ ₹ 2P = ₹ 26P
Given maturity value = ₹ 30,000
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 27

Question 5.
Rishabh has recurring deposit account in a post office for 3 years at 8% p.a. simple interest. If he gets ₹ 9,990 as interest at the time of maturity, find:
(i) The monthly installment.
(ii) The amount of maturity.
Solution:
Let Installment per month = ₹ P
Number of months(n) = 36
Rate of interest(r)= 8% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 28
Given interest = ₹ 9,990
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 29
(ii) Maturity value = ₹ (2,250 × 36) + ₹ 9,990 = ₹ 90,990

Question 6.
Gopal has a cumulative deposit account and deposits ₹ 900 per month for a period of 4 years he gets ₹ 52,020 at the time of maturity, find the rate of interest.
Solution:
Installment per month(P) = ₹ 900
Number of months(n) = 48
Let rate of interest(r)= r% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 30
Maturity value= ₹ (900 × 48) + ₹ (882)r
Given maturity value = ₹ 52,020
Then ₹ (900 × 48) + ₹ (882)r = ₹ 52,020
⇒ 882r = ₹ 52,020 – ₹ 43,200
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 31

Question 7.
Deepa has a 4-year recurring deposit account in a bank and deposits ₹ 1,800 per month. If she gets ₹ 1,08,450 at the time of maturity, find the rate of interest.
Solution:
Installment per month(P) = ₹ 1,800
Number of months(n) = 48
Let rate of interest(r)= r% p.a.
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 32
Maturity value = ₹ (1,800 x 48) + ₹ (1,764)r
Given maturity value = ₹ 1,08,450
Then ₹ (1,800 x 48) + ₹ (1764)r = ₹ 1,08,450
⇒ 1764r = ₹ 1,08,450 – ₹ 86,400
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 33

Question 8.
Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is of 8% per annum and Mr. Britto gets Rs. 8,088 from the bank after 3 years, find the value of his monthly instalment.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 34

Question 9.
Shahrukh opened a Recurring Deposit Acoount in a bank and deposited Rs. 800 per month for 1 \(\frac { 1 }{ 2 }\) years. If he received Rs. 15,084 at the time of maturity, find the rate of interest per annum.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 35

Question 10.
Katrina opened a recurring deposit account with a Nationalised Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly installment is ₹ 1,000, find the :
(i) interest earned in 2 years
(ii) maturity value
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 36

Question 11.
Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs. 1200 as interest at the time of maturity, find
(i) the monthly installment
(ii) the amount of maturity
Solution:
Interest, I = Rs. 1,200
Time, n = 2 years = 2 × 12 = 24 months
Rate, r = 6%
(i) To find: Monthly instalment, P
Now,
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) q11
So, the monthly instalment is Rs. 800.

(ii) Total sum deposited = P × n = Rs. 800 × 24 = Rs. 19,200
∴ Amount of maturity = Total sum deposited + Interest on it
= Rs. (19,200 + 1,200)
= Rs. 20,400

Question 11.
Peter has a recurring deposit account in Punjab National Bank at Sadar Bazar, Delhi for 4 years at 10% p.a. He will get ₹ 6,370 as interest on maturity. Find :
(i) monthlyinstallment,
(ii) the maturity value of the account.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Banking (Recurring Deposit Accounts) - 37

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations

Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 5 Quadratic Equations

Quadratic Equations Exercise 5A – Selina Concise Mathematics Class 10 ICSE Solutions

Find which of the following equations are quadratic:

Solution 1(i)
(3x – 1)2 = 5(x + 8)
⇒ (9x2 – 6x + 1) = 5x + 40
⇒ 9x2 – 11x – 39 =0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.

Solution 1(ii)
5x2 – 8x = -3(7 – 2x)
⇒ 5x2 – 8x = 6x – 21
⇒ 5x2 – 14x + 21 =0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.

Solution 1(iii)
(x – 4)(3x + 1) = (3x – 1)(x +2)
⇒ 3x2 + x – 12x – 4 = 3x2 + 6x – x – 2
⇒ 16x + 2 =0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.

Solution 1(iv)
x2 + 5x – 5 = (x – 3)2
⇒ x2 + 5x – 5 = x2 – 6x + 9
⇒ 11x – 14 =0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.

Solution 1(v)
7x3 – 2x2 + 10 = (2x – 5)2
⇒ 7x3 – 2x2 + 10 = 4x2 – 20x + 25
⇒ 7x3 – 6x2 + 20x – 15 = 0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.

Solution 1(vi)
(x – 1)2 + (x + 2)2 + 3(x +1) = 0
⇒ x2 – 2x + 1 + x2 + 4x + 4 + 3x + 3 = 0
⇒ 2x2 + 5x + 8 = 0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.

Question 2(i)
Is x = 5 a solution of the quadratic equation x2 – 2x – 15 = 0?
Solution:
x2 – 2x – 15 = 0
For x = 5 to be solution of the given quadratic equation it should satisfy the equation.
So, substituting x = 5 in the given equation, we get
L.H.S = (5)2 – 2(5) – 15
= 25 – 10 – 15
= 0
= R.H.S
Hence, x = 5 is a solution of the quadratic equation x2 – 2x – 15 = 0.

Question 2(ii).
Is x = -3 a solution of the quadratic equation 2x2 – 7x + 9 = 0?
Solution:
2x2 – 7x + 9 = 0
For x = -3 to be solution of the given quadratic equation it should satisfy the equation
So, substituting x = 5 in the given equation, we get
L.H.S =2(-3)2 – 7(-3) + 9
= 18 + 21 + 9
= 48
≠ R.H.S
Hence, x = -3 is not a solution of the quadratic equation 2x2 – 7x + 9 = 0.

Question 3.
If \(\sqrt{\frac{2}{3}}\) is a solution of equation 3x2 + mx + 2 = 0, find the value of m.
Solution:
For x = \(\sqrt{\frac{2}{3}}\) to be solution of the given quadratic equation it should satisfy the equation
So, substituting x = \(\sqrt{\frac{2}{3}}\) in the given equation, we get
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations 3

Question 4.
\(\frac{2}{3}\) and 1 are the solutions of equation mx2 + nx + 6 = 0. Find the values of m and n.
Solution:
For x =  \(\frac{2}{3}\) and x = 1 to be solutions of the given quadratic equation it should satisfy the equation
So, substituting x =  \(\frac{2}{3}\) and x = 1 in the given equation, we get
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations 1
Solving equations (1) and (2) simultaneously,
4m  + 6n + 54 = 0 …..(1)
m + n  + 6 = 0 ….(2)
(1) – (2) × 6
⇒ -2m + 18 = 0
⇒ m = 9
Substitute in (2)
⇒ n = -15

Question 5.
If 3 and -3 are the solutions of equation ax2 + bx – 9 = 0. Find the values of a and b.
Solution:
For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the equation
So, substituting x = 3 and x = -3 in the given equation, we get
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations 2
Solving equations (1) and (2) simultaneously,
9a + 3b – 9 = 0 …(1)
9a – 3b – 9 = 0 …(2)
(1) + (2)
⇒ 18a – 18 = 0
⇒ a = 1
Substitute in (2)
⇒ b = 0

Quadratic Equations Exercise 5B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Without solving, comment upon the nature of roots of each of the following equations :
(i) 7x2 – 9x +2 =0
(ii) 6x2 – 13x +4 =0
(iii) 25x2 – 10x +1=0
(iv) x2 + 2√3x – 9=0
(v) x2 – ax – b2 =0
(vi) 2x2 +8x +9=0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 1Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 2

Question 2.
Find the value of p, if the following quadratic equation has equal roots : 4x2 – (p – 2)x + 1 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 3

Question 3.
Find the value of ‘p’, if the following quadratic equations have equal roots : x2 + (p – 3)x + p = 0
Solution:
x2 + (p – 3)x + p = 0
Here, a = 1, b = (p – 3), c = p
Since, the roots are equal,
⇒ b2– 4ac = 0
⇒ (p – 3)2– 4(1)(p) = 0
⇒p2 + 9 – 6p – 4p = 0
⇒ p2– 10p + 9 = 0
⇒p2-9p – p + 9 = 0
⇒p(p – 9) – 1(p – 9) = 0
⇒ (p -9)(p – 1) = 0
⇒ p – 9 = 0 or p – 1 = 0
⇒ p = 9 or p = 1

Question 4.
The equation 3x2 – 12x + (n – 5)=0 has equal roots. Find the value of n.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 4

Question 5.
Find the value of m, if the following equation has equal roots : (m – 2)x2 – (5+m)x +16 =0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 5

Question 6.
Find the value of p for which the equation 3x2– 6x + k = 0 has distinct and real roots.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 6

Quadratic Equations Exercise 5C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Solve : x² – 10x – 24 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 7

Question 2.
Solve : x² – 16 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 8

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 9
Solution:

Question 4.
Solve : x(x – 5) = 24
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 11

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 11
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 13

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 14
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 15

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 16
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 17

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 18
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 19

Question 9.
Solve : (2x – 3)² = 49
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 20

Question 10.
Solve : 2(x² – 6) = 3(x – 4)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 21

Question 11.
Solve : (x + 1)(2x + 8) = (x + 7)(x + 3)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 22

Question 12.
Solve : x² – (a + b)x + ab = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 23

Question 13.
(x + 3)² – 4(x + 3) – 5 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 24

Question 14.
4(2x – 3)² – (2x – 3) – 14 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 25

Question 15.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 26
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 27

Question 16.
2x2 – 9x + 10 = 0, When
(i) x∈ N
(ii) x∈ Q
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 28

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 29
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 30

Question 18.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 31
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 32

Question 19.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 33
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 34

Question 20.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 35
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 36

Question 21.
Find the quadratic equation, whose solution set is :
(i) {3, 5} (ii) {-2, 3}
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 37

Question 22.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 38
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 39

Question 23.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 40
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 41

Question 24.
Find the value of x, if a + 1=0 and x2 + ax – 6 =0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 42

Question 25.
Find the value of x, if a + 7=0; b + 10=0 and 12x2 = ax – b.
Solution:
If a + 7 =0, then a = -7
and b + 10 =0, then b = – 10
Put these values of a and b in the given equation
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 43

Question 26.
Use the substitution y= 2x +3 to solve for x, if 4(2x+3)2 – (2x+3) – 14 =0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 44

Question 27.
Without solving the quadratic equation 6x2 – x – 2=0, find whether x = 2/3 is a solution of this equation or not.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 44

Question 28.
Determine whether x = -1 is a root of the equation x2 – 3x +2=0 or not.
Solution:
x2 – 3x +2=0
Put x = -1 in L.H.S.
L.H.S. = (-1)– 3(-1) +2
= 1 +3 +2=6 ≠ R.H.S
Then x = -1 is not the solution of the given equation.

Question 29.
If x = 2/3 is a solution of the quadratic equation 7x2+mx – 3=0; Find the value of m.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 46

Question 30.
If x = -3 and x = 2/3 are solutions of quadratic equation mx+ 7x + n = 0, find the values of m and n.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 47

Question 31.
If quadratic equation x2 – (m + 1) x + 6=0 has one root as x =3; find the value of m and the root of the equation.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 48

Question 32.
Given that 2 is a root of the equation 3x² – p(x + 1) = 0 and that the equation px² – qx + 9 = 0 has equal roots, find the values of p and q.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 49

Question 33.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 50
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 51

Question 34.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 52
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 52

Question 35.
If -1 and 3 are the roots of x2 + px + q = 0, find the values of p and q.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 54

Quadratic Equations Exercise 5D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 55
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 56
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 57
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 58
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 59
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 60
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 61
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 62

Question 2.
Solve each of the following equations for x and give, in each case, your answer correct to one decimal place :
(i) x2 – 8x+5=0
(ii) 5x2 +10x – 3 =0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 63

Question 3(i).
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
(i) 2x2 – 10x +5=0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 64

Question 3(ii).
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
4x + 6/x + 13 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 65

Question 3(iii).
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
x2 – 3x – 9 =0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 66

Question 3(iv).
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
x2 – 5x – 10 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 67

Question 4.
Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places :
(i) 3x2 – 12x – 1 =0
(ii) x2 – 16 x +6= 0
(iii) 2x2 + 11x + 4= 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 68

Question 5.
Solve:
(i) x4 – 2x2 – 3 =0
(ii) x4 – 10x2 +9 =0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 69

Question 6.
Solve :
(i) (x2 – x)2 + 5(x2 – x)+ 4=0
(ii) (x2 – 3x)2 – 16(x2 – 3x) – 36 =0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 70

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 70
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 72
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 73
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 74

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 75
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 76

Question 9.
Solve the following equation and give your answer correct to 3 significant figures:
5x² – 3x – 4 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 77

Question 10.
Solve for x using the quadratic formula. Write your answer correct to two significant figures.
(x – 1)2 – 3x + 4 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 78

Question 11.
Solve the quadratic equation x² – 3 (x+3) = 0; Give your answer correct to two significant figures.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations ex 5d q11

Quadratic Equations Exercise 5E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 79
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 80

Question 2.
Solve: (2x+3)2=81
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 81

Question 3.
Solve: a²x² – b² = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 82

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 83
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 84

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 85
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 86

Question 6.
Solve: 2x4 – 5x² + 3 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 87

Question 7.
Solve: x4 – 2x² – 3 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 88

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 89
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 90

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 91
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 92

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 93
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 94

Question 11.
Solve : (x² + 5x + 4)(x² + 5x + 6) = 120
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 95

Question 12.
Solve each of the following equations, giving answer upto two decimal places.
(i) x2 – 5x -10=0 (ii) 3x2 – x – 7 =0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 96.

Question 13.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 97
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 98

Question 14.
Solve :
(i) x2 – 11x – 12 =0; when x ∈ N
(ii) x2 – 4x – 12 =0; when x ∈ I
(iii) 2x2 – 9x + 10 =0; when x ∈ Q
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 99

Question 15.
Solve : (a + b)²x² – (a + b)x – 6 = 0; a + b ≠ 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 100

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 101
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 102

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 103
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 104

Question 18.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 105
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 106
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 107

Question 19.
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 108
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 109

Question 20.
Without solving the following quadratic equation, find the value of ‘m’ for which the given equation has real and equal roots.
x² + 2(m – 1)x + (m + 5) = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 110

Quadratic Equations Exercise 5F – Selina Concise Mathematics Class 10 ICSE Solutions

Solution 1(i)
Given: (x + 5)(x – 5)=24
⇒ x2 – 52 = 24   …. since (a – b)(a + b) = a2 – b2
⇒ x2 – 25 = 24
⇒ x2 = 49
⇒ x = ± 7

Solution 1(ii)
Given: 3x2 – 2\(\sqrt{6}\)x + 2 = 0
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 112
Solution 1(iii)
Given: 3\(\sqrt{2}\)x2 – 5x – \(\sqrt{26}\) = 0
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 113
Question 2.
One root of the quadratic equation 8x2 + mx + 15 is 3/4. Find the value of m. Also, find the other root of the equation.

Solution:
Given quadratic equation is  8x2 + mx + 15 = 0   …. (i)
One of the roots of (i) is \(\frac{3}{4}\), so it satisfies (i)
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 114
So, the equation (i) becomes 8x2 – 26x + 15 = 0
⇒ 8x2 – 20x – 6x + 15 = 0
⇒ 4x(2x – 5) -3(2x – 5) = 0
⇒ (4x – 3)(2x – 5) = 0
⇒ x = \(\frac{3}{4}\) or x = \(\frac{5}{2}\)
⇒ x = \(\frac{3}{4}, \frac{5}{2}\)
Hence, the other root is \(\frac{5}{2}\)

Question 3.
One root of the quadratic equation x2 – 3x – 2ax – 6a = 0 is -3, find its other root.

Solution:
Given quadratic equation is …. (i)
One of the roots of (i) is -3, so it satisfies (i)
⇒ x2 – 3x – 2ax – 6a = 0
⇒ x(x + 3) – 2a(x + 3) = 0
⇒ (x – 2a)(x + 3) = 0
⇒ x = -3, 2a
Hence, the other root is 2a.

Question 4.
If p – 15 = 0 and 2x2 + 15x + 15 = 0;find the values of x.

Solution:
Given i.e p – 15 = 0 i.e. p = 15
So, the given quadratic equation becomes
2x2 + 15x + 15 = 0
⇒ 2x + 10x + 5x + 15 = 0
⇒ 2x(x + 5) + 5(x + 5)
⇒ (2x + 5)(x + 5) = 0
⇒ x = -5, \(-\frac{5}{2}\)
Hence, the values of x are -5 and \(-\frac{5}{2}\)

Question 5.
Find the solution of the equation 2x2 -mx – 25n = 0; if m + 5 = 0 and n – 1 = 0.

Solution:
Given quadratic equation is 2x2 -mx – 25n = 0 ….. (i)
Also, given and m + 5 = 0 and n – 1 = 0
⇒ m = -5 and n = 1
So, the equation (i) becomes
2x2 + 5x + 25 = 0
⇒ 2x + 10x – 5x – 25 = 0
⇒ 2x(x + 5) -5(x + 5) = 0
⇒ (x + 5)(2x – 5) = 0
⇒ x = -5, \(\frac{5}{2}\)
Hence, the solution of given quadratic equation are x and \(\frac{5}{2}\)

Question 6.
If m and n are roots of the equation \(\frac{1}{x}-\frac{1}{x-2}=3\) where x ≠ 0 and x ≠ 2; find m × n.

Solution:
Given quadratic equation is \(\frac{1}{x}-\frac{1}{x-2}=3\)
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 115
Since, m and n are roots of the equation, we have
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 116

Question 7.
Solve, using formula :
x2 + x – (a + 2)(a + 1) = 0
Solution:
Given quadratic equation is x2 + x – (a + 2)(a + 1) = 0
Using quadratic formula,
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 117

Question 8.
Solve the quadratic equation 8x2 – 14x + 3 = 0
(i) When x ∈ I (integers)
(ii) When x ∈ Q (rational numbers)
Solution:
Given quadratic equation is 8x2 – 14x + 3 = 0
⇒ 8x2 – 12x – 2x + 3 = 0
⇒ 4x(2x – 3) – (2x – 3) = 0
⇒ (4x – 1)(2x – 3) = 0
⇒ x = \(\frac{3}{2}\) or x = \(\frac{1}{4}\)
(i) When x ϵ I, the equation 8x2 – 14x + 3 = 0 has no roots
(ii) When x ϵ Q the roots of 8x2 – 14x + 3 = 0 are
x = \(\frac{3}{2}\) x = \(\frac{1}{4}\)

Question 9.
Find the value of m for which the equation (m + 4 )2 + (m + 1)x + 1 = 0 has real and equal roots.
Solution:
Given quadratic equation is (m + 4 )2 + (m + 1)x + 1 = 0
The quadratic equation has real and equal roots if its discriminant is zero.
⇒ D = b2 – 4ac = 0
⇒ (m + 1)2 -4(m + 4)(1) = 0
⇒ m2 + 2m + 1 – 4m – 16 = 0
⇒ m2 – 2m – 15 = 0
⇒ m2 – 5m + 3m – 15 = 0
⇒ m(m – 5) +3(m =5) = 0
⇒ (m – 5)(m + 3) = 0
⇒ m = 5 or m = -3

Question 10.
Find the values of m for which equation 3x2 + mx + 2 = 0 has equal roots. Also, find the roots of the given equation.
Solution:
Given quadratic equation is 3x2 + mx + 2 = 0 …. (i)
The quadratic equation has equal roots if its discriminant is zero
⇒ D = b2 – 4ac = 0
⇒ m2 – 4(2)(3) = 0
⇒ m2 = 24
⇒ m = \(\pm 2 \sqrt{6}\)
When m = \(2 \sqrt{6}\), equation (i) becomes
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 118
When m = \(-2 \sqrt{6}\), equation (i) becomes
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 119
∴ x= \(-\frac{\sqrt{6}}{3}, \frac{\sqrt{6}}{3}\)

Question 11.
Find the value of k for which equation 4x2 + 8x – k = 0 has real roots.

Solution:
Given quadratic equation is 4x2 + 8x – k = 0 …. (i)
The quadratic equation has real roots if its discriminant is greater than or equal to zero
⇒ D = b2 – 4ac ≥ 0
⇒ 82 – 4(4)(-k) ≥ 0
⇒ 64 + 16k ≥ 0
⇒ 16k ≥ -64
⇒ k ≥ -4
Hence, the given quadratic equation has real roots for k ≥ -4

Question 12.
Find, using quadratic formula, the roots of the following quadratic equations, if they exist
(i) 3x2 – 5x + 2 = 0
(ii) x2 + 4x + 5 = 0
Solution:
(i) Given quadratic equation is 3x2 – 5x + 2 = 0
D = b2 – 4ac = (-5)2 – 4(3)(2) = 25 – 24 = 1
Since D > 0, the roots of the given quadratic equation are real and distinct.
Using quadratic formula, we have
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 120
⇒ x = 1 or x = \(\frac{2}{3}\)

(ii) Given quadratic equation is x2 + 4x + 5 = 0
D = b2 – 4ac = (4)2 – 4(1)(5) = 16 – 20 = – 4
Since D < 0, the roots of the given quadratic equation does not exist.

Solution 13:
(i) Given quadratic equation is \(\frac{1}{18-x}-\frac{1}{18+x}=\frac{1}{24}\)
Selina Concise Mathematics Class 10 ICSE Solutions Quadratic Equations - 121
⇒ 48x = 324 – x2
⇒ x2 + 48x – 324 = 0
⇒ x2 + 54x – 6x – 324 = 0
⇒ x(x + 54) -6(x + 54) = 0
⇒ (x + 54)(x – 6) = 0
⇒ x = -54 or x = 6
But as x > 0, so x can’t be negative.
Hence, x = 6.
(ii) Given quadratic equation is \((x-10)\left(\frac{1200}{x}+2\right)=1260\)
⇒ (x – 10)\(\left(\frac{1200+2 x}{x}\right)\) = 1260
⇒ (x – 10)(1200 + 2x) = 1260x
⇒ 1200x + 2x2 – 12000 – 20x = 1260x
⇒ 2x2 – 12000 – 80x = 0
⇒ x2 – 40x – 6000 = 0
⇒ x2 – 100x + 60x – 6000 = 0
⇒ (x – 100)(x – 60) = 0
⇒ x = 100 or x = -60
But as x < 0, so x can’t be positive.
Hence, x = -60.

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Selina Concise Mathematics Class 10 ICSE Solutions Probability

Selina Concise Mathematics Class 10 ICSE Solutions Probability

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Mathematics Chapter 25 Probability. You can download the Selina Concise Mathematics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Mathematics for Class 10 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability

Probability Exercise 25(A) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A coin is tossed once. Find the probability of:
(i) getting a tail
(ii) not getting a tail
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 1

Question 2.
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 2

Question 3.
In a single throw of a die, find the probability of getting a number:
(i) greater than 4.
(ii) less than or equal to 4.
(iii) not greater than 4.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 3

Question 4.
In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will not be an even number.
(iii) will be an odd number.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 4

Question 5.
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will:
(i) be a black card.
(ii) not be a red card.
(iii) be a red card.
(iv) be a face card.
(v) be a face card of red colour.
Solution:
Total number of cards = 52
Total number of outcomes = P(s) = 52
There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are black. So, there are 26 red cards and 26 black cards.
(i) Number of black cards in a deck = 26
P(E) = favourable outcomes for the event of drawing a black card = 26
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 5
(iv) There are 52 cards in a deck of cards, and 12 of these cards are face cards (4 kings, 4 queens, and 4 jacks).
P(E) = 12
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 6

Question 6.
(i) If A and B are two complementary events then what is the relation between P(A) and P(B)?
(ii) If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?
Solution:
(i) Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1.
P(A) + P(B) = 1
(ii) P(A) = 0.46
Let P(B) be the probability of not happening of event A
We know,
P(A) + P(B) = 1
P(B) = 1 – P(A)
P(B) = 1 – 0.46
P(B) = 0.54
Hence the probability of not happening of event A is 0.54

Question 7.
In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of:
(i) winning of Geeta
(ii) not winning of Ritu
Solution:
(i) Winning of Geeta is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(winning of Geeta) = 1
P(winning of Geeta) = 1 – P(winning of Ritu)
P(winning of Geeta) = 1 – 0.73
P(winning of Geeta) = 0.27
(ii) Not winning of Ritu is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(not winning of Ritu) = 1
P(not winning of Ritu) = 1 – P(winning of Ritu)
P(not winning of Ritu) = 1 – 0.73
P(not winning of Ritu) = 0.27

Question 8.
In a race between Mahesh and John, the probability that John will lose the race is 0.54. Find the probability of:
(i) winning of Mahesh
(ii) winning of John
Solution:
(i) But if John looses, Mahesh wins
Hence, probability of John losing the race = Probability of Mahesh winning the race since it is a race between these two only
Therefore, P(winning of Mahesh) = 0.54
(ii) P(winning of Mahesh) + P(winning of John) = 1
0.54 + P(winning of John) = 1
P(winning of John) = 1 – 0.54
P(winning of John) = 0.46

Question 9.
(i) Write the probability of a sure event
(ii) Write the probability of an event when impossible
(iii) For an event E, write a relation representing the range of values of P(E)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 7
The number of elements in ‘E’ can’t be less than ‘0’ i.e. negative and greater than the number of elements in S.

Question 10.
In a single throw of die, find the probability of getting:
(i) 5
(ii) 8
(iii) a number less than 8
(iv) a prime number
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 8
(ii) There are only six possible outcomes in a single throw of a die. If we want to find probability of 8 to come up, then in that case number of possible or favourable outcome is 0 (zero)
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 9

Question 11.
A die is thrown once. Find the probability of getting:
(i) an even number
(ii) a number between 3 and 8
(iii) an even number or a multiple of 3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 10

Question 12.
Which of the following cannot be the probability of an event?
(i) 3/5
(ii) 2.7
(iii) 43%
(iv) -0.6
(v) -3.2
(vi) 0.35
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 11

Question 13.
A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball
(ii) a black ball
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 12

Question 14.
A single letter is selected at random from the word ‘Probability’. Find the probability that it is a vowel.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 13

Question 15.
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 14
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 15

Probability Exercise 25(B) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:
(i) an even number
(ii) a multiple of 3
(iii) an even number and a multiple of 3
(iv) an even number or a multiple of 3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 16
(iv) From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10
Favorable number of events = n(E) = 7
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 17

Question 2.
Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is:
(i) a multiple of 5
(ii) a multiple of 6
(iii) between 40 and 60
(iv) greater than 85
(v) less than 48
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 18

Question 3.
From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 3
(ii) 5
(iii) 3 and 5
(iv) 3 or 5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 19

Question 4.
A die is thrown once. Find the probability of getting a number:
(i) less than 3
(ii) greater than or equal to 4
(iii) less than 8
(iv) greater than 6
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 20

Question 5.
A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 21

Question 6.
A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 22

Question 7.
If two coins are tossed once, what is the probability of getting:
(i) both heads.
(ii) at least one head.
(iii) both heads or both tails.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 23

Question 8.
Two dice are rolled together. Find the probability of getting:
(i) a total of at least 10.
(ii) a multiple of 2 on one die and an odd number on the other die.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 24

Question 9.
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is:
(i) a spade(v) Jack or queen
(ii) a red card(vi) ace and king
(iii) a face card(vii) a red and a king
(iv) 5 of heart or diamond(viii) a red or a king
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 25
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 26

Question 10.
A bag contains 16 colored balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is:
(i) red(v) green or red
(ii) not red(vi) white or green
(iii) white(vii) green or red or white
(iv) not white
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 27
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 28
Question 11.
A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is:
(i) white(iii) not green
(ii) red(iv) red or white
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 29

Question 12.
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is:
(i) a red card
(ii) a black card
(iii) a spade
(iv) an ace
(v) a black ace
(vi) ace of diamonds
(vii) not a club
(viii) a queen or a jack
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 30
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 31

Question 13.
Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is:
(i) a multiple of 4 or 6
(ii) a multiple of 3 and 5
(iii) a multiple of 3 or 5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 32

Question 14.
In a single throw of two dice, find the probability of:
(i) a doublet
(ii) a number less than 3 on each dice
(iii) an odd number as a sum
(iv) a total of at most 10
(v) an odd number on one dice and a number less than or equal to 4 on the other dice.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 33
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 34

Probability Exercise 25(C) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A bag contains 3 red balls, 4 blue balls and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is:
(i) yellow
(ii) red
(iii) not yellow
(iv) neither yellow nor red
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 35

Question 2.
A dice is thrown once. What is the probability of getting a number:
(i) greater than 2?
(ii) less than or equal to 2?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 36

Question 3.
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:
(i) a face card
(ii) not a face card
(iii) a queen of black card
(iv) a card with number 5 or 6
(v) a card with number less than 8
(vi) a card with number between 2 and 9
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 37

Question 4.
In a match between A and B:
(i) the probability of winning of A is 0.83. What is the probability of winning of B?
(ii) the probability of losing the match is 0.49 for B. What is the probability of winning of A?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 38

Question 5.
A and B are friends. Ignoring the leap year, find the probability that both friends will have:
(i) different birthdays?
(ii) the same birthday?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 39

Question 6.
A man tosses two different coins (one of Rs 2 and another of Rs 5) simultaneously. What is the probability that he gets:
(i) at least one head?
(ii) at most one head?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 40

Question 7.
A box contains 7 red balls, 8 green balls and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball is:
(i) white
(ii) neither red nor white.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 41

Question 8.
All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting:
(i) a black face card
(ii) a queen
(iii) a black card
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 42

Question 9.
In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start. What is the probability that the music will stop within the first 15 seconds?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 43

Question 10.
In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that:
(i) it is acceptable to a trader who accepts only a good shirt?
(ii) it is acceptable to a trader who rejects only a shirt with major defects?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 44

Question 11.
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 8
(ii) 13
(iii) less than or equal to 12
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 45

Question 12.
Which of the following cannot be the probability of an event?
(i) 3/7
(ii) 0.82
(iii) 37%
(iv) -2.4
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 46

Question 13.
If P(E) = 0.59; find P(not E)
Solution:
P(E) + P(not E) = 1
0.59 + P(not E) = 1
P(not E) = 1 – 0.59 = 0.41

Question 14.
A bag contains a certain number of red balls. A ball is drawn. Find the probability that the ball drawn is:
(i) black
(ii) red
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 47

Question 15.
The probability that two boys do not have the same birthday is 0.897. What is the probability that the two boys have the same birthday?
Solution:
P(do not have the same birthday)+P(have same birthday) = 1
0.897 + P(have same birthday) = 1
P(have same birthday) = 1 – 0.897
P(have same birthday) = 0.103

Question 16.
A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be:
(i) not red?
(ii) neither red nor green?
(iii) white or green?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 48

Question 17.
A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin:
(i) will be a Re 1 coin?
(ii) will not be a Rs 2 coin?
(iii) will neither be a Rs 5 coin nor be a Re 1 coin?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 49

Question 18.
A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.
If the outcomes are equally likely, find the probability that the pointer will point at:
(i) 6
(ii) an even number
(iii) a prime number
(iv) a number greater than 8
(v) a number less than or equal to 9
(vi) a number between 3 and 11
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 50
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 51

Question 19.
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
(i) a queen of red color
(ii) a black face card
(iii) the jack or the queen of the hearts
(iv) a diamond
(v) a diamond or a spade
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 52

Question 20.
From a deck of 52 cards, all the face cards are removed and then the remaining cards are shuffled. Now one card is drawn from the remaining deck. Find the probability that the card drawn is:
(i) a black card
(ii) 8 of red color
(iii) a king of black color
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 53

Question 21.
Seven cards:- the eight, the nine, the ten, jack, queen, king and ace of diamonds are well shuffled. One card is then picked up at random.
(i) What is the probability that the card drawn is the eight or the king?
(ii) If the king is drawn and put aside, what is the probability that the second card picked up is:
a) an ace? b) a king?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 54

Question 22.
A box contains 150 bulbs out of which 15 are defective. It is not possible to just look at a bulb and tell whether or not it is defective. One bulb is taken out at random from this box. Calculate the probability that the bulb taken out is:
(i) a good one
(ii) a defective one
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 55

Question 23.
(i) 4 defective pens are accidentally mixed with 16 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is drawn at random from the lot. What is the probability that the pen is defective?
(ii) Suppose the pen drawn in (i) is defective and is not replaced. Now one more pen is drawn at random from the rest. What is the probability that this pen is:
a) defective
b) not defective?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 56

Question 24.
A bag contains 100 identical marble stones which are numbered 1 to 100. If one stone is drawn at random from the bag, find the probability that it bears:
(i) a perfect square number
(ii) a number divisible by 4
(iii) a number divisible by 5
(iv) a number divisible by 4 or 5
(v) a number divisible by 4 and 5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 57

Question 25.
A circle with diameter 20 cm is drawn somewhere on a rectangular piece of paper with length 40 cm and width 30 cm. This paper is kept horizontal on table top and a die, very small in size, is dropped on the rectangular paper without seeing towards it. If the die falls and lands on paper only, find the probability that it will fall and land:
(i) inside the circle
(ii) outside the circle
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 58

Question 26.
Two dice (each bearing numbers 1 to 6) are rolled together. Find the probability that the sum of the numbers on the upper-most faces of two dice is:
(i) 4 or 5
(ii) 7, 8 or 9
(iii) between 5 and 8
(iv) more than 10
(v) less than 6
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 59

Question 27.
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at most two heads
(iv) all tails
(v) at least one tail
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 60

Question 28.
Two dice are thrown simultaneously. What is the probability that:
(i) 4 will not come up either time?
(ii) 4 will come up at least once?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 61

Question 29.
Cards marked with numbers 1, 2, 3 ……… 20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is:
(i) a prime number
(ii) divisible by 3
(iii) a perfect square
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 62

Question 30.
Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on:
(i) the same day
(ii) consecutive day
(iii) different days
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 63

Question 31.
A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball; find the number of black balls in the box.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 64

Question 32.
From a pack of 52 playing cards, all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is
(i) A face card (King, Jack or Queen)
(ii) An even numbered red card?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 65

Question 33.
A die has 6 faces marked by the given numbers as shown below:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 66
The die is thrown once. What is the probability of getting
(i) a positive integer?
(ii) an integer greater than -3?
(iii) the smallest integer?
Solution:
Given that the die has 6 faces marked by the given numbers as below:
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 67
When a die is rolled, total number of possible outcomes – 6
(i) For getting a positive integer, the favourable outoomes are: 1, 2, 3
⇒ Number of favourable outcomes – 3
⇒ Required probability = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)

(ii) For getting an integer greater than -3, the favourable outcomes
are: -2,-1, 1, 2, 3
⇒ Number of favourable outcomes – 5
⇒ Required probability = \(\frac { 5 }{ 6 } \)

(iii) For getting a smallest integer, the favourable outoomes are: -3
⇒ Number of favourable outcomes = 1
⇒ Required probability = \(\frac { 1 }{ 6 } \)

Question 34.
A bag contains 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is:
(i) a green ball
(ii) a white or a red ball.
(iii)Neither a green ball nor a white ball
Solution:
Number of white balls = 5
Number of red balls = 6
Number of green balls = 9
∴ Total number of balls = 5 + 6 + 9 = 20
Selina Concise Mathematics Class 10 ICSE Solutions Probability image - 68

Question 35.
A game of numbers has cards marked with 11, 12, 13, ….., 40. A card is drawn at random. Find the probability that the number on the card drawn is:
(i) A perfect square
(ii) Divisible by 7.
Solution:
Total number of outcomes = 30
(i) The perfect squares from 11 to 40 are 16, 25 and 36. So, the number of possible outcomes = 3 Hence, the probability that the number on the card drawn is a perfect square
= \(=\quad \frac { Number\quad of\quad possible\quad outcomes }{ Total\quad number\quad of\quad outcomes } =\frac { 3 }{ 30 } =\frac { 1 }{ 10 } \)
(ii) Among the given numbers, 14, 21, 28 and 35 are divisible by 7. So, the number of possible outcomes = 4 Hence, the probability that the number on the card drawn is divisible by 7
= \(\frac{\text { Number of possible autoomes }}{\text { Total number of outoomes }}=\frac{4}{30}=\frac{2}{15}\)

Question 36.
Sixteen cards are labelled as a, b, c, … , m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is:
i. a vowel
ii. a consonant
iii. none of the letters of the word median?
Solution:
Here, Total number of all possible outcomes = 16
i. a, e, i and o are the vowels.
Number of favourable outcomes = 4
∴ Required Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{4}{16}=\frac{1}{4}\)

ii. Number of consonants = 16 – 4 (vowels) = 12
∴ Number of favourable outcomes = 12
∴ Required Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{12}{16}=\frac{3}{4}\)

iii. Median contains 6 letters.
∴ Number of favourable outcomes = 16 – 6 = 10
∴ Required Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{10}{16}=\frac{5}{8}\)

Question 37.
A box contains a certain number of balls. On each of 60% balls, letter A is marked. On each of 30% balls, letter B is marked and on each of remaining balls, letter C is marked. A ball is drawn from the box at random. Find the probability that the ball drawn is:
i. marked C
ii. A or B
iii. neither B nor C
Solution:
A box contains,
60% balls, letter A is marked.
30% balls, letter B is marked.
10% balls, letter C is marked.
i. Total number of all possible outcomes = 100
Number of favourable outcomes = 10
∴ Required Probability = \(\frac{\text { Number of favou rable outcomes }}{\text { Total number of all possible outcomes }}=\frac{10}{100}=\frac{1}{10}\)

ii. The probability that the ball drawn is marked A = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{60}{100}=\frac{6}{10}\) … (1)
The probability that the ball drawn is marked B = \(\frac{\text { Number of favou rable outcomes }}{\text { Total number of all possible outcomes }}=\frac{30}{100}=\frac{3}{10}\) … (2)
∴ Required Probability = \(\frac{6}{10}+\frac{3}{10}=\frac{9}{10}\)
iii. The probability that the ball drawn is neither B nor C
= 1 – [P(B) + P(C)]
= 1 – \(\left[\frac{3}{10}+\frac{1}{10}\right]\)
= 1 – \(\frac{4}{10}\)
= \(\frac{6}{10}\)
= \(\frac{3}{5}\)

Question 38.
A box contains a certain number of balls. Some of these balls are marked A, some are marked B and the remaining are marked C. When a ball is drawn at random from the box P(A) = \(\frac{1}{3}\) and P(B) = \(\frac{1}{4}\). If there are 40 balls in the box which are marked C, find the number of balls in the box.
Solution:
P(C) = 1 – [P(A) + P(B)] = \(1-\left[\frac{1}{3}+\frac{1}{4}\right]=1-\frac{7}{12}=\frac{5}{12}\)
Probability = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}\)
Given that 40 balls in the box are marked C.
⇒ \(\frac{5}{12}=\frac{40}{\text { Total number of all possible outcomes }}\)
⇒ Total number of all possible outcomes = \(\frac{40 \times 12}{5}=96\)
∴ the number of balls in the box is 96.

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency

Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency (Mean, Median, Quartiles and Mode)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 24 Measures of Central Tendency

Measures of Central Tendency Exercise 24A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the mean of the following set of numbers:
(i) 6, 9, 11, 12 and 7
(ii) 11, 14, 23, 26, 10, 12, 18 and 6
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 1

Question 2.
Marks obtained (in mathematics) by 9 student are given below:
60, 67, 52, 76, 50, 51, 74, 45 and 56
(a) find the arithmetic mean
(b) if marks of each student be increased by 4; what will be the new value of arithmetic mean.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 2

Question 3.
Find the mean of the natural numbers from 3 to 12.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 3

Question 4.
(a) Find the mean of 7, 11, 6, 5, and 6
(b) If each number given in (a) is diminished by 2, find the new value of mean.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 4

Question 5.
If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’
Solution:
No. of terms = 5
Mean = 8
Sum of numbers = 8 x 5 = 40 .(i)
But, sum of numbers = 6+4+7+a+10 = 27+a ..(ii)
From (i) and (ii)
27+a = 40
a = 13

Question 6.
The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’.
Solution:
No. of terms = 5 and mean = 8
Sum of numbers = 5 x 8 = 40 ..(i)
but sum of numbers = 6+y+7+x+14 = 27+y+x .(ii)
from (i) and (ii)
27 + y + x = 40
x + y = 13
y = 13 – x

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 6

Question 8.
If 69.5 is the mean of 72, 70, ‘x’, 62, 50, 71, 90, 64, 58 and 82, find the value of ‘x’.
Solution:
No. of terms = 10
Mean = 69.5
Sum of the numbers = 69.5 x 10 = 695 ……….(i)
But sum of numbers = 72+70+x+62+ 50+71+90+64+58+82
= 619 + x ……(ii)
from (i) and (ii)
619 + x = 695
x = 76

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 7
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 8

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 9
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 10

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 11
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 12

Question 12.
If the mean of the following distribution is 3, find the value of p.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 13
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 14
Question 13.
In the following table, ∑f = 200 and mean = 73. Find the missing frequencies f1, and f2.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 15
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 16

Question 14.
Find the arithmetic mean (correct to the nearest whole-number) by using step-deviation method.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 17
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 18

Question 15.
Find the mean (correct to one place of decimal) by using short-cut method.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 19
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 20

Measures of Central Tendency Exercise 24B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 21
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 22

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 23
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 24

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 25
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 26
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 27

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 28
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 29

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 30
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 31

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 32
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 33

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 34
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 35

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 36
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 37

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 38
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 39

Question 10.
Calculate the mean of the distribution, given below, using the short cut method :
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 40
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 41

Question 11.
Calculate the mean of the following distribution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 42
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 43

Measures of Central Tendency Exercise 24C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A student got the following marks in 9 questions of a question paper.
3, 5, 7, 3, 8, 0, 1, 4 and 6.
Find the median of these marks.
Solution:
Arranging the given data in descending order:
8, 7, 6, 5, 4, 3, 3, 1, 0
The middle term is 4 which is the 5th term.
Median = 4

Question 2.
The weights (in kg) of 10 students of a class are given below:
21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24.
Find the median of their weights.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 122

Question 3.
The marks obtained by 19 students of a class are given below:
27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28. Find:
(i) median
(ii) lower quartile
(iii) upper quartile
(iv) interquartile range
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 44

Question 4.
From the following data, find:
(i) Median
(ii) Upper quartile
(iii) Inter-quartile range
25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 45

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 46
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 47

Question 6.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 49

Question 7.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 51

Question 8.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 53

Question 9.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 55

Question 10.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 57

Question 11.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 59
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 60

Question 12.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 62

Measures of Central Tendency Exercise 24D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the mode of the following data:
(i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6
(ii) 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8
Solution:
(i) Mode = 7
Since 7 occurs 4 times
(ii) Mode = 11
Since it occurs 4 times

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 63
Solution:
Mode is 122 cm because it occur maximum number of times. i.e. frequency is 18.

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 64
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 65

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 66
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 67

Question 5.
Find the median and mode for the set of numbers:
2, 2, 3, 5, 5, 5, 6, 8 and 9
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 68

Question 6.
A boy scored following marks in various class tests during a term; each test being marked out of 20.
15, 17, 16, 7, 10, 12, 14, 16, 19, 12 and 16
(i) What are his modal marks?
(ii) What are his median marks?
(iii) What are his total marks?
(iv) What are his mean marks?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 69

Question 7.
Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks.
0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7 and 8.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 70

Question 8.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 72

Measures of Central Tendency Exercise 24E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 74
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 75
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 76

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 77
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 78

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 79
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 80

Question 4.
The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m-1 and median q. Find p and q.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 81

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 82
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 83

Question 6.
The marks of 20 students in a test were as follows:
2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20.
Calculate:
(i) the mean (ii) the median (iii) the mode
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 84

Question 7.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 86
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 87

Question 8.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 89

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 90
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 91

Question 10.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 93

Question 11.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 95
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 96

Question 12.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 98

Question 13.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 100
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Question 14.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 103

Question 15.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 105

Question 16.
The median of the observations 11, 12, 14, (x – 2) (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Solution:
Data in ascending order:
11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47
Total number of observations = n = 9 (odd)
⇒ Median – \(\left(\frac{n+1}{2}\right)^{t h}\) term = \(\left(\frac{9+1}{2}\right)^{t h}\) term =5th term
Given, median = 24
⇒ 5th term = 24
⇒ x + 4 = 24
⇒ x = 20
Thus, the observation are as follows:
11, 12, 14, 18, 24, 29, 32, 38, 47
∴ Mean = \(\frac{\sum x}{n}=\frac{11+12+14+18+24+29+32+38+47}{9}=\frac{225}{9}=25\)

Question 17.
The number 6, 8, 10, 12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 106

Question 18.
(Use a graph paper for this question). The daily pocket expenses of 200 students in a school are given below :
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 107
Draw a histogram representing the above distribution and estimate the mode from the graph.
Solution:
Histogram is as follows:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 108
In the highest rectangle which represents modal class draw two lines AC and BD intersecting at E.
From E, draw a perpendicular to x-axis meeting at L.
Value of L is the mode. Hence, mode = 21.5

Question 19.
The marks obtained by 100 students in a mathematics test are given below :
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 109
Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm = 10 units on both the axes.
Use the ogive to estimate :
(i) Median
(ii) Lower quartile
(iii) Number of students who obtained more than 85% marks in the test.
(iv) Number of students failed, if the pass percentage was 35.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 110
The ogive is as follows:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 111
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 112

Question 20.
The mean of following numbers is 68. Find the value of ‘x’.
45, 52, 60, x, 69, 70, 26, 81 and 94.
Hence, estimate the median.
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 113

Question 21.
The marks of 10 students of a class in an examination arranged in ascending order is as follows:
13, 35, 43, 46, x, x + 4, 55, 61, 71, 80
If the median marks is 48, find the value of x. Hence, find the mode of the given data.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 114

Question 22.
The daily wages of 80 workers in a project are given below.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 116
Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = Rs. 50 on x – axis and 2 cm = 10 workers on y – axis). Use your ogive to estimate.
i. the median wages of the workers.
ii. thelower quartile wage of workers.
iii. the number of workers who earn more than Rs. 625 daily.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 117Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 118
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 119

Question 23.
The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to:
i. Frame a frequency distribution table.
ii. To calculate mean.
iii. To determine the Modal class.
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 115
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 120
Selina Concise Mathematics Class 10 ICSE Solutions Measures of Central Tendency image - 121

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles)

Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 19 Constructions (Circles)

Constructions Circles Exercise 19 – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Draw a circle of radius 3 cm. Mark a point P at a distance of 5 cm from the centre of the circle drawn. Draw two tangents PA and PB to the given circle and measure the length of each tangent.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 1

  1. Draw a circle with centre O and radius 3 cm.
  2. From O, take a point P such that OP = 5 cm
  3. Draw a bisector of OP which intersects OP at M.
  4. With centre M, and radius OM, draw a circle which intersects the given circle at A and B.
  5. Join AP and BP.
    AP and BP are the required tangents.
    On measuring AP = BP = 4 cm

Question 2.
Draw a circle of diameter of 9 cm. Mark a point at a distance of 7.5 cm from the centre of the circle. Draw tangents to the given circle from this exterior point. Measure the length of each tangent.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 2

  1. Draw a circle of diameter 9 cm, taking O as the centre.
  2. Mark a point P outside the circle, such that PO = 7.5 cm.
  3. Taking OP as the diameter, draw a circle such that it cuts the earlier circle at A and B.
  4. Join PA and PB.
    Thus, PA and PB are required tangents. PA = PB = 6 cm

Question 3.
Draw a circle of radius 5 cm. Draw two tangents to this circle so that the angle between the tangents is 45º.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 3

  1. Draw a circle with centre O and radius BC = 5 cm
  2. Draw arcs making an angle of 180º- 45º = 135º at O such that ∠AOB = 135º
  3. AT A and B, draw two rays making an angle of 90º at each point which meet each other at point P, outside the circle.
  4. AP and BP are the required tangents which make an angle of 45º with each other at P.

Question 4.
Draw a circle of radius 4.5 cm. Draw two tangents to this circle so that the angle between the tangents is 60º.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 4

  1. Draw a circle with centre O and radius BC = 4.5 cm
  2. Draw arcs making an angle of 180º – 60º = 120º at O such that ∠AOB = 120º
  3. AT A and B, draw two rays making an angle of 90º at each point which meet each other at point P, outside the circle.
  4. AP and BP are the required tangents which make an angle of 60º with each other at P.

Question 5.
Using ruler and compasses only, draw an equilateral triangle of side 4.5 cm and draw its circumscribed circle. Measure the radius of the circle.
Solution:
Steps of construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 5

  1. Draw a line segment BC = 4.5 cm
  2. With centers B and C, draw two arcs of radius 4.5 cm which intersect each other at A.
  3. Join AC and AB.
  4. Draw perpendicular bisectors of AC and BC intersecting each other at O.
  5. With centre O, and radius OA or OB or OC draw a circle which will pass through A, B and C.
    This is the required circumcircle of triangle ABC.
    On measuring the radius OA = 2.6 cm

Question 6.
Using ruler and compasses only.
(i) Construct triangle ABC, having given BC = 7 cm, AB – AC = 1 cm and ∠ABC = 45°.
(ii) Inscribe a circle in the ∆ABC constructed in (i) above. Measure its radius.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 6
i) Construction of triangle:

  • Draw a line segment BC = 7 cm
  • At B, draw a ray BX making an angle of 45º and cut off BE = AB – AC = 1 cm
  • Join EC and draw the perpendicular bisector of EC intersecting BX at A.
  • Join AC.
    ∆ABC is the required triangle.

ii) Construction of incircle:

  • Draw angle bisectors of ∠ABC and ∠ACB intersecting each other at O.
  • From O, draw perpendiculars OL to BC.
  • O as centre and OL as radius draw circle which touches the sides of the ∆ABC. This is the required in-circle of ∆ABC.
    On measuring, radius OL = 1.8 cm

Question 7.
Using ruler and compasses only, draw an equilateral triangle of side 5 cm. Draw its inscribed circle. Measure the radius of the circle.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 7

  1. Draw a line segment BC = 5 cm
  2. With centers B and C, draw two arcs of 5 cm radius each which intersect each other at A.
  3. Join AB and AC.
  4. Draw angle bisectors of ∠B and ∠C intersecting each other at O.
  5. From O, draw OL ⊥ BC.
  6. Now with centre O and radius OL, draw a circle which will touch the sides of ∆ABC
    On measuring, OL = 1.4 cm

Question 8.
Using ruler and compasses only,
(i) Construct a triangle ABC with the following data:
Base AB = 6 cm, BC = 6.2 cm and ∠CAB – 60°
(ii) In the same diagram, draw a circle which passes through the points A, B and C and mark its centre as O.
(iii) Draw a perpendicular from O to AB which meets AB in D.
(iv) Prove that AD = BD
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 8

  1. Draw a line segment AB = 6 cm
  2. At A, draw a ray making an angle of 60º with BC.
  3. With B as centre and radius = 6.2 cm draw an arc which intersects AX ray at C.
  4. Join BC.
    ∆ABC is the required triangle.
  5. Draw the perpendicular bisectors of AB and AC intersecting each other at O.
  6. With centre O, and radius as OA or OB or OC, draw a circle which will pass through A, B and C.
  7. From O, draw OD ⊥ AB.
    Proof: In right ∆OAD and ∆OBD
    OA = OB (radii of same circle)
    Side OD = OD (common)
    ∴ ∆OAD ≅ ∆OBD (RHS)
    ⇒ AD = BD (CPCT)

Question 9.
Using ruler and compasses only construct a triangle ABC in which BC = 4 cm, ∠ACB = 45° and perpendicular from A on BC is 2.5 cm. Draw a circle circumscribing the triangle ABC.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 9

  1. Draw a line segment BC = 4 cm.
  2. At C, draw a perpendicular line CX and from it, cut off CE = 2.5 cm.
  3. From E, draw another perpendicular line EY.
  4. From C, draw a ray making an angle of 45º with CB, which intersects EY at A.
  5. Join AB.
    ∆ABC is the required triangle.
  6. Draw perpendicular bisectors of sides AB and BC intersecting each other at O.
  7. With centre O, and radius OB, draw a circle which will pass through A, B and C.
    Measuring the radius OB = OC = OA = 2 cm

Question 10.
Perpendicular bisectors of the sides AB and AC of a triangle ABC meet at O.
(i) What do you call the point O?
(ii) What is the relation between the distances OA, OB and OC?
(iii) Does the perpendicular bisector of BC pass through O?
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 10

  1. O is called the circumcentre of circumcircle of ∆ABC.
  2. OA, OB and OC are the radii of the circumcircle.
  3. Yes, the perpendicular bisector of BC will pass through O.

Question 11.
The bisectors of angles A and B of a scalene triangle ABC meet at O.
i) What is the point O called?
ii) OR and OQ are drawn perpendiculars to AB and CA respectively. What is the relation between OR and OQ?
iii) What is the relation between angle ACO and angle BCO?
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 11

  1. O is called the incentre of the incircle of ∆ABC.
  2. OR and OQ are the radii of the incircle and OR = OQ.
  3. OC is the bisector of angle C
    ∴ ∠ACO = ∠BCO

Question 12.
i) Using ruler and compasses only, construct a triangle ABC in which AB = 8 cm, BC = 6 cm and CA = 5 cm.
ii) Find its incentre and mark it I.
iii) With I as centre, draw a circle which will cut off 2 cm chords from each side of the triangle.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 12

  1. Draw a line segment BC = 6 cm.
  2. With centre B and radius 8 cm draw an arc.
  3. With centre C and radius 5 cm draw another arc which intersects the first arc at A.
  4. Join AB and AC.
    ∆ABC is the required triangle.
  5. Draw the angle bisectors of ∠B and ∠A intersecting each other at I. Then I is the incentre of the triangle ABC
  6. Through I, draw ID ⊥ AB
  7. Now from D, cut off \(D P=D Q=\frac{2}{2}=1 \mathrm{cm}\)
  8. With centre I, and radius IP or IQ, draw a circle which will intersect each side of triangle ABC cutting chords of 2 cm each.

Question 13.
Construct an equilateral triangle ABC with side 6 cm. Draw a circle circumscribing the triangle ABC.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 13

  1. Draw a line segment BC = 6 cm
  2. With centers B and C, draw two arcs of radius 6 cm which intersect each other at A.
  3. Join AC and AB.
  4. Draw perpendicular bisectors of AC, AB and BC intersecting each other at O.
  5. With centre O, and radius OA or OB or OC draw a circle which will pass through A, B and C.
    This is the required circumcircle of triangle ABC.

Question 14.
Construct a circle, inscribing an equilateral triangle with side 5.6 cm.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 14

  1. Draw a line segment BC = 5.6 cm
  2. With centers B and C, draw two arcs of 5.6 cm radius each which intersect each other at A.
  3. Join AB and AC.
  4. Draw angle bisectors of ∠B and ∠Cintersecting each other at O.
  5. From O, draw OL ⊥ BC.
  6. Now with centre O and radius OL, draw a circle which will touch the sides of ∆ABC
    This is the required circle.

Question 15.
Draw a circle circumscribing a regular hexagon of side 5 cm.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 15

  1. Draw a regular hexagon ABCDEF with each side equal to 5 cm and each interior angle 120º.
  2. Join its diagonals AD, BE and CF intersecting each other at O.
  3. With centre as O and radius OA, draw a circle which will pass through the vertices A, B, C, D, E and F.
    This is the required circumcircle.

Question 16.
Draw an inscribing circle of a regular hexagon of side 5.8 cm.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 16

  1. Draw a line segment AB = 5.8 cm
  2. At A and B, draw rays making an angle of 120° each and cut off AF = BC = 5.8 cm
  3. Again F and C, draw rays making an angle of 120° each and cut off FE = CD = 5.8 cm.
  4. Join DE. Then ABCDEF is the regular hexagon.
  5. Draw the bisectors of ∠A and ∠B intersecting each other at O.
  6. From O, draw OL ⊥ AB
  7. With centre O and radius OL, draw a circle which touches the sides of the hexagon.
    This is the required in circle of the hexagon.

Question 17.
Construct a regular hexagon of side 4 cm. Construct a circle circumscribing the hexagon.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 17

  1. Draw a circle of radius 4 cm with centre O
  2. Since the interior angle of regular hexagon is 60o, draw radii OA and OB such that ∠AOB = 60°
  3. Cut off arcs BC, CD, EF and each equal to arc AB on given circle
  4. Join AB, BC, CD, DE, EF, FA to get required regular hexagon ABCDEF in a given circle.
    The circle is the required circum circle, circumscribing the hexagon.

Question 18.
Draw a circle of radius 3.5 cm. Mark a point P outside the circle at a distance of 6 cm from the centre. Construct two tangents from P to the given circle. Measure and write down the length of one tangent.
Solution:
Steps of Construction:
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 18

  1. Draw a line segment OP = 6 cm
  2. With centre O and radius 3.5 cm, draw a circle
  3. Draw the midpoint of OP
  4. With centre M and diameter OP, draw a circle which intersect the circle at T and S
  5. Join PT and PS.
    PT and PS are the required tangents. On measuring the length of PT = PS = 4.8 cm

Question 19.
Construct a triangle ABC in which base BC = 5.5 cm, AB = 6 cm and m∠ABC =120˚.
i. Construct a circle circumscribing the triangle ABC.
ii. Draw a cyclic quadrilateral ABCD so that D is equidistant from B and C.
Solution:
Steps of Construction:
i.
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 19
a. Draw a line BC = 5.4 cm.
b. Draw AB = 6 cm, such that m ∠ABC = 120°.
c. Construct the perpendicular bisectors of AB and BC, such that they intersect at O.
d. Draw a circle with O as the radius.
ii.
(e) Extend the perpendicular bisector of BC, such that
it intersects the circle at D.
(f) Join BD and CD.
(g) Here BD = DC.

Question 20.
Using a ruler and compasses only :
(i) Construct a triangle ABC with the following data: AB = 3.5 cm, BC = 6 cm and ∠ABC = 120°.
(ii) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
(iii) Measure ∠BCP.
Solution:
Steps of constructions:

1. Draw a line segment BC = 6 cm.
At B, draw a ray BX making an angle of 120° with BC.
With B as centre and radius 3.5 cm, cut-off AB = 3.5 cm.
Join AC
Thus, ABC is the required triangle.

2. Draw perpendicular bisector MN of BC which cuts BC at point o.
With O as centre and radius = OB, draw a circle.
Draw angle bisector of ∠ABC which meets the cirde at point P.
Thus, point P is equidistant from AB and BC

3. On measuring, ∠BCP = 30°

Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 20

Question 21.
Construct a ∆ABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle.
Solution:
Steps of construction:

  1. Draw BC = 6.5 cm.
  2. With B as centre, draw an arc of radius 5.5 cm.
  3. With C as oentre, draw an arc of radius 5 cm.
    Let this arc meets the previous arc at A.
  4. Join AB and AC to get ∆ABC.
  5. Draw the bi sectors of ∠ABC and ∠ACB.
    Let these bisectors meet each other at O.
  6. Draw ON ⊥ BC.
  7. With O as centre and radius ON, draw a inarcle that touches all the sides of ∆ABC
  8. By measurement, radius ON = 1.5 cm

Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 21

Question 22.
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence :
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Solution:
Steps of construction:

  1. Draw AB = 5.5 cm
  2. Construct ∠BAR = 1050
  3. With centre A and radius 6 cm, aut off arc on AR at C.
  4. Join BC. ABC is the required triangle.
  5. Draw angle bisector BD of ∠ABC, which is the loss of points equidistant from BA and BC.
  6. Draw perpendicular bisector EF of BC, which is the locus of points equidistant from B and C.
  7. BD and EF intersect each other at point P.
    Thus, P satisfies the above two lod.
    By measurement, PC = 4.8 cm

Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 22
Question 23.
Construct a regular hexagon of side 5 cm. Hence construct all its lines of symmetry and name them. (2016)
Solution:
Steps of construction:

  1. Draw AF measuring 5 cm using a ruler.
  2. With A as the centre and radius equal to AF, draw an arc above AF.
  3. With F as the centre, and same radius cut the previous arc at Z
  4. With Z as the centre, and same radius draw a circle passing through A and F.
  5. With A as the centre and same radius, draw an arc to cut the circle above AF at B.
  6. With B as the centre and same radius, draw an arc to cut the circle at C.
  7. Repeat this process to get remaining vertices of the hexagon at D and E.
  8. Join consecutive arcs on the circle to form the hexagon.
  9. Draw the perpendicular bisectors of AF, FE and DE.
  10. Extend the bisectors of AF, FE and DE to meet CD, BC and AB at X, L and O respectively.
  11. Join AD, CF and EB.

These are the 6 lines of symmetry of the regular hexagon.
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 23

Question 24.
Draw a line AB = 5 cm. Mark a point C on AB such that AC = 3 cm. Using a ruler and a compass only, construct:
(i) A circle of radius 2.5 cm, passing through A and C.
(ii) Construct two tangents to the circle from the external point B. Measure and record the length of the tangents.   (2016)
Solution:
Steps for construction:

  1. Draw AB = 5 cm using a ruler.
  2. With A as the centre cut an arc of 3 cm on AB to obtain C.
  3. With A as the centre and radius 2.5 cm, draw an arc above AB.
  4. With same radius, and C as the centre draw an arc to cut the previous arc and mark the intersection as O.
  5. With O as the centre and radius 2.5 cm, draw a circle so that points A and C lie on the circle formed.
  6. Join OB.
  7. Draw the perpendicular bisector of OB to obtain the mid-point of OB, M.
  8. With the M as the centre and radius equal to OM, draw a circle to cut the previous circle at points P and Q.
  9. Join PB and QB. PB and QB are the required tangents to the given circle from exterior point B.

Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 24
QB = PB = 3 cm
That is, length of each tangent is 3 cm.

Solution 25.
Steps of construction :

  1. Draw a line AB = 7 cm
  2. Taking P as centre and same radius, draw an arc of a circle which intersects AB at M.
  3. Taking M as centre and with the same radius as before drawn an arc intersecting previously drawn arc, at point N.
  4. Draw the ray AX passing through N, then ∠XAB = 60°
  5. Taking A as centre and radius equal to 5 cm, draw an arc cutting AX at C.
  6. Join BC
  7. The required triangle ABC is obtained.
  8. Draw angle bisector of ∠CAB and ∠ABC
  9. Mark their intersection as O
  10. With O as center, draw a circle with radius OD

Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 25

Solution 26.
Steps for construction :

  1. Draw BC = 6.8 cm.
  2. Mark point D where BD = DC = 3.4 cm which is mid-point of BC.
  3. Mark a point A which is intersection of arcs AD = 4.4 cm and AB = 5 cm from a point D and B respectively.
  4. Join AB, AD and AC. ABC is the required triangle.
  5. Draw bisectors of angle B and angle C which are ray BX and CY where I is the incentre of a circle.
  6. Draw incircle of a triangle ABC.

Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 26

Solution 27.
Steps for construction :

  1. Draw concentric circles of radius 4 cm and 6 cm with centre of O.
  2. Take point P on the outer circle.
  3. Join OP.
  4. Draw perpendicular bisectors of OP where M is the midpoint of OP.
  5. Take a distance of a point O from the point M and mark arcs from M on the inner circle it cuts at point A and B respectively.
  6. Join PA and PB.

We observe that PA and PB are tangents from outer circle to inner circle are equal of a length 4.5 cm each.
Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 27

Solution 28.
Steps for construction :

  1. Draw BC = 7.2 cm.
  2. Draw an angle ABC = 90°using compass.
  3. Draw BD perpendicular to AC using compass.
  4. Join BD.
  5. Draw perpendicular bisectors of AB and BC which intersect at I, where I is the circumcentre of a circle.
  6. Draw circumcircle using circumcentre I. we get radius of a circle is 4.7 cm.

Selina Concise Mathematics Class 10 ICSE Solutions Constructions (Circles) image - 28

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere

Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere (Surface Area and Volume)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume)

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The height of a circular cylinder is 20 cm and the radius of its base is 7 cm. Find :
(i) the volume
(ii) the total surface area.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere ex 20a q1

Question 2.
The inner radius of a pipe is 2.1 cm. How much water can 12 m of this pipe hold?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 1
Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 3

Question 4.
How many cubic meters of earth must be dug out to make a well 28 m deep and 2.8 m in diameter? Also, find the cost of plastering its inner surface at Rs 4.50 per sq meter.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 4

Question 5.
What length of solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 5

Question 6.
A cylinder has a diameter of 20 cm. The area of curved surface is 100 sq cm. Find:
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 6

Question 7.
A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all round. Find the weight, in kilogram, of 2 metres of the pipe if 1 cm3 of the metal weights 7.7 g.
Solution:
Inner radius of the pipe = r =\(\frac{5}{2}\) = 2.5 cm
External radius of the pipe = R = Inner radius of the pipe + Thickness of the pipe
= 2.5 cm + 0.5 cm
= 3 cm
Length of the pipe = h = 2 m= 200 cm
Volume of the pipe = External Volume – Internal Volume
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 7
Since 1cm3 of the metal weights 7.7 9,
∴ Weight of the pipe = (1728.6 × 7.7)g = \(\left(\frac{1728.6 \times 7.7}{1000}\right)\) kg = 13.31 kg

Question 8.
A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm 10.5 cm. Find the rise in level of the water when the solid is submerged.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 8

Question 9.
A cylindrical container with internal radius of its base 10 cm, contains water up to a height of 7 cm. Find the area of wetted surface of the cylinder.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 9

Question 10.
Find the total surface area of an open pipe of length 50 cm, external diameter 20 cm and internal diameter 6 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 10

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 153
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 11

Question 12.
The radius of a solid right circular cylinder increases by 20% and its height decreases by 20%. Find the percentage change in its volume.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 12

Question 13.
The radius of a solid right circular cylinder decreases by 20% and its height increases by 10%. Find the percentage change in its :
(i) volume
(ii) curved surface area
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 13
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 14

Question 14.
Find the minimum length in cm and correct to nearest whole number of the thin metal sheet required to make a hollow and closed cylindrical box of diameter 20 cm and height 35 cm. Given that the width of the metal sheet is 1 m. Also, find the cost of the sheet at the rate of Rs. 56 per m.
Find the area of metal sheet required, if 10% of it is wasted in cutting, overlapping, etc.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 15

Question 15.
3080 cm3 of water is required to fill a cylindrical vessel completely and 2310 cm3 of water is required to fill it upto 5 cm below the top. Find :
(i) radius of the vessel.
(ii) height of the vessel.
(iii) wetted surface area of the vessel when it is half-filled with water.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 16
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 17

Question 16.
Find the volume of the largest cylinder formed when a rectangular piece of paper 44 cm by 33 cm is rolled along it :
(i) shorter side.
(ii) longer side.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 18

Question 17.
A metal cube of side 11 cm is completely submerged in water contained in a cylindrical vessel with diameter 28 cm. Find the rise in the level of water.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 19Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 19

Question 18.
A circular tank of diameter 2 m is dug and the earth removed is spread uniformly all around the tank to form an embankment 2 m in width and 1.6 m in height. Find the depth of the circular tank.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 20

Question 19.
The sum of the inner and the outer curved surfaces of a hollow metallic cylinder is 1056 cm2 and the volume of material in it is 1056 cm3. Find its internal and external radii. Given that the height of the cylinder is 21 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 21
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 22

Question 20.
The difference between the outer curved surface area and the inner curved surface area of a hollow cylinder is 352 cm2. If its height is 28 cm and the volume of material in it is 704 cm3;find its external curved surface area.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 23

Question 21.
The sum of the heights and the radius of a solid cylinder is 35 cm and its total surface area is 3080 cm2, find the volume of the cylinder.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 24

Question 22.
The total surface area of a solid cylinder is 616 cm2. If the ratio between its curved surface area and total surface area is 1 : 2; find the volume of the cylinder.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 25

Question 23.
A cylindrical vessel of height 24 cm and diameter 40 cm is full of water. Find the exact number of small cylindrical bottles, each of height 10 cm and diameter 8 cm, which can be filled with this water.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 26

Question 24.
Two solid cylinders, one with diameter 60 cm and height 30 cm and the other with radius 30 cm and height 60 cm, are metled and recasted into a third solid cylinder of height 10 cm. Find the diameter of the cylinder formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 28

Question 25.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 151Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 29
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 30

Question 26.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 31
The given figure shows a solid formed of a solid cube of side 40cm and a solid cylinder of radius 20 cm and height 50 cm attached to the cube as shown.
Find the volume and the total surface area of the whole solid (Take π = 3.14)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 32

Question 27.
Two right circular solid cylinders have radii in the ratio 3 : 5 and heights in the ratio 2 : 3, Find the ratio between their :
(i) curved surface areas.
(ii) volumes.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 33

Question 28.
A dosed cylindrical tank, made of thin ironsheet, has diameter = 8.4 m and height 5.4 m. How much metal sheet, to the nearest m2, is used in making this tank, if \(\frac{1}{15}\) of the sheet actually used was wasted in making the tank?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 34

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the volume of a cone whose slant height is 17 cm and radius of base is 8 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 35

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 153
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 36

Question 3.
The circumference of the base of a 12 m high conical tent is 66 m. Find the volume of the air contained in it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 37

Question 4.
The radius and height of a right circular cone are in the ratio 5:12 and its volume is 2512 cubic cm. Find the radius and slant height of the cone. (Take π = 3.14)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 38

Question 5.
Two right circular cones x and y are made, x having three times the radius of y and y having half the volume of x. Calculate the ratio between the heights of x and y.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 39

Question 6.
The diameters of two cones are equal. If their slant heights are in the ratio 5:4, find the ratio of their curved surface areas.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 40

Question 7.
There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 41

Question 8.
A heap of wheat is in the form of a cone of diameter 16.8 m and height 3.5 m. Find its volume. How much cloth is required to just cover the heap?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 42

Question 9.
Find what length of canvas, 1.5 m in width, is required to make a conical tent 48 m in diameter and 7 m in height. Given that 10% of the canvas is used in folds and stitching. Also, find the cost of the canvas at the rate of Rs. 24 per meter.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 43
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 44

Question 10.
A solid cone of height 8 cm and base radius 6 cm is melted and re-casted into identical cones, each of height 2 cm and diameter 1 cm. Find the number of cones formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 45

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 46
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 47

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 48
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 49

Question 13.
A vessel, in the form of an inverted cone, is filled with water to the brim. Its height is 32 cm and diameter of the base is 25.2 cm. Six equal solid cones are dropped in it, so that they are fully submerged. As a result, one-fourth of water in the original cone overflows. What is the volume of each of the solid cones submerged?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 50

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 51
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 52

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 53
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 54

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 55
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 56

Question 3.
A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one. How many such balls can be made?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 57

Question 4.
How many balls each of radius 1 cm can be made by melting a bigger ball whose diameter is 8 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 58

Question 5.
8 metallic sphere; each of radius 2 mm, are melted and cast into a single sphere. Calculate the radius of the new sphere.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 59

Question 6.
The volume of one sphere is 27 times that of another sphere. Calculate the ratio of their:
(i) radii
(ii) surface areas
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 59

Question 7.
If the number of square centimeters on the surface of a sphere is equal to the number of cubic centimeters in the volume, what is the diameter of the sphere?
Solution:
atics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 61

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 154Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 155

Question 9.
The internal and external diameters of a hollow hemi-spherical vessel are 21 cm and 28 cm respectively. Find:
(i) internal curved surface area
(ii) external curved surface area
(iii) total surface area
(iv) volume of material of the vessel.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 62
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 63

Question 10.
A solid sphere and a solid hemi-sphere have the same total surface area. Find the ratio between their volumes.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 64

Question 11.
Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted and recasted into a single solid sphere. Taking ∏ = 3.1, find the surface area of solid sphere formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 65

Question 12.
The surface area of a solid sphere is increased by 21% without changing its shape. Find the percentage increase in its:
(i) radius
(ii) volume
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 66
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 67

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 68

Question 2.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 69

Question 3.
The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid right circular cone of height 32 cm. find the diameter of the base of the cone.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 70

Question 4.
Total volume of three identical cones is the same as that of a bigger cone whose height is 9 cm and diameter 40 cm. find the radius of the base of each smaller cone, if height of each is 108 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 71

Question 5.
A solid rectangular block of metal 49 cm by 44 cm by 18 cm is melted and formed into a solid sphere. Calculate the radius of the sphere.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 72

Question 6.
A hemi-spherical bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into conical shaped small containers each of diameter 3 cm and height 4 cm. How many containers are necessary to empty the bowl?
Solution:
Radius of hemispherical bowl = 9 cm
Volume = \(\frac{1}{2} \times \frac{4}{3} \pi r^{3}=\frac{2}{3} \pi 9^{3}=\frac{2}{3} \pi \times 729=486 \pi \mathrm{cm}^{2}\)
Diameter each of cylindrical bottle = 3 cm
Radius = \(\frac{3}{2}\)cm, and height = 4 cm
∴ Volume of bottle = \(\frac{1}{3} \pi \pi^{2} n=\frac{1}{3} \pi \times\left(\frac{3}{2}\right)^{2} \times 4=3 \pi\)
∴ No. of bottles = \(\frac{486 \pi}{3 \pi}=162\)

Question 7.
A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone if it is completely filled.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 73

Question 8.
A solid cone of radius 5 cm and height 8 cm is melted and made into small spheres of radius 0.5 cm. Find the number of spheres formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 74

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 75
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 76

Question 10.
A solid metallic cone, with radius 6 cm and height 10 cm, is made of some heavy metal A. In order to reduce weight, a conical hole is made in the cone as shown and it is completely filled with a lighter metal B. The conical hole has a diameter of 6 cm and depth 4 cm. Calculate the ratio of the volume of the metal A to the volume of metal B in the solid.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 77
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 78

Question 11.
A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 3 cm and height 8 cm. Find the number of cones.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 79

Question 12.
The surface area of a solid metallic sphere is 2464 cm2. It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate :
(i) the radius of the sphere.
(ii) the number of cones recast. (Take π = \(\frac{22}{7}\))
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 80

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 81

Question 2.
A buoy is made in the form of a hemisphere surmounted by a right circular cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 m and its volume is two-third the volume of hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two decimal places.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 82

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 83
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 84

Question 4.
The cubical block of side 7 cm is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 85

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the top which is open is 5 cm. It is filled with water. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 86

Question 6.
A hemispherical bowl has negligible thickness and the length of its circumference is 198 cm. find the capacity of the bowl.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 87

Question 7.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 88

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 89
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 90

Question 9.
A solid hemisphere of diameter 28 cm is melted and recast into a number of identical solid cones, each of diameter 14 cm and height 8 cm. Find the number of cones so formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 91

Question 10.
A cone and a hemisphere have the same base and same height. Find the ratio between their volumes.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 92

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20F – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base are removed. Find the volume of the remaining solid.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 93

Question 2.
From a solid cylinder whose height is 16 cm and radius is 12 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume and total surface area of the remaining solid.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 94
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 95

Question 3.
A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 80 m, calculate the total area of canvas required. Also, find the total cost of canvas used at Rs 15 per meter if the width is 1.5 m
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 96

Question 4.
A circus tent is cylindrical to a height of 8 m surmounted by a conical part. If total height of the tent is 13 m and the diameter of its base is 24 m; calculate:
(i) total surface area of the tent
(ii) area of canvas, required to make this tent allowing 10% of the canvas used for folds and stitching.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 97
Height of the cylindrical part = H = 8 m
Height of the conical part = h = (13 – 8)m = 5 m
Diameter = 24 m → radius = r = 12 m
Slant height of the cone = l
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 98
Slant height of cone = 13 m
(i) Total surface area of the tent = 2πrh + πrl = πr(2h + l)
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 99
(ii)Area of canvas used in stitching = total area
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 100

Question 5.
A cylindrical boiler, 2 m high, is 3.5 m in diameter. It has a hemispherical lid. Find the volume of its interior, including the part covered by the lid.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 101

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 102
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 156
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 104

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 105
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 106

Question 8.
A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm x 10.5 cm. Find the rise in level of the water when the solid is submerged.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 107

Question 9.
Spherical marbles of diameter 1.4 cm are dropped into beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 108

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 109
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 110

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 111
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 112

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 113
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 114

Question 13.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 115
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 116
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 117

Question 14.
A cylindrical can, whose base is horizontal and of radius 3.5 cm, contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate:
(i) the total surface area of the can in contact with water when the sphere is in it;
(ii) the depth of water in the can before the sphere was put into the can.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 118
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 119

Question 15.
A hollow cylinder has solid hemisphere inward at one end and on the other end it is closed with a flat circular plate. The height of water is 10 cm when flat circular surface is downward. Find the level of water, when it is inverted upside down, common diameter is 7 cm and height of the cylinder is 20 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 120

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20G – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
What is the least number of solid metallic spheres, each of 6 cm diameter, that should be melted and recast to form a solid metal cone whose height is 45 cm and diameter is 12 cm?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 121

Question 2.
A largest sphere is to be carved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere. (Answer correct to the nearest integer)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 122

Question 3.
A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in identical cones of height 12 cm and diameter 6 cm having a semi-spherical shape on the top. Find the number of cones required.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 123

Question 4.
A solid is in the form of a cone standing on a hemisphere with both their radii being equal to 8 cm and the height of cone is equal to its radius. Find in terms of π, the volume of the solid.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 124

Question 5.
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of wire.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 125

Question 6.
Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 126

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 127
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 128

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 129
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 130

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 131
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 132

Question 10.
A cylindrical water tank of diameter 2.8m and height 4.2m is being fed by a pipe of diameter 7 cm through which water flows at the rate of 4m/s. Calculate, in minutes, the time it takes to fill the tank.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 133

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 134
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 135

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 136
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 137
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 138

Question 13.
A solid, consisting of a right circular cone, standing on a hemisphere, is placed upright, in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of the cone is 4 cm. Give your answer to the nearest cubic centimeter.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 139
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 140

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 141
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 142

Question 15.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 143
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 144

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 145
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 146
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 148

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 149
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 150

Question 18.
Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 157

Question 19.
A certain number of metallic cones, each of radius 2 cm and height 3 cm, are melted and recast into a solid sphere of radius 6 cm. Find the number of cones used.
Solution:
Let the number of cones melted be n.
Let the radius of sphere be rs = 6 cm
Radius of cone be rc = 2 cm
And, height of the cone be h = 3 cm
Volume of sphere = n (Volume of a metallic cone)
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 158

Question 20.
A conical tent is to accommodate 77 persons. Each person must have 16m3 of air to breathe. Given the radius of the tent as 7m, find the height of the tent and also its curved surface area.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 159

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation

Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation (Histograms, Frequency Polygon and Ogives)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 23 Graphical Representation (Histograms, Frequency Polygon and Ogives)

Graphical Representation Exercise 23 – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 2
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 3
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 4
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 5

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 6
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 7
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 8

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 9
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 10
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 11

Question 4.
Construct a frequency distribution table for the number given below, using the class intervals 21-30, 31-40 … etc.
75, 67, 57, 50, 26, 33, 44, 58, 67, 75, 78, 43, 41, 31, 21, 32, 40, 62, 54, 69, 48, 47, 51, 38, 39, 43, 61, 63, 68, 53, 56, 49, 59, 37, 40, 68, 23, 28, 36, 47
Use the table obtained to draw:
(i) a histogram (ii) an ogive
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 12
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 13

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 14
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 15

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 16
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 17
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 18

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 19
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 20

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 21
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 22

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 23
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 24
Selina Concise Mathematics Class 10 ICSE Solutions Graphical Representation image - 25

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances

Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances

Heights and Distances Exercise 22A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The height of a tree is √3 times the length of its shadow. Find the angle of elevation of the sun.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 1

Question 2.
The angle of elevation of the top of a tower from a point on the ground and at a distance of 160 m from its foot, is found to be 60o. Find the height of the tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 2

Question 3.
A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68o with the ground. Find the height, upto which the ladder reaches.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 3

Question 4.
Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30o and 38o respectively. Find the distance between them, if the height of the tower is 50 m.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 4

Question 5.
A kite is attached to a string. Find the length of the string, when the height of the kite is 60 m and the string makes an angle 30o with the ground.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 5

Question 6.
A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45o, (ii) 60o. Find the height of the tower in each case.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 6

Question 7.
The upper part of a tree, broken over by the wind, makes an angle of 45o with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. What was the height of the tree before it was broken?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 7

Question 8.
The angle of elevation of the top of an unfinished tower at a point distance 80 m from its base is 30o. How much higher must the tower be raised so that its angle of elevation at the same point may be 60o?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 8

Question 9.
At a particular time, when the sun’s altitude is 30o, the length of the shadow of a vertical tower is 45 m. Calculate
(i) the length of the tower.
(ii) the length of the shadow of the same tower, when the sun’s altitude is
(a) 45o (b) 60o
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 9

Question 10.
Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes angle 32o24′ with the pole and when it is turned to rest against another pole, it makes angle 32o24′ with the road. Calculate the width of the road.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 10

Question 11.
Two climbers are at points A and B on a vertical cliff face. To an observer C, 40m from the foot of the cliff, on the level ground, A is at an elevation of 48o and B of 57o. What is the distance between the climbers?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 11

Question 12.
A man stands 9 m away from a flag-pole. He observes that angle of elevation of the top of the pole is 28o and the angle of depression of the bottom of the pole is 13o. Calculate the height of the pole.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 12

Question 13.
From the top of a cliff 92 m high, the angle of depression of a buoy is 20o. Calculate, to the nearest metre, the distance of the buoy from the foot of the cliff.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 13

Heights and Distances Exercise 22B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 14
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 15

Question 2.
Find the height of a tree when it is found that on walking away from it 20 m, in a horizontal line through its base, the elevation of its top changes from 60o to 30o.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 16

Question 3.
Find the height of a building, when it is found that on walking towards it 40 m in a horizontal line through its base the angular elevation of its top changes from 30o to 45o.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 17

Question 4.
From the top of a light house 100 m high, the angles of depression of two ships are observed as 48o and 36o respectively. Find the distance between the two ships(in the nearest metre) if:
(i) the ships are on the same side of the light house.
(ii) the ships are on the opposite sides of the light house.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 18

Question 5.
Two pillars of equal heights stand on either side of a roadway, which is 150 m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60o and 30o ; find the height of the pillars and the position of the point.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 19

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 20
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 21

Question 7.
The angle of elevation of the top of a tower is observed to be 60o. At a point, 30 m vertically above the first point of observation, the elevation is found to be 45o. Find:
(i) the height of the tower,
(ii) its horizontal distance from the points of observation.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 22

Question 8.
From the top of a cliff, 60 metres high, the angles of depression of the top and bottom of a tower are observed to be 30o and 60o. Find the height of the tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 23

Question 9.
A man on a cliff observes a boat, at an angle of depression 30o, which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be 60o. Assuming that the boat sails at a uniform speed, determine:
(i) how much more time it will take to reach the shore.
(ii) the speed of the boat in metre per second, if the height of the cliff is 500 m.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 24

Question 10.
A man in a boat rowing away from a lighthouse 150 m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60o to 45o. Find the speed of the boat.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 25

Question 11.
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60o. When he moves 40 m away from the bank, he finds the angle of elevation to be 30o. Find:
(i) the height of the tree, correct to 2 decimal places,
(ii) the width of the river.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 26

Question 12.
The horizontal distance between two towers is 75 m and the angular depression of the top of the first tower as seen from the top of the second, which is 160 m high, is 45o. Find the height of the first tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 27

Question 13.
The length of the shadow of a tower standing on level plane is found to be 2y metres longer when the sun’s altitude is 30o than when it was 45o. Prove that the height of the tower is y(√3 + 1) metres.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 28

Question 14.
An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60o. After 10 seconds, its elevation is observed to be 30o; find the uniform speed of the aeroplane in km per hour.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 29

Question 15.
From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be 30o and 45respectively. Find the distances of the two stones from the foot of the hill.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 30

Heights and Distances Exercise 22C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 31

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 32

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 33
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 34

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 35
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 36

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 37
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 38

Question 5.
The radius of a circle is given as 15 cm and chord AB subtends an angle of 131o at the centre C of the circle. Using trigonometry, calculate:
(i) the length of AB;
(ii) the distance of AB from the centre C.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 39

Question 6.
At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is 5/12. On walking 192 metres towards the tower, the tangent of the angle is found to be 3/4. Find the height of the tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 40

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 41
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 42

Question 8.
With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m. The man’s eye s 2 m above the ground. He observes the angle of elevation of C, the top of the pole, as xo , where tan xo = 2/5. Calculate:
(i) the distance AB in metres;
(ii) angle of elevation of the top of the pole when he is standing 15 metres from the pole. Give your answer to the nearest degree.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 43

Question 9.
The angles of elevation of the top of a tower from two points on the ground at distances a and b metres from the base of the tower and in the same line are complementary. Prove that the height of the tower is √ab metre.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 44

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 45
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 46

Question 11.
A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 47

Question 12.
A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60o. When he moves 50 m away from the bank, he finds the angle of elevation to be 30o. Calculate:
(i) the width of the river;
(ii) the height of the tree.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 48

Question 13.
A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole is 60o and the angle of elevation of the top of the pole, as seen from the foot of the tower is 30o. Find:
(i) the height of the tower ;
(ii) the horizontal distance between the pole and the tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 49

Question 14.
A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole, the angle of elevation of the top of the tower is 60o and the angle of depression of the bottom of the tower is 30o. Find:
(i) the height of the tower, if the height of the pole is 20 m;
(ii) the height of the pole, if the height of the tower is 75 m.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 50
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 51

Question 15.
From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30o and the angle of depression of its image in the water of the lake is observed to be 60o. Find the actual height of the bird above the surface of the lake.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 52

Question 16.
A man observes the angle of elevation of the top of a building to be 30o. He walks towards it in a horizontal line through its base. On covering 60 m, the angle of elevation changes to 60o. Find the height of the building correct to the nearest metre.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 53
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 54

Question 17.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships, on the same side of a light house in a horizontal line with its base, are 30° and 40° respectively. Find the distance between the two ships. Give your answer corrected to the nearest metre.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 55

Question 18.
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 56
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 57
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 58

Question 19.
An aeroplane, at an altitude of 250 m, observes the angles of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 59

Question 20.
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Solution:
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Question 21.
The angles of depression of two ships A and B as observed from the top of a light house 60m high, are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Heights and Distances image - 62

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Trigonometrical Identities

Selina Concise Mathematics Class 10 ICSE Solutions Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 21 Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)

Trigonometrical Identities Exercise 21A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
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Solution:
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Question 2.
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Solution:
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Question 3.
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Solution:
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Question 4.
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Solution:
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Question 5.
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Solution:
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Question 6.
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Solution:
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Question 7.
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Solution:
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Question 8.
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Solution:
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Question 9.
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Solution:
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Question 10.
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Solution:
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Question 11.
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Solution:
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Question 12.
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Solution:
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Question 13.
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Solution:
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Question 14.
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Solution:
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Question 15.
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Solution:
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Question 16.
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Solution:
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Question 17.
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Solution:
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Question 18.
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Solution:
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Question 19.
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Solution:
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Question 20.
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Solution:
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Question 21.
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Solution:
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Question 22.
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Solution:
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Question 23.
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Solution:
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Question 24.
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Solution:
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Question 25.
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Solution:
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Question 26.
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Solution:
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Question 27.
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Solution:

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Question 28.
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Solution:
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Question 29.
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Solution:
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Question 30.
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Solution:
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Question 31.
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Solution:
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Question 32.
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Solution:
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Question 33.
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Solution:
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Question 34.
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Solution:
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Question 35.
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Solution:
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Question 36.
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Solution:
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Question 37.
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Solution:
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Question 38.
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Solution:
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Question 39.
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Solution:
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Question 40.
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Solution:
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Question 41.
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Solution:
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Question 42.
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Solution:
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Question 43.
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Solution:
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Question 44.
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Solution:
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Question 45.
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Solution:
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Question 46.
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Solution:
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Question 47.
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Solution:
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Question 48.
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Solution:
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Trigonometrical Identities Exercise 21B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
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Solution:
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Question 2.
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Solution:
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Question 3.
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Solution:
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Question 4.
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Solution:
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Question 5.
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Solution:
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Question 6.
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Solution:
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Question 7.
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Solution:
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Trigonometrical Identities Exercise 21C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
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Solution:
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Question 2.
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Solution:
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Question 3.
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Solution:
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Question 4.
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Solution:
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Question 5.
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Solution:
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Question 6.
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Solution:
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Question 7.
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Solution:
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Question 8.
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Solution:
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Question 9.
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Solution:
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Question 10.
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Solution:
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Question 11.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Trigonometrical Identities image - 137

Question 12.
Without using trigonometrical tables, evaluate:
cosec2 57° – tan233° + cos 44° cosec46° – \(\sqrt{2}\)cos45° – tan260°
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Trigonometrical Identities image - 138

Trigonometrical Identities Exercise 21D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
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Solution:
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Question 2.
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Solution:
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Question 3.
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Solution:
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Question 4.
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Solution:
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Question 5.
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Solution:
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Question 6.
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Solution:
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Trigonometrical Identities Exercise 21E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
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Solution:
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Question 2.
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Solution:
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Question 3.
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Solution:
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Question 4.
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Solution:
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Question 5.
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Solution:
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Question 6.
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Solution:
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Question 7.
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Solution:
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Question 8.
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Solution:
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Question 9.
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Solution:
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Question 10.
\(\frac{\cot A-1}{2-\sec ^{2} A}=\frac{\cot A}{1+\tan A}\)
Solution:
To prove that : \(\frac{\cot A-1}{2-\sec ^{2} A}=\frac{\cot A}{1+\tan A}\)
Selina Concise Mathematics Class 10 ICSE Solutions Trigonometrical Identities image - 180
Hence proved.

Question 11.
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Solution:
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Question 12.
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Solution:
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Question 13.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Trigonometrical Identities
Selina Concise Mathematics Class 10 ICSE Solutions Trigonometrical Identities image - 190

Question 14.
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Solution:
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Question 15.
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Solution:
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Question 16.
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Solution:
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Question 17.
Evaluate without using trigonometric tables,
sin2 28° + sin2 62° + tan2 38° – cot2 52° + \(\frac{1}{4}\)sec2 30°
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Trigonometrical Identities image - 198

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line

Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line

Equation of a Line Exercise 14A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find, which of the following points lie on the line x – 2y + 5 = 0:
(i) (1, 3) (ii) (0, 5)
(iii) (-5, 0) (iv) (5, 5)
(v) (2, -1.5) (vi) (-2, -1.5)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 1

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 3

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 4
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 5

Question 4.
For what value of k will the point (3, -k) lie on the line 9x + 4y = 3?
Solution:
The given equation of the line is 9x + 4y = 3.
Put x = 3 and y = -k, we have:
9(3) + 4(-k) = 3
27 – 4k = 3
4k = 27 – 3 = 24
k = 6

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 6
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 7

Question 6.
Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2)?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 8

Question 7.
(i) The line y = 3x – 2 bisects the join of (a, 3) and (2, -5), find the value of a.
(ii) The line x – 6y + 11 = 0 bisects the join of (8, -1) and (0, k). Find the value of k.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 9

Question 8.
(i) The point (-3, 2) lies on the line ax + 3y + 6 = 0, calculate the value of a.
(ii) The line y = mx + 8 contains the point (-4, 4), calculate the value of m.
Solution:
(i) Given, the point (-3, 2) lies on the line ax + 3y + 6 = 0.
Substituting x = -3 and y = 2 in the given equation, we have:
a(-3) + 3(2) + 6 = 0
-3a + 12 = 0
3a = 12
a = 4
(ii) Given, the line y = mx + 8 contains the point (-4, 4).
Substituting x = -4 and y = 4 in the given equation, we have:
4 = -4m + 8
4m = 4
m = 1

Question 9.
The point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3. Does P lie on the line x – 5y + 15 = 0?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 10

Question 10.
The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1: 2. Does the line x – 2y = 0 contain Q?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 11

Question 11.
Find the point of intersection of the lines:
4x + 3y = 1 and 3x – y + 9 = 0. If this point lies on the line (2k – 1)x – 2y = 4; find the value of k.
Solution:
Consider the given equations:
4x + 3y = 1 ….(1)
3x – y + 9 = 0 ….(2)
Multiplying (2) with 3, we have:
9x – 3y = -27 ….(3)
Adding (1) and (3), we get,
13x = -26
x = -2
From (2), y = 3x + 9 = -6 + 9 = 3
Thus, the point of intersection of the given lines (1) and (2) is (-2, 3).
The point (-2, 3) lies on the line (2k – 1)x – 2y = 4.
(2k – 1)(-2) – 2(3) = 4
-4k + 2 – 6 = 4
-4k = 8
k = -2

Question 12.
Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.
Solution:
We know that two or more lines are said to be concurrent if they intersect at a single point.
We first find the point of intersection of the first two lines.
2x + 5y = 1 ….(1)
x – 3y = 6 ….(2)
Multiplying (2) by 2, we get,
2x – 6y = 12 ….(3)
Subtracting (3) from (1), we get,
11y = -11
y = -1
From (2), x = 6 + 3y = 6 – 3 = 3
So, the point of intersection of the first two lines is (3, -1).
If this point lie on the third line, i.e., x + 5y + 2 = 0, then the given lines will be concurrent.
Substituting x = 3 and y = -1, we have:
L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 5 – 5 = 0 = R.H.S.
Thus, (3, -1) also lie on the third line.
Hence, the given lines are concurrent.

Equation of a Line Exercise 14B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 12
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 13

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 14
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 15

Question 3.
Find the slope of the line passing through the following pairs of points:
(i) (-2, -3) and (1, 2)
(ii) (-4, 0) and origin
(iii) (a, -b) and (b, -a)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 16

Question 4.
Find the slope of the line parallel to AB if:
(i) A = (-2, 4) and B = (0, 6)
(ii) A = (0, -3) and B = (-2, 5)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 17

Question 5.
Find the slope of the line perpendicular to AB if:
(i) A = (0, -5) and B = (-2, 4)
(ii) A = (3, -2) and B = (-1, 2)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 18

Question 6.
The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 19

Question 7.
The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 20

Question 8.
Without using the distance formula, show that the points A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 21

Question 9.
Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 22

Question 10.
(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 23
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 24

Question 11.
Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 25

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 26
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 27

Question 13.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 28
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 29

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 30
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 31

Question 15.
A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find:
(i) the slope of the altitude of AB,
(ii) the slope of the median AD, and
(iii) the slope of the line parallel to AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 32

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 33
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 34

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 35
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 36

Question 18.
The points (-3, 2), (2, -1) and (a, 4) are collinear. Find a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 37

Question 19.
The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 38

Question 20.
Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.
Which segment appears to have the steeper slope, AB or AC?
Justify your conclusion by calculating the slopes of AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 39

Question 21.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 40
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 41

Equation of a Line Exercise 14C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the equation of a line whose:
y-intercept = 2 and slope = 3.
Solution:
Given, y-intercept = c = 2 and slope = m = 3.
Substituting the values of c and m in the equation y = mx + c, we get,
y = 3x + 2, which is the required equation.

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 42
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 43

Question 3.

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 45

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 141
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 46

Question 5.
Find the equation of the line passing through:
(i) (0, 1) and (1, 2) (ii) (-1, -4) and (3, 0)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 47

Question 6.
The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find:
(i) the gradient of PQ;
(ii) the equation of PQ;
(iii) the co-ordinates of the point where PQ intersects the x-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 48

Question 7.
The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find:
(i) the equation of AB;
(ii) the co-ordinates of the point where the line AB intersects the y-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 49

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 50
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 51

Question 9.
In ΔABC, A = (3, 5), B = (7, 8) and C = (1, -10). Find the equation of the median through A.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 52

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 53
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 54

Question 11.
Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x – y = 4.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 55

Question 12.
In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A.
Also, find the equation of the line through vertex B and parallel to AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 56

Question 13.
A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 57

Question 14.
Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 58

Question 15.
Find the equation of the line, whose:
(i) x-intercept = 5 and y-intercept = 3
(ii) x-intercept = -4 and y-intercept = 6
(iii) x-intercept = -8 and y-intercept = -4
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 59

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 142
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 60

Question 17.
Find the equation of the line with x-intercept 5 and a point on it (-3, 2).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 61

Question 18.
Find the equation of the line through (1, 3) and making an intercept of 5 on the y-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 62

Question 19.
Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 63

Question 20.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 64
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 65

Question 21.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 66
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 67

Question 22.
A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC, Find:
(i) the co-ordinates of the centroid of triangle ABC.
(ii) the equation of a line, through the centroid and parallel to AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 68

Question 23.
A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point P in AC; such that AP: CP = 2: 3.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 69

Equation of a Line Exercise 14D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the slope and y-intercept of the line:
(i) y = 4
(ii) ax – by = 0
(iii) 3x – 4y = 5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 70

Question 2.
The equation of a line x – y = 4. Find its slope and y-intercept. Also, find its inclination.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 71

Question 3.
(i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0?
(ii) Is the line x – 3y = 4 perpendicular to the line 3x – y = 7?
(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1?
(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 72
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 73

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 74
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 75

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 76
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 77

Question 6.
(i) Lines 2x – by + 3 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
(ii) Lines mx + 3y + 7 = 0 and 5x – ny – 3 = 0 are perpendicular to each other. Find the relation connecting m and n.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 78

Question 7.
Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 79

Question 8.
The equation of a line AB is 2x – 2y + 3 = 0.
(i) Find the slope of the line AB.
(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 80

Question 9.
The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 81

Question 10.
If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 82

Question 11.
The line through A(-2,3) and B(4,b) is perpendicular to the line 2x – 4y =5. Find the value of b.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 83

Question 12.
Find the equation of the line through (-5, 7) and parallel to:
(i) x-axis (ii) y-axis
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 84

Question 13.
(i) Find the equation of the line passing through (5, -3) and parallel to x – 3y = 4.
(ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 85

Question 14.
Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 86

Question 15.
Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 87

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 88
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 89

Question 17.
B (-5, 6) and D (1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 90

Question 18.
A = (7, -2) and C = (-1, -6) are the vertices of square ABCD. Find the equations of diagonal BD and of diagonal AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 91

Question 19.
A (1, -5), B (2, 2) and C (-2, 4) are the vertices of triangle ABC, find the equation of:
(i) the median of the triangle through A.
(ii) the altitude of the triangle through B.
(iii) the line through C and parallel to AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 92

Question 20.
(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.
(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 93

Question 21.
The line 4x – 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A.
Determine the equation of the line through A and perpendicular to 4x – 3y + 12 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 94

Question 22.
The point P is the foot of perpendicular from A (-5, 7) to the line whose equation is 2x – 3y + 18 = 0. Determine:
(i) the equation of the line AP
(ii) the co-ordinates of P
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 95

Question 23.
The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC.
If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 96

Question 24.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 97
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 98

Question 25.
Find the value of a for which the points A(a, 3), B(2, 1) and C(5, a) are collinear. Hence, find the equation of the line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 99

Equation of a Line Exercise 14E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5. Find its co-ordinates of point P.
Also, find the equation of the line through P and parallel to 3x + 5y = 7.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 100

Question 2.
The line segment joining the points A(3, -4) and B (-2, 1) is divided in the ratio 1: 3 at point P in it. Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y + 4 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 101
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 102
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 103
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 104
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 105
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 106
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 107
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 108
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 109
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 110

Question 3.
A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.
Solution:

Question 4.
Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other.
Solution:

Question 5.
Solution:

Question 6.
(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.
Solution:

Question 7.
Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
Solution:

Question 8.
A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.
Solution:

Question 9.
A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.
Solution:

Question 10.
Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x – 2y = 1.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 111

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 112
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 113

Question 12.
O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:
(i) the equation of median of triangle OAB through vertex O.
(ii) the equation of altitude of triangle OAB through vertex B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 114

Question 13.
Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.
Does the line 3x = y + 1 bisect the line segment joining the two given points?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 115

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 143Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 116

Question 15.
Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:
(i) perpendicular to the line 2x – y + 7 = 0
(ii) parallel to it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 117

Question 16.
The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find:
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point, where the perpendicular through A, as obtained in (i), meets BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 118

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 119
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 120

Question 18.
P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 121

Question 19.
A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 122

Question 20.
A line AB meets the x-axis at point A and y-axis at point B. The point P (-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:
(i) the co-ordinates of A and B.
(ii) the equation of line through P and perpendicular to AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 123

Question 21.
A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from the positive side of y-axis. Find the equation of the line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 124

Question 22.
Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 125

Question 23.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 126
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 127

Question 24.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 146
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 147

Question 25.
The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 157
Question 26.
Points A and B have coordinates (7, -3) and (1, 9) respectively. Find:
(i) the slope of AB.
(ii) the equation of the perpendicular bisector of the line segment AB.
(iii) the value of ‘p’ if (-2, p) lies on it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 149

Question 27.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 150
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 158
Question 28.
The equation of a line 3x + 4y – 7 = 0. Find:
(i) the slope of the line.
(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 151
Question 29.
ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find:
(i) Co-ordinates of A
(ii) Equation of diagonal BD
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 152

Question 30.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 153
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 154
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 155

Question 31.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 156
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 128

Question 32.
Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 129

Question 33.
A straight line passes through the points P(-1, 4) and Q(5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid- t point of the line segment AB. Find:
(i) the equation of the line.
(ii) the co-ordinates of points A and B.
(iii) the co-ordinates of point M
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 130

Question 34.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 131
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 132

Question 35.
A line through point P(4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 133

Question 36.
Find the equation of line through the intersection of lines 2x – y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 134

Question 37.
Find the equation of the line through the Points A(-1, 3) and B(0, 2). Hence, show that the points A, B and C(1, 1) are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 135

Question 38.
Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2), find :
(i) the co-ordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 136

Question 39.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 137
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 138

Question 40.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 139

Question 41.

i. Since A lies on the X-axis, let the co-ordinates of A be (x, 0).
Since B lies on the Y-axis, let the co-ordinates of B be (0, y).
Let m = 1 and n = 2
Using Section formula,
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 140
⇒ Slope of line perpendicular to AB = m = -2
P = (4, -1)
Thus, the required equation is
y – y1 = m(x – x1)
⇒ y – (-1) = -2(x – 4)
⇒ y + 1 = -2x + 8
⇒ 2x + y = 7

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions)

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions)

Locus and Its Constructions Exercise 16A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Given— PQ is perpendicular bisector of side AB of the triangle ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 1
Prove— Q is equidistant from A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 2
Construction: Join AQ
Proof: In ∆AQP and ∆BQP,
AP = BP (given)
∠QPA = ∠QPB (Each = 90 )
PQ = PQ (Common)
By Side-Angle-Side criterian of congruence, we have
∆AQP ≅ ∆BQP (SAS postulate)
The corresponding parts of the triangle are congruent
∴ AQ = BQ (CPCT)
Hence Q is equidistant from A and B.

Question 2.
Given— CP is bisector of angle C of ∆ ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 3
Prove— P is equidistant from AC and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 4

Question 3.
Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX at point Y.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 5
Prove—
(i) X is equidistant from AB and AC.
(ii) Y is equidistant from A and C.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 6

Question 4.
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
Solution:
Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 7
Steps of Construction:
i) Draw a line segment BC = 6.3 cm
ii) With centre B and radius 4.2 cm, draw an arc.
iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
iv) Join AB and AC.
∆ABC is the required triangle.
v) Again with centre B and C and radius greater than \(\frac{1}{2} \mathrm{BC}\) draw arcs which intersects each other at L and M.
vi) Join LM intersecting AC at D and BC at E.
vii) Join DB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) 12
Hence, D is equidistant from B and C.

Question 5.
In each of the given figures: PA = PH and QA = QB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) 4
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of points equidistant from two given fixed points.
Solution:
Construction: Join PQ which meets AB in D.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 9
Proof: P is equidistant from A and B.
∴ P lies on the perpendicular bisector of AB.
Similarly, Q is equidistant from A and B.
∴ Q lies on perpendicular bisector of AB.
∴ P and Q both lie on the perpendicular bisector of AB.
∴ PQ is perpendicular bisector of AB.
Hence, locus of the points which are equidistant from two fixed points, is a perpendicular bisector of the line joining the fixed points.

Question 6.
Construct a right angled triangle PQR, in which ∠ Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 10
Steps of Construction:
i) Draw a line segment QR = 4.5 cm
ii) At Q, draw a ray QX making an angle of 90°
iii) With centre R and radius 8 cm, draw an arc which intersects QX at P.
iv) Join RP.
∆PQR is the required triangle.
v) Draw the bisector of ∠PQR which meets PR in T.
vi) From T, draw perpendicular PL and PM respectively on PQ and QR.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 11
Hence, T is equidistant from PQ and QR.

Question 7.
Construct a triangle ABC in which angle ABC = 75°. AB = 5 cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P ; prove that P is equidistant from B and C ; and also from AC and BC.
Hence P is equidistant from AC and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 12
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 13

Question 8.
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 14

Question 9.
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A.
Prove that –
(i) point A is equidistant from all the three sides of the triangle.
(ii) AM bisects angle LMN.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 15
Construction: Join AM
Proof:
∵ A lies on bisector of ∠N
∴A is equidistant from MN and LN.
Again, A lies on bisector of ∠L
∴ A is equidistant from LN and LM.
Hence, A is equidistant from all sides of the triangle LMN.
∴ A lies on the bisector of ∠M

Question 10.
Use ruler and compasses only for this question:
(i) construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
Solution:
Steps of construction:
(i) Draw line BC = 6 cm and an angle CBX = 60o. Cut off AB = 3.5. Join AC, triangle ABC is the required triangle.
(ii) Draw perpendicular bisector of BC and bisector of angle B
(iii) Bisector of angle B meets bisector of BC at P.
⇒ BP is the required length, where, PB = 3.5 cm
(iv) P is the point which is equidistant from BA and BC, also equidistant from B and C.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 16
PB=3.6 cm

Question 11.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point E Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 17
Prove that:
(i) F is equidistant from A and B.
(ii) F is equidistant from AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 18
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 20

Question 12.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 21
Since P lies on the bisector of angle B,
therefore, P is equidistant from AB and BC …. (1)
Similarly, P lies on the bisector of angle C,
therefore, P is equidistant from BC and CD …. (2)
From (1) and (2),
Hence, P is equidistant from AB and CD.

Question 13.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 22
Steps of construction:
(i) Draw a line segment AB of 6 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
(iii) Take any point on LM say P.
(iv) Join PA and PB.
Since, P lies on the right bisector of line AB.
Therefore, P is equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.

Question 14.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 23
Steps of Construction:
i) Draw a ray BC.
ii) Construct a ray RA making an angle of 75° with BC. Therefore, ABC= 75°.
iii) Draw the angle bisector BP of ∠ABC.
BP is the required locus.
iv) Take any point D on BP.
v) From D, draw DE ⊥ AB and DF ⊥ BC
Since D lies on the angle bisector BP of ∠ABC
D is equidistant from AB and BC.
Hence, DE = DF
Similarly, any point on BP is equidistant from AB and BC.
Therefore, BP is the locus of all points which are equidistant from AB and BC.

Question 15.
Draw an angle ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC ; and also equidistant from A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 24
Steps of Construction:
i) Draw a line segment BC = 5 cm
ii) At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.
iii) Draw the angle bisector of ∠ABC.
iv) Draw the perpendicular bisector of AB which intersects the angle bisector at P.
P is the required point which is equidistant from AB and BC, as well as from A and B.

Question 16.
In the figure given below, find a point P on CD equidistant from points A and B.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 25
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 26
Steps of Construction:
i) AB and CD are the two lines given.
ii) Draw a perpendicular bisector of line AB which intersects CD in P.
P is the required point which is equidistant from A and B.
Since P lies on perpendicular bisector of AB; PA = PB.

Question 17.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 27
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 28
Steps of Construction:
i) In the given triangle, draw the angle bisector of ∠BAC.
ii) Draw the perpendicular bisector of BC which intersects the angle bisector at P.
P is the required point which is equidistant from AB and AC as well as from B and C.
Since P lies on angle bisector of ∠BAC,
It is equidistant from AB and AC.
Again, P lies on perpendicular bisector of BC,
Therefore, it is equidistant from B and C.

Question 18.
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ ABC = 60°. Locate by construction the point P such that :
(i) P is equidistant from B and C.
(ii) P is equidistant from AB and BC.
(iii) Measure and record the length of PB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 29
Solution:
Steps of Construction:
1) Draw a line segment AB = 7 cm.
2) Draw angle ∠ABC = 60° with the help of compass.
3) Cut off BC = 8 cm.
4) Join A and C.
5) The triangle ABC so formed is the required triangle.
i) Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.
ii) Draw the angle bisector of ∠ABC. Any point situated on this angular bisector is equidistant from lines AB and BC.
The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.
P is the required point which is equidistant from AB and AC as well as from B and C.
On measuring the length of line segment PB, it is equal to 4.5 cm.

Question 19.
On a graph paper, draw lines x = 3 and y = -5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 30
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 3 which is parallel to y-axis
And draw another line m, y = -5, which is parallel to x-axis
These two lines intersect each other at P.
Now draw the angle bisector p of angle P.
Since p is the angle bisector of P, any point on P is equidistant from l and m.
Therefore, this line p is equidistant from l and m.

Question 20.
On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distance from the given line is always equal to 3 units.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 31
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 6 which is parallel to y-axis
Take points P and Q which are at a distance of 3 units from the line l.
Draw lines m and n from P and Q parallel to l
With locus = 3, two lines can be drawn x = 3 and x = 9.

Locus and Its Constructions Exercise 16B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Describe the locus of a point at a distance of 3 cm from a fixed point.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 32
The locus of a point which is 3 cm away from a fixed point is circumference of a circle whose radius is 3 cm and the fixed point is the centre of the circle.

Question 2.
Describe the locus of a point at a distance of 2 cm from a fixed line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 33
The locus of a point at a distance of 2 cm from a fixed line AB is a pair of straight lines l and m which are parallel to the given line at a distance of 2 cm.

Question 3.
Describe the locus of the centre of a wheel of a bicycle going straight along a level road.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 34
The locus of the centre of a wheel, which is going straight along a level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.

Question 4.
Describe the locus of the moving end of the minute hand of a clock.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 35
The locus of the moving end of the minute hand of the clock will be a circle where radius will be the length of the minute hand.

Question 5.
Describe the locus of a stone dropped from the top of a tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 36
The locus of a stone which is dropped from the top of a tower will be a vertical line through the point from which the stone is dropped.

Question 6.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) 31
The locus of the runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge will be the circumference of a circle whose radius is equal to the radius of the inner circular track plus 1.5 m.

Question 7.
Describe the locus of the door handle as the door opens.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 38
The locus of the door handle will be the circumference of a circle with centre at the axis of rotation of the door and radius equal to the distance between the door handle and the axis of rotation of the door.

Question 8.
Describe the locus of a point inside a circle and equidistant from two fixed points on the circumference of the circle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 39
The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.

Question 9.
Describe the locus of the centers of all circles passing through two fixed points.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 40
The locus of the centre of all the circles which pass through two fixed points will be the perpendicular bisector of the line segment joining the two given fixed points.

Question 10.
Describe the locus of vertices of all isosceles triangles having a common base.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 41
The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.

Question 11.
Describe the locus of a point in space which is always at a distance of 4 cm from a fixed point.
Solution:
The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to 4 cm.

Question 12.
Describe the locus of a point P so that:
AB2 = AP2 + BP2, where A and B are two fixed points.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 42
The locus of the point P is the circumference of a circle with AB as diameter and satisfies the condition AB2 = AP2 + BP2.

Question 13.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
i) AB and BC
ii) B and D.
Solution:
i)
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 43
The locus of the point in a rhombus ABCD which is equidistant from AB and BC will be the diagonal BD.
ii)
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 44
The locus of the point in a rhombus ABCD which is equidistant from B and D will be the diagonal AC.

Question 14.
The speed of sound is 332 meters per second. A gun is fired. Describe the locus of all the people on the Earth’s surface, who hear the sound exactly one second later.
Solution:
The locus of all the people on Earth’s surface is the circumference of a circle whose radius is 332 m and centre is the point where the gun is fired.

Question 15.
Describe:
i) The locus of points at distances less than 3 cm from a given point.
ii) The locus of points at distances greater than 4 cm from a given point.
iii) The locus of points at distances less than or equal to 2.5 cm from a given point.
iv) The locus of points at distances greater than or equal to 35 mm from a given point.
v)The locus of the centre of a given circle which rolls around the outside of a second circle and is always touching it.
vi) The locus of the centers of all circles that are tangent to both the arms of a given angle.
vii) The locus of the mid-points of all chords parallel to a given chord of a circle.
viii) The locus of points within a circle that are equidistant from the end points of a given chord.
Solution:
i) The locus is the space inside of the circle whose radius is 3 cm and the centre is the fixed point which is given.
ii) The locus is the space outside of the circle whose radius is 4 cm and centre is the fixed point which is given.
iii) The locus is the space inside and circumference of the circle with a radius of 2.5 cm and the centre is the given fixed point.
iv) The locus is the space outside and circumference of the circle with a radius of 35 mm and the centre is the given fixed point.
v) The locus is the circumference of the circle concentric with the second circle whose radius is equal to the sum of the radii of the two given circles.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 45
vi) The locus of the centre of all circles whose tangents are the arms of a given angle is the bisector of that angle.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 46
vii) The locus of the mid-points of the chords which are parallel to a given chords is the diameter perpendicular to the given chords.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 47
viii) The locus of the points within a circle which are equidistant from the end points of a given chord is the diameter which is perpendicular bisector of the given chord.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 48
Question 16.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 49
Draw a line XY parallel to the base BC from the vertex A.
This line is the locus of vertex A of all the triangles which have the base BC and length of altitude equal to AD.

Question 17.
In the given figure, obtain all the points equidistant from lines m and n ; and 2.5 cm from O.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 50
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 51
Draw an angle bisector PQ and XY of angles formed by the lines m and n. From O, draw arcs with radius 2.5 cm, which intersect the angle bisectors at a, b, c and d respectively.
Hence, a, b, c and d are the required four points.

Question PQ.
By actual drawing obtain the points equidistant from lines m and n and 6 cm from the point P, where P is 2 cm above m, m is parallel to n and m is 6 cm above n.
Solution:
Steps of construction:
i) Draw a linen.
ii) Take a point Lonn and draw a perpendicular to n.
iii) Cut off LM = 6 cm and draw a line q, the perpendicular bisector of LM.
iv) At M, draw a line m making an angle of 90°.
v) Produce LM and mark a point P such that PM = 2 cm.
vi) From P, draw an arc with 6 cm radius which intersects the line q, the perpendicular bisector of LM, at A and B.
A and B are the required points which are equidistant from m and n and are at a distance of 6 cm from P.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 52

 

Question 18.
A straight line AB is 8 cm long. Draw and describe the locus of a point which is:
(i) always 4 cm from the line AB
(ii) equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B. Describe the figure AXBY.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 53
(i) Draw a line segment AB = 8 cm.
(ii) Draw two parallel lines l and m to AB at a distance of 4 cm.
(iii) Draw the perpendicular bisector of AB which intersects the parallel lines l and m at X and Y respectively then, X and Y are the required points.
(iv) Join AX, AY, BX and BY.
The figure AXBY is a square as its diagonals are equal and intersect at 90°.

Question 19.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respec¬tively. Draw and describe the locus of a point which is :
(i) equidistant from BA and BC.
(ii) 4 cm from M.
(iii) 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 54
i) Draw an angle of 60° with AB = BC = 8 cm
ii) Draw the angle bisector BX of ∠ABC
iii) With centre M and N, draw circles of radius equal to 4 cm, which intersects each other at P. P is the required point.
iv) Join MP, NP
BMPN is a rhombus since MP = BM = NB = NP = 4 cm

Question 20.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
(i) the locus of the centers of all circles which touch AB and AC.
(ii) the locus of the centers of all circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 55
Steps of Construction:
i) Draw a line segment BC = 4.5 cm
ii) With B as centre and radius 6 cm and C as centre and radius 5 cm, draw arcs which intersect each other at A.
iii) Join AB and AC.
ABC is the required triangle.
iv) Draw the angle bisector of ∠BAC
v) Draw lines parallel to AB and AC at a distance of 2 cm, which intersect each other and AD at O.
vi) With centre O and radius 2 cm, draw a circle which touches AB and AC.

Question 21.
Construct a triangle ABC, having given AB = 4.8 cm. AC = 4 cm and ∠ A = 75°. Find a point P.
(i) inside the triangle ABC.
(ii) outside the triangle ABC.
equidistant from B and C; and at a distance of 1.2 cm from BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 56
Steps of Construction:
i) Draw a line segment AB = 4.8 cm
ii) At A, draw a ray AX making an angle of 75°
iii) Cut off AC = 4 cm from AX
iv) Join BC.
ABC is the required triangle.
v) Draw two lines l and m parallel to BC at a distance of 1.2 cm
vi) Draw the perpendicular bisector of BC which intersects l and m at P and P’
P and P’ are the required points which are inside and outside the given triangle ABC.

Question PQ.
O is a fixed point. Point P moves along a fixed line AB. Q is a point on OP produced such that OP = PQ. Prove that the locus of point Q is a line parallel to AB.
Solution:
P moves along AB, and Qmoves in such a way that PQ is always equal to OP.
But Pis the mid-point of OQ
Now in ∆OQQ’
P’and P” are the mid-points of OQ’ and OQ”
Therefore, AB||Q’Q”
Therefore, Locus of Q is a line CD which is parallel to AB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 57

 

Question 22.
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 58
Steps of Construction:
i) Draw a ray BC.
ii) At B, draw a ray BA making an angle of 75° with BC.
iii) Draw a line l parallel to AB at a distance of 2 cm
iv) Draw another line m parallel to BC at a distance of 1.5 cm which intersects line l at P.
P is the required point.

Question 23.
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 59
Steps of Construction:
i) Draw a line segment AB = 5.6 cm
ii) From A and B, as centers and radius 9.2 cm, make two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw two lines n and m parallel to BC at a distance of 2 cm
v) Draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively.
P and Q are the required points which are equidistant from AB and AC.
On measuring the distance between P and Q is 4.3 cm.

Question 24.
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 60
Steps of Construction:
i) Draw a line segment AB = 6 cm
ii) With A and B as centers and radius 9 cm, draw two arcs which intersect each other at C.
iii) Join AC and BC.
iv) Draw the perpendicular bisector of BC.
v) With A as centre and radius 4 cm, draw an arc which intersects the perpendicular bisector of BC at P.
P is the required point which is equidistant from B and C and at a distance of 4 cm from A.

Question 25.
Ruler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
(ii) Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
(iv) Mark the point Q, in your construction, which would make A QBC equal in area to A ABC, and isosceles.
(v) Measure and record the length of CQ.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 61
Steps of Construction:
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle 60 degree and cut off BA=9 cm.
(iii) Join AC. ABC is the required triangle.
(iv) Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from B and C.
(v) Through A, draw a line m || BC.
(vi) The perpendicular bisector of BC and the parallel line m intersect each other at Q.
(vii) Then triangle QBC is equal in area to triangle ABC. m is the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.
On measuring CQ = 8.4 cm.

Question 26.
State the locus of a point in a rhombus ABCD, which is equidistant
(i) from AB and AD;
(ii) from the vertices A and C.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 62

Question 27.
Use a graph paper for this question. Take 2 cm = 1 unit on both the axes.
(i) Plot the points A(1,1), B(5,3) and C(2,7).
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 63+
Steps of Construction:
i) Plot the points A(1, 1), B(5, 3) and C(2, 7) on the graph and join AB, BC and CA.
ii) Draw the perpendicular bisector of AB and angle bisector of angle A which intersect each other at P.
P is the required point.
Since P lies on the perpendicular bisector of AB.
Therefore, P is equidistant from A and B.
Again,
Since P lies on the angle bisector of angle A.
Therefore, P is equidistant from AB and AC.
On measuring, the length of PA = 5.2 cm

Question 28.
Construct an isosceles triangle ABC such that AB = 6 cm, BC=AC=4cm. Bisect angle C internally and mark a point P on this bisector such that CP = 5cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 64
Steps of Construction:
i) Draw a line segment AB = 6 cm.
ii) With centers A and B and radius 4 cm, draw two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw the angle bisector of angle C and cut off CP = 5 cm.
v) A line m is drawn parallel to AB at a distance of 5 cm.
vi) P as centre and radius 5 cm, draw arcs which intersect the line m at Q and R.
vii) Join PQ, PR and AQ.
Q and R are the required points.

Question PQ.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 65

Steps of Construction:
i) Draw a circle with radius = 4 cm.
ii) Take a point A on it.
iii) A as centre and radius 6 cm, draw an arc which intersects the circle at B.
iv) Again A as centre and radius 5 cm, draw an arc which intersects the circle at C.
v) Join AB and AC.
vi) Draw the perpendicular bisector of AC, which intersects AC at Mand meets the circle at E and F.
EF is the locus of points inside the circle which are equidistant from A and C.
vii) Join AE, AF, CE and CF.
Proof:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 66
Similarly, we can prove that CF = AF
Hence EF is the locus of points which are equidistant from A and C.
ii) Draw the bisector of angle A which meets the circle at N.
Therefore. Locus of points inside the circle which are equidistant from AB and AC is the perpendicular bisector of angle A.

Question 29.
Plot the points A(2,9), B(-1,3) and C(6,3) on a graph paper. On the same graph paper, draw the locus of point A so that the area of triangle ABC remains the same as A moves.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 67
Steps of construction:
i) Plot the given points on graph paper.
ii) Join AB, BC and AC.
iii) Draw a line parallel to BC at A and mark it as CD.
CD is the required locus of point A where area of triangle ABC remains same on moving point A.

Question 30.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
(i) Complete the rectangle ABCD such that:
(a) P is equidistant from A B and BC.
(b) P is equidistant from C and D.
(ii) Measure and record the length of AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 68
i) Steps of Construction:
1) Draw a line segment BC = 5 cm
2) B as centre and radius 4 cm draw an arc at an angle of 45 degrees from BC.
3) Join PC.
4) B and C as centers, draw two perpendiculars to BC.
5) P as centre and radius PC, cut an arc on the perpendicular on C at D.
6) D as centre, draw a line parallel to BC which intersects the perpendicular on B at A.
ABCD is the required rectangle such that P is equidistant from AB and BC (since BD is angle bisector of angle B) as well as C and D.
ii) On measuring AB = 5.7 cm

Question 31.
Use ruler and compasses only for the following questions. All constructions lines and arcs must be clearly shown.
(i) Construct a ∆ABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are at distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
Solution:
i. Steps of construction:

  1. Draw BC = 6.5 cm using a ruler.
  2. With B as center and radius equal to approximately half of BC, draw an arc that cuts the segment BC at Q.
  3. With Q as center and same radius, cut the previous arc at P.
  4. Join BP and extend it.
  5. With B as center and radius 5 cm, draw an arc that cuts the arm PB to obtain point A.
  6. Join AC to obtain ΔABC.

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 69

ii. With A as center and radius 3.5 cm, draw a circle.
The circumference of a circle is the required locus.

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 70
iii. Draw CH, which is bisector of Δ ACB. CH is the required locus.

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 71
iv. Circle with center A and line CH meet at points X and Y as shown in the figure. xy = 8.2 cm (approximately)

More Resources for Selina Concise Class 10 ICSE Solutions

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Selina Concise Mathematics Class 10 ICSE Solutions Circles

Selina Concise Mathematics Class 10 ICSE Solutions Circles

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles

Circles Exercise 17A – Selina Concise Mathematics Class 10 ICSE Solutions

Circles Class 10 Question 1.
In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC. Show your steps of working.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 2

Question 2.
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
(i) Prove that AC is a diameter of the circle.
(ii) Find ∠ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 4

Question 3.
Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of:
(i) ∠ OCA,
(ii) ∠OAC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 6

Question 4.
In each of the following figures, O is the centre of the circle. Find the values of a, b, and c.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 7
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 8

Question 5.
In each of the following figures, O is the centre of the circle. Find the value of a, b, c and d.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 9
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 10
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 11

Question 6.
In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O1 and O2 are the centres of two circles.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 12
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 13

Question 7.
In the figure given beow, find :
(i) ∠ BCD,
(ii) ∠ ADC,
(iii) ∠ ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 14
Show steps of your workng.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 15

Question 8.
In the given figure, O is centre of the circle. If ∠ AOB = 140° and ∠ OAC = 50°; find :
(i) ∠ ACB,
(ii) ∠ OBC,
(iii) ∠ OAB,
(iv) ∠CBA
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 16
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 17

Question 9.
Calculate :
(i) ∠ CDB,
(ii) ∠ ABC,
(iii) ∠ ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 18
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 19

Question 10.
In the figure given below, ABCD is a eyclic quadrilateral in which ∠ BAD = 75°; ∠ ABD = 58° and ∠ADC = 77°. Find:
(i) ∠ BDC,
(ii) ∠ BCD,
(iii) ∠ BCA.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 20
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 21

Question 11.
In the following figure, O is centre of the circle and ∆ ABC is equilateral. Find :
(i) ∠ ADB
(ii) ∠ AEB
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 22
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 23

Question 12.
Given—∠ CAB = 75° and ∠ CBA = 50°. Find the value of ∠ DAB + ∠ ABD
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 24
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 25

Question 13.
ABCD is a cyclic quadrilateral in a circle with centre O.
If ∠ ADC = 130°; find ∠ BAC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 26
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 27

Question 14.
In the figure given below, AOB is a diameter of the circle and ∠ AOC = 110°. Find ∠ BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 28
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 29

Question 15.
In the following figure, O is centre of the circle,
∠ AOB = 60° and ∠ BDC = 100°.
Find ∠ OBC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 30

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 31

Question 16.
ABCD is a cyclic quadrilateral in which ∠ DAC = 27°; ∠ DBA = 50° and ∠ ADB = 33°.
Calculate :
(i) ∠ DBC,
(ii) ∠ DCB,
(iii) ∠ CAB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 32
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 33

Question 17.
In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°. Find the number of degrees in:
(i) ∠DCE;
(ii) ∠ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 34
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 35

Question 17 (old).
In the figure given below, AB is diameter of the circle whose centre is O. Given that:
∠ ECD = ∠ EDC = 32°.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 36
Show that ∠ COF = ∠ CEF.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 37

Question 18.
In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 38
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 39

Question 19.
In the following figure,
(i) if ∠BAD = 96°, find BCD and
(ii) Prove that AD is parallel to FE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 40
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 41

Question 20.
Prove that:
(i) the parallelogram, inscribed in a circle, is a rectangle.
(ii) the rhombus, inscribed in a circle, is a square.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 42
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 43

Question 21.
In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 44
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 45

Question 22.
Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 46

Question 23.
The figure given below, shows a circle with centre O. Given: ∠ AOC = a and ∠ ABC = b.
(i) Find the relationship between a and b
(ii) Find the measure of angle OAB, if OABC is a parallelogram.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 47
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 48

Question 24.
Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD as the centre O is equal to twice the angle APC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 49

Question 24 (old).
ABCD is a quadrilateral inscribed in a circle having ∠A = 60°; O is the centre of the circle. Show that: ∠OBD + ∠ODB = ∠CBD + ∠CDB
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 50

Question 25.
In the figure given RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°
Calculate:
(i) ∠RNM;
(ii) ∠NRM.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 51
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 52

Question 26.
In the figure given alongside, AB || CD and O is the centre of the circle. If ∠ ADC = 25°; find the angle AEB. Give reasons in support of your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 53
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 54

Question 27.
Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at Cand D. Prove that AC is parallel to BD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 55
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 56

Question 28.
ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 57

Question 29.
AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find:
(i) ∠PRB
(ii) ∠PBR
(iii) ∠BPR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 58
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 59

Question 30.
In the given figure, SP is the bisector of angle RPT and PQRS is a cyclic quadrilateral. Prove that: SQ = SR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 60
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 61

Question 31.
In the figure, O is the centre of the circle, ∠AOE = 150°, DAO = 51°. Calculate the sizes of the angles CEB and OCE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 62
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 63

Question 32.
In the figure, P and Q are the centres of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 64
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 65

Question 33.
The figure shows two circles which intersect at A and B. The centre of the smaller circle is O and lies on the circumference of the larger circle. Given that ∠APB = a°. Calculate, in terms of a°, the value of:
(i) obtuse ∠AOB
(ii) ∠ACB
(iii) ∠ADB.
Give reasons for your answers clearly.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 66
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 67

Question 34.
In the given figure, O is the centre of the circle and ∠ ABC = 55°. Calculate the values of x and y.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 68
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 69

Question 35.
In the given figure, A is the centre of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that ∠BCD = 2∠ABE
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 70
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 71

Question 36.
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°; calculate:
(i) ∠ DAB,
(ii) ∠BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 72
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 73

Question 37.
∠ In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠ EAB = 63°; calculate:
(i) ∠EBA,
(ii) BCD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 74
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 75

Question 38.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°; calculate:
(i) ∠ DAB,
(ii) ∠ DBA,
(iii) ∠ DBC,
(iv) ∠ ADC.
Also, show that the ∆AOD is an equilateral triangle.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 76
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 77

Question 39.
In the given figure, I is the incentre of the ∆ ABC. Bl when produced meets the circumcirle of ∆ ABC at D. Given ∠BAC = 55° and ∠ ACB = 65°, calculate:
(i) ∠DCA,
(ii) ∠ DAC,
(iii) ∠DCI,
(iv) ∠AIC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 78
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 79

Question 40.
A triangle ABC is inscribed in a circle. The bisectors of angles BAC, ABC and ACB meet the circumcircle of the triangle at points P, Q and R respectively. Prove that:
(i) ∠ABC = 2 ∠APQ
(ii) ∠ACB = 2 ∠APR
(iii) ∠QPR = 90° – \(\frac{1}{2}\)BAC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 80
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 81

Question 40 (old).
The sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E; the sides DA and CB are produced to meet at F. If ∠BEC = 42° and ∠BAD = 98°; calculate:
(i) ∠AFB,
(ii) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 82
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 83

Question 41.
Calculate the angles x, y and z if: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 84
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 85

Question 42.
In the given figure, AB = AC = CD and ∠ADC = 38°. Calculate:
(i) Angle ABC
(ii) Angle BEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 86
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 87

Question 43.
In the given figure, AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, and r in terms of x.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 88
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 89

Question 44.
In the given figure, AC is the diameter of circle, centre O. CD and BE are parallel. Angle AOB = 80° and angle ACE = 10°. Calculate:
(i) Angle BEC;
(ii) Angle BCD;
(iii) Angle CED.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 90
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 91

Question 45.
In the given figure, AE is the diameter of circle. Write down the numerical value of ∠ABC + ∠CDE. Give reasons for your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 92
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 93

Question 46.
In the given figure, AOC is a diameter and AC is parallel to ED. If ∠CBE = 64°, calculate ∠DEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 94
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 95

Question 47.
Use the given figure to find
(i) ∠BAD
(ii) ∠DQB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 96
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 97

Question 48.
In the given figure, AOB is a diameter and DC is parallel to AB. If ∠ CAB = x°; find (in terms of x) the values of:
(i) ∠COB
(ii) ∠DOC
(iii) ∠DAC
(iv) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 98
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 99

Question 49.
In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130°. Find:
(i) ∠DAB
(ii) ∠DBA
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 100
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 101

Question 50.
In the given figure, PQ is the diameter of the circle whose centre is O. Given ∠ROS = 42°; calculate ∠RTS.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 102
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 103

Question 51.
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°; calculate
(i) ∠RPQ
(ii) ∠STP.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 104
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 105

Question 52.
AOD = 60°; calculate the numerical values of:
AB is the diameter of the circle with centre O. OD is parallel to BC and ∠AOD = 60°; calculate the numerical values of:
(i) ∠ABD,
(ii) ∠DBC,
(iii) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 106
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 107

Question 53.
In the given figure, the centre of the small circle lies on the circumference of the bigger circle. If ∠APB = 75° and ∠BCD = 40″; find:
(i) ∠AOB,
(ii) ∠ACB,
(iii) ∠ABD,
(iv) ∠ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 108
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 109

Question 54.
In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°; find:
(i) ∠BCD,
(ii) ∠ACB.
Hence, show that AC is a diameter.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 110
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 111

Question 55.
In a cyclic quadrilateral ABCD, ∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5; find each angle of the quadrilateral.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 112

Question 56.
The given figure shows a circle with centre O and ∠ABP = 42°. Calculate the measure of
(i) ∠PQB
(ii) ∠QPB + ∠PBQ
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 195
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 196

Question 57.
In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other. If ∠ MAD =x and ∠BAC = y.
(i) express ∠AMD in terms of x.
(ii) express ∠ABD in terms of y.
(iii) prove that : x = y
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 197
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 198

Question 61 (old).
In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that ∠APB = 60°.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 199

Circles Exercise 17B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.
Prove it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 200

Question 2.
In the following figure, AD is the diameter of the circle with centre 0. chords AB, BC and CD are equal. If ∠DEF = 110°, calculate:
(i) ∠ AFE,
(ii) ∠FAB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 201

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 202

Question 3.
If two sides of a cycli-quadrilateral are parallel; prove thet:
(i) its other two side are equal.
(ii) its diagonals are equal.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 203

Question 4.
The given figure show a circle with centre O. also, PQ = QR = RS and ∠PTS = 75°. Calculate:
(i) ∠POS,
(ii) ∠ QOR,
(iii) ∠PQR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 204
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 205

Question 5.
In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of:
(i) ∠ AOB,
(ii) ∠ ACB,
(iii) ∠ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 206
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 207

Question 6.
In a regular pentagon ABCDE, inscribed in a circle; find ratio between angle EDA and angel ADC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 208

Question 7.
In the given figure. AB = BC = CD and ∠ABC = 132°, calculate:
(i) ∠AEB,
(ii) ∠ AED,
(iii) ∠COD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 209
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 210

Question 8.
In the figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If angle AOB = 108°, find:
(i) ∠ CAB,
(ii) ∠ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 211
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 212

Question 9.
The figure shows a circle with centre O. AB is the side of regular pentagon and AC is the side of regular hexagon. Find the angles of triangle ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 213
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 214

Question 10.
In the given figire, BD is a side of a regularhexagon, DC is a side of a regular pentagon and AD is adiameter. Calculate:
(i) ∠ ADC
(ii) ∠BAD,
(iii) ∠ABC
(iv) ∠ AEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 215
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 216
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 217

Circles Exercise 17C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In the given circle with diametre AB, find the value of x.

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 218
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 219
∠ABD = ∠ACD = 30° (Angle in the same segment)
Now in ∆ ADB,
∠BAD + ∠ADB + ∠DBA = 180° (Angles of a A)
But ∠ADB = 90° (Angle in a semi-circle)
∴ x + 90° + 30° = 180° ⇒ x + 120° = 180°
∴ x= 180° – 120° = 60° Ans.

Question 1.
In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 220
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 221

Question 2.
In the given figure, ABC is a triangle in which ∠ BAC = 30° Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose centre is O.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 222
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 223

Question 3.
Prove that the circle drawn on any one a the equalside of an isoscele triangle as diameter bisects the base.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 224

Question 3 (old).
The given figure show two circles with centres A and B; and radii 5 cm and 3cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 225
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 226

Question 4.
In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠ CBE = 65°, calculate ∠DEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 227
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 228

Question 5.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 229

Question 6.
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 230
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 231

Question 7.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Provet that the points B, C, E and D are concyclic.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 232

Question 7 (old).
Chords AB and CD of a circle intersect each other at point P such that AP = CP.
Show that: AB = CD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 233
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 234

Question 8.
In the given rigure, ABCD is a cyclic eqadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ ADC = 92°, ∠ FAE = 20°; determine ∠ BCD. Given reason in support of your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 235
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 236

Question 9.
If I is the incentre of triangle ABC and Al when produced meets the cicrumcircle of triangle ABC in points D. if ∠ BAC = 66° and ∠ = 80o.calculate:
(i) ∠ DBC
(ii) ∠ IBC
(iii) ∠ BIC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 237
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 238
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 239

Question 10.
In the given figure, AB = AD = DC = PB and ∠ DBC = x°. Determine, in terms of x:
(i) ∠ ABD,
(ii) ∠ APB.
Hence or otherwise, prove thet AP is parallel to DB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 240
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 241

Question 11.
In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠ CQE are supplementary.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 242
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 243

Question 12.
In the given, AB is the diameter of the circle with centre O.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 244
If ∠ ADC = 32°, find angle BOC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 245

Question 13.
In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ prouduced meet at point A: whereas sides PQ and SR produced meet at point B.
If ∠A: ∠B = 2 : 1;find angles A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 246
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 247

Question 17 (old).
If the following figure, AB is the diameter of a circle with centre O and CD is the chord with lengh equal radius OA.
If AC produced and BD produed meet at point p; show that ∠APB = 60°
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 248
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 249
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 250

Question 14.
In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 251
If the bisector of angle A meet BC at point E and the given circle at point F, prove that:
(i) EF = FC
(ii) BF =DF
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 252

Question 15.
ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point e; whereas sides BC and AD produced meet at point F. I f ∠ DCF : ∠F : ∠E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 253
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 254

Question 16.
The following figure shows a cicrcle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, Find the perimetre of the cyclic quadrilateral PORS.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 256

Question 17.
In the following figure, AB is the diameter of a circle with centre O. If chord AC = chord AD.prove that:
(i) arc BC = arc DB
(ii) AB is bisector of ∠ CAD.
Further if the lenghof arc AC is twice the lengthof arc BC find :
(a) ∠ BAC
(b) ∠ ABC
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 258
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 259

Question 18.
In cyclic quadrilateral ABCD; AD = BC, ∠ = 30° and ∠ = 70°; find;
(i) ∠ BCD
(ii) ∠BCA
(iii) ∠ABC
(iv) ∠ ADC
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 260
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 261

Question 19.
In the given figure, ∠ACE = 43° and ∠ = 62°; find the values of a, b and c.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 262
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 263

Question 20.
In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠ BAC = 25°
Find
(i) ∠ CAD
(ii) ∠ CBD
(iii) ∠ ADC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 264
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 265

Question 21.
ABCD is a cyclic quadrilateral of a circle with centre o such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle..if AD and BC produced meet at P, show that APB =60°
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 266

Question 22.
In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 267
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 268

Question 23.
In the figure, given below, AD = BC, ∠ BAC = 30° and ∠ = 70° find:
(i) ∠ BCD
(ii) ∠ BCA
(iii) ∠ ABC
(iv) ∠ADC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 269
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 270

Question 24.
In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED (2016)
Solution:
i. AD is parallel to BC, i.e., OD is parallel to BC and BD is transversal.
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Selina Concise Mathematics Class 10 ICSE Solutions Circles - 272

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 273

Question 25.
In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of
i. ∠BCD
ii. ∠BOD
iii. ∠OBD
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 274
Solution:
∠DAE and ∠DAB are linear pair
So,
∠DAE + ∠DAB = 180°
∴∠DAB = 110°

Also,
∠BCD + ∠DAB = 180°……Opp. Angles of cyclic quadrilateral BADC
∴∠BCD = 70°
∠BCD = \(\frac { 1 }{ 2 }\) ∠BOD…angles subtended by an arc on the centre and on the circle
∴∠BOD = 140°

In ΔBOD,
OB = OD……radii of same circle
So,
∠OBD =∠ODB……isosceles triangle theorem
∠OBD + ∠ODB + ∠BOD = 180°……sum of angles of triangle
2∠OBD = 40°
∠OBD = 20°

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