## ICSE Class 9 Maths Sample Question Paper 10 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Evaluate without using tables :
sin 38° sin 52° – cos 38° cos 52°.
sin 38° sin 52° – cos 38° cos 52° = sin 38° sin (90° – 38°) – cos 38° cos (90° – 38°)
= sin 38° cos 38° – cos 38° sin 38°
= 0.

(b) In the figure, CD is a diameter which meets the chord AB at E such that AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle

Given : CD is a diameter, AE = BE = 4 cm, CE = 3 cm.
∠OEB = 90° (∵ E is the mid-point of BA).
Let the radius OB = OC = r cm.
Then, OE = OC – CE = (r – 3) cm.

In ΔOBE, by Pythagoras theorem,
OB2 = OE2 + BE2
⇒ r2 = (r – 3)2 + 42
⇒ r2 = r2 – 2.r.3 + 9 + 16
⇒ 6r = 25
⇒ $$r=\frac{25}{6}=4 \frac{1}{6}$$
The required radius = $$4 \frac{1}{6}$$ cm.

(c) If a = 1 – √3, find the value of $$\left(a-\frac{1}{a}\right)^{3}$$

Question 2.
If $$a=\frac{1}{a-5}$$,Find
(i) $$a-\frac{1}{a}$$
(ii) $$a+\frac{1}{a}$$
(iii) $$a^{2}-\frac{1}{a^{2}}$$

(b) The water bills (in ₹) of 32 houses in a locality are given below.
80, 48, 52, 78, 103, 85, 37, 94, 72, 73, 66, 52, 92, 85, 78, 81, 64, 60, 75, 78, 108, 63, 71, 54, 59, 75, 100, 103, 35, 89, 95, 73.
Construct a frequency distribution table with a class size of 10.
Here, minimum value = 35
Maximum value = 108
Class size =10.

(c) Factorize : (a2 – b2) (c2 – d2) – 4 abcd.

Question 3.
(a) Show that:

(b) In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that:
(i) ΔACE ≅ ΔDBE
(ii) AE = DF.

Given: AB = CD,CE = BF and ∠ACE= ∠DBF.
AB = CD
AB + BC = BC + CD (Adding BC on both sides)
⇒ AC = BD.
(i) Now, in ΔACE and ΔDBF
AC = BD (Proved above)
∠ACE = ∠DBF (Given)
CE = BF (Given)
ΔACE ≅ ΔDBF (SAS axiom)
Hence Proved.

(ii) AE = DF (c.p.c.t.)
Hence Proved.

(c) If log10y = x, find the value of 102x in terms of y.
Given:
log10 y = x
⇒ y = 10x
Or 10x = y
∴ 102x = (10x)2 = y2

Question 4.
(a) Given, 5 cos A -12 sin A = 0, find the value of $$\frac{\sin A+\cos A}{2 \cos A-\sin A}$$

(b) Solve by the substitution method :
2x – $$\frac{3}{4}$$y – 3 = 0; 5x – 2y – 7 = 0

(c) What sum of money will amount to ₹ 3630 in 2 years at 10% p. a. compound interest ?

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) If x= $$7-4 \sqrt{3}$$,find the value of $$\sqrt{x}+\frac{1}{\sqrt{x}}$$

(b) In the given figure, X and Y are mid-points of sides AB and AC respectively of ΔABC. If
= 6 cm, AB = 7.4 cm and AC = 6.4 cm, then find the perimeter of the trapezium XBCY.

Given : BC = 6 cm, AB = 7.4 cm and AC = 6.4 cm X,
∴ Y are mid-points of AB and AC, respectively
∵ By mid-points theory.

(c) The mean weight of 8 students is 46.5 kg. Three more students having weight 41.7 kg, 52.8 kg and 51.8 kg joined the group. What is the new mean weight of the students ?
Given : No. of students = 8, mean weight = 46.5 kg.
∴ Total weight of 8 students = 8 x 46.5 kg = 372 kg
∵ 3 more students joined the group,
∴ Total weight of 11 students = 372 + 41.7 + 52.8 + 51.8 = 518.3 kg.
∴ New mean of 11 students = $$\frac{518.3}{11}$$ = 47.12 kg.

Question 6.
(a) Factorize : (x2 – 4x) {x2 – 4x – 1} – 20.
(x2 – 4x) (x2 – 4x – 1) – 20
Let x2 – 4x = a
⇒ a (a – 1) – 20 = a2 – a – 20
⇒ a2 – (5 – 4) a – 20
⇒ a2 -5a + 4a -20
⇒ a (a – 5) + 4 (a – 5)
⇒ (a – 5) (a + 4)
Putting the value of a = x2 – 4x, we get
(x2 – 4x – 5) (x2 – 4x + 4)
= (x2 – 5x + x -5) (x2 – 1. x .2 +22)
= {x {x – 5) + 1 (x – 5)} (x – 2)2
= (x -5) (x + 1) (x – 2)2.

(b) The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them saves ₹ 200 per month, find their monthly incomes.
The ratio of incomes = 9 : 7 and the ratio of expenditures = 4 : 3
Let their incomes be 9x and 7x and their expenditures be 4y and 3y.
:. Their savings are 9x – 4y and 7x – 3y respectively.
According to the question,
9x – = 200 ———- (i)
7x – = 200 ———- (ii)
Multiplying equation (i) by 3 and equation (ii) by 4, we get

Their incomes are ₹ 1800 and ₹ 1400.

(c) The centre of a circle is C (2α -1,3α +1) and it passes through the point A (- 3, – 1). If the diameter of the circle is of length 20 units, find the value (s) of a.
Given points are C(2α – 1, 3α+1) and A(-3,-1)
Diameter 20 units.

Question 7.
(a) Insert 3 rational numbers between $$\frac{1}{4}$$ and $$\frac{1}{2}$$.

(b) In the figure, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm and
∠ABD = ∠BCD = 90°. Calculate the length of AB.

Given : AD = 13 cm, DC = 12 cm, BC = 3 cm and ∠ABD = ∠BCD = 90°. (Pythagoras theorem)
In ΔBCD, BD2 = BC2 + CD2
= 32 + 122 = 9 + 144 = 153
In ΔBCD, AD2 = BD2 + AB2
132= 153 + AB2
169 = 153 + AB2
AB2 = 169 – 153 = 16
AB = √16 = 4 cm.

(c) Prove that $$: \frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}=1$$

Question 8.
(a) A sum amounts to ₹ 9680 in 2 years and to ₹ 10648 in 3 years compounded annually. Find the sum and the rate of interest per annum.

(b) A triangle is formed by the lines x + 2y – 3 = 0, 3x – 2y + 7 = 0 and y + 1=0. Find graphically :
(i) The coordinates of vertices of the triangle.

Let the lines intersect at the points A, B and C.
(i) The coordinates of vertices of triangle are A (- 1, 2), B (- 3, – 1) and C (5, – 1).
(ii) From A, draw AM ⊥ BC.
Area of Δ ABC
= $$\frac{1}{2}$$ × BC × AM
= $$\frac{1}{2}$$ × 8 × 3
= 12 sq. units.

(c) Solve : $$\frac{8}{x+3}-\frac{3}{2-x}=2$$

Question 9.
(a) In the figure, ABCD is a parallelogram and P is any point on BC. Prove that area of ΔABP + area of ΔDPC = area of ΔAPD.

Given : ABCD is a parallelogram
We know, triangles on same base and between same parallels are equal.
∴ Area of AAPD = Area of AABD ……. (i)
and Area of AABP = Area of ABDP ……. (ii)
Also, since diagonal of a parallelogram divides it into two triangles of equal area
∴ Area of ΔBCD = Area of ΔABD …… (iii)
⇒ Area of ΔBDP + Area of ΔDPC = Area of ΔABD
⇒ Area of ΔABP + Area of ΔDPC = Area of ΔAPD [Using (i) and (ii)]
Hence Proved.

(b) Factorize : 1+a+b+c+ab+bc+ca+abc.
1+a+b+c+ab+bc+ca+abc=(1+a)+(b+ab)+(c+ca)+(bc+abc)
=(1+a)+b(1+a)+c(1+a)+bc(1+a)
=(1+a)(1+b+c+bc)
=(1+a)((1+b)+c(1+b)
= (1+ a) (1+ b) (1+ c).

(c) The area of cross-section of a pipe is 5.4 cm2 and water is pumped out of it at the rate of 27 km/h. Find in litres the volume of water which flows out of the pipe in one minute.

Question 10.
(a) If 4 cos2 x° – 1 = 0 and 0 ≤ x°≤ 90°,find the value of cos2 x° sin2 x°.

(b) Draw the parallelogram ABCD in which AB = BC = 4.8 cm and AC = 7.5 cm. Find the angle between the two diagonals. What special name can you give it to this parallelogram?
Given : AB = BC = 4.8 cm, AC = 7.5cm

Steps of construction:
(1) Draw BC = 4.8 cm.
(2) From B and C, draw arcs of length 4.8 cm and 7.5 cm, respectively which intersect at A.
(3) From A and C, draw arcs each of length 4.8 cm which intersect at D.
(4) Join BA, AD and CD to complete the required parallelogram.
(5) Join AC and BD which intersect at O.
∴ ∠AOD=90°
The special name of this parallelogram is rhombus as all sides are equal.

(c) Elavuate $$\frac{5^{10+n} \times 25^{3 n-4}}{5^{7 n}}$$

Question 11.
(a) Prove that: $$x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}=1$$

(b) In the figure, ABCD is a rhombus in which the diagonal DB is produced to E. If ∠ABE = 160°, then find x, y and z.

(c) If a + b – c = 4 and a2 + b2 + c2 = 38, find the value of ab – bc – ca.

## ICSE Class 9 Maths Sample Question Paper 9 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Find xy, if x + y = 6 and x – y = 4.
Given: x + y =6,x – y=4.
Now, 4xy =(x+y)2-(x-y)2
= 62 – 42
=20
xy=$$\frac{20}{4}$$=5.

(b) Find the mean of first 10 prime numbers.
First lo prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
∴ ∑ =129,n=10
Mean = $$\frac{\sum x}{n}=\frac{129}{10}=12.9$$

(c) If x = acosθ + bsinθ and y = asinθ – bcosθ, prove that x2 + y2 = a2 + b2.

Question 2.
(a) Simplify : $$\left(\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}\right)\left(\frac{1}{2+\sqrt{3}}+\frac{1}{2-\sqrt{3}}\right)$$

(b) The area of a trapezium is 540 cm2. If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the length of parallel sides.
Given : Area of trapezium = 510 cm2
Perpendicular distance between parallel lines = 18 cm
Ratio of parallel sides = 7 : 5
Let the length of parallel sides be 7x and 5x.
Area of trapezium = $$\frac{1}{2}$$ (Sum of parallel sides) x Perpendicular distance

7x = 7 x 5 = 35
and 5x = 5 x 5 = 25
Hence, the length of parallel sides are 35 cm and 25 cm.

(c) Solve : (3x + 1) (2x + 3) = 3.
(3x + 1) (2x + 3) = 3
6 x 2 + 9x + 2x + 3 – 3 = O
6 x 2 +11x = 0
x (6x + 11) = 0
x=0 or 6x+11=0
x = 0 or x=$$\frac{-11}{6}$$
x = 0 or $$\frac{-11}{6}$$

Question 3.
(a) Express $$\log _{10}\left(\frac{a^{3} c^{2}}{\sqrt{b}}\right)$$
A ladder 13 cm long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the in terms of log10 a, log10 b and log10 c.

(b) A ladder 13 cm long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.
Let AC be the ladder and BC be the height of the wall.
Then, AC = 13m, AB= 15m, ∠B=90°
∴ By Pythagoras theorem,

(c) Factorize : a4 + b4 – 11a2 b2.

Question 4.
(a) Simplify: $$\sqrt{\frac{1}{4}}+(0.04)^{-1 / 4}-(8)^{2 / 3}$$

(b) In the figure, ABCD is a trapezium in which DA || CB. AB has been produced to E. Find the angles of the trapezium.

Given: DA || CB
x + + 100 = 180° (Co-interior angles) … (i)
⇒ x+2y = 170°
Also, x + 25° = y (Corresponding angles) … (ii)
x – y= – 25°
Subtracting equation (ii) from equation (i), we get
3y = 195°
= $$y=\frac{195^{\circ}}{3}=65^{\circ}$$
Putting y = 65° in equation (ii), we get
X – 65° = – 25°
x = 65 – 25°=40°
∠A = x + 25° = 40°+ 25° = 65°
∠B = 180° – y= 180 – 650= 115°
∠C = 2y + 10° =2 x 65° + 10° = 1400
∠D = x = 40°

(c) Calculate the compound interest for the second year on ₹ 8000 when invested for 3 years at 10% p.a.

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) Solve for x : 25x – 1 = 52x – 1 – 100.

(b) In a ΔABC, ∠A = 80°, ∠B = 40° and bisectors of ∠B and ∠C meet at O. Find ∠BOC.

(c) Without using tables, evaluate :

Question 6.
(a) If x = 3 + 2√2 , find the value of $$x^{3}-\frac{1}{x^{3}}$$

(b) Construct a rectangle each of whose diagonals measures 6 cm and the diagonals intersect at an angle of 45°.

Given: Each diagonal = 6 cm, diagonals intersect at 45°.
Steps of construction:
(1) Draw diagonal AC = 6 cm.
(2) Bisect AC at O
(3) At O, draw ∠COX = 45° and extend XO to Y.
(4) From O, cut-off XY at D and B such that OD = OB = 3 cm,
i.e., BD = 6 cm.
(5) Join A, B, C, D to get the required rectangle ABCD.

(c) KM is a straight line of 13 units. If K has the coordinates (2, 5) and M has the coordinates (x, – 7), find the possible values of x.

Question 7.
(a) If $$x=\frac{\sqrt{7}+1}{\sqrt{7}-1} \text { and } y=\frac{\sqrt{7}-1}{\sqrt{7}+1}$$find the value of $$\frac{x^{2}+x y+y^{2}}{x^{2}-x y+y^{2}}$$

(b) In ΔABC, AC = 3 cm. If M is the mid-point of AB and a straight line through M parallel to AC cuts BC at N, what is the length of MN ?

Given : AC = 3 cm, M is the mid-point of AB, MN || AC.
Since, M the mid-point of AB and MN || AC
Therefore, by mid-point theorem, we have
MN = $$\frac{1}{2}$$AC= $$\frac{1}{2}$$ x 3cm = 1.5cm

(c) If θ is an acute angle and tan θ = $$\frac{5}{12}$$, find the value of cosθ + cot θ.

Question 8.
(a) Factorize : 1 – 2ab – (a2 + b2)

(b) In the given figure, AOC is a diameter of a circle with centre O and arc A×B = $$\frac{1}{2}$$ arc BYC. Find ∠BOC.

(c) In a class of 90 students, the marks obtained in a weekly test were as under.

Construct a combined histogram and frequency polygon.

Question 9.
(a) Divide ₹ 1,95,150 between A and B so that the amount that A receives in 2 years is the same as that of B receives in 4 years. The interest is compounded annually at the rate of 4% p.a.

A’s share = 1,01,400.
and B’s share = (1,95,150 – 1,01,400) = ₹ 93,750.

(b) Solve simultaneously : $$\frac{3}{x+y}+\frac{2}{x-y}=3 ; \frac{2}{x+y}+\frac{3}{x-y}=\frac{11}{3}$$

(c) If a = cz, b = ax and c = by, prove that xyz = 1.

Question 10.
(a) If sin (A + B) = 1 and cos (A – B) = $$\frac{\sqrt{3}}{2}$$, 0° < A + B ≤ 90°, A > B, then find A and B.
sin (A + B) = 1 = sin 90°
A + B = 90°
cos (A – B) =$$\frac{\sqrt{3}}{2}$$ = cos 30°
A – B = 30°
Adding (i) and (ii), we get
A + B + A – B = 90° + 30°
⇒ 2A = 120°
A = $$\frac{120^{\circ}}{2}$$ = 60°
Putting A= 60° in (i), we get
60° + B = 90°
⇒ B = 90° – 60° = 30°
∴ A = 60°, B = 30°

(b) The sum of the digits of a two-digit number is 5. The digit obtained by increasing the digit in ten’s place by unity is one-eighth of the number. Find the number.

Let the digit at ten’s place be  and that at unit’s place be y.
∴ The number = 10x + y
By 1st condition,
x + y = 5
By 2nd condition,
x + 1 = $$\frac{1}{8}$$ (10x + y)
⇒ 8 (x + 1) = 10x + y
⇒ 8x + 8 = 10x + y
⇒ 2x + y = 8
Subtracting (i) from (ii), we get
x = 3.
Putting x = 3 in (i), we get
3 + y = 5
⇒ y = 5 – 3 = 2
The required number = 10x + y = 10 x 3 + 2 = 32

(c) Given, log10 x = a and log10 y =
(i) Write down 10a-1 in terms of x
(ii) Write down 102b in terms of y.
(iii) If log10 P = 2a-b, express P in terms of x and y.

Question 11.
(a) Draw the graph of the equations 2x – 3y = 7 and x + 6y = 11 and find their solutions.

The two lines intersect each other at the point (5, 1).
∴ x = 5, y = 1

(b) In the figure, area of ΔABD = 24 sq. units. If AB = 8 units, find the height of ΔABC.
Given : Area of A ABD = 24 sq. units, AB = 8 units, DC || AB.
Area of ΔABC = Area of ΔABD
(∵ They are on same base and between same parallels)
= 24 sq. units.
⇒ $$\frac{1}{2}$$ x AB x Height of ΔABC = 24
⇒ $$\frac{1}{2}$$ x 8 x Height of ΔABC = 24
⇒ Height of Δ ABC = $$\frac{24}{4}$$ = 6 units.

(c) If x – y = 8 and xy = 20, evaluate : (i) x + y (ii) x2 – y2.
Given x – y – 8, xy = 20.
(i) (x + y)2 = (x – y)2 + 4 xy = 82 + 4 x 20 = 64 + 80 = 144
x + y = ± √144 = ± 12
x2 – y2 = (x + y) (x – y)
= (± 12) x 8
= ± 96.

## ICSE Class 9 Maths Sample Question Paper 8 with Answers

Section – A [40-Marks]
(Attempt all questions from this Section)

Question 1.
(a) The mean of 100 observations was found to be 30. If two observations were wrongly taken as 32 and 12 instead of 23 and 11, find the correct mean.
Here, n = 100, $$\bar{x}$$= 30
∴ Incorrect = Σx= $$\bar{x}$$ n = 30 x 100 = 3000.
∴ Correct Σ x = 3000 – (32 + 12) + (23 + 11)
= 3000 – 44 + 34 = 2990
∴ Correct mean = $$\frac{2990}{100}=29.9$$

(b) Determine the rate of interest for a sum that becomes $$\frac{216}{125}$$ times of itself in 3 years, compounded annually.
Let principal be ? P and rate of interest be r% p. a. So,

(c) Without using tables, find the value of :

Question 2.
(a) If $$x=\frac{3+\sqrt{7}}{2}$$ find the value of $$4 x^{2}+\frac{1}{x^{2}}$$

(b) In the given figure, ABCD is a rectangle with sides AB = 8 cm and AD = 5 cm. Compute : (i) area of parallelogram ABEF, (ii) area of ΔEFG.

Given : AB 8 cm, AD = 5 cm.
(i) Area of parallelogram ABEF = Area of rectangle ABCD
(∵ they are on same base and between same parallels)
= (8 x 5) cm2 = 40 cm2
Area of ΔEFG = $$\frac{1}{2}$$ x Area of parallelogram ABEF
(∵ both are on same base and between same parallels)
= $$\frac{1}{2}$$ x 40 cm2 20 cm2

(c) Without using tables, find the value of :
$$\frac{(b+c)^{2}}{b c}+\frac{(c+a)^{2}}{c a}+\frac{(a+b)^{2}}{a b}$$

Question 3.
(a) Solve for x : 2x +3 + 2x+1 = 320.

(b) In the given figure, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of lengths 24 cm and 18 cm respectively.

Given: radius = OA = OC = 15 cm, AB || CD.
Let AB = 24 cm, CD = 18 cm.
We know perpendicular drawn from centre to the chord, bisects the chord
∴ M and N are mid-point of sides AB and CD respectively.
AM= $$\frac{1}{2}$$ AB= $$\frac{1}{2}$$ x24=12cm.

(c) Factorize : 4 (2a – 3)2 – 3 (2a – 3) (a – 1) – 7 (a – 1)2.
Given expression is, 4 (2a – 3)2 – 3 (2a – 3) (a – 1) – 7 (a – 1)2
Let 2a – 3 = x and a – 1 = y
The expression becomes
= 4a2 – 3xy – 7y2 = 4x2 – (7 – 4) xy – 7y2
= 4x2 – 7xy + 4xy – 7y2
= x (4x – 7y) + y (4x – 7y)
= (4x – 7y) (x + y)
Substituting values of x and y, we have
= {4 (2a – 3) – 7 (a – 1)} {2a – 3 + a – 1)
= (8a -12 -7a + 7) (3a – 4) = (a – 5) (3a – 4).

Question 4.
(a) Solve for x : log (x + 5) + log (x – 5) = 4 log 2 + 2 log 3.
log {x + 5) + log {x – 5)
=4 log 2 + 2 log 3 log (x + 5) + log (x – 5)
= log 24 + log 32 log (x + 5) + log (x – 5)
= log 16 + log 9 log [(x + 5) (x – 5)]
= log (16 x 9) log(x2 – 25) – log 144
⇒ x2 – 25 = 144
⇒ x2 = 144 + 25 = 169
⇒ x = √169 = 13

(b) Solve simultaneously : $$2 x+\frac{x-y}{6}=2 ; x-\frac{(2 x+y)}{3}=1$$

(c) If 8 cot 915, find the value of: $$\frac{(2+2 \sin \theta)(1-\sin \theta)}{(1+\cos \theta)(2-2 \cos \theta)}$$

Question 5.
(a) The cost of enclosing a rectangular garden with a fence all around at the rate of ₹ 15 per metre is ₹ 5400. If the length of the garden is 100 m, find the area of the garden.
Total cost of fendng = ₹ 5400
Rate = ₹ 15 per metre
Perimeter = $$\frac{5400}{15}$$ = 360 m
Length, l= 100 m
2 (l + b) = 360
b = 180 $$\frac{360}{2}$$ 100 = 80 m
Area = i x b = 100 m x 80 m = 8000 m2

(b) If 4 sin2 x° – 3 = 0 and x° is an acute angle, find (i) sin x° (ii) x°.
Given: 4 sin2 x° – 3 = O
(i) 4 sin2 x° =3
sin2 x° = $$\frac{3}{4}$$
sin x°= $$\frac{\sqrt{3}}{2}$$

(ii) Now sin x°= $$\frac{\sqrt{3}}{2}$$
⇒ sin x°= sin 60°
⇒ x° = 60°

(c) Draw a frequency polygon from the following data :

 Age (in years) 25-30 30-35 35-40 40-45 45-50 No. of doctors 40 60 50 35 20

Question 6.
(a) If $$\frac{(x-\sqrt{24})(\sqrt{75}+\sqrt{50})}{\sqrt{75}-\sqrt{50}}=1$$ find the value of x.

(b) From the given figure, find the values of a and b.

∠DBC = ∠ADB = a° (Alternate angles)
Now, a + 28° = 75° (Exterior angle is equal to sum of interior opposite angles)
⇒ a = 75° – 28° = 47°.
Also, ∠ABC + ∠BAD = 180° (Co-interior angles)
⇒ a + b + 90° = 180°
⇒ 47° + b + 90° = 180°
⇒ b = 180° – 137° = 43°
a = 47°, b = 43°

(c) Show that the points A (2, – 2), B (8, 4), C (5, 7) and D (- 1, 1) are the vertices of a
rectangle. Also, find the area of the rectangle.

i.e., opposite sides are equal and diagonals are equal
ABCD is a rectangle.
Hence Proved.
Area of rectangle = AB x BC = 6√2 x 3√2 = 36 sq. units.

Question 7.
(a) By using suitable identity, evaluate : (9.8).
(9.8)1 = (10 – 0.2)3 = 103 – 3 x 102 x 0.2 + 3 x 10 x (0.2)2 – (0.2)3
= 1000 – 60 + 1.2 – 0.008 = 941.192.

(b) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB
and parallel to BC intersects AC at D. Show that (i) D is mid-point of AC (ii) MD L AC
(iii) CM = MA = $$\frac{1}{2}$$ AB.
Given : M is mid-point of AB, ∠C = 90°, MD||BC.
Join MC.
(i) ∵ M is mid-point of AB and MD|| BC

∴ By the converse of mid-point theorem,
D is mid-point of AC.
Hence Proved.

(ii) ∠BCD + ∠CDM = 1800 (Co-interior angles, MDIIBC)
⇒ 90° + ∠CDM = 180° (∠BCD = 90°)
∠CDM= 180°-90°=90°
MD ⊥ AC. Hence Proved.

(iii) In ΔAMD and ΔCMD,
AD = CD (D is mid-point of AC)
∠ADM = ∠CDM (Each being 90°)
MD = MD (Common side)
∴ ΔAMD ≅ ΔCMD. (SAS axiom)
∴ AM = CM (c.p.c.t.)
Also, AM = $$\frac{1}{2}$$ AB ( M is mid-point of AB)
∴ CM = AM = $$\frac{1}{2}$$ AB.  Hence Proved.

(c) Solve $$x+\frac{1}{x}=2 \frac{1}{2}$$

Question 8.
(a) If : a = b2x, b – c2y and c = a2z, show that 8xyz = 1.

(b) In the given figure, ABC is a right triangle at C. If D is the mid-point of BC, prove that AB2 = 4AD2 – 3AC2.

Given :∠C = 90°, D is mid-point of BC.
In ΔABC, In ΔACD, ⇒ AB2 = AC2 + BC2 AD2
⇒ AC2 + CD2 CD2
⇒ AD2 – AC2 (Pythagoras theorem) … (i) (Pythagoras theorem)
⇒ $$\left(\frac{1}{2} \mathrm{BC}\right)^{2}$$ = AD2 – AC2 (∵ D is mid-point of BC)
⇒ $$\frac{1}{4}$$ BC2 = AD2 – AC2 4
⇒ BC2 = 4AD2 – 4AC2 …(h)
Using (i) and (ii), we have
AB2 = AC2 + 4AD2 – 4AC2
⇒ AB2 = 4AD2 – 3AC2
Hence Proved.

(c) Find the value of x, if tan 3x = sin 45° cos 45° + sin 30°.
Given: tan 3x = sin 45° cos 45° + sin 30°
⇒  tan 3x $$=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2}$$
⇒ tan 3x = $$=\frac{1}{2}+\frac{1}{2}$$
⇒ tan 3x = 1
⇒ tan 3x = tan 45°
⇒ 3x = 45° ,
⇒ 45°
x = $$\frac{45^{\circ}}{3}$$ = 15°

Question 9.
(a) Factorize : 12 – (x + x1) (8 – x – x2).
12 – (x + x2) (8 – x – x2) = 12 – (x + x2) {8 – (x + x2)}
Let  x + x2 = a
⇒ 12 – a (8 – a) = 12 – 8a + a2
⇒ a2 – 8a + 12
⇒ a2 – (6 + 2) a + 12
= a2 – 6a – 2m + 12
= a (a – 6) – 2 (a – 6)
⇒ (a-6) (a- 2)
Substituting a = x + x2, we get
⇒ (x + x2 – 6) (x + x2 – 2)
⇒ (x2 + x – 6) (x2 + x – 2)
⇒ (x2 + 3x – 2x – 6) (x2 + 2x – x – 2)
⇒ {x (x + 3) – 2 (x + 3)} {x (x + 2) – 1 (x + 2)}
⇒ (x + 3) (x – 2) (x + 2) (x – 1)
⇒ (x – 1) (x + 2) (x – 2) (x + 3)

(b) A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Let the speed of the train be x km/h and that of the car be y km/h.
Total distance travelled = 370 km.
Case I : Distance travelled by train = 250 km.
Distance travelled by car = (370 – 250) km = 120 km
∴ Time taken by train $$=\frac{250}{x} \mathrm{~h}$$

(c) Express as a single logarithm :
2 log10 5 – log10 2 + 3log10 4+1
2 log10 5 – log10 2 + 3 log10 4 + 1 = log10 52 – log10 2 + log10 43 + log10 10
$$=\log _{10}\left(\frac{5^{2} \times 4^{3} \times 10}{2}\right)$$
= log10 8000 = log10 (20)3 = 3 log10 20

Question 10.
(a) The value of a car purchased 2 years ago depreciates by 10% every year. Its present value is ₹ 1,21,500. Find the cost price of the car. What will be its value after 2 years ?

(b) Construct a quadrilateral ABCD given that AB = 4.5 cm, ∠BAD = 60°, ∠ABC = 105°, AC = 6.5 cm and AD = 5 cm.
Given : AB 4.5 cm, Z BAD = 60°, Z ABC 105°, AC = 6.5 cm and AD – 5 cm. Steps of
construction :
(1) Draw AB = 4.5 cm.
(2) At A, draw ∠BAX = 60°.
(3) At B, draw ∠ABY = 105°.

(4) From A, cut BY at C such that AC = 6.5 cm.
(5) From A, cut AX at D such that AD = 5 cm.
(6) Join CD.
Hence, ABCD is the required quadrilateral.

(c) Factorize : x9 + y9.
= x9 + y9 = (x3)3 + (y3)3 = (x3 + y3) {(x3)2 – x3y3 + (y3)2}
= (x + y) (x2 – xy + y2) (x6 – x3y3 + y6).

Question 11.
(a) In the given figure, ABCD is a parallelogram. Find the values of x, y and z.

Given : ABCD is a parallelogram.
AB = CD
⇒ 3x – 1 = 2x + 2
⇒ 3x – 2x =2 + 1
⇒ x = 3
Also, ∠D = ∠B = 102° ( ∵ Opposite angles are equal) Exterior
In ΔACD, y = 50° + 102° (∵ angle is equal to sum of interior opposite angles)
= 152°
and ∠A + ∠D = 180°
⇒ z + 50° + 102° = 180°
⇒ z = 180° – 152° = 28°
⇒ x = 3, y = 152° z = 28°

(b) Draw the graph of 3x + 2 = 0 and 2y – 1 = 0 on the same graph sheet. Do these lines intersect ? If yes, find the point of intersection.
Given: 3x+2=0 ……..(i)
and 2y – 1 = 0 ……. (ii)
From (i), 3x = – 2
= $$x=\frac{-2}{3}$$
It is a straight line parallel to Y-axis at $$x=\frac{-2}{3}$$
From (ii), 2y = 1
⇒ y = $$\frac{1}{2}$$
It is a straight line parallel to Y-axis at y = $$\frac{1}{2}$$

(c) Prove that (sin A + cos A)2 + (sin A – cos A)2 = 2.
L. H.S. = (sin A + cos A)2 + (sin A – cos A)2
= sin2A + cos2 A + 2 sin A cos A + sin2A + cos2A – 2 sin A cos A = 2 (sin2A + cos2A)
= 2 x 1
= 2 = R.H.S.

## ICSE Class 9 Maths Sample Question Paper 7 with Answers

Section – A [40-Marks]
(Attempt all questions from this Section)

Question 1.
(a) If, in a ∆ABC, AB = 3 cm, BC = 4 cm and ∠ABC = 90°, find the values of cos C, sin C and
tan C.
Given : AB = 3 cm, BC = 4 cm, ∠ABC = 90°
By Pythagoras theorem,
AC2 = AB2 + BC2 = 32 + 42 = 25

(b) A man purchased an old scooter for ₹ 16,000. If the cost of the scooter after 2 years depreciates to ₹ 14,440, find the rate of depreciation.
Present value (V0) = ₹ 16,000
Value after 2 year (V1) = ₹ 14,440
∴ n =2
Let r be the rate of depreciation.

(c) Prove that √2 + √5 is irrational.
Let us assume that √2 + √5 is a rational number.
Then $$\sqrt{2}+\sqrt{5}=\frac{a}{b}$$
Where a and b co-prime positive integers.
$$\frac{a}{b}-\sqrt{2}=\sqrt{5}$$

Question 2.
(a) If $$x=\frac{1}{x-2 \sqrt{3}}$$ , find the values of (i) x – $$\frac{1}{x}$$ (ii) x + $$\frac{1}{x}$$.

(b) In the given figure, ABC is an equilateral triangle. Find the measures of angles marked by x, y and z.

Given : ABC is an equilateral triangle.
∠ABC = ∠ACB = ∠B AC = 60°.
Now, ∠BAD + ∠ADB = ∠ABC (Ext. angle is equal to sum of int. opp. angles)
⇒ x + 40° = 60°
⇒ x = 60° – 40°
⇒ x = 20°.
Also, ∠CAE + ∠AEC = ∠ACB (Ext. angle is equal to sum of int. opp. angles)
⇒ y + 30° = 60°
⇒ y = 60° – 30°
⇒ y = 30°
and ∠ACE +∠ACB = 180° (Linear Pair)
⇒ z + 60° = 180°
⇒ z = 180° – 60°
⇒ z = 120°

(c) Solve $$\frac{2}{3} x^{2}-\frac{1}{3} x-1=0$$
$$\frac{2}{3} x^{2}-\frac{1}{3} x-1=0$$
$$3 \times \frac{2}{3} x^{2}-3 \times \frac{1}{3} x-3 \times 1=3 \times 0$$
⇒ 2x2 – x – 3 = 0
⇒ 2 x 2 – (3 – 2)x -3=0
⇒ 2x2 – 3x + 2x – 3 = 0
⇒ x (2x – 3) + 1 (2x – 3) = 0
⇒ (2x – 3) (x + 1) = 0
⇒ 2x-3=0 or x + 1= 0
⇒ x= $$\frac{3}{2}$$ or x =-1
⇒ x= $$\frac{3}{2}$$ or -1

Question 3.
(a) Factorize : a3 – b3 – a + b.
a3 -b3 – a + b = (a-b) (a2 + ab + b2) – (a-b) = (a-b) (a2 + ab + b2 – 1).

(b) Draw a histogram to represent the following :

 Class Interval 40 – 48 48-56 56-64 64-72 72 – 80 Frequency 15 25 35 30 10

(c) Prove that $$\sqrt{\frac{1-\sin 30^{\circ}}{1+\sin 30^{\circ}}}=\tan 30^{\circ}$$

Question 4.
(a) Simplify: $$\frac{5^{2(x+6)} \times(25)^{-7+2 x}}{(125)^{2 x}}$$

(b) In the figure, DE||BC. Prove that (i) Area of ΔACD = Area of ΔABE (ii) Area of ΔOBD = Area of ΔOCE.

Given DE || BC
Area of ΔBCD = Area of ΔBCE
(Triangles on same base and between same parallels have equal area) Now, Area of ΔACD + Area of ΔBCD = Area of ΔABE + Area of ΔBCE
⇒ Area of ΔACD = Area of ΔABE (∵ Area of ABCD = Area of ABCE).
Hence Proved.

(ii) Area of ABCD = Area of ABCE [From (i)]
⇒ Area of ABCD – Area of ΔOBC = Area of ΔBCE – Area of ΔOBC
(Subtracting area of ΔOBC from both side)
⇒ Area of ΔOBD = Area of ΔOCE.
Hence Proved

(c) If log10 x + $$\frac{1}{3}$$ log10 y = 1, express y in terms of x.
Given log10 x + $$\frac{1}{3}$$ log10y = 1
log10 x + log10 y1/3 = log10 10
log10 (xy1/3) = log10 10

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) The mean of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. Find the correct mean of the observations.
We know,
Σx =$$\bar{x}$$ x n
Incorrect ∑ x = 35 x 9 = 315
Correct ∑ x =315 – 18 + 81 = 378
Correct mean = $$\frac{378}{9}=42$$

(b) In the given figure, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find (i) radius of the circle (ii) Length of chord CD.

⇒ 169 =144 ÷ CN2
⇒ CN2 = (169 – 144) = 25
⇒ CN= √25 =5
⇒ CD =2 CN (∵ N is mid-point of CD)
⇒ 2 x 5 = 10cm.

(c) If $$x^{2}+\frac{1}{x^{2}}=83$$ find the value of $$x^{3}-\frac{1}{x^{3}}$$

Question 6.
(a) A cumulative frequency distribution is given below. Convert this into a frequency distribution table.

 Marks Below 45 Below 60 Below 75 Below 90 Below 105 Below 120 No. of Students 0 8 23 48 85 116

 Marks No. of Students Class Interval Frequency Below 45 0 0-45 0 Below 60 8 45 – 60 8 (8-0) Below 75 23 60 – 75 15 (23 – 8) Below 90 48 75 – 90 25 (48 – 23) Below 105 85 90 -105 37 (85 – 48) Below 120 116 105 – 120 31 (116 – 85)

(b) Half the perimeter of a garden, whose length is 4 more than its width, is 36 m. Find the dimensions of the garden.
Let length and breadth of the garden be x m and y m respectively.
According to the question,
x = 4 + y …(i)
and x + y = 36 …(ii)
Substituting x = 4 + y in equation (ii), we get
4 + y + y = 36
2y = 36 – 4
y = $$\frac{32}{2}$$ = 16
Substituting y= 16 in equation (i), we get
x = 4 + 16 = 20
∴ Length = 20 m and breadth = 16 m.

(c) If x and y are rational numbers and $$\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=x-y \sqrt{3}$$

Question 7.
(a) Factorize : (x2 + y2 – z2)2 – 4x2y2.
(x2 + y2 – z2)[1] – 4x2y2 = (x2 + y1 – z2)2 – (2xy)2
= (x2 + y2 – z2 + 2xy) (x2 + y2 – z2 – 2xy)
= {(x2 + y2 + 2xy) – z2} {(x2 + y2 – 2xy) – z2}
= {(x + y)2 – (z)2}  – y)2 – (z)2}
= {x + y + z) {x + y – z) {x – y + z) {x – y – z).

(b) Prove that in a right angled triangle, the median drawn to the hypotenuse is half the hypotenuse in length.

(c) Find the value of x if 3 cot2 (x – 5°) = 1.
3 cot2 (x – 5°) =1
1 cot2 (x – 5°) = $$\frac{1}{3}$$
cot (x – 5°) = $$\frac{1}{\sqrt{3}}$$
cot (x – 5°) = cot 60°
x – 5°= 60°
x = 60° + 5°
x = 65°

Question 8.
(a) Solve: $$\frac{x+y}{x y}=2 ; \frac{x-y}{x y}=1$$

(b) Construct a parallelogram ABCD with AB = 5.1 cm, BC = 7 cm and ∠ABC = 75°.
Given: AB =5.1cm, BC = 7cm and ∠ABC = 75°
Steps of construction:
(1) Draw BC=7cm.
(2) At B, draw ∠ XBC = 75°
(3) From B, cut-off BA = 5.1 cm on BX.
(4) From C, draw an arc of radius 5.1 cm.
(5) From A, draw an arc of 7 cm to cut the arc from C at D.
Hence, ABCD is the required parallelogram.

(c) Calculate the distance between A (7, 3) and B on the X-axis whose abscissa is 11.
Given : A (7, 3)
∵ B lies on the X-axis whose abscissa is 11, the coordinates of B are (11, 0)
$$\mathrm{AB}=\sqrt{(11-7)^{2}+(0-3)^{2}}=\sqrt{4^{2}+(-3)^{2}}=\sqrt{16+9}=\sqrt{25}$$
= 5 Units.

Question 9.
(a) A sum of money ₹ 15,000 amounts to ₹ 16,537.50 in x years at the rate of 5% p.a. compounded annually. Find x.

(b) In the given figure, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

Given: ∠PSR = 90°, PQ = 10 cm, QS = 6 cm, RQ = 9 cm.
∴ In Δ PQ2, PQ2 = PS2 + QS2 (Pythagoras theorem)
102 =PS2+62
PS2 = 100 – 36
PS = √64 = 8cm
In ΔPRS, PR2 = PS2 + RS2 (Pythagoras theorem)
PR2 = 8 + (9 + 6)2 = 64 + 225 = 289
PR=√289=17cm.

(c) In the given figure, ACB is a semicircle whose radius is 10.5 cm and C is a point on the semicircle at a distance of 7 cm from B. Find the area of the shaded region.

For semi-circle,
r = 10.5 cm
∴ Area =$$\text { Area }=\frac{1}{2} \pi r^{2}=\frac{1}{2} \times \frac{22}{7} \times(10.5)^{2}=173.25 \mathrm{~cm}^{2}$$
For triangle ABC,
AB2 = BC2 + AC2 (Pythagoras theorem, ∠C = 90°)
(2 x 10.5)2 = 72 ÷ AC2
AC2 =441 – 49 =392
AC = 19.8 cm.
Area = x BC x AC = x 7 x 19.8 = 69.3 cm2
The area of shaded region = (173.25 – 69.3) cm2
= 103.95 cm2

Question 10.
(a) If a2 + b2 + c2 – ab – be – ca = 0, prove that a = b = c.
Given a2 + b2 + c2 – ab – be – ca =0
⇒ 2 (a2 + b2 + c2 – ab – be – ca) = 0
⇒  2a2 + 1b2 + 2c2 -2ab – 2bc – 2ca = 0
⇒ (a2 – 2ab + b2) + (b2 – 2be + c2) + (c2 – 2ca + a2) = 0
⇒ (a – b)2 + (b – c)2 + (c – a)2 =0
The above expression is possible only if
⇒ (a- b)2 = 0 Ab- c)2 = 0, (c – a)2 = 0
a-b =0, b – c = 0, c-a = 0
a = b,b = c, c = a
a = b = c.
Hence Prove.

(b) Solve graphically x + 3y = 6; 2x – 3y = 12 and hence find the value of a, if Ax + 3y = a
x+3y=6 ………. (i)
2x – 3y = 12 ….. (ii)
from (i)  x = 6 – 3y

 X 6 3 0 y 0 1 2

∴ (6, 0), (3, 1), (0, 2)
From (ii),
2x = 3y + 12
x = $$\frac{3 y+12}{2}$$

 X 6 3 0 y 0 1 2

(6, 0), (3, – 2), (0, – 4)
These points are piotted in the graph.

The two lines intersect at the point (6, 0).
∴ x = 6, y = 0
Now 4x + 3 y = a
⇒ 4 x 6 + 3 x 0 = a
24 + 0 =a
⇒ a = 24

(c) Given, 1008 = 2p.3q.7r, find the values of p, q, r and hence evaluate 2p.3q.7-r÷192.

Question 11.
(a) If log $$\frac{x-y}{2}=\frac{1}{2}$$(log x + log y), prove that x2 + y2 = 6xy.

x2 + y2 – 2xy = 4xy
x2 + y2 – 4xy + 2 xy
x2 + y2 = 6 xy.
Hence Proved

(b) In a pentagon ABCDE, AB||ED and ∠B = 140°. Find ∠C and ∠D if ∠C: ∠D = 5:6.

Given : AB||ED, ZB = 140°, ∠C : ∠D = 5:6.
Let  ∠C =5x, ∠D = 6x.
Now,∠A+∠E= 180° (Co-interior angles, AB||ED)
Also, ∠A+ ∠B+ ∠C+ ∠D+ ∠E= (5-2) x 180°
(∠A + ∠E) + ∠B + ∠C + ∠D
= 3 x 180° 180° + 140° + 5x + 6x = 540°
11 x = 540° – 320°
$$x=\frac{220^{\circ}}{11}$$
∠C = 5x = 5 x 20° = 100°
∠D = 6x = 6 x 20° = 120°

(c) Factorize : 4 a3b – 44 a2b + 112
4 a3b  – 44 a2b + 112 ab = 4 ab (a2 – 11a + 28)
= 4 ab {(a2  – (7 + 4) a + 28)}
= 4ab(a2 – 7a – 4a+28)
= 4ab {a (a – 7) – 4(a – 7))
= 4ab (a – 7) (a – 4).

## ICSE Class 9 Maths Sample Question Paper 6 with Answers

Section – A [40-Marks]
(Attempt all questions from this Section)

Question 1.
(a) Prove that log (1 + 2 + 3) = log 1 + log 2 + log 3.
log (1 + 2 + 3) = log 6 = log (1 x 2 x 3)
= log 1 + log 2 + log 3.

(b) In the given figure, CD is a diameter which meets the chord AB in E such that
AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle.

Given : AE = BE = 4 cm, CE = 3 cm
Let r be the radius (OB = OC)
OE = OC – CE = r – 3.

⇒ OB2 = OE2 + BE2 (Pythagoras theorem)
⇒ r2 = (r – 3)2 + 42 r2
⇒ r2 – 6r + 9 + 16
⇒ 6r = 25
$$r=\frac{25}{6}=4 \frac{1}{6} \mathrm{~cm}$$

(c) If ₹ 6,400 is invested at 6 $$\frac{1}{4}$$ % p.a. compound interest, find (i) the amount after 2 years (ii) the interest earned in 2 years.

Question 2.
(a) Evaluate tan x and cos y from the given figure.

In ΔACD, AC2 = AD2 + CD2
132 – 52 + CD2
⇒ CD2 = 169 – 25 = 144
⇒ CD = 12.
In A BCD, BC2 = CD2 + BD2
= 144 + 162 = 144 + 256 = 400
BC =20

(b) ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that the altitudes are equal.

Given: AC=AB
∴ In ΔBEC and ΔCFB,
∠C=∠B (∵ AB = AC)
∠BEC = ∠CFB (Each being a right angle)
BC = BC (Common side)
∴ ΔBFC  ≅ ΔCFB (AAS axiom)
∴ BE = CF (c.p.ct.)
Hence Proved.

(c) The mean of 5 observations is 15. If the mean of first three observations is 14 and that of the last three is 17, find the third observation.
Mean of 5 observations = 15
∴ Sum of 5 observations = 15 x 5 = 75
Mean of first 3 observations = 14
∴ Sum of first 3 observations = 14 x 3 = 42
Mean of last 3 observations = 17
∴ Sum of last 3 observations = 17 x 3 = 51
∴ The third observation = (42 + 51) – 75 = 18.

Question 3.
(a) Factorize : x4 + 4
x4 + 4 = (x4 + 4x2 + 4) – 4x2 = {(x2)2 + 2 .x. 2 + (2)2} – (2x)2
= (x2 + 2)2 – (2x)2 = (x2 + 2 + 2x) (x2 + 2 – 2x)
= (x2 + 2x+ 2) (x2 – 2x+ 2)

(b) Evaluate : $$\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \cdot \tan 60^{\circ}}$$

(c) Simplify:$$(81)^{3 / 4}-3 \times(7)^{0}-\left(\frac{1}{27}\right)^{-2 / 3}$$

Question 4.
(a) If x $$\frac{2}{x}$$ = 5, find the value of $$x^{3}-\frac{8}{x^{3}}$$

(b) If the hypotenuse of a right angled triangle is 6 m more than twice the shortest side and third side is 2 m less than hypotenuse, find the sides of the triangle.
Let the shortest side be x m.
Then, Hypotenuse =(2x+6)cm,thirdside=2x+6-2=(2x+4)m.
∴ (2x + 6)2 = (2x + 4)2 + x2 (Using Pythagoras theorem)
= (2x)2 + 2.2x.6 + 62 = (2x)2 + 2.2xA +42 + x2
= 4x2 + 24x + 36 = 4x2 + 16x + 16 + x2
= 24x – 16x = x2 + 16 – 36
= x2 – 8x – 20 = 0
= x2 – (10 – 2) x – 20 =0
= x2 – 10x + 2x – 20 =0
= x (x – 10) + 2 (x – 10) = 0
= (x – 10) (x + 2) = 0
= x – 10 =0 or x + 2 = 0
= x = 10 or x = – 2
∴ x = 10 (∵ x cannot be negative)
∴ 2x + 6 = 2 x 10 + 6 = 26
and 2x + 4 = 2 x 10 + 4 = 24
Therefore, the sides are 10 m, 26 m and 24 m.

(c) Simplify: $$\frac{3}{\sqrt{6}+\sqrt{3}}-\frac{4}{\sqrt{6}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}$$

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) Solve : log10 6 + log10 (4x + 5) = log10 (2x + 7) +1

(b) 3 men and 4 women can do a piece of work in 14 days while 4 men and 6 women can do it in 10 days. How long would it take 1 woman to finish the work ?
Let 1 man take x days and 1 woman take y days to finish the work.
∴ In 1 day, 1 man does = $$\frac{1}{x}$$ work and 1 woman does = $$\frac{1}{y}$$ work.
So, 3 men and 4 women do=$$3 \times \frac{1}{x}+4 \times \frac{1}{y}=\frac{3}{x}+\frac{4}{y}$$
It is given that 3 men and 4 women finish the work in 14 days.
$$\frac{3}{x}+\frac{4}{y}=\frac{1}{14}$$ ………….(i)
Also, 4 men and 6 women do the work in 10 days.
= $$\frac{4}{x}+\frac{6}{y}=\frac{1}{10}$$ …………(ii)
Multiplying equation (i) by 4 and equation (ii) by 3, we get

∴ One woman finish the work in 140 days.

(c) There are two regular polygons with number of sides equal to (n – 1) and (n + 2). Their exterior angles differ by 6°. Find the value of n
For first polygon,

$$\frac{3}{n^{2}+2 n-n-2}=\frac{1}{60}$$
n2  + n – 2 = 180
n2 + n- 182 = 0
n2  + (14 – 13) n – 182 = 0
n2  + 14n – 13n – 182 = 0
n(n + 14) -13 (n + 14) = 0
(n + 14) (n – 13) = 0
n + 14 = 0 or n – 13 =0
n = -14 = 0  or n = 13 (∵n cannot be negative)
∴ n = 13.

Question 6.
(a) If $$a^{2}+\frac{1}{a^{2}}=7$$, find the value of $$a^{2}-\frac{1}{a^{2}}$$

(b) Construct a trapezium ABCD in which AD || BC, Z B = 60°, AB = 5 cm, BC = 6.2 cm, and CD = 4.8 cm.
Given : AD||BC, ZB = 60°, AB = 5 cm, BC = 6.2 cm, and CD = 4.8 cm.

Steps of construction :
(1) Draw BC = 6.2 cm.
(2) At B, draw ∠CBX = 60° and cut off BA = 5 cm.
(3) At A, draw exterior ∠XAY = 60° such that AY||BC.
(4) From C, cut-off AY at D such that CD = 4.8 cm and join CD.
Hence, ABCD is the required trapezium.

(c) The inner dimensions of a closed wooden box are 2 m, 1.2 m and 0.75 m. The thickness of the wood is 2.5 cm. Find the cost of wood required to make the box if 1m3 of wood costs ₹ 5400.
The inner dimensions of the closed box are 2 m, 1.2 m, 0.75 m.
Inner volume = (2 x 1.2 x 0.75) m3 = 1.8 m3
Thickness of the box = 2.5 cm = 2.5/100m= 0.025 m
∴ Outer dimensions are (2 + 2 x 0.025) m, (1.2 + 2 x 0.025) m, (0.75 + 2 x 0.025) m
i.e. 2.05 m, 1.25, 0.8 m.
∴ Outer volume = (2.05 x 1.25 x 0.8) m3 = 2.05 m3
Volume of wood = (2.05 – 1.8) m3 0.25 m3
Cost of 1 m3 of wood = ₹ 5400
Cost of 0.25 m3 of wood = ₹5400 x 0.25
= ₹ 1350

Question 7.
(a) Solve : 4x2 + 15 =16x
4x2+15 =16x
4x2 – 16x+15 =0
4x2 – (10-t-6)x+15 =0
4x2 10x – 6x+15 =0
= 2x(2x – 5)- 3(2x – 5) =0
(2x – 5)(2x – 3) =0
2x – 5 =0 or 2x – 3=0
2x =5 or 2x=3

(b) Find graphically the vertices of the triangle whose sides have equations
2y – x = 8, 5y – x = 14 and y – 2x = 1.
Given equations are,
2y – x =8 ……….(i)
5y – x =14 …(ii)
and y – 2x =1 …(iii)
From(i), x =2y – 8

∴ (- 6, 1), (- 4, 2), (- 2, 3)
From (ii), x=2 y-8

(- 4, 2), (1, 3), (6, 4)
From (iii) y=2 x+1

∴ (1, 3), (2, 5), (- 1, – 1)
These points are plotted on the graph.

The three lines intersect at point (- 4, 2), (1, 3) and (2, 5) which are the required vertices of triangle formed by them

(c) If 3tan2 θ-1=0′, find cos 2θ, given that θ is acute.
Given:  3tan2 θ-1=0
tan2θ = 1/3
tan θ   =$$\frac{1}{\sqrt{3}}$$]
⇒ tanθ = tan 30°
θ = 30°
cos2θ = cos (2 x 30°) = cos 60° =$$\frac{1}{2}$$

Question 8.
(a) Solve for x : 3(2x + 1) – 2x+2 + 5 = 0.
⇒ 3(2x + 1) – 2x + 2 + 5 =0
⇒ 3.2x + 3 – 2x. 22 + 5 =0
⇒ 3.2x – 4.2x + 8=0
⇒ -2x = – 8
⇒ 2x = 23
⇒ x =3

(b) Find the area of a triangle whose perimeter is 22 cm, one side is 9 cm and the difference of the other two sides is 3 cm.
One side = 9 cm, perimeter = 22 cm.
Let other two sides be a cm and b cm and a > b.
According to the question,
a + b + 9 = 22
⇒ a + b = 13 ………..(i)
and a – b =3 (Given) ………(ii)
Adding equations (i) and (ii), we have
2 a = 16 ⇒ a = 8
Subtracting equation (ii) from equation (i), we have
2b = 10 ⇒ b = 5
The sides are a = 8 cm, b = 5 cm, c = 9 cm

(c) Insert four irrational numbers between 2√3 and 3√2

Question 9.
(a) Form a cumulative frequency distribution table from the following data by exclusive method taking 4 as the magnitude of class intervals.
31, 23, 19, 29, 20, 16, 10, 13, 34, 38, 33, 28, 21, 15, 18, 36, 24, 18, 15, 12, 30, 27, 23, 20, 17, 14, 32, 26, 25, 18, 29, 24, 19, 16, 11, 22, 15, 17, 10, 25.

(b) Solve simultaneously : $$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{4}{y}=-\frac{1}{2}$$

(c) The diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that area of ΔOAD = area of ΔOBC. Prove that ΔBCD is a trapezium.

Given: Area of ΔOAD = area of ΔOBC.
Draw DM ⊥ AB,CN ⊥AB.
∴ DM || CN (∵Both DM, CN are perpendicular to AB)
Now,
Area of ΔOAD = Area of ΔOBC
Area of ΔOAD + Area of ΔOAB = Area of ΔOBC + Area of ΔOAB

Question 10.
(a) If the interest is compounded half yearly, calculate the amount when the principal is ₹ 7400, the rate of interest is 5% p. a. and the duration is one year.

(b) Find the coefficient of x2 and x in the product of (x – 2) (x – 3) (x – 4).
Given :  (x -2) (x – 3) {x – 4)
Here,  a = – 2, b = – 3, c = -4
Coefficient of x2 = a + b + c = (- 2) + (- 3) + (- 4) = – 9
Coefficient of x = ab + be + ca = (- 2) (- 3) + (- 3) (- 4) + (- 4) (- 2)
= 6 + 12 + 8 = 26

(c) If the figure given, ABCD is a trapezium in which AB || DC. P is the mid-point of AD and PR || AB. Prove that PR = $$\frac{1}{2} (AB + CD)$$.

Question 11.
(a) Factorize : a3 + 3a2b + 3ab2 + 2b3.
a3 + 3a2b + 3 ab2 + 2b3 = (a3 + 3 a2b + 3ab2 + b3) + b3
= (a + b)3 + (b)3 = (a + b + b) {(a + b)2 – (a + b)b + b2}
= (a + 2b) (a2 + 2ab + b2 – ab – b2 + b2)
= (a + 2b) {a2 + ab + b2)

(b) In the point A (2, – 4) is equidistant from the points P (3, 8) and Q (- 10, y), find the values of y.
(c) Simplify: $$\sqrt[a b]{\frac{x^{a}}{x^{b}}} \cdot b \sqrt[x]{\frac{x^{b}}{x^{c}}} \cdot \sqrt[c a]{\frac{x^{c}}{x^{a}}}$$