## ICSE Class 9 Maths Sample Question Paper 5 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) At what rate % p.a. will sum of ₹ 4000 yield ₹ 1324 as compound interest in 3 years ?
(a) Given : P = ₹ 4,000, C.I. = ₹  1,324, n = 3 years.
Let r be the rate % p.a.
Now A = P + C.I. = ₹ (4,000 + 1,324) = ₹ 5,324.

(b) If x = 2 + √3 , prove that x2 – 4x + 1 = 0.

(c) How many times will the wheel of a car having radius 28 cm, rotate in a journey of 88 km
Given : r = 28 cm and
Distance = 88 km = 88 x 1,000 x 100 cm = 88,00,000 cm
Now, Distance covered in 1 rotation = Circumference of wheel

Question 2.
(a) Factorize : $$x^{2}+\frac{1}{x^{2}}-11$$

(b) From the adjoining figure, find the value of x.

In ΔACD,
AC = CD  (Given)
⇒ ∠ADC = ∠CAD (Angles opposite to equal sides)
Now, ∠ADC + ∠CAD + ∠ACD = 180° (Sum of angles in a triangle is 180°)
⇒ 2∠ADC = 180° – 56°
⇒ ∠ADC $$\frac{124^{\circ}}{2}$$
∠ADC = $$\frac{124^{\circ}}{2}$$ = 62°

In ΔABD, AD = BD (Given)
∠ABD = ∠ BAD (Angles opposite to equal sides)
(Exterior angle is equal to sum of interior opposite angles)
2∠ABD =62°
∠ABD$$\frac{62^{\circ}}{2}$$=31°
In ΔABC, ∠A + ∠B + ∠ C = 180° (Sum of angles in a triangle is 180°)
x°+31°+56°=180°
x° = 180°- 87°= 93°

(c) Simplify : $$\frac{5 .(25)^{n+1}-25 .(5)^{2 n}}{5 \cdot(5)^{2 n+3}-(25)^{n+1}}$$

Question 3.
(a) If θ = 30°, verify that cos 3 θ = 4 cos3 θ -3 cos θ .
L.H.S. = cos 3θ = cos (3 x 30°) = cos 90° = 0 ………..(1)
L.H.S. = 4cos3 θ -3cosθ= 4 cos3 30° – 3 cos 30°

From (i) and (ii)
L.H.S. = R.H.S.
Hence Proved.

(b) Solve by cross multiplication method :
x – 3y – 7 = 0; 3x – 3y = 15.

(c) Prove that: $$\log \frac{11}{5}+\log \frac{14}{3}-\log \frac{22}{15}=\log 7$$

= log 11 – log 5 + log 14 – log 3 – log 22 + log 15
= log 11 – log 5 + log (2 x 7)- log 3 – log (2 x 11) + log (3 x 5)
= log 11 – log 5 + log 2 + log 7 – log 3 – log 2 – log 11 + log 3 + log 5
= log 7 = R.H.S
Hence Proved.

Question 4.
(a) If a2 – 3a – 1 = 0, find the value of a + $$a^{2}+\frac{1}{a^{2}}$$
(a)

(b) Construct a combined histogram and frequency polygon for the following data :

(c) Of two unequal chords of a circle, prove that longer chord is nearer to the centre of the circle.
Given : AB > CD, OM⊥AB, ON ⊥ CD.
Join OA and OC.

We know perpendicular drawn from the centre to the chord bisects the chord.
∴ AM= $$\frac{1}{2}$$AB
and CN = $$\frac{1}{2}$$CD
Now, AM > CN(∵ AB > CD)
In ΔOAM OA2 = AM2 + OM2 (Pythagoras theorem)
In ΔOCN OC2 = CN2 + ON2 (Pythagoras theorem)
AM2 + OM2 =CN2 +ON2(∵ OA = OC, Radii)
⇒ OM2 – ON2 = – (AM2 – CN2)
⇒ OM2 – ON2 <O (∵AM > CN)
⇒ OM2 < ON2
⇒ OM < ON
i.e., longer chord is nearer to the centre.
Hence Proved

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

Question 5.
(a) Factorize : $$\frac{y^{6}}{343}+\frac{343}{y^{6}}$$

(b) The diagonals AC and DB of a parallelogram intersect at O. If P is the mid-point of AD,
prove that (i) PO || AB (ii) PO = $$\frac{1}{2}$$CD.
Given : In parallelogram ABCD, diagonals AC and BD intersect at O. P is the mid-point of AD.
(i) Since diagonals of a parallelogram bisect each other
∴ O is the mid-point of DB
P and O are mid-points of sides AD and BD respectively
By mid-point theorem,

(ii) Also, by mid-point theorem

(c) If θ is acute and 3sin θ = 4cos θ, find the value of 4sin2 θ – 3cos2 θ + 2.

Question 6.
(a) If the points A (4,3) and B (x, 5) are on the circle with centre C (2, 3), find the value of x
Given : A (4, 3), B (x, 5), C (2, 3).
∵ C (2, 3) is the centre,
∴ AC = BC (Redii)
AC2 = BC2
(4 – 2)2 + (3 – 3)2 =(x- 2) + (5 – 3)2
= 4 + 0 =(x – 2)2 + 4
(x-2)2 =0
x – 2 =0
x =2.

(b) ABCD is a trapezium with AB | | CD, and diagonals AC and BD meet at O.
Prove that area of ΔDAO = area of ΔOBC.

Given : AB||CD, diagonals AC and BD meet at O.
AB||DC
∴  Area of ΔABD = Area of ΔABC (Triangles on same base and between same parallels are equal in area)
∴ Area of ΔDAO + Area of ΔOAB = Area of ΔOBC + Area of Δ OAB (Addition area axiom)
⇒ Area of ΔDAO = Area of Δ OBC.
Hence Proved.

(c) Simplify: $$\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}$$

Question 7.
(a) If x + y = 10 and x2 + y1 = 58, find the value of x3 + y3.
Given :
x + y =10, x2 + y2 = 58.
(x + y)2 = x2 + y2 + 2xy
⇒ 102 = 58 + 2xy
2xy = 100 – 58
⇒ xy = $$\frac{42}{2}$$ = 21
x3 + y3 = (x + y)3 – 3xy (x + y)
= 103 – 3 x 21 x 10
= 1000 – 630 = 370.

(b) The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Let the larger supplementary angle be
Then, smaller supplementary angle = 180° – x According to the question,
x – (180° – x) = 18°
⇒ x – 180° + x =18°
⇒ 2x = 18° + 180°
⇒ x= $$\frac{198^{\circ}}{2}$$ = 99°
∴ 180° –  x = 1800 99° = 81°
The supplementary angles are 990 and 81°.

(c) The mean of 5 numbers is 20. If one number is excluded, mean of the remaining numbers becomes 23. Find the excluded number.
Mean of 5 numbers = 20
∴Sum of 5 numbers = 20 x 5 = loo.
1f one number is excluded,
Then, Mean of 4 numbers = 23
Sum of 4 numbers = 23 x 4 = 92
The excluded number = 100 – 92 = 8.

Question 8.
(a) Solve for x : 9 x 3X = (27)2x-5

(b) In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other two sides is
4 : 3, find the sides.
Given: Hypotenuse = 20 cm
and ratio of the other two sides = 4:3
Let the other two sides be 4x and 3x.
∴ By Pythagoras theorem,

4x – 4 x 4 = 16
3x = 3 x 4 = 12
∴ The required sides are 16 cm and 12 cm.

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

Question 9.
(a) In what time will a sum of ₹ 8000 becomes ₹ 9261 at the rate of 10% p. a., if the interest is compounded semi-annually?
Given : P = X 8,000, A = ? 9,261, r = 10% p.a.
Let n be the number of years.
∵ C.I. is compounded semi-annually,

(b) Construct a regular hexagon of side 2.2 cm.
Each side = 2.2 cm.

Steps of construction :
(1) Draw AB 2.2 cm
(2) At A and B, draw angle of 120°.
(3) From A and B, cut-off arcs of 2.2 cm each.
(4) At C, draw 120° and cut it off at D so that CD = 2.2 cm.
(5) At D, draw 120° and cut-off DE = 2.2 cm.
(6) Join EF.
Then, ABCDEF is the required hexagon.

(c) Solve graphically : 2x – 3y + 2 = 4x + 1=3x-y + 2
Given: 2x – 3y + 2 = 4x + 1=3x-y + 2
∴ 2x – 3y + 2 = 4x + 1 and 4x + 1= 3x-y + 2
⇒ 4x – 2x = – 3y + 2 – 1 and 4x-3x = -y + 2- 1
⇒  2x = 1 – 3y

The two lines intersect at the point (2, – 1).
x =2, y = -1

Question 10.
(a) Express (x2 -5x + 7) (x2 + 5x – 7) as a difference of two squares.
(x2 – 5x + 7) (x2 + 5x – 7) = {x2 – (5x – 7)} {x2 + (5x – 7)}
= (x2)2 – (5x – 7)2

(b) Simplify: $$(64)^{2 / 3}-\left(\frac{1}{81}\right)^{-1 / 4}+8^{2 / 3} \cdot\left(\frac{1}{2}\right)^{-1} \cdot 3^{0}$$

(c) In a pentagon ABCDE, BC | | ED and ∠B: ∠A: ∠E = 5:3:4. Find ∠B.

Given: ∠B: ∠A: ∠E =5:3:4
Let Now, ∠B = 5x, ∠A = 3x, ∠E = 4x.
∠C + ∠D = 180°(Co-interior angles; BC||ED)
Sum of angles in a figure with number of sides ‘n’ = (n – 2) x 180°
In pentagon, sum of angles = (5 – 2) x 180°
= 3 x 180° = 540°
∴ ∠ A + ∠B + ∠C + ∠D + ∠E = 540°
⇒ 3x + 5x + 180° + 4x = 540°
⇒ 12x = 540° – 180°
⇒ $$x=\frac{360^{\circ}}{12}$$
⇒ x = 30°
∴ ∠B = 5x = 5 x 30° = 150°

Question 11.
(a) If p + q = 10 and pq = 21, find 3 (p2 + q2).
p + q = 10, pq = 21.
∴ p2 + q2 = (p + q)2 – 2pq = 102 – 2 x 21 = 100 – 42 = 58
3 (p2 + q2) = 3 x 58
= 174.

(b) Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm
Let length of each of equal sides be a and that of base be b.
b  = 6 cm(Given)
and Perimeter   = 16 cm
⇒ a + b  = 16
⇒ 2a   + 6  = 16
⇒ 2a = 16 – 6

(c) Prove that : $$\frac{1}{1+\tan ^{2} \theta}+\frac{1}{1+\cot ^{2} \theta}=1$$

= cos2 θ + sin2 θ=1
Hence Proved.

## ICSE Class 9 Maths Sample Question Paper 4 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Expand : $$\left(\frac{2}{3} x-\frac{3}{2 x}-1\right)^{2}$$

(b) A person invests ₹ 10,000 for two years at a certain rate of interest, compounded annu­ally. At the end of one year, this sum amounts to ₹ 11,200. Calculate :
(i) The rate of interest p. a.
(ii) The amount at the end of second year.

(c) Factorize : 64x6 – 729y

Question 2.
(a) If a2 + b2 = 7ab, prove that 2 log (a + b) = log 9 + log a + log b.
Given : a2 + b2 .= 7ab
⇒ a2 + b2 – 9ab – 2ab
⇒ a2 + b2 + 2ab = 9 ab
⇒ (a + b)2 = 9 ab
Taking log of both sides, we get
log (a + b)2 = log (9ab)
⇒ 2 log (a + b) = log 9 + log a + log b.

(b) Prove that: $$\tan ^{2} \theta-\frac{1}{\cos ^{2} \theta}+1=0$$
To prove:
$$\tan ^{2} \theta-\frac{1}{\cos ^{2} \theta}+1=0$$
Consider L.H.S. = $$\tan ^{2} \theta-\frac{1}{\cos ^{2} \theta}+1$$
= tan2 θ- sec2 θ +1
= (1 + tan2 θ) – sec2 θ (∵ 1 + tan2 0 = sec2 0)
= sec2 θ – sec2 θ = θ= R.H.S.
Hence Proved

(c) In the given figure, two chords AB and CD of a circle intersect at P. If AB = CD, prove that arc AD = arc CB.
AB = CD
Minor arc AB = Minor arc CD
Minor arc AB – minor arc BD = Minor arc CD – Minor arc BD
⇒ arc AD = arc CB.
Hence Proved.

Question 3.
(a) In the given figure, ∠ BCD = ∠ADC and ∠BCA = ∠ADB.
Show that: (i)ΔACD ≅ ΔBDC (ii) BC = AD (iii) ∠A = ∠B.

Given: ∠BCD = ∠ADC
and ∠BCA = ∠ADB
=> ∠BCA + ∠BCD = ∠ADB + ∠BCD
=> ∠BCA + ∠BCD = ∠ADB + ∠ADC (v ∠BCD = ∠ADC)
=> ∠ACD = ∠BDC
In ΔACD and ΔBDC
(i) ∠ADC = ∠BCD (Given)
CD = CD (Common side)
⇒ ∠ACD = ∠BDC (Proved above)
∴ ΔACD ≅ ΔBDC (ASA axiom)
(ii) ∴BC = AD (c.p.c.t.)
(iii) ∴ ∠A = ∠B. (c.p.c.t.)
Hence Proved.

(b) $$a=\frac{2-\sqrt{5}}{2+\sqrt{5}} \text { and } b=\frac{2+\sqrt{5}}{2-\sqrt{5}}, \text { find } a^{2}-b^{2}$$

(c) Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, – 1) are the vertices of a square ABCD.

Question 4.
(a) The mean height of 10 girls in a class is 1.38 m and the mean height of 40 boys is 1.44 m. Find the mean height of 50 students of the class.
Given : Mean height of 10 girls = 1.38 m
∴ Sum of heights of 10 girls = 1.38 x 10 = 13.8 m
and Mean height of 40 boys = 1.44 m
∴ Sum of heights of 40 boys = 1.44 x 40 = 57.6 m.
∴ Sum of heights of 50 students = 13.8 +57.6 = 71.4 m.
∴ Mean heights of 50 students $$\frac{71.4}{50}$$
=1.428m

(b) In the given figure, AABC is a right triangle with ∠C = 90° and D is mid-point of side BC. Prove that AB2 = 4AD2 – 3AC2.

Given ∠C = 90°, D is the mid-point of BC.
∴ CD = BD
∴  In ΔABC,
AB2 = AC2 + BC(Pythagoras theorem)
= AC2 + (2CD)2 (∵ CD = BD = $$\frac{1}{2}$$ BC)
AB2 = AC2 + 4 CD2  …(i)
In ΔACD,
AD2 = AC2 + CD2  (Pythagoras theorem)
⇒ CD2 = AD2 – AC2 …(ii)
From equations (i) and (ii), we get
AB2 = AC2 + 4 (AD2 – AC2)
= AC2 + 4AD2 – 4AC2
= 4AD2 – 3AC2
Hence Proved.

(c) The following observation have been arranged in ascending order.
3, 6, 7, 10, x, x + 4, 19, 20, 25, 28 . If the median of the data is 13, find the value of x.
Given : Numbers in ascending order : 3, 6, 7, 10, x, x + 4, 19, 20, 25, 28.
Median = 13
Here, n = 20.

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

Question 5.
(a) Factorize : 5x2 + 17xy – 12y2.
5x2 + 17xy – 12y2 = 5x2 + (20 – 3) xy – 12y2
= 5x2 + 20xy – 3xy – 12y2
= 5x(x + 4y) – 3y (x + 4y)
= (x + 4y) (5x – 3y).

(b) If twice the son’s age in years is added to the father’s age, the sum is 70. But if twice the  father’s age is added to the son’s age, the sum is 95. Find the ages of father and son.
Let father’s age be x years and that of son’s be y years.
By 1st condition, x + 2y = 70 …(i)
By 2nd condition, 2x + y = 95 …(ii)
Multiplying equation (ii) by 2, we get
4x + 2y = 190
Subtracting equation (iii) from equation (i), we get

⇒ 40 +2y = 70
⇒ 2y = 70 – 40
⇒ $$y=\frac{30}{2}=15$$
∴ Father’s age is 40 years and son’s age is 15 years.

(c) In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.
Given: E, F are mid-points of sides AC and AB, respectively.

We know, the line joining the mid-points of any two sides of a triangle is parallel to the third side
FE || BC
⇒ FQ || BP
Now, since F is mid-point of AB and FQ || BP
∴ By converse of mid-point theorem,
⇒ Q is the mid-point of AP.
AQ = QP. Hence Proved.

Question 6.
(a) If a + $$\frac{1}{a} = p,$$ prove that $$a^{3}+\frac{1}{a^{3}}=p\left(p^{2}-3\right).$$

(b) The side of a square exceeds the side of another square by 3 cm and the sum of the areas of the two squares is 549 cm2. Find the perimeters of the squares.
Let the side of one square be x cm
Then, side of other square = (x + 3) cm.
Area of two squares are x2 cm2 and (x + 3)2 cm2, respectively.
According to the question,
x2 + (x + 3)2 = 549 ⇒ x2 + x2 + 6x + 9 = 549
⇒ 2x2 + 6x – 540 = 0
⇒ x2 + 3x – 270 = 0
⇒ x2 + (18 – 15)x – 270 =0
⇒ x2 + 15x – 15x – 270 =0
⇒ x (x + 18) – 15 (x + 18) =0
⇒ (x + 18) (x – 15) =0
⇒ x = – 18 or 15
∴ x – 15      (∵ x cannot be negative)
∴ x+ 3 =15+ 3 = 18
∴ Perimeter of one square = 4 x 15 = 60 cm
and Perimeter of other square = 4 x 18 = 72 cm

(c) Simplify :
$$\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\cos \left(90^{\circ}-\theta\right)}{\sec \left(90^{\circ}-\theta\right)}-3 \tan ^{2} 30^{\circ}$$

Question 7.
(a) If log10 a = b, express 102b3 in terms of a.

(b) ΔABC and ΔDBC are on the same base BC with A, D on opposite sides of BC. If area of ΔABC = area of ΔDBC, prove that BC bisects AD.

(c) If cos θ + sec θ = 2, show that cos8 θ + sec8 θ = 2.

Question 8.
(a) If each interior angle is double the exterior angle, find the number of sides.

(b) Solve by the substitution method :
5x + 4y – 4 = 0; x – 20 = 12y.
Given: 5x+4y – 4 =0 ………..(i)
and x – 20 =12y ………… (2)
From (ii), x = 12y + 20 ……. (3)
Putting x = 12 + 20 in equation (i), we have
5(12y+20)+4y – 4 =0
⇒ 60y+100+4y – 4=0
⇒ 64y = – 96
⇒ $$y=-\frac{96}{64}=-\frac{3}{2}$$
From (iii), x = 12 x $$\left(\frac{-3}{2}\right)$$ +20 = 2
x = 2,y = $$\frac{-3}{2}$$

(c) If x = 2 + √3 , find the value of x – $$\frac{1}{x}$$

Question 9.
(a) Evaluate : $$x^{2 / 3} \cdot y^{-1} \cdot z^{1 / 2}$$ when x – 8, y = 4 and z = 25

(b) Construct a ΔABC is which base AB 5 cm, ∠A = 30° and AC – BC – 2.5 cm.

Steps of construction :
(1) Draw base AB = 5 cm.
(2) Draw ∠BAX = 30°
(3) From AX, cut-off AD = 2.5 cm
(4) Join BD.
(5) Draw the perpendicular bisector of BD to cut AX at C.
(6) Join BC.
Thus, ABC in the required triangle.

(c) Simplify: $$\left(2 x-\frac{1}{2 x}\right)^{2}-\left(2 x+\frac{1}{2 x}\right)\left(2 x-\frac{1}{2 x}\right)$$

Question 10.
(a) Simplify: $$\left(a^{m-n}\right)^{m+n} \cdot\left(a^{n-l}\right)^{n+l} \cdot\left(a^{l-m}\right)^{l+m}$$

(b) If area of a semi-circular region is 1232 cm2, find its perimeter.

(c) If x = 15°, evaluate : 8 sin cos 4x. sin 6x.
Given: x=15°
∴ 8 sin 2xcos 4x.sin 6x = 8 sin (2 x 15°). cos (4 x 15°) . sin (6 x 15°)
= 8 sin 30° . cos 60° . sin 90°
$$=8 \times \frac{1}{2} \times \frac{1}{2} \times 1=2$$

Question 11.
(a) A farmer increases his output of wheat in his farm every year by 8%. This year he pro­duced 2187 quintals of wheat. What was his yearly produce of wheat 2 years ago?

(b) Draw the graph of the equation 3x – y = 4.

(c) Evaluate : (99.9)2 – (0.1)1
(99.9)2 – (0.1)2 = (99.9 + 0.1)
(99.9 – 0.1) = 100 x 99.8
= 9980.

## ICSE Class 9 Maths Sample Question Paper 3 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Factorize : 8(a – 2b)2 – 2a + 4b – 1
8 (a – 2b)2 -2a + 4b – 1 = 8 (a- 2b)2 -2(a – 2b) – 1
a – 2b = x
Then, the expression becomes
= 8x2 – 2x – 1
= 8x2 – 4x + 2x – 1
= 4x (2x – 1) + 1 (2x – 1)
= (2x – 1) (4x + 1)
= {2 (a- 2b) – 1} {4 (a – 2b) + 1} (Putting x-a-2b)
= (2a – 4b – 1) (4a- 8b + 1)

(b) The mean of 6 observations is 17.5. If five of them are 14, 9, 23, 25 and 10, find the sixth observation.
Mean of 6 observations is 17.5
5 observations are 14, 9, 23, 25, 10
Let 6th observation be x

(c) If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ-1.

Question 2.
(a) A man borrowed ₹15,000 for 2 years. The rate of interest for the two successive years are 8% and 10% respectively. If he repays ₹6,200 at the end of first year, find the outstanding amount at the end of the second year.
(a) For 1st year : P = ₹ 15,000, R = 8% p.a.
$$\mathrm{I}=\frac{15000 \times 8 \times 1}{100}$$ = ₹ 16,200
A = P + I = 15,000 + 1,200 =₹16,200
Amount of money repaid = ₹ 6,200
For 2nd year : P = 16,200 – 6,200 = ₹10,000, R = 10% p.a.
$$\mathrm{I}=\frac{10,000 \times 10 \times 1}{100}$$ = = ₹1,000
∴ A =P+I=10,000+1,000=11,000
∴ The amount outstanding at the end of 2nd year = 11, 000.

(b) Solve for x if log2 (x2 – 4) = 5.
Given: log2(x2 – 4) = 5
x2 – 4 =25
x2 =32 + 4
X = ±√36= ±6.

(c) In the following figure, AB is a 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.

(i) Given: chord AC = chord AD
⇒ arc AC = arc AD …(i)
Also, arc ACB = arc ADB (AB is a diameter) …(ii)
Subtracting (i) from (ii),
arc ACB – arc AC = arc ADB – arc AD
⇒ arc BC = arc DB. Hence Proved.

(ii) ∵ arc BC = arc BD (Proved above)
∴ ∠BAC = ∠DAB
⇒ AB is bisector of ∠CAD.
Hence Proved.

Question 3.
(a) Prove that: $$\frac{2^{n}+2^{n-1}}{2^{n+1}-2^{n}}=\frac{3}{2}$$

(b) Given that 16 cot A = 12, find the value of $$\frac{\sin A+\cos A}{\sin A-\cos A}$$

(c) If the altitudes from two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles.

Question 4.
(a) If a + b + 2c = 0, prove that a3 + b3 + 8c3 – 6abc = 0.
Given : a + b + 2c = 0
⇒ a + b = – 2c
Cubing both sides, we get
(a + b)3 = (- 2c)3
⇒ a3 + b3 + 3ab (a + b) = – 8c3
⇒ a3 + b3 + 3ab (- 2c) = – 8c3
⇒ a3 + b3 + 8c3 – 6abc = 0.
Hence Solved.

(b) Express $$0.1 \overline{34}$$ in the form $$\frac{p}{q}$$ ,p, q ∈ Z and q ≠ 0.
Let x = $$0.1 \overline{34}$$ = 0.1343434…
Multiplying both sides of (i) by 10, we get
10x = 1.343434 ………(i)
Multiplying both sides of (ii) by 100, we get
1000x = 134.3434 ………….(ii)
Subtracting (ii) from (iii), we get
1000x – 10x= 134.3434 … – 1.3434 ……………

(c) If the sides are in the ratio 5 : 3 : 4, prove that it is a right angled triangle.
Ratio of sides = 5:3:4
Let the length of sides be 5x, 3x, 4x.
Here,(3x)2 + (4x)2 = 9x2 + 16x2 = 25x2 = (5x)2
∴ By Pythagoras theorem, the triangle is right angled
Hence Proved.

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

Question 5.
(a) On what sum of money will the difference between compound interest and simple inter­est for 2 years be equal to ₹25 if the rate of interest charged for both is 5% p.a. ?

(b) Show by distance formula that the points A (-1, -1), B (2, 3) and C (8,11) are collinear.
Given points are A (-1, -1), B (2, 3), C (8, 11).
Now

AB + BC = 5 + 10 = 15
⇒ AB + BC = AC
The points are collinear.
Hence Proved.

(c) Factorize : a6 – 26a3 – 27.
a6 – 26a3 – 27 = a6 – (27 – 1) a3-27 = a6 – 27a3 + a3 – 27
= a3 (a3 – 27) + 1 (a3 – 27) = {a3 – 27) (a3 + 1)
= (a3 – 33) (a3 + 13)
= (a – 3) (a2 + 3a + 9) (a + 1) (a2 – a + 1)

Question 6.
(a) Simplify: $$\frac{\left(x^{a+b}\right)^{2} \cdot\left(x^{b+c}\right)^{2} \cdot\left(x^{c+a}\right)^{2}}{\left(x^{a} \cdot x^{b} \cdot x^{c}\right)^{4}}$$

(b) In the given figure AABC, D is the mid-point of AB, E is the mid-point of AC. Calculate :
(i) DE, if BC = 8 cm.
(ii) ∠ADE, if ∠DBC = 125°.

Given : D is mid-point of AB, E is the mid-point of AC, BC = 8 cm, ∠DBC = 125°.
The line joining the mid-points of any two sides of a triangle is parallel to the third and is equal the half of it.

(c) If a and b are rational numbers, find the values of a and b :
$$\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}$$

Question 7.
(a) Draw a histogram from the following data :

 Weight (in kg) 40-44 45-49 50-54 55-59 60-64 65-69 No. of students 2 8 12 10 6 4

(b) Solve:
83x – 67y = 383
67x – 83y = 367.
83x – 67y = 383 ………….(i)
67a – 83y = 367 ………….(ii)
Adding (i) and (ii), we get
150x – 150y = 750
x – y = 5 ………. (iii)
Subtracting (ii) from (i), we get
16x + 16y = 16
x + y = 1 ……… (iv)
Adding (iii) and (iv), we get
2x = 6
X = 3
Putting x = 3 in (iv), we get
3 + y = 1
⇒ y = 1 – 3 = -2.
x = 3, y = – 2

(c) In the following figure, area of parallelogram AFEC is 140 cm2. State, giving reason, the area of (i) parallelogram BFED (ii) ABFD.

Given : Area of parallelogram AFEC = 140 cm2.
(i)  Area of parallelogram BFED = Area of parallelogram AFEC
( ∵ They are on same base and between same parallels)
= 140 cm2

(ii) Area of Δ BFD = $$\frac{1}{2}$$ x Area of parallelogram BFED
(∵ They are on same base and between same parallels)
$$\frac{1}{2}$$ x 140 cm2 = 70 cm2.

Question 8.
(a) If log10 a = m and log10 b = n, express $$\frac{a^{3}}{b^{2}}$$ in terms of m and n

(b) Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Hence, find the area of the region bounded by these lines and X-axis.

(c) Factorize : $$8 x^{3}-\frac{1}{27 y^{3}}$$
Answre:

Question 9.
(a) If $$\frac{x^{2}+1}{x}=2 \frac{1}{2}$$ find the values of $$(i) x-\frac{1}{x} (ii) x^{3}-\frac{1}{x^{3}}$$

(b) In the following figure, OAB is a quadrant of a circle. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of the shaded portion.

Given : OA = 3.5 cm, OD = 2 cm
Area of shaded region = Area of quadrant AOB – Area of ΔAOD.

= 9.625 – 3.5 = 6.125 cm2.

(c) If a + b + c = 9 and ab + be + ca = 40, find the value of a2 + b2 + c2.
Given: a + b + c = 9 ab + bc + ca = 40
We know, (a+b+c)2 =a2+ b2 + c2 +2(ab+bc+ca)
(9)2 =a2+ b2 + c2+ 2 x 40
a2+ b2 + c2 = 81 – 80 = 1

Question 10.
(a) Solve: $$(\sqrt{2})^{2 x+4}=8^{x-6}$$

(b) Construct a rhombus whose diagonals are 5 cm and 6.8 cm.
Given, diagonals are 5 cm and 6.8 cm.
Steps of construction :
(1) Draw AC = 5 cm.
(2) Draw perpendicular bisector PQ of AC which intersect AC at O.
(3) From POQ, cut-off OB = OD = $$\frac{6.8}{2}$$ = 3.4 cm.
(4) Join the points A, B, C, D.
Then, ABCD is the required rhombus.

(c) In a quadrilateral ABCD, AO and BO are the bisectors of ∠A and ∠B respectively.
Prove that ∠AOB = $$\frac{1}{2}$$ (∠C + ∠D).
Given, AO and BO are bisectors of ∠A and ∠B respectively.

Question 11.
(a) Find the value of log5√5 (125).

(b) The sum of a two-digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number.
Let the digits in tens and units place be x and y respectively.
The number = 10x + y
The number obtained by reversing digits = 10y + x By 1st condition,
(10 + y) + (10y + x) = 165
⇒ 11x + 11y = 165
⇒ x + y =15 …(ii)
By 2nd condition, x-y =3 …(iii)
or y-x =3 …(iv)
Adding (ii) and (iii), we get 2x = 18
⇒ x =9
Putting x = 9 in (ii), we get y = 15 – 9 = 6
Again, adding (ii) and (iv), we get
2 y =18
⇒ y =9
Putting y = 9 in (ii), we get x = 15 – 9 = 6.
Substituting these values in (i), we get
The number = 10 x 9 + 6 or 10 x 6 + 9 = 96 or 69

(c) If the area of an equilateral triangle is 81√3 cm2, find its perimeter.
Given : Area of equilateral triangle = 81√3 cm2
Let the length side of each of equilateral triangle be a cm.

## ICSE Class 9 Maths Sample Question Paper 2 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) If x – 3 – 2√2, find the value of x2 + -y.
Given =

(b) Factorize : 9x2 – 4 (y + 2x)2
9x2 -4(y + 2x)2 = (3x)2 – {2 (y + 2x)}2
= (3x)2 – (2y + 4x)2
= (3x + 2y + 4x) (3x – 2y – 4x)
= (7x + 2y) (-x -2y)
= – (x + 2y) (7x + 2y).

(c) The area enclosed between the concentric circles is 770 cm2. Given that the radius of the outer circle is 21 cm, calculate the radius of the inner circle.
Given : The area enclosed between the circles = 770 cm2
Radius of outer circle (R) = 21 cm.
Let radius of inner circle be r.

Question 2.
(a) A man invests ₹46,875 at 4% p.a. compound interest for 3 years. Calculate :
(i) The interest for the first year.
(ii) The amount at the end of the second year.
(iii) The interest for the third year.

(b) If ax = by = cz and b2 = ac, Prove that $$y=\frac{2 x z}{x+z}$$

(c) In the following figure, AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Given: AD = BC,
∠OAD = ∠OBC = 90°
In Δ OAD and ΔOBC,
ΔOAD = ΔOBC (Given)
Δ AOD =Δ BOC (Vertically opposite angles)
∴ ΔOAD ≅ ΔOBC (SAS axiom)
∴ OA = OB (c.p.c.t.)
Hence, CD bisects AB. Hence Proved.

Question 3.
(a) Solve the following equations by cross multiplication method :
3x – 7y = – 10, – 2x + y = 3.

(b) Find the value of :
2√2 cos 45° cos 60° + 2√3 sin 30° tan 60° – cos 0°
2√2 cos 45° cos 60° + 2√3 sin 30° tan 60° – cos 0°
$$=2 \sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2}+2 \sqrt{3} \times \frac{1}{2} \times \sqrt{3}-1$$

(c) Construct a frequency polygon for the following frequency distribution using a graph sheet.

 Marks 40 – 50 50-60 60 – 70 70-80 80 – 90 90 – 100 No. of Students 5 8 13 9 7 5

Use 1 cm – 10 marks and 1 cm = 5 students.

Question 4.
(a) Express as a single logarithm :
2 log 3 – $$\frac{1}{2}$$ log 64 + log 16.
(b) If $$x+\frac{1}{x}=3$$,evaluate $$x^{3}+\frac{1}{x^{3}}$$
(c) Prove that the line joining mid-points of two parallel chords of a circle passes through the centre of the circle.

(c) Given: AB || CD, M and N are mid-points of sides AB and CD respectively.
Construction: Join OM, ON and draw a straight line parallel to AB and CD.
Since, line segment joining the mid-point of the chord with centre of the circle is perpendicular to the chord

∴ OM ⊥ AB and ON ⊥ CD
⇒ ∠AMO = 90° and ∠NOE = 90°
Now, ∠MOE = 90° (Co-interior angles, OE || AB)
∠NOE = 90° (Co-interior angles, OE || CD)
∠MOE + ∠NOF =90° + 900 = 1800
So, MON is a straight line passing through the centre of the circle.
Hence Proved.

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

Question 5.
(a) Find a point on the Y-axis which is equidistant from the points A (6, 5) and B (- 4, 3).
Given : A (6, 5), B (- 4, 3).
Let the point on the Y-axis be P (0, b).
According to the question,
AP = BP
⇒ AP2 = BP2
⇒ (6 – 0)2 + (5 – b)2 = (- 4 – 0)2 + (3 – b)2
⇒ 36 + 25 – 10b + V- = 16 + 9 – 6fo + b2
⇒ – 10b + 6b = 25 – 61
⇒  -4b =-36
⇒ b = 9
Required Point = (0,9)

(b) In the following figure, area of parallelogram ABCD is 29 cm2. Calculate the height of parallelogram ABEF, if AB = 5.8 cm.

Area of parallelogram ABCD = 29 cm2, AB = 5.8 cm.
Area of parallelogram ABEF = 29 cm2 (area of parallelograms on same base are equal)
⇒ AB x Height = 29 cm2
⇒ $$\text { Height }=\frac{29 \mathrm{~cm}^{2}}{\mathrm{AB}}=\frac{29 \mathrm{~cm}^{2}}{5.8 \mathrm{~cm}}=5 \mathrm{~cm}$$

(c) A sum of money doubles itself at compound interest in 15 years. In how many years will it become eight times ?

Question 6.
(a) Construct the quadrilateral ABCD, given that AB = 5 cm, BC = 2.5 cm, CD = 6 cm,
∠BAD = 90° and the diagonal AC = 5.5 cm.
Given : AB = 5 cm, BC = 2.5 cm, CD = 6 cm, Z BAD 90°,
AC = 5.5 cm.
Steps of construction :
(1) Draw AB 5 cm.
(2) At A, draw ∠BAP = 90°.

(3) From B and A, draw arcs of lengths 2.5 cm and 5.5 cm, respectively which intersect at C.
(4) From C, cut-off AP at D such that CD = 6 cm.
Thus, ABCD is the required quadrilateral.

(b) Factorize : (a + 1) (a + 2) (a + 3) (a + 4) – 3.
(a + 1) (a + 2) (a + 3) (a + 4) – 3 = (a + 1) (a + 4) (a + 2) (a + 3) – 3
= (a2 + 5a + 4) (a2 + 5a + 6) – 3
= (p + 4) (p + 6) – 3  (Putting a2 + 5a = p)
= p2 + 6p + 4p + 24 – 3
= p2 + 10p + 21
= p2 + (7 + 3) p + 21
= p2 + 7p + 3p + 21
= p2 (p + 7) + 3 (p + 7) = (p + 7) (p + 3)
= (a2 + 5a + 7) (a2 + 5a + 3) (∵ p = a2 + 5a)

(c) In the following figure, D and E are mid-points of the sides AB and AC respectively. If BC = 6 cm and ∠B = 72°, compute (i) DE (ii) ∠ADE.

Given : BC= 5.6 cm, ∠B = 72° and D, E are mid-points of sides AB, AC, respectively, (i)
The line joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

Question 7.
(a) Evaluate without using tables :
$$\left(\frac{\cos 47^{\circ}}{\sin 43^{\circ}}\right)^{2}+\left(\frac{\sin 72^{\circ}}{\cos 18^{\circ}}\right)^{2}-2 \cos ^{2} 45^{\circ}$$

(b) Solve:
$$\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27}$$

(c) In ΔABC, ∠ACB = 90°, AB = c unit, BC = a unit,
AC = b unit, C perpendicular to AB and CD = p unit.
Prove that $$\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$$

Question 8.
(a) If $$x^{2}+\frac{1}{x^{2}}=27$$,find the values of :
(i) $$x+\frac{1}{x}$$
(ii) $$x-\frac{1}{x}$$
(iii) $$x^{2}-\frac{1}{x^{2}}$$

(b) If 1 is added to the numerator of a fraction, it becomes $$\frac{1}{5}$$. If 1 is subtracted from the denominator, it becomes $$\frac{1}{7}$$. Find the fraction.

(c) Find the mean and median of the numbers :
41, 39, 52, 48, 54, 62, 46, 52, 40, 96, 42, 40, 98, 60, 52.
41, 39, 52, 48, 54, 62, 46, 52, 40, %, 42, 40, 98, 60, 52.
∴ ∑ x =822, n=15
∴ $$\text { Mean }=\frac{\sum x}{n}=\frac{822}{15}=54.8$$
Rearranging in ascending order, we get
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98

Question 9.
(a) The volume of a cuboidal block of silver is 10368 cm3. If its dimensions are in the ratio 3:2:1, find :
(i) Dimensions of the block.
(ii) Cost of gold polishing its entire surface at ₹0.50 per cm2.
(a) (i) Ratio of dimensions = 3:2:1
Let its length, breadth and height be 3x cm, 2x cm and x cm respectively.
Volume of block = 3× x 2x × x = 10368
⇒ 6x3 = 10368
$$x^{3}=\frac{10368}{6}=1728 \Rightarrow x=12$$
Length (l) =3x = 3 × 12=36cm
Breadth (b) = 2x = 2 × 12=24cm
Height (h) =x = 12 cm

(ii) Total surface area =2(lb+lh+bh)=2(36 x 24+36 x 12+24x 12)
= 2 (864 + 432 + 288)
= 2 x 1584 = 3168 cm2
∵ Rate of gold polishing = ₹ 0.50 = ₹$$\frac{1}{2}$$
Total cost of gold polishing of entire surface

(b) Factorize : 2 – y (7 – 5y).
2 – y (7 – 5y) = 2-7y + 5y2
= 2 – (5 + 2) y + 5y2 = 2 – 5y – 2y + 5y2
= 1 (2 – 5y) – y (2 – 5y) = (2 – 5y) (1 – y)

(c) Solve graphically : x – 2y = 1; x + y – 4.
x – 2y =1 ……….(i)
x + y = 4 ……………(ii)
from (i)

∴ The points are (1, 0), (3, 1), (5, 2)

From (ii)

The points are (4, 0), (3, 1), (2, 2)
These points are plotted on the graph.

The two straight lines intersect at (3, 1)
∴ x=3,y=1

Question 10.
(a) From the adjoining figure, find the values of :
(i) cot2 x – cosec2 x
(ii) $$\tan ^{2} y-\frac{1}{\cos ^{2} y}$$
(a) In AABD,
AB2 = AD2 + BD2 (By Pythagoras theorem)
= 42 + 32 = 16 + 9 = 25
AB = √25 = 5.

(b) $$\text { If } \frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}, \text { prove that }: a^{a} . b^{b} \cdot c^{c}=1 \text { . }$$
$$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k(\text { say })$$
log a =k(b – c); log b = k(c-a); log c = k(a-b)
Now, a log a + b log c + c log c = ak (b – c) + bk (c – a) + ck (a – b)
⇒ log ab + log bb + log cc = kab – kac + kbc – kab + kac – kbc
⇒ log (aa. bb . cc) = 0
⇒ log (aa. bb . cc)= log 1
⇒ aa. bb. cc = 1.
Hence Proved.

(c) Prove that √2 is not a rational number.

Let us assume that√2 is a rational number.
If $$\sqrt{2}=\frac{p}{q}, p, q \in \mathrm{I}$$ have no comman factor and C”q≠ 0.
$$2=\frac{p^{2}}{q^{2}} \Rightarrow p^{2}=2 q^{2} \Rightarrow p^{2}$$ C”is an even integer
⇒ p is an even integer
⇒ p = 2m, where m∈I
⇒ p2 = 4m2 ⇒ 2y2 = 4m2 ⇒ q2 = 2m2
⇒ q2 is an even integer
⇒ y is an even integer.
Thus, p and q are both even integers and therefore, have a common factor 2 which contradicts that p and q have no common factor.
√2 is not a rational number.
Hence Proved.

Question 11.
(a) Simplify: $$\left(a+\frac{1}{a}\right)^{2}-\left(a-\frac{1}{a}\right)^{2}$$

(b) In the given figure, ABCD is a trapezium. Find the values of x and y.

(c) Simplify: $$\frac{(25)^{3 / 2} \times(243)^{3 / 5}}{(16)^{5 / 4} \times(8)^{4 / 3}}$$

## ICSE Class 9 Maths Sample Question Paper 1 with Answers

Max Marks :80
[2 Hours]

General Instructions

• Answers to this Paper must be written on the paper provided separately.
• You will not be allowed to write during the first 15 minutes.
• This time is to be spent in reading the question paper.
• The time given at the head of this Paper is the time allowed for writing the answers.
• Section A is compulsory. Attempt any four questions from Section B.
• The intended marks for questions or parts of questions are given in brackets [ ].

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

Question 1.
(a) Rationalize the denominator : $$\frac{14}{5 \sqrt{3}-\sqrt{5}}$$ [3]

(b) Factorize the given expression completely : 6×2 + 7x – 5 [3]
6 x2+ 7x-5 = 6x2 + (10 – 3)* – 5
– 6x2 + 10x- 3x – 5
= 2x(3x + 5) – 1(3x + 5)
= (3x + 5) (2x – 1).

(c) In the given figure, AB = $$\frac{1}{2}$$ BC, where BC = 14 cm. Find : [4]
(i) Area of quadrilateral AEFD
(ii) Area of ΔABC
(iii) Area of semicircle
Hence find the area of shaded region. Use 7π = $$\left(\text { Use } \pi=\frac{22}{7}\right)$$

Question 2.
(a) Mr. Ravi borrows ₹ 16,000 for 2 years. The rate of interest for the two successive years are 10% and 12% respectively. If he repays ₹ 5,600 at the end of first year, find the amount outstanding at the end of the second year. [3]

(b) Simplify: $$\left(\frac{8}{27}\right)^{-\frac{1}{3}} \times\left(\frac{25}{4}\right)^{\frac{1}{2}} \times\left(\frac{4}{9}\right)^{0}+\left(\frac{125}{64}\right)^{\frac{1}{3}}$$ [3]

(c) In the given figure, ABCD is a parallelogram. AB is produced to P, such that AB = BP and PQ is drawn parallel to BC to meet AC produced at Q.
Given AB = 8 cm, AD = 5 cm, AC = 10 cm,
(i) Prove that point C is mid-point of AQ.
(ii) Find the perimeter of quadrilateral BCQP. [4]

(a) Here, P = ₹ 16000
For first year: R = 10% , T = 1 year
∴ $$\text { Interest }=\frac{16000 \times 10 \times 1}{100}= 1600$$
Amount = ₹ (16000 + 1600) = ₹ 17600
∴ Amount repaid = ₹ 5600.

For Second Year :
P = (17600 – 5600) = 12000, R = 12% , T = 1 year
∴ Intrest = $$\frac{12000 \times 12 \times 1}{100}$$ = ₹1440
∴ Amount =(12000+1440) = ₹13440

(b)

(c) Given : ABCD is parallelogram, AB = BP, AB = 8 cm, AD = 5 cm, AC = 10 cm.
(i) ∵ AB = BP
∴ B is mid-point of AP
Also, BC || PQ (Given)
AC = CQ (By mid-point theorem)
∴ C is mid-point of AQ. Hence Proved.

(ii) BP = AB = 8 cm (Given)
BC = AD = 5 cm (∵ ABCD is a parallelogram)
CQ = AC = 10 cm [From part, (i)]
PQ = 2 BC = 2×5 = 10 cm(By mid-point theorem)
∴ Perimeter of quadralateral BCQP = BP + PQ + CQ + BC

Question 3.
(a) Solve following pairs of linear equations using cross-multiplication method : [3]
5x – 3y = 2
4x + 7y = – 3

(b) Without using tables, evaluate : [3]
$$4 \tan 60^{\circ} \sec 30^{\circ}+\frac{\sin 31^{\circ} \sec 59^{\circ}+\cot 59^{\circ} \cot 31^{\circ}}{8 \sin ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}$$

(c) Construct a frequency polygon for the following frequency distribution, using a graph sheet. [4]

 Marks 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 7 18 26 37 20 6

Use : 1 cm = 10 marks, 1 cm = 5 students

Question 4.
(a) Evaluate : 3 log 2 – $$\frac{1}{3}log 27 + log 12 – log 4 + 3 log 5$$. [3]

(b) If x –$$\frac{1}{x}$$ =3, evaluate x3 – $$\frac{1}{x^{3}}$$[3]

(c) In the given diagram, O is the centre of the circle and AB is parallel to CD. AB = 24 cm
and distance between the chords AB and CD is 17 cm. If the radius of the circle is 13 cm, find the length of the chord CD.

Section – B [40 Marks]
(Attempt any four questions from this Section)

Question 5.
(a) Find the coordinates of the points on Y-axis which are at a distance of 5√2 units from
the point (5, 8). [3]
(a) Let the coordinates of the point on Y-axis be (0, y).
Distance = 5 √2
⇒$$\sqrt{(0-5)^{2}+(y-8)^{2}}=5 \sqrt{2}$$
Squaring both sides, we get
(0-5)2 + (y-8)2 = (5-√2)2
⇒ 25 + y2 – 2 y-8 + 64 = 50
⇒ y2 – 16y + 89 – 50 = 0
⇒ y2 – 16y + 39 = 0
⇒ y2 – (13 + 3)y + 39 = 0
⇒ y2-13y-3y+ 39 =0
⇒ y(y – 13) – 3(y – 13) = 0
⇒ (y – 13) (y – 3) = 0
⇒ y-13=0 or y-3 = 0
⇒ y = 13 or y = 3.
.’. The required point is (0,13) or (0, 3).

(b) In the given figure, BC is parallel to DE. Prove that area of ΔABE = Area of ΔACD. [3]

Given: BC || DE
∴ Area of ΔBCE = Area of ΔBCD
(Triangles, on same base and between the same parallels are equal in area)
⇒ Area of ΔBCE + Area of ΔABC = Area of ΔBCD + Area of ΔABC
(Adding area of ΔABC to both sides) .
⇒ Area of ΔABE = Area of ΔACD. Hence Proved.

(c) A stun of ₹ 12,500 is deposited for 1 $$\frac{1}{2}$$ years, compounded half-yearly. It amounts to ₹ 13,000 at the end of first half year. Find : [4]
(i) The rate of interest
(ii) The final amount. Give your answer correct to the nearest rupee.
P = ₹ 12,500, A = ₹ 13,000, T = – year.
∴ Interest for $$\frac{1}{2}$$ year
= ₹ (13000 – 12500) = ₹ 500.
(i) Let R be the rate of interest.
∴ $$\mathrm{R}=\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{500 \times 100}{12500 \times \frac{1}{2}}=8$$
∴ The rate of interest = 8 % p.a.

(ii) Now, n = 1 $$\frac{1}{2}$$years =$$\frac{3}{2}$$years.
C.I. is calculated half-yearly,
$$\mathrm{R}=\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{500 \times 100}{12500 \times \frac{1}{2}}=8$$

Question 6.
(a) Construct a parallelogram ABCD in which AB = 6.4 cm, AD = 5.2 cm and the
perpendicular distance between AB and DC is 4 cm. [3]
(a) Given : AB = 6.4 cm, AD = 5.2 cm,
Perpendicular distance between AB and DC is 4 cm.

Steps of construction :
(1) Draw a line segment XY and take any point P on it.
(2) At P, draw a perpendicular PZ and cut-off PD = 4 cm.
(3) From D, cut-off XY at A such that DA = 5.2 cm.
(4) From A, cut-off XY at B such that AB = 6.4 cm.
(5) From B and D, draw arcs of 5,2 cm and 6.4 cm radii respectively which intersect at C.
(6) Join AD, BC and CD to obtain the required parallelogram ABCD.

(b) Factorize : 4a2 – 9b2 – 16c2 + 24be [3]
4a2 – 9b2 – 16c2 + 24 be =4a2– (9b2 – 14bc + 16c2)
= (2a)2 – {(3b)2 – 2-3b-4c + (4c)2}
= (2a)2 – (3b – 4c)2
= (2a + 3b – 4c) (2a – 3b + 4c).

(c) In the given diagram, ABCD is a parallelogram, ΔAPD and ΔBQC are equilateral triangles.
Prove that: . [4]
(i) ∠PAB = ∠QCD
(ii) PB = QD

Given : ABCD is parallelogram, ΔAPD and ΔBQC are equilateral triangles.
(i) ∠DAB = ZBCD (Opp. angles of a || gm are equal)
⇒ ∠DAB + ∠PAD = ∠BCD + ∠BCQ (∠PAD = ∠BCQ = 60°)
⇒ ∠PAB = ∠DCQ. Hence Proved.

(ii) In ΔPAB and ΔQCD,
AB DC (Opp. sides of aIgm are equal)
∠PAB = ∠QCD [From (i)
AP = CQ (∵AP=AD=BC=CQ)
∠PAB ≅ ΔQCD (SAS axiom)
PB = QD (c.p.c.t.)
Hence Proved.

Question 7.
(a) Solve for x : sin2 x + cos2 30° = $$\frac{5}{4}$$; where 0° ≤ x ≤ 90° [3]

(b) Evaluate for x :$$\left(\sqrt{\frac{5}{3}}\right)^{x-8}=\left(\frac{27}{125}\right)^{2 x-3}$$ [3]

(c) In the given figure, triangle ABC is a right angle triangle with ∠B = 90° and D is mid­point of side BC. Prove that AC2 = AD2 + 3 CD2. [4]

Question 8.
(a) In the given figure, ∠ABC = 66°, ∠DAC = 38°. CE is perpendicular to AB and AD is perpendicular to BC. Prove that CP > AP. [3]

Given: ∠ABC = 66°, ∠DAC = 38°, CE ⊥AB, AD ⊥ BC.
In ∠ABD, ∠BAD + ∠ABD = ∠ADC (Exterior angle is equal to sum
of interior opposite angles)
In ∠SACE, ∠ACE + ∠AEC + ∠CAE = 180° (Sum of angles in a triangle is 180°)
∠ACE + 90° + (24° + 38°) = 180°
∠ACE + 152° = 180°
∠ACE = 180° – 152° = 28°.
Now, ∠CAP > ∠ACP ( 38°> 28°)
CP > AP (In a triangle, greater angle has greater side opposite to it)
Hence Proved.

(b) Mr. Mohan has ₹ 256 in the form of ₹ 1 and ₹ 2 coins. If the number of ₹ 2 coins are three more than twice the number of ₹ 1 coins, find the total value of ₹ 2 coins. [3]
Total amount = ₹ 256
Let the no. of ₹ 1 coins be x and that of ₹ 2 coins be y.
∴ Value of x coins = ₹ 1 × x = ₹  x
Value of y coins = ₹ 2 x y = ₹ 2y.
∴ x + 2y = 256
Also, y = 3 + 2x
Using equation (ii) in (i), we have
Also, y=3+2x
Using equation (ii) in (i), we have
⇒ x+2(3+2x)= 256
⇒ x+6+4x= 256
⇒ 5x =256 – 6
⇒ x=$$\frac{250}{5}$$=50.
Putting the value of x in equation (ii), we get
y =3+2x 50 =3+ 100 = 103.
∴ Total value of ₹ 2 coins = ₹ 2y
=₹ 2x 103
=₹ 206.

(c) Find (i) mean and (ii) median for the following observations : [4]
10, 47, 3, 9, 17, 27, 4, 48, 12, 15
Given observations are 10, 47, 3, 9, 17, 27, 4, 48, 12, 15.
Here, n 10
(i) Σx = 192
$$\text { Mean }=\frac{\Sigma x}{n}=\frac{192}{10}=19.2$$

(ii) Rearranging the observations in ascending order, we have
3, 4, 9, 10, 12, 15, 17, 27, 47, 48

Question 9.
(a) Three cubes are kept adjacently, edge to edge. If the edge of each cube is 7 cm, find total surface area of the resulting cuboid. [3]
Given : Length of each side of cube = 7 cm
For cuboid, 7cm
l= (7 + 7 + 7) cm = 21 cm
b = 7 cm, h = 7 cm.
We know, Total surface area = 2 (lb + bh + Ih)
= 2 (21 x 7 + 7 x 7 + 21 x 7)
= 2 (147+ 49 + 147) = 2 x 343 = 686 cm2

(b) In the given figure, arc AB = twice (arc BC) and ∠AOB = 80°. Find : [3]
(i) ∠BOC
(ii) ∠OAC

(i) Given: Arc AB = 2 (arc BC),∠AOB =80°
∠AOB=2∠BOC
∠BOC = $$\frac{1}{2}$$ ∠AOB
$$\frac{1}{2}$$ × 80°
=40°

(ii) In ΔAOC
OA = OC (Radii)
⇒ ∠OCA = ∠OAC (Angles opposite to equal
sides are equal)
Now, ∠OAC + ∠AOC + ∠OCA = 180° (Angle sum property)
∠OAC + (∠AOB + ∠BOC) + ∠OAC = 180° (∠OAC= ∠OCA)
= 2∠OAC + (80° + 40°) = 180°
2∠OAC + 120° = 180°
2∠OAC = 180° – 120° = 60°
∴ ∠OAC =$$\frac{60^{\circ}}{2}$$ 3o°

(c) Solve graphically the following system of linear equations (use graph sheet): [4]
x – 3y = 3
2x + 3y = 6
Also, find the area of the triangle formed by these two lines and the Y-axis.
x – 3y = 3 …………….. (i)
2x + 3y = 6 ………. (ii)
from equation (i)
x = 3y + 3

 X 3 0 -3 y 0 -1 -2

The points are (3, 0), (0, – 1), (- 3, – 2).
From equation (ii),
⇒ 2x = 6 – 3y
⇒ $$x=\frac{6-3 y}{2}$$

 X 3 0 -3 y 0 2 4

The points are (3, 0), (0, 2), (- 3, 4).
These points are plotted on the graph.

The two lines intersect at the point (3, 0).
∴ x=3,y=0
Triangle formed by the lines (i), (ii) and Y-axis is ABC.

Question 10.
(a) Each interior angle of a regular polygon is 135°. Find : [3]
(i) The measure of each exterior angle.
(ii) Number of sides of the polygon.
(iii) Name the polygon.
(a) Given: Each interior angle = 135°
(i) Exterior angle = 180° – 135° = 45°

(iii) The polygon is a regular octagon.

(b) If log 4 = 0.6020, find the value of log 80. [3]
Given : log 4 = 0.6020
⇒ log 22 = 0.6020
⇒ 2 log 2 = 0.6020
⇒ log 2
$$=\frac{0.6020}{2}$$
Now, log 80 = log (8 x 10) = log 8 + log 10
= log 23 + log 10 = 3 log 2 + log 10
= 3 x 0.3010 + 1 = 1.9030

(c) Evaluate x and y from the figure diagram. [4]

Question 11.
(a) ΔABC is an isosceles triangle such that AB = AC. D is a point on side AB such that
BC = CD. Given ∠BAC = 28°. Find the value of ∠DCA. [3]
(b) Prove that opposite angles of a parallelogram are equal. [3]
(c) The cross-section of a 6 m long piece of metal is shown in the figure. Calculate : [4]
(i) The area of the cross-section
(ii) The volume of the piece of metal in cubic centimetres.

(a) Given : AB = AC, BC = CD, ∠BAC = 28°

Since, AB = AC
∠ABC = ∠ACB. (Equal sides have equal angles opposite to them)
∠ABC + ∠ACB + ∠BAC = 1800 (Sum of angleš in a triangle is 1800)
∠ABC + ∠ABC + 28° = 180°
2∠ABC =180°-28°
∠ABC= $$\frac{152^{\circ}}{2}$$=76°
∠BDC = ∠CBD = 76°
Now, ∠ACD + ∠CAD = ∠BDC (Exterior angle is equal to sum of interior opposite angles)
∠ACD + 28° = 76°
∠ACD = 76° – 28° = 480

(b) Given : A parallelogram ABCD.

To prove:∠A = ∠C and ∠B = ∠D.
Proof: AB II DC, AD II BC ( ABCD is a parallelogram)
∠A + ∠D = 1800 (Co-interior angles) …(i)
and ∠D + ∠C = 180° (Co-interior angles) …(ii)
From (i) and (ii),∠A + ∠D =∠D + ∠C
∠A=∠C
Similarly,∠B = ∠D.Hence Proved.

(c) In triangle, length of equal sides (a) = 5 cm, base (b) = 8 cm.
In rectangle, Length (L) = 8 cm, Breadth (B) = 6.5 cm.
(i) The area of cross-section = Area of rectangle + Area of triangle

(ii)  Length of metal = 6 m = 600 cm.
Volume = Area of cross-section x Length
= 64 cm2 x 600 cm
= 38400 cm3.