2
PERIMETER AREA OF PLANE
FIGURES, VOLUME AND
SURFACE AREA OF SOLIDS
Z
B
1. Find the area of the triangle whose sides are 13 cm, 20 cm and 21
cm. Also, find the altitude of the triangle corresponding to the largest
side.
Ans. Let a = 13, b = 20 cm and c = 21 cm.
a + b + c 13 + 20 + 21
54
=
cm =
cm = 27 cm
Now, s =
2
2
2
Thus, area of the triangle = s ( s − a ) ( s − b) ( s − c)
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= 27(27 − 13) (27 − 20) (27 − 21) cm 2
= 27 × 14 × 7 × 6 cm 2
= 3 × 3 × 3 × 2 × 7 × 7 × 2 × 3 cm 2
= 2 × 3 × 3 × 7 cm2 = 126 cm2
Here, largest side = 21 cm
Area of the triangle = 126 cm2
126 × 2
1
cm
× b × h = 126
⇒
⇒ h=
21
2
⇒ h = 12 cm
Hence, altitude of the triangle to the largest side = 12 cm.
2. Find the area of the triangle whose sides are 50 cm, 48 cm, and 14
cm. Find the height of the triangle corresponding to the side
measuring 48 cm.
Ans.Let a = 50 cm, b = 48 cm and c = 14 cm.
Now,
Math Class VIII
s=
a + b + c 50 + 48 + 14
112
=
cm =
cm = 56 cm
2
2
2
1
Question Bank
Thus, area of the triangle
= s ( s − a ) ( s − b) ( s − c )
= 56(56 − 50) (56 − 48) (56 − 14) cm 2
= 56 × 6 × 8 × 42 cm2
= 2 × 2 × 2 × 7 × 2 × 3 × 2 × 2 × 2 × 2 × 3 × 7 cm 2
Z
B
= 2 × 2 × 2 × 2 × 3 × 7 = cm2 = 336 cm2
Area of the triangle = 336 cm2
⇒
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1
× b × h = 336
2
⇒
1
× 48 × h = 336
2
336 × 2
cm ⇒ h = 14 cm
48
Hence, the required height is 14 cm.
3. Find the area of a triangular field whose sides are 17 m, 19 m and
32 m. Find the altitude of the triangle corresponding to the small-
⇒ h=
est side. (Given 15 = 3.873 )
Ans.Let a = 17 m, b = 19 m and c = 32 m
17 + 19 + 32
68
m=
m = 34 m
2
2
Thus, area of the triangle
then
s=
= s ( s − a ) ( s − b) ( s − c )
= 34(34 − 17) (34 − 19) (34 − 34) m 2 = 34 × 17 × 15 × 2 m 2
= 2 × 17 × 17 × 3 × 5 × 2 m 2 = 2 × 17 3 × 5 m 2
= 34 15 m 2 = 34 × 3.873 m2 = 131.68 m2
Here smallest side (b) = 17 m
Area of the triangle 131.68
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1
1
× b × h = 131.68 ⇒
× 17 × h = 131.68
2
2
131.68 × 2
h=
⇒
⇒ h = 15.49 m ⇒ h = 15.49 m
17
Hence, the required height is 15.49 m
4. Find the area of an isosceles triangle in which each of the equal
sides measures 30 cm and the third side is 48 cm long.
Ans.In ∆ ABC,
A
AB = AC = 30 cm, BC = 48 cm
AD ⊥ BC is drawn which bisects BC at D
48
24 cm
24 cm C
= 24 cm
Then,
BD = DC =
B
2
D
48
cm
In right ∆ ABC, we have
AB2 = AD2 + BD2 ⇒ (30)2 = AD2 + (24)2
⇒ AD2 = (30)2 – (24)2 = 900 – 576 = 324 = (18)2 ∴ AD = 18 cm
1
Hence, area of triangle ABC = × base × altitude
2
1
= × 48 × 18 = 432 cm 2
2
5. The base and the height of a triangle are in the ratio 5 : 3 and its
area is 43.2 m2. Find the base and the height of the triangle.
Ans.Let base of the triangle be 5x and height of the triangle be 3x
Area of the triangle = 43.2 m2
(given)
⇒
cm
30
cm
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30
Z
B
1
1
× b × h = 43.2 ⇒ × 5 x × 3 x = 43.2
2
2
43.2 × 2
15 2
x = 43.2
⇒
⇒ x2 =
15
2
⇒ x2 = 5.76
⇒ x = 5.76 ⇒ x = 2.4 m
Hence, base of the triangle = (5 × 2.4) m = 12 m
Height of the triangle = (3 × 2.4) m = 7.2 m.
⇒
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6. Two sides of a triangle are 6 cm and 8 cm. If height of the triangle
corresponding to 6 cm side is 4 cm ; find :
(a) area of the triangle.
(b) height of the triangle corresponding to 8 cm side.
Ans.
BC = 6 cm
height
AD = 4 cm
area of triangle ABC
1
1
= base × height = × BC × AD
2
2
A
E
1
2
= × 6 cm × 4 cm = 12 cm
2
8 cm
Again, area of triangle
4 cm
1
ABC = × AC × BE
2
C
B
D
1
6 cm
12 = × 8 × BE
2
12 × 2
BE =
Thus
8
BE = 3 cm
∴ (a) 12 cm2 (b) 3 cm.
7. Two sides of a triangle are 6.4 m and 4.8 m. If height of the triangle corresponding to 4.8 m side is 6 m ; find :
(i) area of the triangle.
(ii) height of the triangle corresponding to 6.4 m side.
Ans. ABC is the triangle in which BC = 4.8 m
A
AC = 6.4 m and AD = 6 m
E
Area of triangle
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B
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ABC = × BC × AD
2
1
= × 4.8 × 6 = 14.4 m2
2
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6.4 m
6m
B
D
C
4.8 m
Question Bank
BE is height of triangle corresponding to 6.4 m
1
1
AC × BE = 14.4,
× 6.4 × BE = 14.4
Thus
2
2
14.4 × 2 14.4
BE =
=
= 4.5 m
6.4
3.2
∴
(i) 14.4 m2 (ii) 4.5 m.
8. The area of an equilateral triangle is 144 3 cm 2 ; find its perimeter.
Ans. Let each side of an equilateral triangle be x cm.
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Area of equilateral triangle =
3
(side)2
4
3 2
x = 144 3
(given)
4
4
x 2 = 144 3 ×
⇒
⇒ x2 = 144 × 4
3
x2 = 576 ⇒ x = 576 = 24 cm.
⇒
⇒ Each side = 24 cm
Hence, perimeter of equilateral triangle = 3 × side = 3(24) = 72 cm.
9. The area of an equilateral triangle is numerically equal to its
perimeter. Find its perimeter correct to 2 decimal places.
Ans. Let each side of the equilateral triangle be x
=
3 2
x
4
Perimeter of equilateral triangle = 3x
Area of equilateral triangle = Perimeter of equilateral triangle
∴ Area of equilateral triangle =
3 2
x = 3x
4
⇒ x2 = 3x ×
Math Class VIII
4
3x × 4 × 3 3x × 4 × 3
=
=
= 4x 3
3
3
3× 3
5
Question Bank
⇒ x 2 = 3(4 x) ⇒ x = 4 3
Hence, perimeter = 12 3 units
= 12 × (1.732) = 20.784 = 20.78 units.
10. A field is in the shape of a quadrilateral ABCD in which side
AB = 18 m, side AD = 24 m, side BC = 40 m, DC = 50 m and ∠A
= 90°. Find the area of the field.
D
C
50 m
Ans. Since ∠ A = 90°
∴ By Pythagoras theorem,
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B
24 m
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In ∆ ABD,
BD2 = AB2 + AD2
2
2
2
BD = AB + AD = 18 + 24
2 A
18 m
40 m
90°
B
= 324 + 576 = 900 = 30 m.
1
Area of ∆ ABD = (18) × (24) = (18) × (12) = 216 m2
2
In ∆ BCD, sides are 30, 40, 50
⇒ By Pythagoras theorem ∠CBD = 90°
[∴ DC2 = BD2 + BC 2, ∴ (50)2 = (30)2 + (40)2
1
Thus, area of ∆ BCD = × (40) × (30) = 600 m 2
2
Hence, area of quadrilateral ABCD = Area of ∆ ABD + area of
∆ BCD = 216 + 600 = 816 m2.
11. The lengths of the sides of a triangle are in the ratio 4 : 5 : 3 and its
perimeter is 96 cm. Find its area.
Ans.Let the sides of the triangle ABC be 4x, 5x and 3x
Let AB = 4x, AC = 5x and BC = 3x
Perimeter of ∆ ABC = 4x + 5x + 3x, 12x = 96
(Given)
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A
96
x=
x=8
12
∴ Sides are BC = 3(8) = 24 cm,
AB = 4 (8) = 32 cm,
AC = 5 (8) = 40 cm
5x
4x
B
∴
3x
Z
B
(AC)2 = (AB)2 + (BC)2
C
[∵ (5x)2 = (3x)2 + (4x)2]
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1
1
Hence, area of ∆ ABC = (BC) (AB) = × 24 × 32
2
2
= 12 × 32 = 384 cm2
12. Calculate the area of the quadrilateral PQRS shown in the adjoining figure, it being given that PR = 8 cm, RQ = 17 cm,
∠ RPQ = 90°, RS = 3 cm and ∠ PRS = 90°.
S
6 cm R
8 cm
17
cm
Q
P
Ans. In ∆ PQR,
Using Pythagoras theorem, = PQ2 = QR2 – PR2
PQ = (QR) 2 − (PR) 2 = (17)2 − (8) 2
= 289 − 64 = 225 = 15 cm
1
1
2
Thus, area of ∆ PQR = × PQ × PR = × 15 × 8 cm = 60 cm2
2
2
1
1
Area of ∆ PRS = × SR × PR = × 6 × 8 = 24 cm2
2
2
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Hence, area of the quadrilateral PQRS
= area of ∆ PQR + area of ∆ PRS
= (60 + 24) cm2 = 84 cm2.
13. Find the area of a quadrilateral ABCD whose diagonal AC is 25
cm long and the lengths of perpendiculars from opposite vertices
B and D on AC are BE = 3.6 cm and DF = 2.4 cm.
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B
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B
F
A
Ans.Area of the quadrilateral ABCD
= area of ∆ ABC + area of ∆ ACD
1
1
= × AC × BE + × AC × DF
2
2
1
= × AC × (EB + DF)
2
1
= × 25 × (3.6 + 2.4) cm 2
2
1
= × 25 × 6 cm = 75 cm 2 .
2
14. Find the area of the quadrilateral ABCD, given in the adjoining
figure in which AB = 28 cm. BC = 78 cm, CD = 112 cm, BD = 50
cm and DA = 30 cm.
(Given 26 = 5.099)
C
m
112 c
30
50
78
cm
D
cm
cm
A 28 cm
B
Ans. In ∆ ABD
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50 + 30 + 28 108
=
= 54 cm
2
2
Thus, area of ∆ ABD
s=
= 54(54 − 50) (54 − 30) (54 − 28) cm2
= 54 × 4 × 24 × 26 cm2
= 2 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 3 × 2 × 13 cm 2
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B
= 22 × 22 × 22 × 2 × 32 × 32 × 13 cm 2
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= 2 × 2 × 2 × 3 × 3 13 × 2 cm
2
2
= 72 26 cm = 72 × 5.0990 = 367.13 cm2
In ∆ DBC
112 + 78 + 50 240
s=
=
= 120 cm
2
2
Thus area of ∆ DBC
= 120(120 − 112) (120 − 78) (120 − 50) cm2
= 120 × 8 × 42 × 70 cm2
= 2 × 2 × 3 ×10 × 2 × 2 × 2 × 2 × 3 × 7 × 7 ×10 cm2
= 22 × 22 × 22 × 32 × 72 × 102
= 2 × 2 × 2 × 3 × 7 × 10
= 1680 cm2
Area of the quadrilateral ABCD
= area of ∆ ABD + area of ∆ DBC
= (367.13 + 1680) cm2
= 2047.13 cm2.
15. In the adjoining figure, ABCD is a square and
E, F are mid-points of the sides BC, CD respectively. If CE = 7 cm., find the area of the unshaded region.
Math Class VIII
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D
F
C
E
A
B
Question Bank
Ans. ABCD is a square
∴ AB = BC = CD = DA, CD = BC
1
1
CD = BC
⇒ CF = CE.
2
2
CF = CE = 7 cm.
⇒
∴
1
49
×7×7 =
cm 2
2
2
Area of square = (Side)2 = 14 × 14 = 196 cm2
Thus area of shaded region =
Z
B
49 343
=
= 171.5 cm 2
Area of unshaded portion = 196 −
2
2
16. In the adjoining figure, find the area of the path which is 1.9 m
wide all inside the rectangle.
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14 m
22 m
Ans.Length of rectangle ABCD = 22 m
Breadth of rectangle ABCD = 14 m
C
D
E
1.9 m
A
(22 – 3.8)
22 m
G
14 m
(14 – 3.8)
1.9 m
H
F
B
Area of rectangle ABCD = 22 × 14 = 308 m2
Length of rectangle EFGH = 22 – 2 × 1.9 = 18.2 m
Breadth of rectangle EFGH = 14 – 2 × 1.9 = 10.2 m
Area of rectangle EFGH = 18.2 × 10.2 = 185.64 m2
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Question Bank
Hence, area of path = area of rectangle ABCD – area of rectangle
EFGH (308 – 185.64) m2 = 122.36 m2
17. Find the area and the perimeter of the following figures. All angles
are right angles and all measurements are in centimetres.
7
8
2
3
2
3
5
Z
B
3
12
9
2
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5
2
2
7
(ii)
8
(i)
Ans. (i) Area of rectangle ABCD
= 8 × 2 = 16 cm2
Area of rectangle IJKL = 5 × 2 = 10 cm2
[∵ L = 9 – 2 – 2 = 5 cm. B = 8 – 3 – 3 = 2 cm.]
8
A
B
2
D
3
9
5
E
H
3
C
3
F
5
3
K
8
2
G
Area of the figure = 16 + 16 + 10 = 42 cm2.
Perimeter of figure = sum of all the sides
= 8 + 2 + 3 + 5 + 3 + 2 + 8 + 2 + 3 + 5 + 3 + 2 = 46 cm.
(ii) Area of rectangle AEHG = 12 × 2 = 24 cm2
Area of rectangle EFCB = 5 × 2 = 10 cm2
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Question Bank
Area of rectangle OLMN = 3 × 2 = 6 cm2
Area of rectangle HIJK = 5 × 2 = 10 cm2
Total area of figure = 24 + 10 + 6 + 10 = 50 cm2
A E
D
12
7
5
F
3
O 3 N
2
L 3 M
3
5
K
B
2
C
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7
J
2
I
Perimeter of figure = Sum of all the sides
= 7 + 2 + 5 + 3 + 3 + 2 + 3 + 3 + 5 + 2 + 7 + 12 = 54 cm.
18. The inner dimensions of a closed wooden box are 2 m by 1.2 m by
0.75 m. The thickness of the wood is 2.5 cm. Find the cost of
wood required to make the box if 1 m3 of wood costs Rs. 5400.
Ans.Thickness of wood = 2.5 cm = 0.025 m
∴ Inner volume = 2 × 1.2 × 0.75 = 1.8 m3
External length = 2 + 0.05 = 2.05 m
External breadth = 1.2 + 0.05 = 1.25 m
External weight = 0.75 + 0.05 = 0.8 m
External volume = l × b × h
= 2.05 × 1.25 × 0.8 = 2.05 m3
Hence, volume of wood required to make box
= (2.05 – 1.8) m3 = 0.25 m3
Cost of wood = 0.25 × Rs 5400 = Rs 1350.
19. In the diagram, ABCD is a rectangle of size 18 cm by 10 cm. In
∆ BEC, ∠ E = 90° and EC = 8 cm. Find the area enclosed by the
pentagon by the pentagon ABECD.
Math Class VIII
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Question Bank
C
D
8 cm
E
A
10 cm
B
18 cm
Ans.Area of rectangle ABCD
= 18 × 10 = 180 cm2
In ∆ BEC,
BC2 = CE2 + BE2
(By Pythagoras theorem)
(10)2 = 82 + BE2
∴
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BE2 = 100 – 64 = 36 ⇒ BE = 36 ⇒ BE = 6 cm.
1
2
Hence, area of right ∆ BEC = × 6 × 8 = 24 cm
2
Area of pentagon ABECD = Area of rectangle ABCD – area of
∆ BEC = (180 – 24) cm2 = 156 cm2.
20. In the adjoining figure, ABCDE is a pentagon which is symmetrical about the line passing through A and the mid-point M of the
side CD. All measurements are in centimetres. Find the perimeter
and the area enclosed by the pentagon correct to two decimal places.
(Given 10 = 3.16)
A
5.5
E
B
4
C
Math Class VIII
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13
4.5
D
Question Bank
Ans. Join BE
A
5.5
B
5.5
4.5
4.5
E
4
4.5
C
M
4.5
D
In ∆ ABE , AB = 5.5 cm, AE = 5.5 cm
BE = 4.5 + 4.5 = 9 cm.
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s=
5.5 + 5.5 + 9
= 10
2
Area of ∆ ABE = 10(10 − 5.5) (10 − 5.5) (10 − 9)
= 10 × 4.5 × 4.5 × 1
= 4.5 × 10 = 4.5 × 3.16 = 14.23 cm2
Area of rectangle BCDE = l × b = 9 cm × 4 cm = 36 cm2
Total area of pentagon = (14.23 + 36) cm2 = 50.23 cm2
Perimeter of pentagon = sum of all the sides
= 5.5 + 4 + 4.5 + 4.5 + 4 + 5.5 = 28 cm.
21. Find the area of the shaded region of each of the following figures,
given that all measurements area in centimetres :
16
4
4
4
16
8
5
3
4
3
5
10
6
24
4
5
20
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Question Bank
Ans.(i) Area of rectangle ABCD = 16 × 4 = 64 cm2
16
A
4
B
E
4
D
12
O 3
G
M 3 H
4
8
N
4
3
C
4
8
8
I
4
F
P
J
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B
4
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Area of rectangle EFGH = 8 × 12 = 96 cm2
Area of rectangle IJKL = 20 × 4 = 80 cm2
1

2
Area of two triangles = 2  × 3 × 4  = 12 cm .
2

Total area of the shaded region = Area of rectangle ABCD +
area of rectangle EFGH + area of rectangle IJKL + area of
∆ MNH + area of ∆ PGO
= 64 + 96 + 80 + 6 + 6 = 252 cm2
1
1
2
(ii) Area of ∆ ABC = × b × h = × 10 × 5 = 25 cm
2
2
Area of rectangle ACDI = L × B = 16 × 10 = 60 cm2
16
A
B
10
C
5
C
I
G
H
16
8
24
D
5
F
6
E
5
Area of rectangle HDEG = 8 × 6 = 48 cm2
Area of ∆ EFG =
Math Class VIII
s(s − a) (s − b) (s − c)
5+5+6


where
s
=
=
8,
a
=
6,
b
=
5,
c
=
5


2
15
Question Bank
= 8(8 − 6) (8 − 5)(8 − 5) = 8 × 2 × 3 × 3 = 12 cm 2
Area of the shaded region = (25 + 160 + 48 + 12) cm2 = 245
cm2.
22. A rectangular lawn 75 m by 60 m has two roads each 4 m wide
running in the middle of it, one parallel to length and the other
parallel to breadth. Find the cost of gravelling the roads at Rs. 4.50
per sq. metre.
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4m
4m
60 m
75 m
Ans.Area of the road length wise = (75 × 4) m2 = 300 m2
Area of the road breadth wise = (60 × 4) m2 = 240 m2
Area of the common road = (4 × 4) m2 = 16 m2
∴
Area of the road which to be gravelled
= (300 + 240 – 16) m2 = (540 – 16) m2 = 524 m2
Cost of gravelling l m2 = Rs 4.50
Cost of gravelling 524 m2 = Rs 4.50 × 524 = Rs 2358.
23. The length and breadth of a rectangular park are in the ratio 5 : 2.
A 2.5 m wide path running all around the outside of the park has
an area of 305 m2. Find the dimensions of the park.
Ans.Area of path = 305 m2
Let length of park (l) = 5x, then breadth (b) = 2x
∴
Area = l × b = 5x × 2x = 10 x2
Width of path = 2.5 m
∴
Outer length (L) = 5x + 2 × 2.5 = (5x + 5) m
and breadth (B) = 2x + 2 × 2.5 = 2x + 5 m
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Question Bank
∴
Area of path = L × B – l × b = (5x + 5) (2x + 5) – 5x × 2x
= 10x2 + 25x + 10x + 25 – 10x2
∴
35x + 25 = 305 ⇒ 35x = 280
⇒ x=
280
=8
35
Length of the park = 5x = 5 × 8 = 40 m
and breadth of park = 2x = 2 × 8 = 16 m
24. Find the area of the shaded region of each of the given figures :
∴
A
Z
B
6 cm
F
E
E
5
cm
D 3 cm
C
cm
cm
A
(i)
D
E
B
6 cm
(ii)
C
C
17 cm
G
C
25 cm
A
5cm
H
B
(iii)
E
D
9 cm
17 cm
cm
9 cm
14 cm
20
F
5cm
15
G
H
5
D
B
17 cm
6 cm
8 cm
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B
A
(iv)
Ans. (i) In ∆ BDC
Using Pythagoras Theorem we have
BD = (BC) 2 − (DC) 2 = (15)2 − (9)2
[ ∵ DC = EC – AB = 17 – 8 = 9 cm]
= 225 − 81 = 144 = 12 cm
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1
× BD × DC
2
1
= × 12 × 9 cm 2 = 54 cm 2
2
Area of the rectangle ABDE = AB × BD = 8 × 12 cm2 = 96 cm2
Hence, Area of the shaded region
∴
Area of ∆ BDC =
Z
B
= area of ∆ BDC + area of rectangle ABDE
= (54 + 96) cm2 = 150 cm2
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(ii) In ∆ BDC
Using Pythagoras theorem we have,
BD = (BC) 2 − (CD)2 = (5) 2 − (3)2 = 25 − 9
= 16 = 4 cm
Similarly, AG = 4 cm
1
1
Thus, area of ∆ BDC = × BD × DC = × 4 × 3 cm3 = 6 cm 2
2
2
2
Similarly, area of the ∆ AGH = 6 cm = 6 cm2
Area of rectangle ABEF = AB × BE = 6 × ( 6 + 4) cm2
= 6 × 10 cm2 = 60 cm2
Hence, area of the shaded region
= ( 6 + 6 + 60) cm2 = 72 cm2
(iii) Area of the rectangle ABCD
= AB × BC = 25 × 14 cm2 = 350 cm2
In ∆ AEB , Using Pythagoras theorem, we have
AE = (AB)2 − (EB) 2 = (25) 2 − (20) 2
= 625 − 400 = 225 = 15 cm
Math Class VIII
18
Question Bank
1
1
2
Area of the ∆ AEB , = × AE × EB = × 15 × 20 cm = 150 cm2
2
2
Area of the shaded region
= area of rectangle ABCD – area of ∆ ABE
= (350 – 150) cm2 = 200 cm2
(iv) In ∆ AHB
AH = AG – GH = 17 – (9 + 5) = 17 – 14 = 3 cm
Area
of
1
 2 15 2
2
1
∆ AHB = × AH × HB =  × 3 × 5  cm = cm = 7.5 cm
2
2

2
From C draw perpendicular ar on GF
Then, area of region in two rectangular forms
= [(9 × 5) + (17 – 3) × 5] cm2
= [45 + 14 × 5] cm2
= (45 + 70) cm2 = 115 cm2
Z
B
L
A
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Area of ∆ DEF =
1
1
× DF × FE = × 5 × 3 cm 2
2
2
15
cm 2 = 7.5 cm 2
2
∴ Area of the shaded region
= (7.5 + 115 + 7.5) cm2 = 130 cm2.
25. The parallel sides of a trapezium are 17 cm and 25 cm. If the distance between them is 13 cm, find the area of the trapezium.
Ans. Area of trapezium
C
D
=
Math Class VIII
13 cm
1
= (AB + CD) × h
2
1
= × 42 × 13 = 273 sq. cm 2
2
A
19
25 cm
B
Question Bank
26. The cross-section ABCD of a swimming pool is a trapezium.
A
14 m
1.5 m
D
B
8m
C
Its width AB = 14 m, depth at the shallow end is 1.5 m and at the
deep end is 8 m. Find the area of the cross-section.
Ans.Here, two parallel sides of trapezium are AD and BC and distance
between them is 14 m.
1
1
2
Hence, area of trapezium = (1.5 + 8) × 14 = (9.5 × 14 ) = 66.5 m .
2
2
27. The cross-section of a canal is shown
8m
in the adjoining diagram. If the canal
is 8 m wide at the top and 6 m wide at
the bottom and the area of the crosssection is 16.8 m2, calculate its depth.
Ans.Area
of
trapezium
ABCD
6m
1
= × ( AB + CD ) × h
2
8m
D
C
But, area of trapezium = 16.8 m2
Z
B
L
A
A
N
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∴ 16.8 =
1
× (6 + 8) × h
2
16.8
= 2.4 h = 2.4 m
7
28. From the adjoining diagram, calculate
(i) the area of trapezium ACDE
(ii) the area of parallelogram ABDE
(iii) the area of triangle BCD
⇒ h=
A
Math Class VIII
20
A
E
h
B
6m
7 cm
D
6.5 m
B
13 m
C
Question Bank
Ans.
(i) Area of trapezium ACDE
1
1
× (AC + DE) × h = × (13 + 7) × 6.5 = 65 m 2
2
2
(ii) Area of parallelogram ABDE = b × h = 7 × 6.5 = 45.5 m2
=
1
1
2
(iii) Area of ∆ BCD = × BC× h = × 6 × 6.5 = 19.5 m
2
2
29. Calculate the area enclosed by the following shapes. All measurements are in cm.
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3
4
2
16
10
Z
B
2
7
9
5
7
2
14
(i)
9
7
(ii)
(iii)
Ans. (i) ABCD is a trapezium
C
D
16
10
A
14
B
1
Hence, Area of trapezium ABCD = × (10 + 16) × 14 = 182 cm2
2
Math Class VIII
21
Question Bank
(ii) Area
of
trapezium
1
= (AB + CD) × AD
2
D
ABCD
C
3
4
E
F
2
1
= (5 + 3) × 9 = 36 cm 2
5
5
2
Area of rectangle GAFE = 2 × 5 = 10
2
cm2
A
G
2
Total area of the figure = (36 + 10) cm =
A B
46 cm2
(iii) Area of rectangle ABCD = 9 × 2 = 18 cm2
2
1
Area of rectangle EFGH = 9 × 2 = 18 cm2
Area of parallelogram BIHJ = 2 × 5 = 10
cm2
7
Total area of the figure = (18 + 18 + 10)
2
cm2 = 46 cm2
D C
30. From the adjoining sketch, calculate
15 cm
D
(i) the length AD
(ii) the area of trapezium ABCD
41 c
m
(iii) the area of triangle BCD
Z
B
5
7
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A
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B
E
F
2
7
2 J
H
G
9
C
Ans. (i) In ∆ ABD, we have
A
40 cm
BD2 = AD2 + AB2 ⇒ AD2 = BD2 – AB2 = (41)2 – (40)2
= 1681 – 1600 = 81
B
∴ AD = 81 = 9 cm.
1
(ii) Area of trapezium ABCD = (AB + CD) × AD
2
1
= × (40 + 15) × 9 = 247.5 cm 2
2
Math Class VIII
22
Question Bank
(iii) Area of triangle BCD = Area of trapezium ABCD – Area
1
of ∆ ABD =  247.5 − × 40 × 9  cm 2
2


= (247.5 – 180) cm2 = 67.5 cm2
31. The two parallel sides and the distance between them are in the
ratio 3 : 4 : 2. If the area of the trapezium is 175 cm2 ; find its
height.
Ans.Let the two parallel sides and the distance between them be 3x, 4x
and 2x cm respectively,
Area =
Z
B
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A
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1
× (sum of parallel sides) × (distance between parallel sides)
2
1
= (3 x + 4 x) × 2 x = 175 (given
2
⇒ 7x × x = 175 ⇒ 7x2 = 175 ⇒ x2 = 25 ⇒ x = 5
∴ Height i.e., distance between parallel sides = 2x = 10 cm
32. A parallelogram has sides of 15 cm and 12 cm; if the distance
between the 15 cm sides is 6 cm; find the distance between 12 cm
sides.
15 cm
D
C
Ans. Base AB = 15 cm
Distance between 15 cm sides
12 cm
i.e., Height DP = 6 cm
Q
∴ Area of parallelogram = base ×
6 cm
2
height = AB × DP = 15 × 6 = 90 cm A P 15 cm
B
Let BQ be distance between 12 cm sides
AD × BQ = area of parallelogram ABCD
∴
90
= 7.5 cm
12
33. A parallelogram has sides of 20 cm and 30 cm. If the distance
between its shorter sides is 15 cm; find the distance between the
longer sides.
∴
Math Class VIII
12 × BQ = 90 BQ =
23
Question Bank
Ans. Let ABCD be the parallelogram in which BC = 30 cm and
CD = 20 cm
Distance between shorter sides,
D
A
30 cm
i.e., CQ = 15 cm
Area of parallelo∴
gram = AB × CQ = 20 × 15
20 cm
Q
= 300 cm2
15 c
m
20
cm
Again BC × AP =
Area of parallelogram
B
30 cm
P
C
30 × AP = 300
Z
B
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A
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300
= 10 cm.
30
Hence, distance between larger sides is 10 cm.
34. In a parallelogram ABCD; AB = 10 cm, AD = 24 and BD = 26 cm,
find the area of parallelogram.
B
C
Ans. Firstly, we find area of ∆ ABD
Sides are a = 10 cm, b = 26 cm, 10 cm
26 cm
and c = 24 cm
AP =
s=
=
a+b+c
2
A
24 cm
D
10 + 26 + 24 60
=
= 30
2
2
Thus, area of ∆ ABD = s ( s − a ) ( s − b) ( s − c)
= 30(30 − 10) (30 − 26) (30 − 24)
= 30 × 20 × 4 × 6 = 5 × 6 × 4 × 5 × 4 × 6
= 5 × 5 × 6 × 6 × 4 × 4 = 120 cm2
∵
Diagonal bisects the parallelogram in two equal parts.
Math Class VIII
24
Question Bank
Area (parallelogram ABCD) = 2 × area (∆ ABD)
= 2 × 120 = 240 cm2
35. The adjacent sides of a parallelogram are 21 cm and 28 cm. If its one
diagonal is 35 cm; find the area of
A
D
the parallelogram.
Ans. Firstly, we find area of ∆ ABC.
21 cm
35 cm
Sides are, a = 28 cm, b = 35 cm
and c = 21 cm
a+b+c
28 cm
B
C
s=
2
28 + 35 + 21 84
s=
=
= 42
2
2
Area of ∆ ABD = s ( s − a ) ( s − b) ( s − c)
∴
Z
B
L
A
A
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= 42(42 − 28) (42 − 35) (42 − 21) = 42 × 14 × 7 × 21
= 2 × 21 × 2 × 7 × 7 × 21 = 2 × 2 × 21 × 21 × 7 × 7
= 2 × 21 × 7 = 294 cm2
Thus, diagonal of parallelogram divides it into two equal parts.
Hence, area of parallelogram = 2 × area of ∆ ABC
= 2 × 294 = 588 cm2.
36. PQRS is a parallelogram with PQ = 26 cm and QR = 20 cm. If the
distance between its longer sides is 12.5 cm, find :
(i) the area of the parallelogram ;
(ii) the distance between its shorter sides.
S
R
26 cm
Ans. (i) Case I
Base = (b) = 26 cm
M 12.5 cm
20 cm
Altitude (h) = 12.5 cm
20 cm
∴ Area of PQRS = b × h = 26
× 12.5 cm = 325 cm2
P
Math Class VIII
25
L
26 cm
Q
Question Bank
(ii) Case II
Base (shorter side) = 20 cm
Area 325
=
= 16.25 cm
Base 20
37. ABCD is a parallelogram having adjacent sides AB = 16 cm and BC
= 14 cm. If its area is 168 cm2, find the distance between its longer
sides and that between its shorter sides.
Ans. Area of parallelogram ABCD = 168 cm2
D
16 cm
C
Longer side AB = 16 cm
Shorter side BC = 14 cm
M
14
cm
Let DL ⊥ AB and BM ⊥ AD
14 cm
Hence, height =
Z
B
L
A
A
N
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∴ Length DL =
Area 168
=
= 10.5 cm
Base 16
A
L
B
16 cm
Area Area 168
=
=
= 12 cm. [∵ AD = BC]
AD
BC
14
38. In the adjoining figure, ABCD is a parallelogram in which AB = 28
cm, BC = 26 cm and diagonal AC = 30 cm. Find
(i) the area of parallelogram ABCD ;
(ii) the distance between AB and DC ;
(iii) the distance between CB and DA.
and length of BM =
C
D
cm
M
30
26 cm
A
28 cm
L
B
Ans. In parallelogram ABCD
(i) In ∆ABC, AB = 28 cm, BC = 26 cm and AC = 30 cm
Math Class VIII
26
Question Bank
1
area (ABCD)
2
(∵ A diagonal bisects the parallelogram into two triangles of
equal area)
Now, sides of triangle ABC are 28 cm, 26 cm, 30 cm
Area of triangle ABC =
a + b + c 28 + 26 + 30 84
=
=
= 42
2
2
2
∴
s=
∴
Area of ∆ ABC = s ( s − a ) ( s − b) ( s − c)
Z
B
L
A
A
N
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= 42(42 − 28) (42 − 26) (42 − 30)
= 42 × 14 × 16 × 12 = 42 × 14 × 16 × 12
= 2 × 3× 7 × 7 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
= 22 × 22 × 22 × 22 × 32 × 7 2
= 2 × 2 × 2 × 2 × 3 × 7 = 336 cm2
Hence, area of parallelogram ABCD = 2 × 336 = 672 cm2
(ii) Distance between AB and DC
Area 672
=
cm = 24 cm
AB
28
(iii) Distance between CB and DA
Area 672
CM =
=
= 25.84 cm
AD
26
39. The parallel sides of a trapezium are 20 cm and 10 cm, Its nonparallel sides are both equal, each being 13 cm. Find the area of
the trapezium.
CL =
10 cm
cm
13
13
Math Class VIII
cm
A
C
13
cm
D
20 cm
E
27
F
B
Question Bank
Ans. Let ABCD is a trapezium
Draw CE || AD and CF ⊥ AB
CE = AD = 13 cm
∴
and AE = DC = 10 cm
EB = AB – AE = 20 – 10 = 10 cm
s=
a + b + c 13 + 13 + 10 36
=
=
= 18
2
2
2
Z
B
∴ Area of ∆ ECB = s ( s − a ) ( s − b) ( s − c )
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A
A
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= 18(18 − 13) (18 − 13) (18 − 10)
= 18 × 5 × 5 × 8 = 2 × 3 × 3 × 5 × 5 × 2 × 2 × 2
= 2 × 2 × 3 × 5 = 60 cm2
Again, area of ∆ ECB = 60 cm 2
1
× 10 × CF = 60
2
60
CF
=
5
×
CF
=
60
⇒
⇒
⇒ CF = 12 cm.
5
1
Hence, area of trapezium ABCD = × (AB + CD) × CF
2
1
1
= (20 + 10) × 12 = × 30 × 12 = 180 cm 2
2
2
40. Find the area of the figure ABCDEFGH, given below, it being given
that AC = 17 m, BC = 8m, EF = 9m, CD = 6 m, GL ⊥ EF and GL =
3.6 m.
⇒
1
× FB × CF = 60
2
FL
H
⇒
9m
3.6 m
G 6m
m
17
A
Math Class VIII
E
D
C
8m
B
28
Question Bank
Ans. Area of the trapezium EFGD
1
1
= (EF + GD) × GL = × (9 + 6) × 3.6 m2
2
2
1
= × 15 × 3.6 m 2 = 27 m2
2
Using Pythagoras theorem, we have
Z
B
2
2
AB = (AC)2 − (BC) 2 = (17) − (8)
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= 289 − 64 = 225 = 15 cm.
Area of rectangle ABCH = AB × BC
= 15 × 8 cm2 = 120 cm2
Hence, area of the figure ABCDEFGH = area of trapezium EFGD +
area of rectangle ABCH
= (27 + 120) cm2 = 147 cm2
41. Find the area of the shaded region in the figure given below :
3.5 m
3m
2.5 m
4.6 m
6m
Ans.From figure,
Area of the upper rectangular part = 3.5 × 3 = 10.5 m2
In trapezium part, distance between parallel sides
= (4.6 – 3) m = 1.6 m
Math Class VIII
29
Question Bank
1
1
2
Area of trapezium = × (6 + 3.5) × 1.6 = × 9.5 × 1.6 = 7.6 cm
2
2
Area of lower rectangular part = (6 × 2.5) m2 = 15 m2
Hence, Area of the shaded region
= (10.5 + 7.6 + 15) m2 = 33.1 m2
42. Find the area of figure ABCDE, it being given that :
Z
B
AE || BD, AF ⊥ BD, CG ⊥ BD
AE = 12 cm, BD = 16 cm, AF = 6.5 cm and CG = 8.5 cm
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A
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E
A
B
G
F
D
C
Ans. This figure consists of two figures one is a triangle and other is a
trapezium
1
1
2
∴ Area of ∆ BCD = BD × CG = × 16 × 8.5 cm = 68 cm2
2
2
Area of trapezium ABDE
=
1
1
(BD + AE) × AF = × (16 + 12) × 6.5 cm 2
2
2
1
× 28 × 6.5 = 91 cm 2
2
Hence, area of the figure ABCDE = Area of ∆BCD + Area of the
pezium ABDE = 68 + 91 = 159 cm2.
43. Find the area of the field ABCDEFA in which BP ⊥ AD,
CR ⊥ AD, FQ ⊥ AD, ES ⊥ AD, and AP = 20 m, AQ = 35 m,
AR = 58 m, AS = 65 m, AD = 75 m, BP = 15 m, CR = 20 m, ES = 15
m and RQ = 10m
=
Math Class VIII
30
Question Bank
E
F
A
P
R
Q
S
D
B
C
Ans. This figure of a field contains triangles and trapeziums
Z
B
1
1
2
Area of ∆ APB = AP × BP = × 20 × 15 = 150 m
2
2
1
Area of ∆ RCD = × RD × RC
2
1
1
= (AD − AR) × RC = × (75 − 58) × 20
2
2
1
= × 17 × 20 = 170 m 2
2
1
Area of ∆ ESD = SD × ES
2
1
1
= (AD- AS) × ES = (75 − 65) × 15
2
2
1
= × 10 × 15 = 75 m 2
2
1
1
2
Area of ∆ AQF = AQ × FQ = × 35 × 10 = 175 m
2
2
Area of trapezium BPRC
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1
1
= (BP + CR) × PR = (15 + 20) × (AR − AP)
2
2
=
Math Class VIII
1
1
× (15 + 20) × (58 − 20) = × 35 × 38 = 665 m 2
2
2
31
Question Bank
and area of trapezium QFES
1
1
= ( FQ + ES ) × QS = ( FQ + ES ) × ( AS − AQ )
2
2
1
1
2
= (10 + 15) × (65 − 35) = × 25 × 30 = 375 m
2
2
Hence, area of the total field ABCDEFA
= (150 + 170 + 75 + 175 + 665 + 375) m2
= 1610 m2.
44. Cotton thread is wound on a reel which has diameter of 2.8 cm.
There are 1200 turns of the thread on the reel. What is the total
length, in metre of the thread on the reel ?
Z
B
L
A
A
N
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22
× 1.4 = 8.8 cm.
7
Length taken by thread in one turn = 8.8 cm.
⇒ Total Length taken by thread in 1200 turns
Ans.Circumference of reel = 2πr 2 ×
10560
m =105.6 m
100
45. How many times will the wheel of a car rotate in a journey of 88
km, given that the diameter of the wheel is 56 cm.
Ans. Distance covered by wheel of car = 88 km = 88 × 1000 × 100 cm.
Diameter of wheel is 56 cm
= 8.8 × 1200 = 10560 cm
56 cm
= 28 cm.
2
⇒ Distance covered by wheel of car in one turn = 2πr
22
2 × × 28 = 176 cm.
7
Number of times wheel rotates
∴
radius =
=
Math Class VIII
88 × 1000 × 100
= 1000 × 50 = 50,000
176
32
Question Bank
46. A Tonga is being driven at 11 km/hr. If each wheel of the Tonga is 1.4
m in diameter, find the number of revolutions made by each wheel
per minute.
Ans. Distance covered by wheel of Tonga in one revolution = 2πr
22 1.4
×
= 4.4 m
7
2
Distance covered by tonga in 60 minutes = 11 km. = 11,000 m.
11,000
Distance covered by tonga in 1 minute =
60
Number of revolutions in one minute
= 2×
Z
B
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11000
11000 10 250 125
2
=
÷ 4.4 =
×
=
=
= 41
60
60
44
6
3
3
47. From a square cardboard of side 21cm, a circle of maximum area is
21 cm
cut out. Find the area of the cardboard left.
Ans. Circle of maximum area can be cut out from a
square if diameter of circle is equal to side of the
square.
d
2
2
∴ Area of square = (21 × 21) cm = 441 cm
21 
22 21 21 693 
∵
r
=
× × =


2
7 2 2
2 
Area of remaining cardboard
Area of circle =
693  2  882 − 693  2 189

=  441 −
cm 2 = 94.5 cm2
 cm = 
 cm =
2 
2
2



48. A wire is in the form of a square of side 27.5 cm. It is straightened
and bent into the shape of a circle. Find the area of the circle.
Ans.Perimeter of square = Circumference of circle
4 × side = 2πr ⇒ 4 × 27.5 = 2 ×
Math Class VIII
33
22
×r
7
Question Bank
⇒ r=
4 × 27.5 × 7 2 × 2.5 × 7
= 17.5 cm
=
2 × 22
2
 22

Area of circle =  × 17.5 × 17.5  cm 2 = 962.5 cm2.
 7

49. Three cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and
recast into a single cube. Find the cost of coating the surface of the
new cube with gold at the rate of Rs. 3.50 per square centimetre?
Ans. Let a cm be the edge of new cube
According to the given condition
a3 = 33 + 43 + 53 = 27 + 64 + 125 = 216 cm3
Z
B
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a = 3 216 ⇒ a = 6 cm.
Suface area of new cube = 6 × (side)2 = 6 × (6)2 = 216 cm2
Cost of coating the surface of new cube = Rs 3.50 × 216 = Rs 756.
50. In the adjoining figure, the area enclosed between the concentric circles
is 770 cm2. If the radius of the outer circle is 21 cm, calculate the
radius of the inner circle.
Ans. Radius of outer circle (R) = 21 cm.
Radius of inner circle (r) = r cm.
Area of shaded protion = 770 cm2
⇒
⇒
⇒
⇒
π( R 2 − r 2 ) = 770
22 2 2
(21 − r ) = 770
7
7
441 − r 2 = 770 ×
= 35 × 7 = 245
22
⇒ r2 = 196
r 2 = 441 − 245
r2 = 196
⇒
⇒ r = 196
r = 14 cm.
⇒
51. Find the area of a flat circular ring formed by two concentric circles
Math Class VIII
34
Question Bank
(circles with same centre whose radii are 9 cm and 5 cm
Ans.
9
r 1=
cm
r2=5 cm
Z
B
External radius, r1 = 9 cm
internal radius, r2 = 5 cm
∴
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A
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N
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IN
πr12
Area of ring =
21 cm
− πr22
=
π(r12
− r22 )
2
= π(9 − 52 )
22
22
[81 − 25] =
× 56 = 176 cm2.
7
7
52. Find the area of the shaded portion in each the following diagrams :
4.5
=
7 cm
2.5 m
(i)
Ans. (i)
(ii)
Radius of circle (r) = 7 cm
Side of square = 7 + 7 = 14 cm
Area of circle = πr 2
22
× 7 × 7 = 154 cm2
7
Area of square = 14 × 14
Area of shaded portion
= 196 – 154 = 42 cm2
(ii) Radii of concentric circles are
=
Math Class VIII
35
7 cm
7 cm
Question Bank
Hence, area of shaded portion = πr12 − πr22
= π[r12 − r22 ] =
r1 =
4.
5c
m
r1 = 4.5 m, r2 = 2.5 m
22
22 
r =2.5cm
× 20.25 − 6.25 =
(4.5) 2 − 2(2.5) 2 

7
7 
22
× 14 = 44 cm 2
7
53. The shape of a park is a rectangle bounded by semicircles at the ends,
each of radius 17.5 m, as shown in the adjoining figure. Find the
area and the perimeter of the park.
=
Z
B
L
A
A
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35 m
35 m
40 m
Ans. Diameter of the part = 35 cm
i.e. radius of the circular part = 17.5 m
∴ Area of two semi-circular parts
1
22
= 2 × πr 2 = πr 2 =
× (17.5)2 m 2
2
7
22
× 17.5 × 17.5 m 2 = 962.5 m 2
7
Now, area of the rectangular part = (40 × 35) m2 = 1400 m2
∴ Area of the park = (962.5 + 1400) m2 = 2362.5 m2.
Now, Kcircumference of a semi-circle
=
1
22
× 2πr = πr =
× 17.5 m = 55 m
2
7
Hence, perimeter of the park = 40 + 55 + 40 + 55 = 190 m
54. Find the area of the space enclosed by two concentric circles of radii
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17 cm and 11cm.
Ans. Here, R = 17 cm and r = 11 cm
Required area = π[(17)2 − (11) 2 ] cm 2
22
22
× (17 + 11) (17 − 11) cm 2 =
× 28 × 6 cm 2 = 528 cm 2
7
7
17 cm
Hence, area of the space = 528 cm2
55. The diameter of a circular park is 52 m. On
its outside, there is path 4 m wide, running
O 11 cm
around it. Find the cost of turfing the path at
Rs. 2.50 per square metre.
Ans. Diameter of the park = 52 m. So, Radius
52
of the park =
= 26 m
2
Width of path = 4 m
Here, r = 26 m and R = (26 + 4) m = 30 m
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Area of the path = π[ R 2 − r 2 ] = π[(30) 2 − (26) 2 ] m 2
=
22
(30 + 26) (30 − 26) m 2
7
22
× 56 × 4 m 2 = 704 m2
7
Now, cost of turfing 1m2 = Rs 2.50
Cost of turfing 704 m2 = Rs 2.50 × 704 = Rs 1760.
56. The sum of diameters of two circles is 2.8 m and the difference of
their circumferences is 0.88 m. Find the radii of two circles.
Ans. Let diameter of two circles be d1 and d2
d1 + d2 = 2.8 m
∴
or
2r1 + 2r2 = 2.8
(∵ d = 2r)
r1 + r2 = 1.4
...(i)
Also,
2πr1 − 2πr2 = 0.88 m
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2π(r1 − r2 ) = 0.88 m
1 7
r1 − r2 = 0.88 × ×
= 0.14
...(ii)
2 22
(i) and (ii), we have
2r1 = 1.54. r1 = 0.77 m or 77 cm.
r2 = 1.4 – r2 = 1.4 – 0.77
= 0.63 m or 63 cm.
57. Calculate the length of the boundary and the area of the shaded
region in the following diagrams. All measurements are in
centimetres.
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10
28
7
(i) Unshaded part is a semi-circle
(ii) Four semi-circles on a square
Ans. (i) Length of boundary
1
= AB + BC + CD + (2πr )
2
+
A
D
22 7
7/2
7/2
= 10 + 7 + 10 + ×
7 2
= 10 + 7 + 10 + 11 = 38 cm.
10 ABCD – area 10
Area of shaded portion = area of rectangle
of
semi-circle
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1 22 7 7
77 C
= 10 × 7 − (πr 2 ) = 70 − × × B× = 707 −
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2 7 2 2 Question
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38
= 70 – 19.25 = 50.75 cm2
(ii) Length of one semi-circle = πr
22
× 7 = 22 cm.
7
Length of 4 semi-circles
= 22 × 4 = 88 cm.
=
7 2
Area of four semi-circles = Area of two circles = 2πr
2 × 22
=
× 7 × 7 = 308 cm 2 .
14
14 14
7
Area of square = 14 × 14 = 196 cm2
Area of shaded portion = 308 + 196 = 504 cm2. 7
58. In the adjoining figure, ABCD is a square of side 14 cm. A, B, C and
D are centres of circular arcs of radius 7 cm. Find the perimeter and
the area of the shaded region.
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D
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A
B
Ans. Perimeter of shaded portion
+ FG
+ GH
+ EH
= Perimeter of circle = 2πr
= EF
= 2×
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× 7 = 44 cm.
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D
7
7
E
C
7
7
F
H
7
7
A
7
G
7
2
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B
B
Area of square = 14 × 14 = 196 cm
Area of four quadrants or area of on circle
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22
= πr =
× 7 × 7 = 154 cm 2
7
Area of shaded portion = (196 – 154) cm2 = 42 cm2.
59. The boundary of the shaded region in the given figure consists of
three semi-circles, the smaller being equal. If the diameter of the
larger one is 28 cm, find
(i) the length of the boundary
(ii) the area of the shaded region
Ans. Length of boundary
2
length of AB + 2(length of AQ )
=
1
1
 28 
 28  
× 2π   + 2  × 2π   
2
 2 
 4 
2
Q
= π × 14 + π × 14
π (14 × 14) =
A
P
B
22
× 28 = 22 × 2 + 22 × 2 = 88 cm.
7
Area of the shaded region =
=
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1
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2
1 22
× × 14 × 14 = 308 cm 2
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60.Find the volume, the total surface area and the lateral surface area of
the cuboid having
(i) length = 24 cm, breadth = 16 cm and height = 7.5 cm
(ii) length = 10 m, breadth = 35 cm and height = 1.2 m
Ans. (i) Here l = 24 cm, b = 16 cm, h = 7.5 cm
∴ Volume of the cuboid
= 2 (lb + bh + hl)
= 2 (24 × 16 + 16 × 7.5 + 7.5 × 24) cm2
= 2 (384 + 120 + 180) cm2
= 2 × 684 cm2 = 1368 cm2
Lateral surface area of the cuboid = 2 (l + b) × h
= 2 [24 + 16] × 7.5 cm2
= 2 × 40 × 7.5 cm2 = 600 cm2
(ii) Here, l = 10 m, b = 35 cm and h = 1.2 m
∴ Volume of the cuboid = l × b × h
= 10 × 0.35 × 1.2 m2 = 4.2 m3
Total surface area of the cuboid
= 2 [lb + bh + hl]
= 2 [10 × 0.35 + 0.35 × 1.2 + 1.2 × 10] m2
= 2 [3.5 + 0.42 + 12] m2 = 2 × 15.92 m2 = 31.84 m2
Lateral suface area of the cuboid
= 2 (l + b) × h = 2 [10 + 0.35] × 1.2 m2
= 2 × 10.35 × 1.2 m2 = 24.84 m2
61. A wall of length 13.5 m, width 60 cm and height 1.6 m is to be
constructed by using bricks, each of dimensions 22.5 cm by 12
cm 8 cm. How many bricks will be needed ?
Ans.Length of the wall (l) = 13.5 m
= 13.5 × 100 cm = 1350 cm
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Breadth of the wall (b) = 60 cm
Height of the wall (h) = 1.6 m
= 1.6 × 100 cm = 160 cm
Volume of the wall = l × b × h
= (1350 × 60 × 160) cm3
Volume of a brick = (22.5 × 12 × 8) cm3
Hence, required number of bricks used
Z
B
1350 × 60 × 160 1350 × 60 × 160 × 10
=
22.5 × 12 × 8
225 × 12 × 8
= 6 × 5 × 20 × 10 = 6000
62. Find the length of each edge of a cube, if its volume is :
(i) 216 cm3
(ii) 1.728 m3
Ans. (i) (Edge)3 = Volume of a cube
(Edge)3 = 216 cm3
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Edge = (3 × 3 × 3 × 2 × 2 × 2)1/ 3
Edge = 3 × 2
⇒
⇒ Edge = 6 cm
(ii) (Edge)3 = Volume of a cube
(Edge)3 = 1.728 m3
∴
⇒
1.728 1728
(Edge) =
=
1000 1000
3
1/ 3
 1728 
⇒ Edge = 

 1000 
1/ 3
 2 × 2 × 2 × 2 × 2 × 2 × 3× 3× 3 
Edge = 
⇒

10 × 10 × 10


2× 2×3
12
Edge =
⇒
⇒ Edge = m
10
10
Edge = 1.2 m
⇒
63. The total surface area of a cube is 216 cm2. Find its volume.
Ans. 6(Edge)2 = Total surface area of a cube.
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6 (Edge)2 = 216 cm2
⇒
(Edge) 2 =
216
6
⇒
(Edge)2 = 36
⇒
⇒ Edge = 6 cm
Edge = 36
Volume of the given cube = (Edge)3
= (6)3 = 6 × 6 × 6 = 216 cm3.
64. A wall 9 m long, 6 m high and 20 cm thick, is to be constructed ;
using bricks of dimensions 30 cm, 15 cm and 10 cm. How many
bricks will be required.
Ans. Length of the wall = 9 m = 9 × 100 cm = 900 cm
Height of the wall = 6 m = 6 × 100 cm = 600 cm
Breadth of the wall = 20 cm
Volume of the wall = (900 × 600 × 20) cm3 = 10800000 cm3
Volume of the one brick = 30 × 15 × 10 cm3 = 4500 cm3
Number of bricks required to construct the wall
Volume of wall
10800000
=
=
= 2400.
Volume of one brick
4500
65. The dining-hall of a hotel is 75 m long; 60 m broad and 16 m high.
It has five-doors 4 m by 3 m each and four windows 3 m by 1.6 m
each. Find the cost of :
(i) papering its walls at the rate of Rs. 12 per m2;
(ii) carpetting its floor at the rate of Rs. 25 per m2.
Ans. Length of the dining hall of a hotel = 75 m
Breadth of the dining hall of a hotel = 60 m
Height of the dining hall of a hotel = 16 m
(i) Area of four walls of the dining hall
= 2 (L + B) × H = 2(75 + 60) × 16
= 2 (135) × 16 = 270 × 16 = 4320 m2
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Area of one door = 4 × 3 m2 = 12 m2
Area of 5 doors = 12 × 5 = 60 m2
Area of one window = 3 × 1.6 = 4.8 m2
Area of 4 windows = 4.8 × 4 = 19.2 m2
Area of the walls to be papered
= 4320 – (60 + 19.2) = 4320 – 79.2 = 4240.8 m2
Cost of papering the walls at the rate of Rs 12 per m2
= 4240.8 × 12 = Rs 50889.60
(ii) Area of floor = L × B = 75 × 60 = 4500 m2
Cost of carpetting the floor at the rate Rs 25 per m2
= 4500 × 25 = Rs 112500
66. Find the volume of wood required to make a closed box of external dimensions 80 cm, 75 cm and 60 cm, the thickness of wall of
the box being 2 cm throughout.
Ans.External length of the closed box = 80 cm
External breadth of the closed box = 75 cm
External Height of the closed box = 60 cm
External volume of the closed box = 80 × 75 × 60 = 360000 cm3
Internal length of the closed box = 80 – 4 = 76 cm
Internal Breadth of the closed box = 75 – 4 = 71 cm
Internal Height of the closed box = 60 – 4 = 56 cm
Internal volume of the closed box = 76 × 71 × 56 cm = 302176
cm3
Volume of wood required to make the closed box = External volume of the closed box – Internal volume of the closed box
= 360000 – 302176 = 57824 cm3.
67. Find the length of the longest pole that can be placed in a room 12
m long, 8 m broad and 9 m high.
Ans.Length of room (l) = 12 m
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Breadth (b) = 8 m
and height = (h) = 9 m
Longest pole = l 2 + b 2 + h2 = (12)2 + (8)2 + (9) 2
= 144 + 64 + 81 = 289 = 17 m.
68. The volume of cuboid is 972 m3. If its length and breadth be 16 m
and 13.5 m respectvely, find its height.
Ans.Here, length of the cuboid (l) = 16 m
Breadth of the cuboid (b) = 13.5 m
Let height of the cuboid be h m
Volume of the cuboid = 972 m3
⇒ l × b × h = 972 ⇒ 16 × 13.5 × h = 972
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972
9
m ⇒ h= m
16 × 13.5
2
⇒ h = 4.5 m
Hence, height of the cuboid = 4.5 m
69. The volume of a cuboid is 1296 m3. Its length is 24 m and its
breadth and height are in the ratio 3 : 2. Find the breadth and
height of the cuboid.
Ans.Length of the cuboid = 24 m
Let breadth of the cuboid be 3x
and height of the cuboid be 2x
Volume of the cuboid = 1296 m2
⇒ l × b × h = 1296 ⇒ 24 × 3x × 2x = 1296 ⇒ 144x2 = 1296
⇒ h=
1296
⇒ x2 = 9 m ⇒ x = 9 ⇒ x = 3
144
Hence, the breadth of the cuboid = 3 × 3 = 9 m
Height of the cuboid = 2 × 3 = 6 m.
70. The surface area of a cuboid is 468 cm2. Its length and breadth are
12 cm and 9 cm respectively. Find its height.
2
⇒ x =
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Ans. Let height of the cuboid be h cm
Surface area of the cuboid = 468 cm2
⇒ 2 [lb + bh + hl] = 468
⇒ 2 [ 12 × 9 + 9 × h + h × 12] = 468
⇒ 2 [ 108 + 9h + 12h] = 468
⇒
⇒
468
2
21h = 234 – 108
21h + 108 =
Z
B
126 ⇒ h = 6 cm
h=
21
Hence, height of the cuboid = 6 cm.
71. A room 9 m long, 6 m broad and 3.6 m high has one door 1.4 m by 2
m and two windows, each 1.6 m by 75 cm. Find :
(i) the area of 4 walls, excluding doors and windows.
(ii) the cost of white-washing the walls from inside at the rate of
Rs. 2.50 per m2;
(iii) the cost of painting its ceiling at Rs. 5 per m2.
Ans. Length of the room (l) = 9 m
Breadth of the room (b) = 6 m
and height of the room (h) = 3.6 m
Area of 4 walls = 2 (l + b) × h
= 2 (9 + 6) × 3.6 = 2 × 15 × 3.6 m2 = 108 m2
Area of the door = 1.4 × 2 m2 = 2.8 m2.
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21h = 126 ⇒
75  2

Area of two windows = 2 1.6 ×
m

100


3

= 2 1.6 ×  m 2 = 2.4 m 2
4

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(i) Area of 4 walls, excluding doors and windows
= (108 – 2.8 – 2.4) m2 = (108 – 5.2) m2 = 102.8 m2.
(ii) Cost of white-washing the walls of 1m2 = Rs 2.50
Cost of white-washing the walls of 102.8 m2 = Rs. 102.8
× 8 250 = Rs 257.
(iii) Area of the ceiling = l × b = (9 × 6) m2 = 54 m2
Cost of painting the ceiling of 1m2 = Rs 5
Cost of painting the ceiling of 54 m2
= Rs 54 × 5 = Rs 270.
72. The length, breadth and height of a cuboid are in the ratio 7 : 6 : 5.
If the surface area of the cuboid is 1926 cm2, find its dimensions.
and also, find the volume of the cuboid.
Ans. Surface area of cuboid = 1926 cm2
Ratio in the length, breadth and height of a cuboid = 7 : 6 : 5
Let length of the cuboid (l) = 7x
Preadth of the cuboid (b) = 6x
and height of the cuboid (h) = 5x
Surface area = 2 (l × b + b × h + h × l)
= 2(7x × 6x + 6x × 5x + 5x × 7x)
= 2(42x2 + 30x2 + 35x2) = 2 × 107 x2
= 241 x2
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1926
= 9 = (3)2 ∴ x = 3
∴
214
Hence, length (l) = 7x = 7 × 3 = 21 cm
Breadth (b) = 6x = 6 × 3 = 18 cm
and height (h) = 5x = 5 × 3 = 15 cm
Then the volume of cuboid = l × b × h = (21 × 18 × 15) cm3
= 5670 cm3
2
214x2 = 1926 ⇒ x =
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73. If the areas of the three adjacent faces of a cuboidal box are 120 cm2,
72 cm2 and 60 cm2 respectively, then find the volume of the box.
Ans. Let l be the length, b be the breadth and h be the height of the cuboid
l × b = 120 cm2 ...(i) b × h = 72 cm2 ...(ii)
∴
h × l = 60 cm2 ...(iii)
Multiplying (i), (ii), (iii) we have
l2 b2 h2 = 120 × 72 × 60 = 518400 ∴
Z
B
lbh = 518400 = 720
Hence, volume of the cuboidal is box 20 cm3.
74. The external dimensions of wooden box, open at the top are 54 cm
by 30 cm by 16 cm. It is made of wood 2 cm thick. Calculate :
(i) the capacity of the box (ii) the volume of wood.
Ans. Internal length of the open box = 54 – 4 = 50 cm
Internal breadth of the open box = 30 – 4 = 26 cm
Internal height of the open box = 16 – 2 = 14 cm.
(i) Capacity of the open box = (50 × 26 × 14) cm3 = 18200 cm3.
(ii) Volume of the wood = (54 × 30 × 16 – 50 × 26 × 14) cm3
= (25920 – 18200) cm3 = 7720 cm3.
75. Three cubes of metal with edges 5 cm, 4 cm and 3 cm respectively
are melted to form a single cube. Find the lateral surface area of
the new cube formed.
Ans.Volume of three cubes
= [ (5)3 + (4)3 + (3)3] cm3 = [125 + 64 + 27] cm3 = 216 cm3
Now, volume of a big single cube = 216 cm3
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i.e. edge of a big single cube = 216 cm = 6 cm
Hence, lateral surface area of the new big cube
= 4 (edge)2 = 4(6)2 cm2 = 4 × 36 cm2 = 144 cm2
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