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Chapter 20: Mensuration
Page No: 229
Exercise 20A

Question 1:
Solution:
(i) Area of a rectangle = product of length and breadth = 24.5 18  441m2.
(ii) Area of a rectangle = product of length and breadth = 12.5  0.8  10 m2.

Question 2:
Solution:
Since the diagonal is 50 m, and one side is 48 m, the other side is given by
502  482  14 m.
So the area of this rectangle is given by the product of length and breadth = 48 14  672 m2.

Question 3:
Solution:
Since the sides are in the ratio, 4:3, let the sides be 4x and 3x. So its area is given by
1728
x2   144  x  12
12
So the sides are of length and bread 48 m and 36 m respectively.
The perimeter (fencing portion) is given by the sum of its sides = 48 + 36 + 48 + 36 = 168 m.
The cost of fencing it = 168  30  Rs.5040

Question 4:
Solution:
3584
Area of the rectangular field = product of length and breadth => breadth =  56 m.
64
Perimeter of the field = 64 + 56 + 64 + 56 = 240 m.
For him to go five rounds, the distance covered = 240  5  1200m .
1200 1
So, the time taken =  hr  12 min .
6000 5

Question 5:
Solution:
Area of the verandah = product of l and b = 40 15  600m 2  60000dm2 .
area of the stone = 6  5  30 dm 2 =
no. of stones required = area of verandah /area of the stone = 60000 / 30  2000
Therefore 2000 stones are required.

Question 6:
Solution:
The area of the room = 13  9  117m2
The area of the room must be equal to the area of the carpet. So,
13  9  117m 2  l  0.75m  l  156m
So the cost of carpeting = 156 105  Rs.16380 .

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Question 7:
Solution:
Let the width of the room be 'b' m
Area of the room = 15b m2
From the cost of the carpet, we determine length of the carpet = 19200 / 80 = 240 m
Area of the carpet = 240 x 0.75
Since the area of the room = area of the carpet => 15b  240  0.75
So, b   240  0.75 /15  12 .
Therefore, width of the room is 12 m.

Question 8:
Solution:
Given length = 5x and breadth = 3x
So the perimeter = 2(5x + 3x) = 16x = 9600/24 = 400
16x = 400 => x = 25
So the length = 125m and breadth = 75m.

Question 9:
Solution:
Given length = l = 10m; breadth = b = 10 m and height = h = 5 m
Length of the largest pole = l 2  b2  h2 = 100  100  25 = 225  15 m
The length of the largest pole = 15 m.

Question 10:
Solution:
Area of square = side × side = 8.5  8.5  72.25 m 2

Question 11:
Solution:
 72  72   5184  2592 cm2
(i) Area of square = 
2 2
 2.4  2.4 
(ii) Area of square =   2.88 m2
2

Question 12:
Solution:
Area of square = (side)² = 16200 m² => side = 16200  90 2 m
Diagonal of square = 2  side  2  90 2  180 m
The length of the diagonal = 180 m.

Question 13:
Solution:
1 hectare = 10000 m2
∴0.5hec = 5000m2
⇒ side of field = 50 2  diagonal  50  2  100m
The length of its diagonal = 100 m.

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Question 14:
Solution:
Given the area of a square plot is 6084 m2. So, the side of the square will be 78 m.
Perimeter = 4 times the side = 312 m.
So, 4 times the perimeter = 1248 m.

Question 15:
Solution:
Perimeter of the square = 40 cms.
Since this is rebent into a rectangle of length 12 cm, its perimeter is 24 + 2b = 40 => b = 8 cm.
So, the area of the square = 100 cm2 and the area of the rectangle = 96 cm2
Thus the square occupies more area by 4 cm2.

Question 16:
Solution:
Area of the walls of the godown = 2h(l + b) = 2 10(50  40)  20  90  1800 m2
Area of the ceiling of the godown = lb = 2000 m2.
So the total area = 1800 + 2000 = 3800 m2.
So the cost of whitewashing at the rate of Rs. 20 per square meter = 3800  20  76000
The cost of whitewashing = Rs. 76000.

Question 17:
Solution:
The dimensions of the room are length = l m, breadth= 10m and Height = h =4 m
2(l+b) h = 168
2(l+10)*4=168
Thus, l+10=21 => l=21-10 = 11 m
Therefore, the length of the room = 11 m.

Question 18:
Solution:
Area of 4 walls = 2 (l+b) h = 77= 2(7.5 + 3.5)h
So, 77  2 11h  h  77 / 22  h  3.5 m
The height of the room is hence 3.5 m.

Question 19:
Solution:
Let the breadth be x and length be 2x
Given height = 4m
Area of four walls (CSA) = 120 m2  2  4  x  2 x 
So, 120= 8 (x + 2x) => 120 = 8x + 16x = 24x
Hence, x = 5 m
Length = 10 m; Breadth = 5 m
So the area of the floor = 50 m2

Question 20:

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Solution:
Area of four wall =2h (l + b) = 2  3.4 8.5  6.5  6.8  15  102 m2
Hence the area of two doors and two windows = 2 1.5  1  2  2  1   3  4  m2  7 m2
Area to be painted = (102 - 7) = 95m2
Therefore the cost of painting =  95  160   Rs. 15200

Page No: 232

Exercise 20 B

Question 1:
Solution:
Since the path is of width 2 m on the inside of the plot, the smaller area will be equal to
(75  4)  (60  4)  3976 sq.m
The area of the path on the outside = 75  60  4500 sq.m
So the area of the path = 4500 – 3976 = 524 m2.
The cost of constructing it at Rs. 125 per m2 =  524 125  Rs. 65500 .

Question 2:
Solution:
The area of the rectangular plot = (95  72)  6840sq.m
Since the inside of the plot is to be constructed a uniform path of width 3.5 m,
The area of the inside rectangle = (95  7)  (72  7)  5720 sq.m
The cost of constructing the path at Rs. 80 per m2 =  6840  5720  80 = Rs. 89600
The cost of laying the grass at 40 per m2 =  5720  40   228800 .
The total cost = Rs. 89600 + Rs. 228800 = Rs. 318400.

Question 3:
Solution:
Length of saree = 5 m = 500 cm; Breadth of saree = 1.3 m = 130 cm
Area of saree with border = 500 130  65000 cm²
Now, length of saree without border = 500 - (25+25) = 450 cm
Breadth of saree without border = 130 - (25+25) = 80 cm
Area without border = 450  80  36000 cm²
Area of border = 65000 - 36000 = 29000 cm²
At the rate of Rs. 1 per 10 cm2, it will cost Rs. 2900.

Question 4:
Solution:
Area of grassy lawn = 38  25 m 2  950 m 2
The rectangle formed by the gravelled path plus grassy lawn has the size (38 + 5) m by (25 + 5)m.
The area of this rectangle = 1290 m2
Area of gravelled path = 340 m2
Cost = 340  120 Rs  Rs 40,800
The cost of gravelling the path = Rs. 40,800.

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Question 5:
Solution:
Area of room = 6  9.5  57 m 2
Area of room including verandah = 12  8.5  102 m 2
So area of verandah = 102 - 57 = 45 m2.
So cost of cementing the floor of this verandah = 45  80  Rs. 3600 .

Question 6:
Solution:
Given that each side of the flower bed is 2.8m long and so the area is 2.8  2.8  7.84m2
A strip of width 30cm is dug all around it.
The sides of the bed will be 2.8  (2  0.3)  3.4m
New area = 3.4  3.4  11.56sqm .
Hence the increase in area = 11.56  7.84  3.72sq.m

Question 7:
Solution:
Let the length be 2x and breadth be 1x. Given the perimeter = 240m.
Therefore, 2  2 x  1x   240m
So, 4x  2x  240m  6x  x  240 / 6  40m
Hence the length = 80m and breadth = 40m
Area of outer park = 80  40  3200 sq.m
Area of inner park = 80 – 4  40  4  76  36  2736 sq.m
Area of path = outer park - inner park = 3200sq.m - 2736sq.m = 464sq.m
Cost of paving the path = 464  80  Rs.37120

Question 8:
Solution:
Since the carpet is laid with a margin of 75 cm from the walls,
The area of the carpet =  22 – 1.5  15.5 – 1.5  287 m2
The area of the strip uncovered = area of the hall – area of the carpet
=  22 15.5  – 287  54m2
Given the width of the carpet = 82 cm, therefore its length = 287 / 0.82  350 m
So the cost of the carpet at the rate of Rs. 60 per m =  350  60   Rs.21000 .

Question 9:
Solution:
Area of the path =  x  5 – x 2  165 => x 2  10 x  25  x 2  165  10 x  25  165
2

So, x  14 .
Area of the lawn = x 2  196m 2 .

Question 10:
Solution:
Given that the area is 305 sq m and let the length of the rectangular park be 5x and the breadth be 2x
Since area = product of length and breadth,

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5x  2x  10x²
The width of the park all around the park = 2.5m
Hence the length and breadth including path  5x  2.5  2.5 m   5 x  5 and breadth of rectangular
park =  2 x  2.5  2.5   2 x  5 
Thus, the area of the park =  5x  2 2 x  5 10 x²  m²  35 x  25 m²
Since it is given that 35x  25  305  35x  280  x  8
Therefore, the length of the park = 5  8  40m and breadth of the park = 2  8  16m

Question 11:
Solution:
Area of the road parallel to the length= 70m  5m  350 sq.m
Area of the road parallel to the breadth= 50m  5m  250 sq.m
Area of the portion of the road overlapping= 5m  5m  25 sq.m
Total area of the road=  350 sq.m  250 sq.m  – 25 sq.m  575 sq.m
Cost of construction = 120  575  Rs. 69, 000

Question 12:
Solution:
The length of the road has length is 115m and width 2m.
Therefore, the area of the road = 115  2  230 m²
The other road has breath 64m and width 2.5m
So the area of this road = 64  2.5  160 m²
The area of the common road = 2  2.5  5 m²
Therefore, the area of the road that needs gravelling = 230  160 – 5  385 m²
The cost of gravelling the roads at Rs. 60 per m2 = 385  60  Rs. 23100 .

Question 13:
Solution:
Area of the longer road = 2  50  100 m2 .
Area of the smaller road = 2.5  40  100 m2 .
Area of the field = 50  40  2000 m 2 .
Hence, the area common to both the roads = 2.5  2  5 m 2 .
So, the total area of the roads = 200  5  195 m 2 .
Therefore, the remaining area of the field = 2000  195  1805 m2 .

Question 14: Calculate the area of the shaded region in each of the figures given below:

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Solution:
(i) The area of the larger rectangle = 43  27  1161 sq.m
The area of the smaller rectangle =  43  3   27 – 2   40  25  1000 sq.m
Therefore, the area of the shaded portion = 1161 – 1000  161 sq.m.

(ii) The area of the larger square = 40  40  1600 sq.m

The area of the rectangular plot of width 3m = 120 sq.m
The area of the rectangular plot of width 2m = 80 sq.m
The area of the rectangle at the centre that is common to both rectangular plots needs to be reduced.
Therefore, 6m is to be reduced from 120  80   200  194 sq.m.
Therefore the area of the shaded portion = 1600 – 194   1406 sq.m.

Question 15:
Solution:
(i) The area of the larger rectangle = 24 19  456sq.m
Area of the smaller rectangle = 16.5  20  330sq.m
Therefore, the area of the shaded portion = 456  330  126sq.m

(ii) The area of the vertical rectangle = 15  3  45 sq.m

The area of the two horizontal rectangles = 2  9  3  54 sq.m.
The area of the smaller horizontal rectangle = 3  5  15sq.m
The total area =  45  54  15  114 sq.m.

Question 16:
Solution:
Area of the largest rectangle =  3.5  0.5  1.75 sq.m
Area of the second largest rectangle =  2.5  0.5   1.25 sq.m
Area of the third smaller rectangle = 1.5  0.5   0.75 sq.m
Area of the smallest rectangle =  0.5  0.5  0.25 sq.m
Hence the area of the shaded portion = 1.75  1.25  0.75  0.25  4 sq.m

Page No: 237

Exercise 20C
Question 1:

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Solution:
Area of a parallelogram = produce of base and height =  32  16.5  528 sq.m

Question 2:
Solution:
Base length = 1.60 m
Height is 0.75 m
So the area = 1.6  0.75   1.2 sq.m.
Question 3:
Solution:
Given the base = 14 dm = 140 cm and height = 6.5 dm = 65 cm.
(i) So the area of the parallelogram = 140  65  9100 sq.cm
(ii) The area of the parallelogram = 1.40  0.65  0.91 sq.m

Question 4:
Solution:
Given the area is 54 cm2 and the base is 15 cm.
54 18
So, the height of the parallelogram =   3.6cm
15 5

Question 5:
Solution:
One side of a parallelogram is 18 cm long and its area is 153 cm2
153 51
The distance of the given side from its opposite side =   8.5cm
18 6

Question 6:
Solution:
Area of the parallelogram = product of its base and height = 18  6.4  115.2 sq.cm
12  AM  115.2  AM  115.2 / 12  9.6 cm

Question 7:
Solution:
Given bigger side as base = 15 cm and smaller side as base = 8 cm
Distance (height) between the longer sides = 4 cm
Area of parallelogram = base  height  15  4  60 cm²
As area of parallelogram = base  height  60  8  height
Therefore, the height = 60 / 8  7.5 cm.
Hence, the distance between the shorter sides is 7.5 cm

Question 8:
Solution:
Given the height = one-third of base => if h  x, b  3 x
Therefore, area = product of base and height = 3x 2  108  x 2  36  x  6.
Therefore, the height is 6 cm and base is 18 cm.

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Question 9:
Solution:
Given base is twice its height => if base = 2x then height = x .
The area of the parallelogram = product of base and height = 512 cm2  2 x 2
So, x 2  256  x  16 cm .
Hence the height = 16 cm and base = 32 cm.

Question 10:
Solution:
(i) Area of the rhombus = product of side and height = 12  7.5  90 sq.cm
(ii) Area of the rhombus = product of side and height =
 2dm  12.6 cm    20  12.6  sq.cm  252 sq.cm
Question 11:
Solution:
(i) Area of the rhombus = half of product of diagonals =  0.5  16  28  224 sq.cm
(ii) Area of the rhombus = half of product of diagonals =  0.5  85  56  sq.cm  2380 sq.cm

Question 12:
Solution:
Area of the rhombus is given by the formula
Area = diagonal  ( side2  (diagonal / 2)2 )  (24  400 144 )  24 16  384sq.cm

Question 13:
Solution:
Area of the rhombus = half of product of diagonals => 148.8  19.2  0.5  diagonal => length of the
148.8
diagonal =  2  15.5cm
19.2

Question 14:
Solution:
Perimeter=4 times the length of the side.
Hence, length of side = 56 / 4  14 cm
Area  base  altitude
Altitude = Area/base = 119 / 14  8.5cm .

Question 15:
Solution:
Area = product of side and height => 441  side x 17.5  side  441 / 17.5  25.2 cm

Question 16:
Solution:
Given, Area of the triangle = Area of the rhombus
Hence 1/ 2  Base  Height  Product of Diagonals / 2

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1 / 2  24.8  16.5  22  x / 2  204.6  22  x / 2
=> 22x  204.62  x  409.2 / 22  18.6 cm .

Page No: 242

Exercise 20D
Question 1:
Solution:
1
Area of a triangle is given by  base  height
2
1
(i) Area =  42  25  525sq.cm
2
1
(ii) Area = 16.8  0.75  6.3sq.m
2
1
(iii) Area =  80  35  1400 sq.cm
2

Question 2:
Solution:
1
Area of a triangle is given by  base  height
2
1 72
That is, 72  16  height  height   9cm
2 8

Question 3:
Solution:
1
Area of a triangle is given by base  height
2
1 224
Hence, 224   28  height  height   16m
2 14

Question 4:
Solution:
1
 base  height
Area of a triangle is given by
2
1 90
Hence, 90  12  base  base   15 cm.
2 6

Question 5:
Solution:
Area of the field = Total cost / rate =  333.18 / 25.6   13.5 hectares

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13.5  10000 m2  135000 m2
Considering that the height = x metres and base = 3x metres.
Hence, 12  3 x  x  135000  x 2  90000  x  300
Therefore, the base = 900 m and height = 300 m.

Question 6:
Solution:
1
The area of a right angled triangle is  base  height
2
Given 1/ 2 14.8 h   129.5
7.4  h   129.5  h  17.5
The other side is 17.5cm.

Question 7:
Solution:
We know that hypotenuse2 = base2 + height2 => (3.7) 2= (1.2) 2 + height2
 13.69  1.44  height 2
 height 2  13.69  1.44
 height = 12.25  3.5m
1 1
Hence the area of the right angled triangle =  base  height =  3.5 1.2  4.2 / 2  2.1m 2
2 2

Question 8:
Solution:
Let the length of the legs be 3x and 4x
We know that area =  base  height = 12 x 2  / 2  6 x 2  1014
1
2
So x  169  x  13
2

Hence the legs are of length 39 cm and 52 cm.

Question 9:
Solution:
Let there is a right angled triangle with sides AC, AB and BC and AC be the longest side of this triangle.
AC ²  AB²  BC ²  1 ²   0.8  ²   BC  ²
=>  BC  ²  1  0.64  0.36
So BC = 0.6 m
1 1
Now, the area of the triangle = base  height =  0.6  0.8  0.24 sq m
2 2
Cost of scarf = 250 0.24  Rs 60

Question 10:
Solution:
3 2
Area of an equilateral triangle is given by the formula a
4

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 3
(i) Area = (182 )  81 3  140.13sq.cm
4
3
(ii) Area = (20 2 )  100 3  173sq.cm
4

Question 11:
Solution:
3 2
Area of an equilateral triangle is given by the formula a  16 3  a 2  64
4
Hence the length of the side a = 8 cm.

Question 12:
Solution:
1 3
 24  h   24  24
Given 2 4
3
h  24  20.76cm
So, 2

Question 13:
Solution:
1
We know that S  (a  b  c ) and area of the triangle = s(s  a)(s  b)(s  c)
2
1 1
(i) S  (a  b  c ) = (13  14  15)  21
2 2
Area of the triangle = s(s  a)(s  b)(s  c) = 21(8)(7)(6)  84 sq.cm

1 1
(ii) S  (a  b  c ) = (52  56  60)  84
2 2
Area of the triangle = s(s  a)(s  b)(s  c) = 84(32)(28)(24)  1344 sq.cm

1 1
(iii) S  (a  b  c ) = (91  98  105)  147
2 2
Area of the triangle = s(s  a)(s  b)(s  c) = 147(56)(49)(42)  4116 sq.cm

Question 14:
Solution:
1 1
S  (a  b  c ) = (33  44  55)  66
2 2
Area of the triangle = s(s  a)(s  b)(s  c) = 66(33)(22)(11)  726 sq.cm sq.cm

1 1
Area of a triangle =  base  height => 726 =  height  44
2 2
Therefore, the height = 726 / 22  33 cm

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Question 15:
Solution:
Let the sides of the triangle be 13x, 14x and 15x
Given perimeter = 84  13x  14x  15x  42x  84  x  2
So, the sides are a = 26 cm, b = 28 cm and c = 30 cm
s  perimeter / 2  84 / 2  42 cm
Area = s(s  a)(s  b)( s  c) = 42  42  26  42  28 42  30  = 336 cm2

Question 16:
Solution:
1
S  (a  b  c ) =  42  34  20  / 2 = 48 cm
2
a = 42 cm b = 34 cm c = 20 cm
Hence, Area = s(s  a)(s  b)(s  c) =  48  6  14  28  cm² = 336 cm²
Height on the base of 42 cm is obtained as follows:
336  1/ 2  42  h  h  16 cm

Question 17:
Solution:
Given the base is 48 cm, and the other two sides measure 30 cm each. Hence
1
S  (a  b  c ) =  48  30  30  / 2  54 cm.
2
Hence, Area = s(s  a)(s  b)(s  c) =  54  6  24  24   432sq.cm cm² = 336 cm²

Question 18:
Solution:
Let x be the side of the triangle
Given that the perimeter = 2x + 12 = 32.
So, x = 10 cm.
Using Pythagoras theorem, the height of the base = 8 cm.
So the area of the triangle = (1/ 2) 12  8 = 48 cm2.

Question 19:
Solution:
Area of quadrilateral is (1/ 2)  d   h1  h2 
Given d = 26, h1 = 12.8 and h2 = 11.2
Hence the area = (1/ 2)  26  12.8  11.2   (1/ 2)  26  24  312 sq.cm.

Question 20:
Solution:
Area of quad. ABCD = (area of triangle ABC) + (area of triangle ACD).
1
For Triangle ABC, S  (a  b  c ) =  26  28  30  / 2  42
2
Hence, Area = s(s  a)(s  b)(s  c) =  42  12  14  16   336sq.cm

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 1
For Triangle ACD, S  (a  b  c ) =  50  40  30  / 2  60
2
Hence, Area = s(s  a)(s  b)( s  c) =  60  30  20  10   600sq.cm
Hence the area of quadrilateral ABCD = 336 + 600 = 936 sq.cm.

Question 21:
Solution:
The area of the rectangle = product of length and breadth = 36  24  864 sq.cm
1 1
The area of the triangle =  base  height   24  15  180sq.cm
2 2
Hence area of the shaded portion = 864 – 180 = 684 sq.cm.

Question 22:
Solution:
Area of rectangle = product of length and breadth = 40  25  1000cm 2
PB = half of 40 = 20 cm
BQ = half of 25 = 12.5 cm
Area of triangle PBQ = (1/ 2)  b  h  (1/ 2)  20 12.5
Area of triangle PBQ= 10 12.5  125 cm2
Since all the four triangles are equal, the area of 4 triangles = 125  4  500 cm 2
Area of shaded region = 1000 – 500 = 500 cm2

Question 23:
Solution:
(i) The area of the rectangle = 18 10  180 sq.cm
1
Area of the triangle DEF = (10  6)  30sq.cm
2
1
Area of the triangle BCE = (10  8)  40sq.cm
2
Hence area of the shaded region = 180 – (30 + 40) = 110 sq.cm
(ii) The area of the square = 20  20  400 sq.cm
1
Area of the triangle LRS = (10  20)  100 sq.cm
2
1
Area of the triangle MRQ = (10  20)  100 sq.cm
2
1
Area of the triangle LMP = (10 10)  50sq.cm
2

Hence area of the shaded region = 400 – (100 + 100 + 50) = 150 sq.cm

Question 24:
Solution:
1
Area of triangle ABD = (5  24)  60 sq.cm
2

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 1
Area of triangle CBD = (8  24)  96 sq.cm
2
Hence area of the quadrilateral ABCD = 60 + 96 = 156 sq.cm.

Page No: 247

Exercise: 20E
22
NOTE: Take   , unless stated otherwise.
7
Question 1:
Solution:
The circumference of a circle is given by 2 r
22
(i) Circumference = 2 r  2   28  176cm.
7
22
(ii) Circumference = 2 r  2  1.4  8.8m.
7

Question 2:
Solution:
The circumference of a circle is given by 2 r
Diameter = twice the radius
22 35
(i) Circumference = 2 r  2    110cm.
7 2
22 4.9
(ii) Circumference = 2 r  2    15.4m.
7 2

Question 3:
Solution:
The circumference of a circle is given by 2 r
Hence the circumference = 2 r  2  3.14 15  94.2cm.

Question 4:
Solution:
The circumference of a circle is given by 2 r
22
Hence, 2 r = 57.2 => 2   r  57.2  r  9.1cm
7

Question 5:
Solution:
The circumference of a circle is given by 2 r
22
Hence, 2 r = 63.8 => 2   r  63.8  r  10.15m
7
Hence diameter = twice the radius = 20.3 m.

Question 6:
Solution:

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Given 2 r = diameter + 30cm.
22 44 30
Hence, 2   r  2r  30  (  2)r  30  r  30  r  7cm.
7 7 7

Question 7:
Solution:
Let the radius of the two circles be 5x and 3x.
22 22
Hence the ratio of their circumferences will be 2   5 x : 2   3x  5 : 3
7 7

Question 8:
Solution:
22
The circumference of the circular field = 2 r = 2   21  132m
7
132 18
At the speed of 8km/hr, the time taken = distance / speed =   59.4sec
8 5

Question 9:
Solution:
Given the inner circumference = 528 m and outer circumference = 616 m.
Hence difference in circumference = (616 - 528) = 88 m.
7
Hence the width of the track = r  88   14m
2  22

Question 10:
Solution:
22
Given the inner circumference of a circular track = 2 r = 2   r  330m  r  52.5m
7
The track is of width 10.5 m. Hence, the radius of the outer circle = 52.5 + 10.5 = 63m.
22
Hence the circumference of the outer circle = 2 r = 2   63  396m
7
The cost of fencing along the outer circle = 396  20  Rs.7920 .

Question 11:
Solution:
22
Circumference of the smaller circle = 2 r = 2   98  616cm
7
22
Circumference of the concentric circle = 2 r = 2  126  792cm
7
Hence the difference in circumference of the two circles = 792 – 616 = 176 cm.

Question 12:
Solution:
Given the side of the equilateral triangle = 8.8cm.
Hence the perimeter = 8.8 + 8.8 + 8.8 = 26.4 = circumference of the circle = 2 r
22
Hence 2   r  26.4  r  4.2cm  d  2r  8.4cm
7

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Question 13:
Solution:
Perimeter of the rhombus = four times the side = 33 4  132cm
Since this is the same as the circumference of the circle,
22
2   r  132  r  21cm
7

Question 14:
Solution:
Perimeter of the rectangle = twice the sum of length and breadth = 2(18.7  14.3)  66cm
This is the circumference of the circle = 2 r = 66 cm
22
Hence the radius is found 2   r  66  r  10.5cm
7

Question 15:
Solution:
22
The circumference of the circle = 2   35  220cm
7
220
Since this is the perimeter of the square, we get 4  lengthofside  220  length   55cm
4

Question 16:
Solution:
Given, the diameter of the well (d) = 140 cm
And the length of the outer edge of the parapet = 616 cm.
Hence width of the parapet =
halfof  diameter of the well  diameter of parapet  
196  140   28 cm
2
Hence, width of the parapet is 28 cm.

Question 17:
Solution:
Diameter of circle = 98 cm
Radius of circle = 49 cm = 0.49 m
22
Circumference of circle = 2 r  2   0.49  3.08
7
Distance covered by one rotation = 3.08 m
Distance covered by 2000 rotation = 3.08  2000  6160 m

Question 18:
Solution:
Given the diameter of the wheel = 70 cm hence its radius = 35 cm.
22
1 revolution covers the circumference of the wheel = 2 r  2   35  2.2m
7

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For 250 revolutions it moves 250  220  55000 cm  550 m

Question 19:
Solution:
Product of number of revolutions (n) and circumference of wheel = Area covered
n  2 r  121 km
22 77
n  2    12100000 cm
7 2
77
n  44   12100000
14
n  242  12100000  x  50000
Hence, the car wheel has to take 50,000 revolutions to cover an area of 121 km.

Question 20:
Solution:
22
Distance covered in 5000 revolutions = 2   r  5000  11000m  r  0.35m  35cm
7
22
The circumference of the wheel = 2   35cm  220cm
7
Hence the diameter of the wheel = 70 cm.

Question 21:
Solution:
22
Distance moved by hour hand in 12 hours = (2   4.2)cm  26.4cm .
7
Distance moved by hour hand in 24 hours =  26.4  2  cm  52.8 cm .
Distance moved by minute hand in 24 hours   44  24  cm  1056 cm .
Hence the sum of the distances covered by their tips in 1 day = 1056cm  52.8cm  1108.8cm

Page No: 252

Exercise: 20F
22
NOTE Take   , unless stated otherwise.
7
Question 1:
Solution:
Area of a circle =  r 2 sq.units
22
(i) Area =  r 2 =  21 21  1386sq.cm
7
22
(ii) Area =  r 2 =  3.5  3.5  38.5sq.m
7

Question 2:
Solution:
Area of a circle =  r 2 sq.units

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 22
(i) Area =  r 2 = 14 14  616 sq.cm
7
22
(ii) Area =  r 2 =  0.7  0.7  1.54 sq.m
7

Question 3:
Solution:
22
Given the circumference = 2   r  264  r  42cm
7
22
Hence its area =  r 2 =  42  42  5544 sq.cm
7

Question 4:
Solution:
22
Given the circumference = 2   r  35.2  r  5.6m
7
22
Hence its area =  r 2 =  5.6  5.6  98.56 sq.m
7

Question 5:
Solution:
22 2
Given the area =  r 2 =  r  616sq.cm  r 2  196  r  14cm
7
22
Hence the circumference = 2  14  88cm
7

Question 6:
Solution:
22 2
Given the area =  r 2 =  r  1386sq.m  r 2  441  r  21m
7
22
Hence the circumference = 2   21  132m
7

Question 7:
Solution:
Let the radii of the two circles be 4x and 5x.
Hence the ratio of their areas =  (4 x)2 :  (5 x)2 = 16: 25.

Question 8:
Solution:
The radius of the circle = length of the string = 21 m.
22
Hence the area =  r 2 =  (21) 2  1386 sq.m
7

Question 9:
Solution:

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Area of the square = a 2  121  a  11cm
Hence the circumference of the circle = four times the length of the side of the square = 44 cm =
22
2   r  r  7cm
7
22
Hence the area of the circle =  r 2 =  (7) 2  154 sq.m
7

Question 10:
Solution:
Radius of the circle = 28 cm. Hence the circumference of the circle = perimeter of the square =
22 176
2   28  176cm => side =  44 cm
7 4
Hence the area of the square formed = 44  44  1936 sq.cm

Question 11:
Solution:
Area of the rectangular sheet = product of length and breadth = 34  24  816 cm²
22 3.5 3.5
Area of each button =  r ²     9.625cm²
7 2 2
Area of 64 buttons = 64  9.625  616cm²
Area of the remaining sheet = area of the sheet - area of 64 buttons = 816 – 616  200cm²

Question 12:
Solution:
Area of the rectangle – area of the circle = product of length and breadth -  r 2
22
= (32  90)  ( 14 14)  2880  616  2264 sq.m
7
The cost of turfing the remaining portion at the rate of Rs. 50 per square metre =
2264  50  Rs.113200

Question 13:
Solution:
Area of the square = 14 14  cm2 = 196 cm2.
1 22
Sum of areas of the 4 quadrants = 4    7  7 cm 2  154 cm 2
4 7
Hence, the area of the shaded region = (196 - 154) cm2 = 42 cm2.

Question 14:
Solution:
Let ABCD be the rectangular field in which AB = 60 m and BC = 40 m.
Let the horse be tied to corner A by a 14 m long rope.
In this case, it can graze through a quadrant of a circle of radius 14 m.
1 22
Therefore, required area = (  14 14)m 2  154 m 2
4 7

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Question 15:
Solution:
Given the diameter of the larger circle is 21 cm. Hence the diameter of the inside circles are 14 cm and 7
cm respectively.
22 22 7 7
The area of the inner circles = (  7  7)  (   )cm 2  192.5cm 2
7 7 2 2
22 21 21
The area of the shaded region is hence [(   )  192.5]cm 2  154cm 2
7 2 2

Question 16:
Solution:
Area of the rectangle = 6  8  48sq.m
Since there are four quarter circles and one full circle inside this rectangle, the sum of their areas is given
 1 22 22 
by  4    2  2)  (  2  2)  m 2  25.14 m 2
 4 7 7 
Hence the area of the remaining plot = 48  25.14  22.86 sq.m

Page No: 253

Exercise 20G

Question 1:
Solution:
(c)
Since the diagonal is 20 cm, and one side is 16 cm, the other side is given by
202  162  12 m.
So the area of this rectangle is given by the product of length and breadth = 16 12  192sq.m m2.

Question 2:
Solution:
(b)
12 12   72 cm2
Area 
2

Question 3:
Solution:
(b)
diagonal 2
Area = 200   diagonal  20cm
2

Question 4:
Solution:
(a)
diagonal 2
Area = 5000m2   diagonal  100m
2

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Question 5:
Solution:
(c)
Perimeter = 240  2  l  b   2  3b  b   b  30m  l  90m

Question 6:
Solution:
(d)
Let the length of each side be a cm. Then, its area = a2 cm2.
5a 25a 2
New length = (125% of a) = cm. New area = cm2.
4 16
25  16
Hence the increase in area = 100  56.25%
16

Question 7:
Solution:
(b)
Let the side of square be a. Length of its diagonal = 2a .
Therefore, required ratio = a2:2a2 = 1:2.

Question 8:
Solution:
(c)
When the perimeters are equal, the area of a square (A) is greater than area of the rectangle (B).

Question 9:
Solution:
(b)
Let the length and breadth be 5x and 3x. Hence 2(8 x)  480  x  30m
Hence the length = 150 m and breadth = 90 m. The area is hence 150  90  13500sq.m

Question 10:
Solution:
(a)
Length of carpet = total cost/rate = (6000/50)m = 120 m.
75
Area of the room = area of the carpet = (120  ) m 2  90 m 2
100
90
 6m
Hence the width of the room = 15

Question 11:
Solution:
(a)
1
We know that S  (a  b  c ) and area of the triangle = s(s  a)(s  b)( s  c)
2

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 1 1
S  (a  b  c ) = (13  14  15)  21
2 2
Area of the triangle = s(s  a)(s  b)(s  c) = 21(8)(7)(6)  84 sq.cm

Question 12:
Solution:
(b)
b  h 12  8
Area of a triangle =   48sq.m
2 2

Question 13:
Solution:
(b)
3 2
Area of an equilateral triangle is given by the formula a = 4 3 sq.cm => a = 4 cm.
4

Question 14:
Solution:
(c)
3 2 3
Area of an equilateral triangle is given by the formula a = (64)  16 3 sq.cm.
4 4

Question 15:
Solution:
(b)
3
Given a  6  a  2 2
2
3 2
Hence the area = a  2 3sq.cm
4

Question 16:
Solution:
(b)
Area = 16  4.5  72 sq.cm

Question 17:
Solution:
(b)
24 18
Area of the rhombus =  216 sq.cm
2

Question 18:
Solution:
(c)
22
r (2   1)  37cm  r  7cm
7

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22
 7  7  154 sq.cm
7

Question 19:
Solution:
(c)
Perimeter of rectangle = 2  l  b   18
Area of four walls = 2  l  b   h
Area of four walls = 18  3  54 m²

Question 20:
Solution:
(a)
Area = 14 * 9 = 126 sq.m = 12600 sq.cm
Hence the number of meters of carpet = 12600 / 63 = 200

Question 21:
Solution:
(c)
l 2  b2  17   289 and 2  l  b   46  1  b   23 .
2

l 2
 b2   l  b 
2
 2lb
529  289
Hence lb   120 sq.cm
2

Question 22:
Solution:
(b)
a2 9 a a 3 4a 4  3 3
  ( ) 2  32     
b 2
1 b b 1 4b 4 1 1

Question 23:
Solution:
(d)
1
(2d ) 2
A1 2 4d 2 4
  2   4 :1
A2 1 2 d 1
d
2

Question 24:
Solution:
(c)
144  width  84  84  7056sq.m  width  49m

Question 25:
Solution:
(d)

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 a2 4
Ratio =   4: 3
3 2 3
a
4

Question 26:
Solution:
(a)
a
a 2   r 2     a : r   :1
r

Question 27:
Solution:
(b)
22 2
 r  154sq.cm  r  7cm
7
3 2 49 3
Area of the triangle = r  .
4 4

Question 28:
Solution:
(c)
d d d 6
Area = 1 2  36sq.cm  1  36  d1  12cm
2 2

Question 29:
Solution:
(d)
d  2d1
Area = 1  144sq.cm  d1  12cm
2
Hence the longer diagonal is of length 24 cm.

Question 30:
Solution:
(c)
22
 r  r  24.64sq.m  r  2.8m
7
22
Hence its circumference = 2 r  2   2.8  17.6m
7

Question 31:
Solution:
(c)
(   r  1 r  1)  (  r  r )  22
Hence,   r 2  2r  1  r 2   22
  2r  1  22  0

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Hence, 2r  6  0  r  3

Question 32:
Solution:
(c)
22
Circumference = 2  1.75m  11m
7
11000
Hence the number of revolutions made for travelling 11 km =  1000
11

Page No: 257

Test paper-20
A.
Question 1:
Solution:
We know that  side  ²   side  ²  diagonal ²
Hence side² = 50²  48² sq.m  196sq.m
Hence side = 14 m
Therefore, the area = 14  48  672 sq.m

Question 2:
Solution:
The total surface area of the room including doors and windows excluding the floor =
2  l  b)h   2 17  6.5  = 221 sq. m
The area of the doors and windows = (2 1.5)  (4 1.5 1)  3  6  9 sq.m
Hence the area of the walls = 221 – 9 = 212 sq.m
The cost of painting the walls = 212  50  Rs.10600

Question 3:
Solution:
1
Area of square = half of product of its diagonals =  64  64  2048sq.cm
2

Question 4:
Solution:
Let x be the side of the square lawn.
Length PQ =  x m  2 m  2 m   x  4 m
Area of PQRS =  x  4  ²   x²  8 x  16  m²
Now, Area of the path = Area of PQRS – Area of the square lawn
Hence, 136  x²  8x  16 – x²  136  8x  16  x  15
Since the side of the lawn = 15 m its area =  Side  ²  15 m  ²  225 m²

Question 5:
Solution:
The area of the rectangular lawn = 30  20  600 sq.m

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Area of the roads = (30  2)  (20  2)  (2  2)  100  4  96sq.m

Question 6:
Solution:
Area of the rhombus is given by the formula
Area = diagonal  ( side2  (diagonal / 2)2 )  (24  169 144)  24  5  120sq.cm

Question 7:
Solution:
Area of the parallelogram = product of base and height
Hence 338  2b 2   b 2  169  b  13m
Hence the altitude is 26 m and base = 13 m.

Question 8:
Solution:
Given b = 24 cm and hypotenuse H = 25 cm
Hence the perpendicular P = H 2  b2  252  242  625  576  7cm
1
Hence the area of the right triangle =  24  7  84 sq.cm
2

Question 9:
Solution:
Given the radius of the wheel = 35 cm.
22
Hence its circumference = 2   35  220cm  2.2m
7
33000
To travel 33 km, that is 33000 m, it will take  15000 revolutions.
2.2

Question 10:
Solution:
22 2
Area of the circle =  r  616  r 2  196  r  14cm
7

B.
Question 11:
Solution:
(a)
22 2
Area of the circle =  r  154sq.cm  r 2  49  r  7cm
7
Diameter = 14 cm.

Question 12:
Solution:
(b)
22
2   r  44  r  7cm
7

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 22
Area of the circle =  7  7  154 sq.cm
7

Question 13:
Solution:
(c)
Diagonal of a square = 2a  14cm  a  7 2cm
 
2
Its area = 7 2  98sq.cm

Question 14:
Solution:
(b)
The area of the square = a2  50sq.cm  a  5 2cm.
Hence the length of its diagonal = 2a  10cm .

Question 15:
Solution:
(a)
The perimeter of the rectangle = 56 m  2  4 x  3x   14 x  x  4 m .
Hence its length = 16 m and breadth = 12 m.
Hence, the area of the field = 16 12  192sq.m

Question 16:
Solution:
(a)
1
We know that S  (a  b  c ) and area of the triangle = s(s  a)(s  b)(s  c)
2
1 1
S  (a  b  c ) = (13  14  15)  21
2 2
Area of the triangle = s(s  a)(s  b)(s  c) = 21(8)(7)(6)  84 sq.cm

Question 17:
Solution:
(a)
3 2 3
Area of the equilateral triangle = a   8  8  16 3sq.cm
4 4

Question 18:
Solution:
(b)
Area of the parallelogram = 14  6.5  91sq.cm

Question 19:
Solution:

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(b)
18  15
Area of the rhombus =  135sq.cm
2

C.
Question 20:
Solution:
d1  d 2
(i) If d1 and d2 be the diagonals of a rhombus, then its area is sq units.
2
(ii) If l, b and h be the length, breadth and height respectively of a room, then area of its 4 walls =
2(l  b)h sq units.
(iii) 1 hectare = 10000 m2.
(iv) 1 acre = 100 m2.
3 2 2
(v) If each side of a triangle is a cm, then its area = a cm .
4

D.
Question 21:
Solution:
(i) F
Area of a triangle = half of (base x height).
(ii) T
Area of a parallelogram = (base x height).
(iii) F
Area of a circle = πr2.
(iv) T
Circumference of a circle = 2πr.

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