Chapter 20: Mensuration
Page No: 229
Exercise 20A
Question 1:
Solution:
(i) Area of a rectangle = product of length and breadth = 24.5 18 441 m2.
(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 2 250 48 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
2 1728 144 12
12x x
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 .5040Rs
Question 4:
Solution:
Area of the rectangular field = product of length and breadth => breadth = 3584
5664
m.
Perimeter of the field = 64 + 56 + 64 + 56 = 240 m.
For him to go five rounds, the distance covered = 240 5 1200m .
So, the time taken = 1200 1
12min6000 5
hr .
Question 5:
Solution:
Area of the verandah = product of l and b = 2 240 15 600 60000m dm .
area of the stone = 26 5 30 dm =
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 = 213 9 117m
The area of the room must be equal to the area of the carpet. So, 213 9 117 0.75 156m l m l m
So the cost of carpeting = 156 105 .16380Rs .
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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 => 15 240 0.75b
So, 240 0.75 /15 12b .
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 = 2 2 2 l b h = 100 100 25 = 225 15 m
The length of the largest pole = 15 m.
Question 10:
Solution:
Area of square = side × side = 28.5 8.5 72.25 m
Question 11:
Solution:
(i) Area of square = 272 72 5184
2592 2 2
cm
(ii) Area of square = 22.4 2.4
2.88 2
m
Question 12:
Solution:
Area of square = (side)² = 16200 m² => side = 16200 90 2 m
Diagonal of square = 2 2 90 2 180 side m
The length of the diagonal = 180 m.
Question 13:
Solution:
1 hectare = 10000 m2
∴0.5hec = 5000m2
⇒ side of field = 50 2 50 2 100diagonal m
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 cm
2
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 11 77 / 22 3.5 h h h 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) = 2120 2 4 2m x 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) = 22 3.4 8.5 6.5 6.8 15 102 m
Hence the area of two doors and two windows = 2 22 1.5 1 2 2 1 3 4 7 m m
Area to be painted = (102 - 7) = 95m2
Therefore the cost of painting = 95 160 . 15200Rs
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 . 65500Rs .
Question 2:
Solution:
The area of the rectangular plot = (95 72) 6840 .sq 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 ²cm
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 ²cm
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 = 2 238 25 950 m m
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 40,800Rs Rs
The cost of gravelling the path = Rs. 40,800.
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Question 5:
Solution:
Area of room = 26 9.5 57 m
Area of room including verandah = 212 8.5 102 m
So area of verandah = 102 - 57 = 45 m2.
So cost of cementing the floor of this verandah = 45 80 . 3600Rs .
Question 6:
Solution:
Given that each side of the flower bed is 2.8m long and so the area is 22.8 2.8 7.84m
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 3.72 .sq m
Question 7:
Solution:
Let the length be 2x and breadth be 1x. Given the perimeter = 240m.
Therefore, 2 2 1 240x x m
So, 4 2 240 6 240/ 6 40x x m x x m
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 .37120Rs
Question 8:
Solution:
Since the carpet is laid with a margin of 75 cm from the walls,
The area of the carpet = 222 – 1.5 15.5 – 1.5 287 m
The area of the strip uncovered = area of the hall – area of the carpet
= 222 15.5 – 287 54m
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 .21000Rs .
Question 9:
Solution:
Area of the path = 2 2 5 – 165x x =>
2 210 25 165 10 25 165x x x x
So, 14x .
Area of the lawn =22 196x m .
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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5 2 10 ²x x x
The width of the park all around the park = 2.5m
Hence the length and breadth including path 5 2.5 2.5 5 5x m x and breadth of rectangular
park = 2 2.5 2.5 2 5x x
Thus, the area of the park = 5 2 2 5 10 ² ² 35 25 ²x x x m x m
Since it is given that 35 25 305 35 280 8x x x
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= 70 5 350 .m m sq m
Area of the road parallel to the breadth= 50 5 250 .m m sq m
Area of the portion of the road overlapping= 5 5 25 .m m sq m
Total area of the road= 350 . 250 . – 25 . 575 .sq m sq m sq m sq m
Cost of construction = 120 575 . 69,000Rs
Question 12:
Solution:
The length of the road has length is 115m and width 2m.
Therefore, the area of the road = 115 2 230 ²m
The other road has breath 64m and width 2.5m
So the area of this road = 64 2.5 160 ²m
The area of the common road = 2 2.5 5 ²m
Therefore, the area of the road that needs gravelling = 230 160 – 5 385 ²m
The cost of gravelling the roads at Rs. 60 per m2 =385 60 . 23100Rs .
Question 13:
Solution:
Area of the longer road =22 50 100 m .
Area of the smaller road =22.5 40 100 m .
Area of the field =250 40 2000 m .
Hence, the area common to both the roads =22.5 2 5 m .
So, the total area of the roads =2200 5 195 m .
Therefore, the remaining area of the field =22000 195 1805 m .
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 = 1 161 – 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 19 456 .sq m
Area of the smaller rectangle = 16.5 20 330 .sq m
Therefore, the area of the shaded portion = 456 330 126 .sq 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 15 .sq 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.
So, the height of the parallelogram = 54 18
3.615 5
cm
Question 5:
Solution:
One side of a parallelogram is 18 cm long and its area is 153 cm2
The distance of the given side from its opposite side = 153 51
8.518 6
cm
Question 6:
Solution:
Area of the parallelogram = product of its base and height = 18 6.4 115.2 .sq cm
12 115.2 115.2 / 12 9.6 AM AM 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 = 15 4 60 ²base height cm
As area of parallelogram = 60 8 base height 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 , 3h x b x
Therefore, area = product of base and height = 2 23 108 36 6.x x x
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 = 2 2512 2cm x
So,2 256 16 x x 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 =
2 12.6 20 12.6 . 252 .dm cm sq cm 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 . 2380 .sq cm sq cm
Question 12:
Solution:
Area of the rhombus is given by the formula
Area = 2 2( ( / 2) ) (24 400 144 ) 24 16 384 .side diagonal sq cmdiagonal
Question 13:
Solution:
Area of the rhombus = half of product of diagonals => 148.8 19.2 0.5 diagonal => length of the
diagonal = 148.8
2 15.519.2
cm
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 17.5 441 / 17.5 25.2 side x side cm
Question 16:
Solution:
Given, Area of the triangle = Area of the rhombus
Hence 1/ 2 / 2 Base Height Product of Diagonals
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1 / 2 24.8 16.5 22 / 2 204.6 22 / 2x x
=> 22 204.6 2 409.2 / 22 18.6 x x cm .
Page No: 242
Exercise 20D
Question 1:
Solution:
Area of a triangle is given by 1
2base height
(i) Area = 1
42 25 525 .2
sq cm
(ii) Area = 1
16.8 0.75 6.3 .2
sq m
(iii) Area = 1
80 35 1400 .2
sq cm
Question 2:
Solution:
Area of a triangle is given by 1
2base height
That is, 1 72
72 16 92 8
height height cm
Question 3:
Solution:
Area of a triangle is given by 1
2base height
Hence, 1 224
224 28 162 14
height height m
Question 4:
Solution:
Area of a triangle is given by 1
2base height
Hence, 1 90
90 12 152 6
base base cm.
Question 5:
Solution:
Area of the field = Total cost / rate = 333.18 / 25.6 13.5 hectares
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2 213.5 10000 135000 m m
Considering that the height = x metres and base = 3x metres.
Hence, 212 3 135000 90000 300x x x x
Therefore, the base = 900 m and height = 300 m.
Question 6:
Solution:
The area of a right angled triangle is 1
2base height
Given 1/ 2 14.8 129.5h
7.4 129.5 17.5h h
The other side is 17.5cm.
Question 7:
Solution:
We know that hypotenuse2 = base
2 + height
2 => (3.7)
2= (1.2)
2 + height
2
21 3.69 1.44 height
2 13.69 1.44height
height = 12.25 3.5m
Hence the area of the right angled triangle = 1
2base height =
213.5 1.2 4.2 / 2 2.1
2m
Question 8:
Solution:
Let the length of the legs be 3x and 4x
We know that area = 1
2base height = 2 212 / 2 6 1014x x
So 2 169 13x x
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.
² ² ² 1 ² 0.8 ² ²AC AB BC BC
=> ² 1 0.64 0.36BC
So BC = 0.6 m
Now, the area of the triangle = 1
2base height =
1 0.6 0.8 0.24
2sq m
Cost of scarf = 250 0.24 60Rs
Question 10:
Solution:
Area of an equilateral triangle is given by the formula 23
4a
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(i) Area = 23(18 ) 81 3 140.13 .
4sq cm
(ii) Area = 23(20 ) 100 3 173 .
4sq cm
Question 11:
Solution:
Area of an equilateral triangle is given by the formula 2 2316 3 64
4a a
Hence the length of the side a = 8 cm.
Question 12:
Solution:
Given
1 324 24 24
2 4h
So,
324 20.76
2h cm
Question 13:
Solution:
We know that 1
( )2
S a b c and area of the triangle = ( )( )( )s s a s b s c
(i) 1
( )2
S a b c = 1
(13 14 15) 212
Area of the triangle = ( )( )( )s s a s b s c = 21(8)(7)(6) 84 sq.cm
(ii) 1
( )2
S a b c = 1
(52 56 60) 842
Area of the triangle = ( )( )( )s s a s b s c = 84(32)(28)(24) 1344 sq.cm
(iii) 1
( )2
S a b c = 1
(91 98 105) 1472
Area of the triangle = ( )( )( )s s a s b s c = 147(56)(49)(42) 4116 sq.cm
Question 14:
Solution:
1 ( )
2S a b c =
1(33 44 55) 66
2
Area of the triangle = ( )( )( )s s a s b s c = 66(33)(22)(11) 726 .sq cm sq.cm
Area of a triangle = 1
2base height => 726 =
144
2height
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 13 14 15 42 84 2x x x x x
So, the sides are a = 26 cm, b = 28 cm and c = 30 cm
/ 2 84 / 2 42 s perimeter cm
Area = ( )( )( )s s a s b s c = 42 42 26 42 28 42 30 = 336 cm2
Question 16:
Solution:
1 ( )
2S a b c = 42 34 20 / 2 = 48 cm
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 16 h h cm
Question 17:
Solution:
Given the base is 48 cm, and the other two sides measure 30 cm each. Hence
1 ( )
2S a b c = 48 30 30 / 2 54 cm.
Hence, Area = ( )( )( )s s a s b s c = 54 6 24 4 432 . 2 sq 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) 1 2d h h
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).
For Triangle ABC, 1
( )2
S a b c = 26 28 30 / 2 42
Hence, Area = ( )( )( )s s a s b s c = 42 12 14 6 336 . 1 sq cm
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For Triangle ACD, 1
( )2
S a b c = 50 40 30 / 2 60
Hence, Area = ( )( )( )s s a s b s c = 60 30 20 0 600 . 1 sq 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
The area of the triangle = 1 1
24 15 180 .2 2
base height sq cm
Hence area of the shaded portion = 864 – 180 = 684 sq.cm.
Question 22:
Solution:
Area of rectangle = product of length and breadth = 240 25 1000cm
PB = half of 40 = 20 cm
BQ = half of 25 = 12.5 cm
Area of triangle PBQ = (1/ 2) (1/ 2) 20 12.5b h
Area of triangle PBQ=210 12.5 125 cm
Since all the four triangles are equal, the area of 4 triangles = 2125 4 500 cm
Area of shaded region = 1000 – 500 = 500 cm2
Question 23:
Solution:
(i) The area of the rectangle = 18 10 180 .sq cm
Area of the triangle DEF = 1
(10 6) 30 .2
sq cm
Area of the triangle BCE = 1
(10 8) 40 .2
sq cm
Hence area of the shaded region = 180 – (30 + 40) = 110 sq.cm
(ii) The area of the square = 20 20 400 .sq cm
Area of the triangle LRS = 1
(10 20) 100 .2
sq cm
Area of the triangle MRQ = 1
(10 20) 100 .2
sq cm
Area of the triangle LMP = 1
(10 10) 50 .2
sq cm
Hence area of the shaded region = 400 – (100 + 100 + 50) = 150 sq.cm
Question 24:
Solution:
Area of triangle ABD = 1
(5 24) 60 .2
sq cm
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Area of triangle CBD = 1
(8 24) 96 .2
sq cm
Hence area of the quadrilateral ABCD = 60 + 96 = 156 sq.cm.
Page No: 247
Exercise: 20E
NOTE: Take 22
7 , unless stated otherwise.
Question 1:
Solution:
The circumference of a circle is given by 2 r
(i) Circumference = 2 r22
2 28 176 .7
cm
(ii) Circumference = 22
2 2 1.4 8.8 .7
r m
Question 2:
Solution:
The circumference of a circle is given by 2 r
Diameter = twice the radius
(i) Circumference = 2 r22 35
2 110 .7 2
cm
(ii) Circumference = 22 4.9
2 2 15.4 .7 2
r m
Question 3:
Solution:
The circumference of a circle is given by 2 r
Hence the circumference = 2 2 3.14 15 94.2 .r cm
Question 4:
Solution:
The circumference of a circle is given by 2 r
Hence, 2 r = 57.2 => 22
2 57.2 9.17
r r cm
Question 5:
Solution:
The circumference of a circle is given by 2 r
Hence, 2 r = 63.8 => 22
2 63.8 10.157
r r m
Hence diameter = twice the radius = 20.3 m.
Question 6:
Solution:
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Given 2 r = diameter + 30cm.
Hence, 22 44 30
2 2 30 ( 2) 30 30 7 .7 7 7
r r r r r cm
Question 7:
Solution:
Let the radius of the two circles be 5x and 3x.
Hence the ratio of their circumferences will be 22 22
2 5 : 2 3 5 :37 7
x x
Question 8:
Solution:
The circumference of the circular field = 2 r = 22
2 21 1327
m
At the speed of 8km/hr, the time taken = distance / speed = 132 18
59.4sec8 5
Question 9:
Solution:
Given the inner circumference = 528 m and outer circumference = 616 m.
Hence difference in circumference = (616 - 528) = 88 m.
Hence the width of the track = 7
88 142 22
r m
Question 10:
Solution:
Given the inner circumference of a circular track = 2 r = 22
2 330 52.57
r m r m
The track is of width 10.5 m. Hence, the radius of the outer circle = 52.5 + 10.5 = 63m.
Hence the circumference of the outer circle = 2 r = 22
2 63 3967
m
The cost of fencing along the outer circle = 396 20 .7920Rs .
Question 11:
Solution:
Circumference of the smaller circle = 2 r = 22
2 98 6167
cm
Circumference of the concentric circle = 2 r = 22
2 126 7927
cm
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
Hence 22
2 26.4 4.2 2 8.47
r r cm d r cm
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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,
222 132 21
7r r cm
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
Hence the radius is found 22
2 66 10.57
r r cm
Question 15:
Solution:
The circumference of the circle = 22
2 35 2207
cm
Since this is the perimeter of the square, we get 220
4 220 554
lengthofside length cm
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 =
196 140
2
28 diameter of the well diameter of parapet cmhalfof
Hence, width of the parapet is 28 cm.
Question 17:
Solution:
Diameter of circle = 98 cm
Radius of circle = 49 cm = 0.49 m
Circumference of circle = 22
2 2 0.49 3.087
r
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.
1 revolution covers the circumference of the wheel = 22
2 2 35 2.2m7
r
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For 250 revolutions it moves 250 220 55000 550 cm m
Question 19:
Solution:
Product of number of revolutions (n) and circumference of wheel = Area covered
2 121 n r km
22 77 2 12100000
7 2n cm
77 44 12100000
14n
242 12100000 50000n x
Hence, the car wheel has to take 50,000 revolutions to cover an area of 121 km.
Question 20:
Solution:
Distance covered in 5000 revolutions = 22
2 5000 11000 0.35 357
r m r m cm
The circumference of the wheel = 22
2 35 2207
cm cm
Hence the diameter of the wheel = 70 cm.
Question 21:
Solution:
Distance moved by hour hand in 12 hours =22
(2 4.2) 26.47
cm cm .
Distance moved by hour hand in 24 hours = 26.4 2 52.8 cm cm .
Distance moved by minute hand in 24 hours 44 24 1056 cm cm .
Hence the sum of the distances covered by their tips in 1 day = 1056 52.8 1108.8cm cm cm
Page No: 252
Exercise: 20F
NOTE Take 22
7 , unless stated otherwise.
Question 1:
Solution:
Area of a circle = 2r sq.units
(i) Area = 2r =
2221 21 1386 .
7sq cm
(ii) Area = 2r =
223.5 3.5 38.5 .
7sq m
Question 2:
Solution:
Area of a circle = 2r sq.units
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(i) Area = 2r =
2214 14 616 .
7sq cm
(ii) Area = 2r =
220.7 0.7 1.54 .
7sq m
Question 3:
Solution:
Given the circumference = 22
2 264 427
r r cm
Hence its area = 2r =
2242 42 5544 .
7sq cm
Question 4:
Solution:
Given the circumference = 22
2 35.2 5.67
r r m
Hence its area = 2r =
225.6 5.6 98.56 .
7sq m
Question 5:
Solution:
Given the area = 2r =
2 222616 . 196 14
7r sq cm r r cm
Hence the circumference = 22
2 14 887
cm
Question 6:
Solution:
Given the area = 2r =
2 2221386 . 441 21
7r sq m r r m
Hence the circumference = 22
2 21 1327
m
Question 7:
Solution:
Let the radii of the two circles be 4x and 5x.
Hence the ratio of their areas = 2(4 )x : 2(5 )x = 16: 25.
Question 8:
Solution:
The radius of the circle = length of the string = 21 m.
Hence the area = 2r =
222(21) 1386 .
7sq m
Question 9:
Solution:
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Area of the square = 2 121 11a a cm
Hence the circumference of the circle = four times the length of the side of the square = 44 cm =
222 7
7r r cm
Hence the area of the circle = 2r =
222(7) 154 .
7sq m
Question 10:
Solution:
Radius of the circle = 28 cm. Hence the circumference of the circle = perimeter of the square =
222 28 176
7cm => side =
17644
4 cm
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 ²cm
Area of each button = 22 3.5 3.5
² 9.625 ²7 2 2
r cm
Area of 64 buttons = 64 9.625 616 ²cm
Area of the remaining sheet = area of the sheet - area of 64 buttons = 816 – 616 200 ²cm
Question 12:
Solution:
Area of the rectangle – area of the circle = product of length and breadth - 2r
= 22
(32 90) ( 14 14) 2880 616 2264 .7
sq m
The cost of turfing the remaining portion at the rate of Rs. 50 per square metre =
2264 50 .113200Rs
Question 13:
Solution:
Area of the square = 14 14 cm2 = 196 cm
2.
Sum of areas of the 4 quadrants =2 21 22
4 7 7 154 4 7
cm cm
Hence, the area of the shaded region = (196 - 154) cm2 = 42 cm
2.
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.
Therefore, required area =2 21 22
( 14 14) 154 4 7
m m
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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.
The area of the inner circles = 2 222 22 7 7
( 7 7) ( ) 192.57 7 2 2
cm cm
The area of the shaded region is hence 2 222 21 21
[( ) 192.5] 1547 2 2
cm cm
Question 16:
Solution:
Area of the rectangle = 6 8 48 .sq m
Since there are four quarter circles and one full circle inside this rectangle, the sum of their areas is given
by 2 21 224 2 2 2 2) 25.14
22) (
74 7m m
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 2 220 16 12 m.
So the area of this rectangle is given by the product of length and breadth = 16 12 192 .sq m m2.
Question 2:
Solution:
(b)
Area 212 12
72 2
cm
Question 3:
Solution:
(b)
Area = 2
200 202
diagonaldiagonal cm
Question 4:
Solution:
(a)
Area = 2
25000 1002
diagonalm diagonal m
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Question 5:
Solution:
(c)
Perimeter = 240 2 2 3 30 90l b b b b m l m
Question 6:
Solution:
(d)
Let the length of each side be a cm. Then, its area = a2 cm
2.
New length = (125% of a) = 5
4
a cm. New area =
225
16
acm
2.
Hence the increase in area = 100 56.25%25 1
1
6
6
Question 7:
Solution:
(b)
Let the side of square be a. Length of its diagonal = 2a .
Therefore, required ratio = a2:2a
2 = 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 ) 480 30x x m
Hence the length = 150 m and breadth = 90 m. The area is hence 150 90 13500 .sq m
Question 10:
Solution:
(a)
Length of carpet = total cost/rate = (6000/50)m = 120 m.
Area of the room = area of the carpet =2 275
(1 20 ) 90 100
m m
Hence the width of the room =
90
156m
Question 11:
Solution:
(a)
We know that 1
( )2
S a b c and area of the triangle = ( )( )( )s s a s b s c
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1 ( )
2S a b c =
1(13 14 15) 21
2
Area of the triangle = ( )( )( )s s a s b s c = 21(8)(7)(6) 84 sq.cm
Question 12:
Solution:
(b)
Area of a triangle = 12 8
48 .2 2
b hsq m
Question 13:
Solution:
(b)
Area of an equilateral triangle is given by the formula 23
4a = 4 3 sq.cm => a = 4 cm.
Question 14:
Solution:
(c)
Area of an equilateral triangle is given by the formula 23
4a =
3(64) 16 3
4 sq.cm.
Question 15:
Solution:
(b)
Given 3
6 2 22
a a
Hence the area = 232 3 .
4a sq cm
Question 16:
Solution:
(b)
Area = 16 4.5 72 .sq cm
Question 17:
Solution:
(b)
Area of the rhombus = 24 18
216 .2
sq cm
Question 18:
Solution:
(c)
22(2 1) 37 7
7r cm r cm
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227 7 154 .
7sq cm
Question 19:
Solution:
(c)
Perimeter of rectangle = 2 18l b
Area of four walls = 2 l b h
Area of four walls = 18 3 54 ²m
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)
22 2 17 289l b and 2 46 1 23l b b .
22 2 2l b l b lb
Hence 529 289
120 .2
lb sq cm
Question 22:
Solution:
(b) 2
2 2
2
9 3 4 4 3 3( ) 3
1 1 4 4 1 1
a a a a
b b b b
Question 23:
Solution:
(d)
22
1
222
1(2 )
4 42 4 :11 1
2
dA d
A dd
Question 24:
Solution:
(c)
144 84 84 7056 . 49width sq m width m
Question 25:
Solution:
(d)
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Ratio = 2
2
44 : 3
3 3
4
a
a
Question 26:
Solution:
(a)
2 2 : :1
aa r a r
r
Question 27:
Solution:
(b)
222154 . 7
7r sq cm r cm
Area of the triangle = 23 49 3
4 4r .
Question 28:
Solution:
(c)
Area = 1 2 11
636 . 36 12
2 2
d d dsq cm d cm
Question 29:
Solution:
(d)
Area = 1 11
2144 . 12
2
d dsq cm d cm
Hence the longer diagonal is of length 24 cm.
Question 30:
Solution:
(c)
2224.64 . 2.8
7r r sq m r m
Hence its circumference = 22
2 2 2.8 17.67
r m
Question 31:
Solution:
(c)
( 1 1 ( ) 22)r r r r
Hence, 2 2 2 1 22r r r
2 1 22 0r
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Hence, 2 6 0 3r r
Question 32:
Solution:
(c)
Circumference = 22
2 1.75 117
m m
Hence the number of revolutions made for travelling 11 km = 11000
100011
Page No: 257
Test paper-20
A.
Question 1:
Solution:
We know that ² ² ²side side diagonal
Hence side² = 50² 48² . 196 .sq m sq 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 ) 2 17 6.5 = 221 .l b h 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 .10600Rs
Question 3:
Solution:
Area of square = half of product of its diagonals = 1
64 64 2048 .2
sq cm
Question 4:
Solution:
Let x be the side of the square lawn.
Length PQ = 2 2 4 x m m m x m
Area of PQRS = 4 ² ² 8 16 ²x x x m
Now, Area of the path = Area of PQRS – Area of the square lawn
Hence, 136 ² 8 16 – ² 136 8 16 15x x x x x
Since the side of the lawn = 15 m its area = ² 15 ² 225 ²Side m m
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 96 .sq m
Question 6:
Solution:
Area of the rhombus is given by the formula
Area = 2 2( ( / 2) ) (24 169 144) 24 5 120 .side diagonal sq cmdiagonal
Question 7:
Solution:
Area of the parallelogram = product of base and height
Hence 2 2338 2 169 13b b b m
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 = 2 2 2 225 24 625 576 7H b cm
Hence the area of the right triangle = 1
24 7 84 .2
sq cm
Question 9:
Solution:
Given the radius of the wheel = 35 cm.
Hence its circumference = 22
2 35 220 2.27
cm m
To travel 33 km, that is 33000 m, it will take 33000
150002.2
revolutions.
Question 10:
Solution:
Area of the circle = 2 222
616 196 147
r r r cm
B.
Question 11:
Solution:
(a)
Area of the circle = 2 222
154 . 49 77
r sq cm r r cm
Diameter = 14 cm.
Question 12:
Solution:
(b)
222 44 7
7r r cm
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Area of the circle = 22
7 7 154 .7
sq cm
Question 13:
Solution:
(c)
Diagonal of a square = 2 14 7 2a cm a cm
Its area = 2
7 2 98 .sq cm
Question 14:
Solution:
(b)
The area of the square = 2 50 . 5 2 .a sq cm a cm
Hence the length of its diagonal = 2 10a cm .
Question 15:
Solution:
(a)
The perimeter of the rectangle = 56 2 4 3 14 4 m x x x x m .
Hence its length = 16 m and breadth = 12 m.
Hence, the area of the field = 16 12 192 .sq m
Question 16:
Solution:
(a)
We know that 1
( )2
S a b c and area of the triangle = ( )( )( )s s a s b s c
1 ( )
2S a b c =
1(13 14 15) 21
2
Area of the triangle = ( )( )( )s s a s b s c = 21(8)(7)(6) 84 sq.cm
Question 17:
Solution:
(a)
Area of the equilateral triangle = 23 38 8 16 3 .
4 4a sq cm
Question 18:
Solution:
(b)
Area of the parallelogram = 14 6.5 91 .sq cm
Question 19:
Solution:
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(b)
Area of the rhombus = 18 15
135 .2
sq cm
C.
Question 20:
Solution:
(i) If d1 and d2 be the diagonals of a rhombus, then its area is 1 2
2
d dsq units.
(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.
(v) If each side of a triangle is a cm, then its area = 23
4a cm
2.
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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