229 Exercise 20A Question 1: Solution: (i) Area of a rectangle

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:


2base height

That is, 1 72

72 16 92 8

height height cm

Question 3:

Solution:


2base height

Hence, 1 224

224 28 162 14

height height m

Question 4:

Solution:


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


(iii) 1

( )2

S a b c = 1

(91 98 105) 1472


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:


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:


Hence the circumference = 2 2 3.14 15 94.2 .r cm

Question 4:

Solution:


Hence, 2 r = 57.2 => 22

2 57.2 9.17

r r cm

Question 5:

Solution:


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

www.vedantu.com 18

(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

www.vedantu.com 21

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


www.vedantu.com 22

1 ( )

2S a b c =

1(13 14 15) 21

2


Question 12:

Solution:

(b)

Area of a triangle = 12 8

48 .2 2

b hsq m

Question 13:

Solution:

(b)


4a = 4 3 sq.cm => a = 4 cm.

Question 14:

Solution:

(c)


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

www.vedantu.com 23

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

www.vedantu.com 27

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


1 ( )

2S a b c =

1(13 14 15) 21

2


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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Documents

229 Exercise 20A Question 1: Solution: (i) Area of a rectangle