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Frank ICSE Mathematics Class 8 Solutions Chapter 18 Mensuration
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Frank ICSE Mathematics Class 8 Math Book, Chapter 18, Mensuration. Here students can easily find step by step solutions of all the problems for Mensuration, Exercise 18.1, 18.2, 18.3, 18.4 and 18.5 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 18 solutions. Here in this post all the solutions are based on latest Syllabus.
Mensuration Exercise 18.1 Solution :
Question no – (1)
Solution :
= r = 7cm
∴ Area = πr2 = 22/7 × 7
= 154 cm2
Perimeter = 2π =2 × 22/7 × 7
= 44 cm
Question no – (2)
Solution :
PR = √(16)2 + (12)2
= √256 + 133
= √400
= 20 cm
∴ Radius = 20/2 = 10 cm
∴ Area = πr2 = (3.14) × 10 × 10
= 314 cm2
Question number – (3)
Solution :
Area of the large semicircle
= 1/2πr2 = 22/7 × (21/2) × (21/2) × 1/2
= 346.5/2 cm2
= 173.25cm2
Area of the small semicircles
= 3πr2 = 3/2 × 22/7 × 7/2
= 57.75 cm2
Total area = 173.25 + 57.75
= 231 cm2
Question no – (4)
Solution :
Length of the wire = (6.6 × 3) cm = 19.8 cm
= 2πr = 19.8
or, 2 × 22/7 × r = 19.8
or, r = 19.8 × 7/2 × 22
= 0.45 × 7
= 3.15 cm
Question no – (5)
Solution :
∴ Perimeter = 2πr = 440
or, 2r = 440 × 7/22 = 140 cm
∴ Area = πr2 = 22/7 × 70 × 70
= 15400 m2
∴ Total cost 15400 × 225
= 3465000 rupees
Question no – (6)
Solution :
The diameter of the circle will be 14 cm
∴ Radius
= 14/2
= 7 cm
∴ Area of the circle
= πr2 = 22/7 × 7 × 7
= 154 cm2
∴ Remaining Area
= (21 × 14) – 154
= 294 – 154
= 140 cm2
Question no – (7)
Solution :
Breadth of the rectangle = diameter of circle = 140 cm
∴ Radius of the circle = 40 cm
∴ Total Area = Area of rectangle + 2 identical semicircles
Area of rectangle + area of circle
= {250 – 80) × (140)} + {22/7 × 40 × 40}
= (170 × 140) + (3.14 ×40 × 40)
= 238000 + 5024
= 18824 cm2
Question no – (8)
Solution :
(i) Diameter = 42m, radius = 21m
∴ Perimeter = 2r + πr = 21 (2 + 22/7)
= 21 × 36/7
= 108 m
Area = πr2/2 = 22/7 × 1/2 × 21
= 693m2
∴ Cost of fencing = (108 × 7)
= 756 rupees
(ii)
∴ Perimeter = 21 + 28 + (22/7 × 35/2)
= 21 + 28 + 55 = 104 m
Area = (1/2 × 21 × 28) + (1/2 × 22/7 × 35/2 × 35/2)
= (14 × 21) + (1925/4)
= 294 + 481.25
= 775.25 m2
∴ Cost of fencing = (104 × 7) = 728 rupees
(iii)
Perimeter = 20 + 20 + 21 + (21/2 × 22/7)
= 20 + 20 + 21 + 33 = 94 m
Area = (21 × 20) + (1/2 × 22/7 × 21/2 × 21/2)
= 420 + 165.75
= 585.75 m2′
Question no – (9)
Solution :
Shaded area – Area of large semicircle –
= 2 (area of smaller semicircle)
= πR2 – 2πr2
= 22/7 [914)2 – 2 (7)2]
= 22/7 × (7)2 × [22 – 2]
= 22 × 7 × 2
= 308 cm2
Question no – (10)
Solution :
Diameter of each circle = 14/2 = 7 cm
(a) Radius of each circle = 7/2 = 3.5 cm
(b) Remaining area = Area of rectangle – 6 (area of circle)
= (21 × 20) – 6 [22/7 × (3.5) × (3.5)]
= 420 – (38.5 × 6)
= 420 – 231
= 189 cm2
Mensuration Exercise 18.2 Solution :
Question no – (2)
Solution :
∴ Area = (12 × 15) + (1/2 × 6 × 15)
= 180 + 45
= 225 cm2
Question no – 3)
Solution :
Sum of the parallel sides = 108/12 cm
= 9 cm
Question no – (4)
Solution :
Suppose initial height = h
Suppose final height = 3h
∴ initial base (sum) = b
∴ initial final base = 3b
initial area = bh/2 final area = 9bh/2
Area will be 9 times of the initial area
Question no – (5)
Solution :
∴ Area = 16/2 × (22.3 + 28.7)
= 16/2 × 51 = 816/2 cm2
= 408cm2
Question no – (6)
Solution :
Distance between parallel sides = Area/(sum of then/2)
= 984/(31.5 + 50.5)/2= 984/82/2
= 24 cm
Question no – (7)
Solution :
∴ Area = 4/2 × (8 + 6)
[80 dm = 8m, 40 dm = 4 m]
= 28m2
Question no – (8)
Solution :
Suppose the parallel sides are – x cm, (x + 6) cm
∴ 9/2 × (x + x + 6) = 270
or, 2x + 6 = 30 × 2
or, x = 27 cm
Length of the
∴ x + 6 = 33 cm
Question no – (9)
Solution :
Suppose the other side = x cm
∴ (x + 42/2) × 26 = 793
or, x + 42 = 30.5 × 200
or, x = 61 – 42 = 19 cm
Question no – (10)
Solution :
Suppose there are x trapeziums
∴ x × 8/2 × (17 + 25)
= 3.78 × 100 × 100
or, x = 225
Question no – (11)
Solution :
∴ Area in m2 = 41 × (67 + 58)/2 × 100 × 100
= 41 × 125/2 × 100 × 100
= 41/160 m2
Question no – (12)
Solution :
Since AB = 57, DC = 27
∴ AE = 57 – 27 = 30 cm
And BC + AD = 134 – (57 + 27) = 50 cm – (i)
And, AD2 – BC2 = 302
or, (AD + BC) (AD – BC) (AD – BC) = 900
or, AD – BC = 900/50
= 18 – (ii)
From (i) and (ii) we get, AD = 34 cm
BC = 16 cm
Mensuration Exercise 18.3 Solution :
Question no – (1)
Solution :
(a) π/2 (10)2 + {(20)2 – (1/2 × 12 × 16)}
= 314/2 + 304
= 157 + 304
= 461 cm2
(b) (14 × 14) + {1/2 × 14 × (19 – 14)}
= 196 + 35
= 231 cm2
(c) (24 × 24) – [2 × 1/2 × 10 × (24 – 0)]
= 567 – 150
= 426 cm2
(d) (6 × 6) + (12 × 6) + {1/2 × 4 × (5 + 12)
= 36 + 72 + 34
= 142 cm2
Question no – (2)
Solution :
(a) 1/2 π(14)2 + 1/2 × 28 × (20 + 30)
= 308 + 700
= 1008 cm2
(b) 1/2 π(21/2)2 + (21 × 21) = 639/2 + 441
= 346.5 + 441
= 787.5 cm2
(c) 1/2 π(7)2 + 1/2 π (14)2 + 1/2 × (4 – 28) × (42 – 7 – 14)
= 77 + 308 + 441
= 826 cm2
(d) π [(11)2 – (4)2] = 22/7 × (121 – 16)
= 22/7 × 105
= 22 × 15
= 330 cm2
Question no – (3)
Solution :
Depth = 112/1/2(16 + 12)
= 112/1/2 × 28 =
112/14
= 8m
Question no – (4)
Solution :
= {2 × 1/2 × (3 + 13) × 5} + (13 × 5)
= 195 + 65
= 260 cm2
Question no – (5)
Solution :
Area of flower beds along = 1/2 × (24 + 24) × 10
= 290 m2
Area of the flower beds along breath = 1/2 × (28 + 18) × 10
= 230 m2
Question no – (6)
Solution :
Perimeter = (6 + 5 + 4 + 5 + 8) cm
= 28 cm
Area = (1/2 × 6 × 8) + [1/2 × 4 × {√62 + 82 + 4}]
= 24 + 28 = 52 cm2
Question no – (7)
Solution :
= [1/2 × 20 × (30 + 10) – [1/2 × π × (10)2]
= 400 – 157
= 243 cm2
Question no – (8)
Solution :
= [1/2 × 7 × 7) + {1/2 × 22/7 × (7√2/2)2}
= 49/2 + 77/2
= 63 cm2
Question no – (9)
Solution :
= (1/2 × 8 × 4) + (1/2 × 8 × 2.5)
= 16 + 10
= 26 cm2
Question no – (10)
Solution :
= (9 × 3) + {1/2 × (9 + 3) × 3}
= 27 + 18
= 45 cm2
Question no – (11)
Solution :
= (1/2 × 12 × 14) + (1/2 × 10 × 12) + {1/2 × (12 + 14) × 8} + (1/2 × 30 × 7}
= 353 cm2
Mensuration Exercise 18.4 Solution :
Question no – (1)
Solution :
Cost = 2 {(30 × 25) + (25 + 18) + (30 + 18)} × 12
= 2 × 170 × 12
= 41760
Question no – (2)
Solution :
Cost of whitewashing walks
= [{2 × (150 + 25)] × 6 + (150 × 25] × 20
= 117000 rupees
Cost of polishing the flower
= (150 × 25 × 40) = 15000 rupees
Question no – (3)
Solution :
Side of the cube = √3750/6
= √625
= 25 m
Question no – (4)
Solution :
Painted are = (8 × 7.5) + (2 × 8 × 6) + (2 × 7.5 × 6)
= 60 + 96 + 90
= 246m2
Question no – (5)
Solution :
(a) 2 {70 × 60) + (60 × 50) + (70 × 50)
= 2 × 10700
= 21400 cm2
(b) 6 × 60 × 60
= 21000 cm2
(c) (2 × 22/7 × 7/2) × 7
= 154 cm2
(d) 6 × 7 × 7
= 294 cm2
Question no – (6)
Solution :
Cost = {(2 × 22/7 × 3.5) × 8} × 130
= 176 × 130
= 22880 rupees
Question no – (7)
Solution :
Required wrapper = 2 × {(50 × 35) + (35 × 10) + (50 × 10)} × 60
= 321000 cm2
= 312000 cm2 = (312000/1000) m2
= 312 m2
Question no – (8)
Solution :
= 2πrh = 2 × 22/7 × 35/2 × 25
= 275 m2
Question no – (9)
Solution :
Area = 2πrh × 900
= (2 × 22/7 × 98/2 × 125) × 900
= (77 × 500 × 900) cm2
= (77 × 5 × 9) m2 = 3456 m2
Question no – (10)
Solution :
= 2πrh = 2 × 22/7 × 350/2 × (750 – 50)
= (22 × 50 × 700) cm2
= 77m2
Question no – (11)
Solution :
Cost = {2 × (7 + 6) × 4 – 7 (7 × 6) × 15
= 2085 rupees
Mensuration Exercise 18.5 Solution :
Question number – (2)
Solution :
Curved surface area,
2πrh = 2 × 22/7 × 8 × 21
= 1056 cm2
Question no – (4)
Solution :
Suppose height = h πr2h = 577.5
or, 22/7 × (7/2)2 × h = 577.5
or, h = 577.5 × 2/11 × 7 = 15 cm
Question no – (5)
Solution :
Suppose height = h = 14 × 11 × h πr2h
or, r2 = 14 × 11 × 7/22
or, r2 = 72
∴ r = 7
Question number – (6)
Solution :
New edge = √33 + 43 + 53
= √27 + 64 + 125
= √216
= 6 cm
Question no – (7)
Solution :
= No of cubical blocks
= 300 × 75 × 602/30 × 10 × 30
= 50
Question no – (8)
Solution :
No of small boxes
= 150 × 75 × 24/12 × 10 × 3
= 750
Question no – (9)
Solution :
Suppose the height = h cm
= 22 × 18 × 14
= 22/7 × (7)2 × h
or, h = 22 × 18 × 14/22 × 7
= 36 cm
Question no – (10)
Solution :
= 2πrh = 4400 – (i) 2πr = 220 – (ii)
Dividing (i) by (ii) we get h = 20 cm
And from (ii) r = 220 × 7/2 × 22 = 35 cm
∴ radius = 35 cm
∴ Volume = πr2h
= 22/7 × 35 × 20
= 77000 cm3
Question no – (11)
Solution :
Curved surface area = 44 × 20
= 880 cm2
Now, 2πrh = 880
or, r = 880 × 7/2 × 22 × 20
= 7 cm
∴ Volume = πr2h
= 22/7 × 7 × 7 × 20
= 3080 cm3
Question no – (12)
Solution :
No of person = 30 × 17 × 6/5
= 612
Question no – (13)
Solution :
Quantity of sort = πr2h
= 22/7 × (7/2)2 × 20
= 770m2
Height of the platform,
= 770/14 × 11
= 5m
Question no – (14)
Solution :
Suppose breadth of room = x m
∴ 2 (2x + x) × 4 = 192
or, x = 8
∴ Breadth = 8m,
length = 16m,
height = 4m
∴ Volume = 8 × 16 × 4
= 512m3
Question no – (15)
Solution :
(i) Volume of can = πr2h
= 22/7 × (14) × 20
= 12320 cm3
(ii) 2πrh = 2 × 22/7 × 14 × 20
= 1760 cm2
Revision Exercise Questions Solution :
Question no – (1)
Solution :
Volume of can,
= 6 × (6)3
= 1296 cm3
Surface area,
= (6)2 × 26
= 936 cm2
Question no – (2)
Solution :
Thickness = 4 × 4 × 4/32 × 8
= 1/4 cm
= 0.25
Question number – (3)
Solution :
= πr2h = 594
or, h = 594 × 7/22 × 1/9
= 21 m
Question no – (4)
Solution :
= No of cubes
= 16 × 8 × 4/2 × 2 × 2
= 64
Question no – (5)
Solution :
Since, r = 3 1/2
or, r = 7/2 πr2
= 22/7 × 7/2
= 77/2 cm2
Area lying outside the circle
= (20 × 15) – 77/2
= 300 – 77/2
= 600 – 77/2
= 523/2
= 261.5 cm2
Question no – (6)
Solution :
Volume of first cake
= πr2h = 22/7 × 7 × 7 × 4
= 616 cm3
Volume of second cake
= 14 × 8 × 5 = 560 cm3
∴ The first cake
= (616 – 560)
= 56 cm3 more volume
Next Chapter Solution :
👉 Chapter 19 👈
