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Joy of Mathematics Class 8 Solutions Chapter 21 Surface Area and Volume of Solids
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 21 Surface Area and Volume of Solids. Here students can easily find step by step solutions of all the problems for Surface Area and Volume of Solids. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 21.1 and 21.2
Surface Area and Volume of Solids Exercise 21.1 Solution
Question no – (1)
Solution :
Here, l = long = 6 dm
B = wide = 3.4 dm
H = high = 3 dm
∴ Surface area of card board = 2(lb + bh + hl) dm²
= 2 (6 × 3.4 + 3.4 × 3 × 6)
= 2 (20.4 + 10.2 + 18)
= 2 (48.6)
= 97.2 dm²
= 97.2 × 100
= 9720 cm²
∴ Surface area of card board in dm units 9.72 dm and in cm unit’s 5720 cm²
Let, the x card board used
Now, x × 5/6 = 9720
=> 5x = 58320
=> x = 11664
∴ 11664 cm² card board is used .
Question no – (2)
Solution :
1st packages = 42 dm × 10 dm × 9 m
Surface area of 1st packages
= 2 (42 × 15 + 15 × 9 × 9 × 42) dm²
= 2 × 1143 dm²
= 2286 dm²
2nd packages = 18 dm × 15 dm × 21 dm
∴ Surface area of 2nd packages
= 2 (18 × 15 + 15 × 18 + 21 × 18) dm²
= 2 × 963 dm²
= 1926 dm²
∴ 2nd packages make less quantity.
∴ 2nd packages make less quantity by,
= (2286 – 1926) dm²
= 360 dm²
Question no – (3)
Solution :
Here, long (i) = 1.8 m = 180 cm
Wide (w) = 1.5 m = 150 cm
High (h) = 1.2 m = 120 cm
∴ Volume of the wooden card
= (180 × 150 × 120 cm³
= 3240000 cm³
And the volume of the cube which wide is 6 cm
= (6)³
= 216
∴ Number of iron cubes that can be fitted in the wooden create
= 3240000/216
= 15000
Question no – (4)
Solution :
Here, long (i) = 5.6 m
Wide (b) = 4.3 m
High (h) = 4 m
∴ Area of the wall = 2h (l + b) cm²
= 2 × 4 × (5.6 + 4.3)
= 8 × 9.9
= 79.2 cm²
∴ the cost price of painting the wall at 65/- per sq. meter
= 79.2 × 65
= 5148 Rs
Question no – (5)
Solution :
Here, length (l) = 8 m
Wide (b) = 6 m
High (h0 = 4 m
∴ Area of the room
= 2h (l + b) m²
= 2 × 4 (8 + 6) m²
= 8 × 14 m²
= 112 m²
∴ Area of two windows
= 1.2 × 1 × 2
= 3 m²
And, area of two doors
= 2.2 × 1.2 × 2
= 5.28 m²
Now, cost of white washing the walls at 3 /- per sq. meter
= (112 × 3) – (3 × 3 + 5.28 × 3)
= 336 – (9 + 15.84)
= 336 – 24.84
= 311.16 Rs
Question no – (6)
Solution :
Let, the height of the room = x m
Length (l) = 4.2 m
Breadth (b) = 3 m
∴ Now, area of the room
= 2h (l + b) m²
= 2x (4.2 + 3) m²
According to the question,
2x (7.2) = 51.2
=> x = 51.2/2 × 7.2
= 512/144
= 3.56
So, the height of the room = 3.56 m
Question no – (7)
Solution :
The flat roof dimensions = 4 × 6 m
Rainfall height = 4.2 cm = 0.042 m
Volume of the rain water collected at roof
= 4 × 6 × 0.042 m³
Let, the depth of the water tank be = x m
4 × 6 × 0.042 = 3.2 × 1.5 × x
=> x = 4 × 6 × 0.042/3.2 × 1.5
=> 1008/4800
= 0.21 m
= 210 mm
So, the depth of the tank = 210 mm
Question no – (8)
Solution :
Here, volume of the lamp = 136 cm³
Volume of the cube which side is 2 cm
= (2 × 2 × 2) cm²
∴ Number of cube = volume of lamp/volume of cube
= 136/2 × 2 × 2
= 17
∴ Weight of one cube
= 207/17
= 12.17 g
Question no – (9)
Solution :
Let , the length of the square based cube = a and breadth of the square based cube = a (since the cuboid has a square base )
∴ Height (h) = 10 cm
∴ Side of the cube = 8 cm
∴ Now, volume of the cube = 8 × 8 × 8 cm³
Volume of 5 cube = 5 × 8 × 8 × 8 cm³
Now, according to the question,
5 × 8 × 8 × 8 = a × a × 10
=> a² = 5 × 8 × 8 × 8/10
=> a = √4 × 8 × 8
=> √266
=> a = 16
∴ side of the square based of the cuboid 16 cm
Question no – (10)
Solution :
Here, the dimensions = 19 m × 16 m × 10 m
∴ Volume of the outer dimension wood
= (13 × 16 × 10) m²
= 2080 m²
∴ Thickness
= 50 cm
= 0.5 m
∴ Volume of the inner wood,
= (13 – 2 × 0.5) × (16 – 2 × 0.5) × (10 – 2 × 0.5) m²
= 1620 m²
∴ Volume of the wood
= (2080-1620) m²
= 460 m²
Question no – (11)
Solution :
Here, the dimension of cuboid = 63 cm × 35 cm × 21 cm
∴ Volume of cuboid,
= (63 × 35 × 21) cm³
= 46,305 cm³
∴ Cube of length 7 cm and its volume
= 7 × 7 × 7 cm³
= 343 cm³
∴ Number of cube,
= 46,305/343
= 135
So, the number of cube 135
Question no – (12)
Solution :
Here, shape of dimension = 80 cm × 50 cm
∴ Area of the sheet
= (80 × 50) cm
= 4000 cm²
Now, new length = 80 – 10 – 10
= 60 cm
And new breadth = 50 – 10 – 10
= 50 – 20
= 30
∴ New area of sheet = 60 × 30 cm²
= 1800 cm²
∴ Surface area of the cuboid
= 2h (l + b)
According to the question,
2 h (l + b) = 1800
=> 2h (60 + 30) = 1800
=>2h = 1800/90
=> h = 10
Now, volume of the cuboid = l × b × h
= 60 × 30 × 10
= 18000 cm³
Question no –(13)
Solution :
Here, length of cuboid (i) = 15 + 15 = 30 cm
Breadth (l) = 15 cm
Height (h) = 15 cm
∴ Surface area of cuboid = 2 (lb + bh + lh)
= 2 (30 × 15 + 15 × 15 + 15 × 15)
= 2 × 1125 cm²
= 2250 cm²
∴ Volume of the cuboid = l × b × h
= 30 × 15 × 15
= 30 × 225
= 6750 cm³
So, the surface area of the cuboid 2250 cm² and volume of the cuboid = 6750 cm³
Question no – (14)
Solution :
Dimension of cuboid = 36 cm × 75 cm × 80 cm
∴ Volume of the cuboid = 36 × 75 × 80 cm³
Let, the sides of the cube be x
∴ Volume of the cube = x³cm³
According of the question,
x³ = 36 × 75 × 80
=> x³ = 21600
=> x = 60
∴ The surface area of the cube
= 6 (60)²
= 6 × 360 cm²
= 2160 cm²
Question no (15)
Solution :
Let, the length, breadth and height of the cuboid are 1 cm, b cm and h cm.
∴ then area of the adjacent rectangular faces are lb, bh, hl, cm²
Now, lb = 80 → (1)
bh = 48 → (2)
hl = 60 → (3)
Multiplying (1), (2) and (3) we get,
lb × bh × hl = 230400
=> (lbh)² = 230400
=> lbh = √230400
=> 480
∴ Area of the cuboid,
= 480/80, 480/48, 480/60,
= 6, 10, 8 cm
Surface Area and Volume of Solids Exercise 21.2 Solution :
Question no – (1)
Solution :
Here, radius of the cylinder = (r) = 70cm
∴ Height of the cylinder (h)
= 7 m
= 700 cm
(a) Curved surface area of the cylinder = 2πrh
= 2 × 22/7 × 70 × 700 (∵ π = 22/7)
= 44 × 7000)
= 308000 cm2
(b) Total surface area of the cylinder
= 2πr (h + r)
= 2 × 22/7 × 70 (700 + 70)
= 2 × 22/7 × 70 × 770
= 44 × 770
= 311080 cm2
(c) Volume of the cylinder = πr2h
= 22/7 × 70 × 70 × 700
= 220 × 4900
= 1078000 cm2
Question no – (2)
Solution :
Let, the radius of the right circular cylinder = r
∴ Height of the cylinder (h) = 20cm
Now, volume of the cylinder = πr2h
According to the question,
πr2h = 770
=> r2 = 770/πh
= 770 × 7/22 × 20
=> r = √7 × 11 × 7/2 × 11 × 2
=> r = 7/2
= 3.5
∴ Radius of the cylinder = 3.5 cm
∴ Curved surface area = 2πrh
= 2 × 22/7 × 35/10 × 20
= 440 cm2
Question no – (3)
Solution :
(a) Let, the radius of the base = r
∴ Circumference of the base = 2πr
∴ 2πr = 88
=> r = 88/2π
= 88 × 7/2 × 22 (∵ π = 22/7)
= 14
∴ Radius (r) = 14 cm
Height (h) = 26 cm
Now,
Total surface area of the cylinder = 2πr = (h + r)
= 2 × 22/7 × 14 (26 + 14)
= 2 × 22/7 × 14 × 40
= 3520 cm2
Volume of the cylinder = πr2h
= 22/7 × 14 × 14 × 26
= 16016 cm3
(b) Let, the radius of the base = r cm
∴ Area of the base = πr2 cm2
∴ πr2 = 346.5
=> r2 = 346.5 × 7/22 (∵ π = 22/7)
=> r2 = 110.25
=> r = 10.5
∴ radius (r) = 10.5 cm
height (h) = 24.5 cm
∴ Total surface area of the cylinder
= 2πr (h + r)
= 2 × 22/7 × 10.5 (24.5 + 10.5)
= 2 × 22/7 × 105/10 × 35
= 2310 cm2
∴ Volume of the cylinder = πr2h
= 22/7 × 10.5 × 10.5 × 24.5
= 8489.25 cm3
Question no – (4)
Solution :
Let, the height of the cylinder = h
∴ diameter f the cylinder = 7 cm
∴ Radius of the cylinder
= 7/2 cm
= 3.5 cm
∴ Volume of the cylinder = πr2h
= 22/7 × 3.5 × 3.5 × h
According to the question,
22/7 × 3.5 × 3.5 × h = 462
=> h = 462 × 7/22 × 3.5 × 3.5
= 12
∴ Thus the height of the cylinder 12 cm.
Question no – (5)
Solution :
Diameter of the cylinder milk tank = 3.5m
∴ Radius of the cylinder milk tank
= 3.5/2 m
= 1.75 m
∴ Height (h) = 7m
∴ Volume of the cylinder milk tank
= π × (1.75)2 × 7 m3
= 22/7 × (1.75)2 × 7 m3
= 67.375m3
= 67375 liter
Question no – (6)
Solution :
Here, diameter of the road roller = 1.26m
∴ Height of the cylinder (h) = 1 m
∴ Area leveled by road roller in one revolution = 2πrh
= 2 × 22/7 × 1.26/100 × 1
= 36 × 22/100 = 792/100
= 7.92
∴ Area leveled in 500 complete revolution
= 7.92 × 500
= 3960
Question no – (7)
Solution :
Here, radius of the cylinder (r) = 35/2 cm
= 17.5 cm
= 0.175 m
∴ Height of the cylinder (h) = 3m
∴ Surface area of the cylinder lamp
= 2πrh
= 2 × 22/7 × 0.175 × 3
= 3.3 m2
∴ Surface area of 1000 lamp
= 3.3 × 1000 m2
= 3300 m2
∴ The total cost price of painting
= 15 × 3300m2
= 49500m2
Question no – (8)
Solution :
Here, length (l) = 22cm
Wide (b) = 12cm
Height of the cylinder (h) = 12cm
∴ Circumference of base of cylinder = 22
∴ 2πr = 22
=> r = 22 × 7/2 × 22
=> r = 3.5cm
∴ Volume of the cylinder = πr2h
= 22/7 × 3.5 × 3.5 × 12
= 462 cm3
Now, circumference of base of cylinder = 12 cm
∴ 2πr = 12
=> r = 12 × 7/2 × 22
= 42/22
= 1.909 cm
and the volume of the cylinder = πr2h
= 22/7 × (1.909)2 × 12
Question no – (9)
Solution :
Here,
Side of the cube = 3.08 m
= 308 cm
= 308000km
∴ Radius of the wire = 1.4cm
Let, the height of the wire = h
Now,
πr2h = 308000
=> h = 308000/π.r2
=> 308000/3.14 × (1.4)2 (π = 3.14)
= 50045 cm
= 5km
Question no – (10)
Solution :
Let, the radius of the cylinder = r and height of the cylinder = h
∴ Curved surface area = 2πrh
∴ Total surface area = 2πr (h + r)
Now, according to the question,
2πrh/2πr(h + r) = 3/5
=> h/h + r = 3/5
=> 5h = 3h + 3r
=> 5h – 3h = 3r
=> 2h = 3r
=> h/r = 3/2
∴ Ratio of the height to the radius of the cylinder = 3 : 2
Question no – (11)
Solution :
Let, the radius of the cylinder 2x and 3x and height of the cylinder 9h and 4h
∴ Volume of the first cylinder (v1) = πr2h
= π × (2x)2 × 9h
∴ Volume of the second cylinder (v2) = π(3x)2 × 4h
∴ v1/v2 = π × 4x × 9h/ π × 9x2 × 4h
∴ v1/v2 = 1/1
∴ Ratio of their volume 1 : 1
Question no – (12)
Solution :
Here, the length of the pipe (h) = 35m
∴ Outer radius of the pipe (R) = 16/2 = 8cm
∴ Inner radius of pipe (r) = 14/2 = 7cm
∴ Volume of the metal = πh (R2 – r2) cm3
= 22/7 × 35 × (64 – 49)cm3
= 22 × 5 × 15 cm3
= 1650 cm3
= 0.165 m3
∴ Volume of the metal 0.165 m3
Question no – (13)
Solution :
Depth of well = 12 m
Radius = 7 m
Volume of earth dug out = πr2h
= 22/7 × 7 × 7 × 12
= 1848 m3
Let, height of the embankment = h m
wide = 7 m
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