Class 8 MAHARASHTRA STATE TEXTBOOK BUREAU Solutions Maths Chapter 16: Surface area and Volume
Surface area and Volume Exercise Ex. 16.1
Solution 1
Given that length = l = 20 cm, breadth = b = 10.5 cm, height = h = 8 cm
Volume of cuboid = l × b × h
= 20 × 10.5 × 8
= 1680 cm2
Volume of box is 1680 cm2.
Solution 2
Given that length = l = 10 cm, breadth = b = 5 cm, Volume of cuboid = 150 cc.
Volume of cuboid = l × b × h
∴ 10 × 5 × h = 150
Thickness of soap bar is 3 cm.
Solution 3
Given that bricks dimensions are l = 25 cm, b = 15 cm and h = 10 cm
Volume of one brick = l × b × h
= 25 × 15 × 10
Given that wall dimensions are l = 6 m = 600 cm, b = 2.5 m = 250 cm and
h = 0.5 m = 50 cm
Volume of wall = l × b × h
= 600 × 250 × 50
The number of bricks required to build a wall
The number of bricks required to build a wall is 2000.
Solution 4
Given that length = l = 10 m, breadth = b = 6 m, h = 3 m.
Volume of cuboid = l × b × h
= 10 × 6 × 3
= 180 m3
Number of water it can hold = 180 × 1000 (1 m3 = 1000 l)
= 1,80,000 L
The capacity of tank is 180 m3 and it can hold 1,80,000 L.
Surface area and Volume Exercise Ex. 16.2
Solution 1
(1) r = 7 cm, h = 10 cm
Curved surface area
= 2Πrh
= 
= 440 cm2
Total surface area
= 2Πr (r + h)
= 
= 
= 748 cm2
(2) r = 1.4 cm, h = 2.1 cm
Curved surface area
= 2Πrh
= 
= 18.48 cm2
Total surface area
= 2Πr (r + h)
= 
= 
= 30.80 cm2
(3) r = 2.5 cm, h = 7 cm
Curved surface area
= 2Πrh
= 
= 110 cm2
Total surface area
= 2Πr (r + h)
= 
= 
= 149.285 cm2
(4) r = 70 cm, h = 1.4 cm
Curved surface area
= 2Πrh
= 
= 616 cm2
Total surface area
= 2Πr (r + h)
= 
= 
= 31416 cm2
(5) r = 4.2 cm, h = 14 cm
Curved surface area
= 2Πrh
= 
= 369.60 cm2
Total surface area
= 2Πr (r + h)
= 
= 
= 480.48 cm2
Solution 2
For cylinder:
d = 50 cm, h = 45 cm, π = 3.14
r =
= 25 cm
Total surface area
= 2Πr (r + h)
= 
= 
= 10, 990 cm2
Solution 3
Curved surface area = 660 sq. cm
h = 21 cm
2Πrh = 660
∴ 
= 660
∴ 44 × 21 r = 660
∴ 
Area of base of cylinder
= Πr2
= 
= 78.57 cm2
Solution 4
Cylindrical open container:
d = 28 cm,
r = 
h = 20 cm,
circular lid of cylindrical container
R = 14 cm, h = 2 cm
Area of sheet required to make cylindrical container open at one end.
= Area of base + Cured surface of cylinder
= ΠR2 + 2ΠRH
= ΠR (R + 2H)
= 
= 
= 2376 cm2
Approximate area of sheet required to make a lid
≈ area of base + Curved surface area of cylinder
≈ Πr2 + 2Πrh
≈ Πr (r + 2h)
≈ 
≈ 
Area of sheet required to make cylinder is 2376 cm2 and approximate area of sheet required to make lid is 792 cm2
Surface area and Volume Exercise Ex. 16.3
Solution 1
(1) r =10.5 cm, h = 8 cm
Volume of cylinder = 𝜋r2h
= 
(2) r = 2.5 m, h = 7 m
Volume of cylinder = 𝜋r2h
= 
(3) r = 4.2 cm, h = 5 cm
Volume of cylinder = 𝜋r2h
= 
(4) r = 5.6 cm, h = 5 cm
Volume of cylinder = 𝜋r2h
= 
Solution 2
d = 1.4 cm, 
, h = 90 cm
volume of cylinder
=
Solution 3
d = 1.6 m, 
volume of cylinder
=
Solution 4
Circumference of cylinder = 132 cm
h = 25 cm
circumference of cylinder = 132 cm
∴ 
∴ 
∴ r = 21 cm
Volume of cylinder = 𝜋r2h
= 
Volume of cylinder is 34650 cm3.