Mathematics Solutions Solutions for Class 8 Math Chapter 17 - Surface Area And Volume

Share with your friends

Mathematics Solutions Solutions for Class 8 Math Chapter 17 Surface Area And Volume are provided here with simple step-by-step explanations. These solutions for Surface Area And Volume are extremely popular among class 8 students for Math Surface Area And Volume Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of class 8 Math Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class 8 Math are prepared by experts and are 100% accurate.

Page No 108:

Question 1:

Find the volume of a box if its length, breadth and height are 20 cm, 10.5 cm and 8 cm respectively.

Answer:

$Volume of the box = length \times breadth \times height = 20 \times 10.5 \times 8 = 1680 {cm}^{2}$

Page No 108:

Question 2:

A cuboid shape soap bar has volume 150 cc. Find its thickness if its length is 10 cm and breadth is 5 cm.

Answer:

Volume of soap bar = 150 cc
$l b h = 150 \Rightarrow 10 \times 5 \times h = 150 \Rightarrow h = \frac{150}{10 \times 5} = 3 cm$

Page No 108:

Question 3:

How many bricks of length 25 cm, breadth 15 cm and height 10 cm are required to build a wall of length 6 m, height 2.5 m and breadth 0.5 m?

Answer:

Let the number of bricks required be n.
Volume of each brick = $25 \times 15 \times 10 = 3750 {cm}^{3}$
Volume of the wall = $6 \times 2.5 \times 0.5 = 7.5 m^{3}$
Number of bricks required(n) = $\frac{Vol of wall}{Vol of brick} = \frac{7.5 \times 1000000}{3750} = 2000$
Thus, 2000 bricks are required to make the wall.

Page No 109:

Question 4:

For rain water harvesting a tank of length 10 m, breadth 6 m and depth 3 m is built. What is the capacity of the tank? How many litre of water can it hold?

Answer:

length = 10 m
breadth = 6 m
depth = 3 m
Volume of the tank = $10 \times 6 \times 3 = 180 m^{3}$
$1 m^{3} = 1000 L \Rightarrow 180 m^{3} = 180000 L$

Page No 110:

Question 1:

In each example given below, radius of base of a cylinder and its height are given. Then find the curved surface area and total surface area.
(1) r = 7 cm, h = 10 cm
(2) r = 1.4 cm, h = 2.1 cm
(3) r = 2.5 cm, h = 7 cm
(4) r = 70 cm, h = 1.4 cm
(5) r = 4.2 cm, h = 14 cm

Answer:

We know that
Curved surface area of a cylinder = $2 π r h$
Total surface area = $2 π r (h + r)$
(1) r = 7 cm, h = 10 cm
Curved surface area of a cylinder = $2 π r h$ = $2 π \times 7 \times 10 = 140 π = 140 \times \frac{22}{7} = 440$ sq cm
Total surface area = $2 π r (h + r)$ = $2 \times \frac{22}{7} \times 7 (10 + 7) = 748$ sq cm
(2) r = 1.4 cm, h = 2.1 cm
Curved surface area of a cylinder = $2 π r h$ = $2 \times \frac{22}{7} \times 1.4 \times 2.1 = 18.48$ sq cm
Total surface area = $2 π r (h + r)$ = $2 \times \frac{22}{7} \times 1.4 (2.1 + 1.4) = 30.8$ sq cm
(3) r = 2.5 cm, h = 7 cm
Curved surface area of a cylinder = $2 π r h$ = $2 \times \frac{22}{7} \times 2.5 \times 7 = 110$ sq cm
Total surface area = $2 π r (h + r)$ = $2 \times \frac{22}{7} \times 2.5 (2.5 + 7) = 149.28$ sq cm
(4) r = 70 cm, h = 1.4 cm
Curved surface area of a cylinder = $2 π r h$ = $2 \times \frac{22}{7} \times 70 \times 1.4 = 616$ sq cm
Total surface area = $2 π r (h + r)$ = $2 \times \frac{22}{7} \times 70 (70 + 1.4) = 31416$ sq cm
(5) r = 4.2 cm, h = 14 cm
Curved surface area of a cylinder = $2 π r h$ = $2 \times \frac{22}{7} \times 4.2 \times 14 = 369.6$ sq cm
Total surface area = $2 π r (h + r)$ = $2 \times \frac{22}{7} \times 4.2 (4.2 + 14) = 480.48$ sq cm

Page No 110:

Question 2:

Find the total surface area of a closed cylindrical drum if its diameter is 50 cm and height is 45 cm. (π = 3.14)

Answer:

Diameter = 50 cm
Radius = 25 cm
Height = 45 cm
Total surface area = $2 π r (h + r)$
$= 2 \times 3.14 \times 25 (25 + 45) = 10990 sq cm$
Thus, the total surface area of the closed cylinderical drum is 10990 sq cm.

Page No 111:

Question 3:

Find the area of base and radius of a cylinder if its curved surface area is 660 sqcm and height is 21 cm.

Answer:

Curves surface area = 660 sq cm
Height = 21 cm
Curved surface area = $2 π r h = 2 \times \frac{22}{7} \times r \times 21 = 660$
$\Rightarrow 132 r = 660 \Rightarrow r = \frac{660}{132} = 5 cm$
Area of base = $π r^{2} = \frac{22}{7} \times 5^{2} = 78.57 sq cm$

Page No 111:

Question 4:

Find the area of the sheet required to make a cylindrical container which is open at one side and whose diameter is 28 cm and height is 20 cm. Find the approximate area of the sheet required to make a lid of height 2 cm for this container.

Answer:

Diameter = 28 cm
Radius = 14 cm
Height = 20 cm
Curved surface area of the cylinder + area of the base = area of the sheet
$\Rightarrow Area of sheet = 2 π r h + π r^{2} \Rightarrow Area of sheet = π r (2 h + r) \Rightarrow Area of sheet = \frac{22}{7} \times 14 (2 \times 20 + 14) \Rightarrow Area of sheet = 2376$
Thus, area of sheet required = 2376 cm²
Area of sheet required to make a lid of height 2 cm will be
$Area of sheet = 2 π r h + π r^{2} = π r (2 h + r) = \frac{22}{7} \times 14 (2 \times 2 + 14) = 44 \times 18 = 792$
Thus, area of sheet used to make the lid is 792 sq cm.

Page No 112:

Question 1:

Find the volume of the cylinder if height (h) and radius of the base (r) are as given below.
(1) r = 10.5 cm, h =8 cm
(2) r = 2.5 m, h = 7 m
(3) r = 4.2 cm, h = 5 cm
(4) r = 5.6 cm, h = 5 cm

Answer:

Volume of the cylinder = $π r^{2} h$
(1) r = 10.5 cm, h =8 cm
Volume = $π r^{2} h$ = $\frac{22}{7} \times 10.5 \times 10.5 \times 8 = 2772$ cubic cm
(2) r = 2.5 m, h = 7 m
Volume = $π r^{2} h$ = $\frac{22}{7} \times 2.5 \times 2.5 \times 7 = 137.5$ cubic cm
(3) r = 4.2 cm, h = 5 cm
Volume = $π r^{2} h$ = $\frac{22}{7} \times 4.2 \times 4.2 \times 5 = 277.2$ cubic cm
(4) r = 5.6 cm, h = 5 cm
Volume = $π r^{2} h$ = $\frac{22}{7} \times 5.6 \times 5.6 \times 5 = 492.8$ cubic cm

Page No 112:

Question 2:

How much iron is needed to make a rod of length 90 cm and diameter 1.4 cm?

Answer:

Total length = 90 cm
Diameter = 1.4 cm
Radius = 0.7 cm
Total iron required = Volume of cylinder
= $π r^{2} h = \frac{22}{7} \times 0.7 \times 0.7 \times 90 = 138.6$ cm³
Thus, 138.6 cm³of iron is required.

Page No 112:

Question 3:

How much water will a tank hold if the interior diameter of the tank is 1.6 m and its depth is 0.7 m?

Answer:

Diameter = 1.6 m
Radius = 0.8 m
Depth = 0.7 m
Volume of the tank = ${πr}^{2} h = \frac{22}{7} \times {(0.8)}^{2} \times 0.7 = 1.408$ m³.
$1 m^{3} = 1000 L 1.408 m^{3} = 1.408 \times 1000 = 1408 L$
Thus, the tank will hold 1408 L.

Page No 112:

Question 4:

Find the volume of the cylinder if the circumference of the cylinder is 132 cm and height is 25 cm.

Answer:

Circumference = 132 cm
$\Rightarrow 2 π r = 132 \Rightarrow 2 \times \frac{22}{7} r = 132 \Rightarrow r = \frac{132 \times 7}{2 \times 22} = 21 cm$
Height = 25 cm
Volume = $π r^{2} h = \frac{22}{7} \times 21 \times 21 \times 25 = 34650$ cm³

View NCERT Solutions for all chapters of Class 8