﻿ Surface Areas and Volumes - Class 10 Maths Chapter 8

# Chapter: Surface Areas and Volumes - Class 10

## Surface Areas and Volumes - Sub Topics

The total area of all the surfaces on a solid shape is known as its surface area. The volume of a shape or object is its capacity to retain something or the space it occupies. This chapter will break down the basics, formulas and real-life uses of surface area and volume. Also covers the Surface area and volume of a combination of solids, the conversion of solid from one shape to another and the frustum of a cone.

• Surface Area
• Volume
• Surface Area of a Combination of Solids
• Conversion of Solid From one Shape to Another
• Frustum of a Cone
• Solved Questions on Surface Area and Volume
• ## Surface Area

The surface area of any given object is the region occupied by the surface of the object. It is of two types:

(i) Total Surface Area
(ii) Curved Surface Area (Lateral Surface Area)

### Total Surface Area

The total surface area constitutes the sum of all the object's surface areas, encompassing both the top and bottom faces along with all other faces.

### Curved Surface Area / Lateral Surface Area

Curved Surface Area applies to curved surfaces of certain 3D objects like cylinders, cones and spheres. It refers to the area that covers the curved part of the object, excluding its top and bottom surfaces. Lateral surface area is the difference between the total surface area and the area of the base and top of the object.

Total and Curved / Lateral Surface Area Formulas for different figures are given in the image below:

## Volume

Volume refers to the measure of the amount of space occupied by a three-dimensional object. It is the capacity of objects to hold something. Volume is only present in three-dimensional objects and not in two-dimensional objects.

Some of the formulas of volume are:

## Surface Area of a Combination of Solids

The surface area of a combination of a solid is equal to the sum of the total surface area of an individual solid, excluding the parts where they overlap. It involves calculating the sum of the individual surface areas of each constituent solid within the combined figure.

For example: If a conical cavity of 4 cm and radius 3 cm is hollowed out from a solid cylinder of height 10 cm and radius 3 cm, then what is the total surface area of the remaining solid? [Use π = 3.14]

a) 263.68 cm2
b) 263.78 cm2
c) 263.76 cm2
d) 263.86 cm2

Explanation: Consider the figure given below

We are given

Height of the cylinder (h1) = 10 cm

Radius of the cylinder (r) = 3 cm

Height of the cone (h2) = 4 cm

Radius of the cone (r) = 3 cm

Let l cm be the slant height of the cone.

Total surface area of the remaining solid = (Area of the top face of the cylinder) + (curved surface area of the cylinder) + (curved surface area of the cone)

We know that

Area of the top face of the cylinder = πr2

Curved surface area of the cylinder = 2πrh1

Curved surface area of the cone = πrl

Slant height of the cone (l) = √[(r2 + (h2)2]

l = √(32 + 42)
l = √(9 + 16)
l = √25
l = 5 cm

Thus

Total surface area of the remaining solid = πr2 + 2πrh1 + πrl
= πr(r + 2h1 + l)
= π (3) (3 + 2(10) + 5)
= 3π(3 + 20 + 5)
= 3π(28)
= 84π
= 84 × 3.14
= 263.76 cm2

Example: If a cone with a radius of 4 cm and a height of 6 cm is placed on top of a hemisphere with the same radius of 4 cm, then what is the total curved surface area of the combined figure? [Use π = 3.14]

a) 191.3 cm2
b) 191.04 cm2
c) 191.4 cm2
d) 191.03 cm2

Explanation: Consider the figure shown below

We know that the curved surface area of a cone = πrl

Curved surface area of a hemisphere = 2πr2

We know that l = √(r2 + h2)

l = √(42 + 62)
l = √(16 + 36)
l = √52
l = 7.21 cm

Curved surface area of the combined figure = (Curved surface area of the cone) + (Curved surface area of the hemisphere)
= πrl + 2πr2
= πr(l + 2r)
= π (4) (7.21 + 2(4)
= π (4) (7.21 + 8)
= π (4) (15.21)
= 60.84 π
= 60.84 × 3.14
= 60.84 × 3.14
= 191.04 cm2

## Volume of a combination of solids

The volume of a combination of solids refers to the total amount of space occupied by multiple interconnected three-dimensional shapes when they are joined or arranged together.

Example: A cone of a radius of 6 cm and a height of 8 cm is placed on top of a cylinder with the same radius of 6 cm and a height of 10 cm. What is the volume of the combination formed? (Use π = 3.14)

a) 1431.82 cm3
b) 1431.88 cm3
c) 1431.86 cm3
d) 1431.84 cm3

Explanation: Consider the figure given below

We know that the volume of the cone = πr2h / 3

Where

h is the height

Volume of a cylinder = πr2h

Where

h is the height

We are given that the radius of the cone = 6 cm and the height of the cone = 8 cm.

Volume of cone = (π × (6)2 × 8) / 3

= 96 π cm3

We are given that the radius of the cylinder = 6 cm and the height of the cylinder = 10 cm.

Volume of cylinder = π × (6)2 × 10

= 360 π cm3

Volume of combination = Volume of cone + Volume of cylinder
= 96 π + 360 π
= 456 π
= 456 × (3.14)
= 1431.84 cm3

## Conversion of Solid From one Shape to Another

The conversion of a solid from one shape to another involves transforming a three-dimensional object from its original form into a different three-dimensional shape while preserving its volume.

Example: If a solid piece of iron in the form of a cuboid of dimensions 98 cm × 66 cm × 48 cm, is moulded to form a solid sphere, then what is the radius of the sphere?

a) 52 cm
b) 42 cm
c) 32 cm
d) 39 cm

Explanation: We know that the volume of a cuboid = l × b × h

Where

l is the length.

h is the height.

Volume of a sphere = 4πr3 / 3

Where

r is the radius of the sphere

Thus, the Volume of the cuboid = 98 × 66 × 48

We know that when one solid is transformed into another solid, its volume remains constant.

Volume of the sphere = Volume of the cuboid

4πr3 / 3 = 98 × 66 × 48

43 × 227 × r3 = 98 × 66 × 48

r3 = $\frac{98×66×48×3×7}{4×22}$

r3 = 98 × 3 × 12 × 3 × 7

r3 = 74088

r = (74088)1/3

r = 42 cm

## Frustum of a Cone

A frustum of a cone is what you get when you cut the tip off a cone. It's like a cone with the top part removed. It’s the part of the cone with its base that is left after the cone is cut by a plane parallel to its base.

### Formulas for Frustum of a Cone

Example: If the diameter of the circular upper surface is 4 cm, the radius of the base of the frustum is 4 cm and the slant height is 5 cm, then what will be the lateral surface area of the frustum?

a) 94.29 cm2
b) 94.26 cm2
c) 94.27 cm2
d) 94.25 cm2

Explanation: Let the radius of the base of the frustum be R cm.

Let the radius of the circular upper surface of the frustum be r cm.

Let the slant height be l cm.

We are given the diameter of the circular upper surface is 4 cm, the radius of the base of the frustum is 4 cm and the slant height is 5 cm

R = 4 cm

r = 2 cm

l = 5 cm

We know that the curved surface area of the frustum = πl(R + r)

Curved surface area of frustum = 227 × 5 × (4 + 2)
= 227 × 5 × 6
= 227 × 30
= 660 / 7
= 94.29 cm2