Chapter – 12

Surface Areas and Volumes

In this post we have given the detailed notes of class 10 Maths Chapter 12 (Surface Areas and Volumes) in English. These notes are useful for the students who are going to appear in class 10 board exams.

Chapter 12: Surface Areas and Volumes

Introduction

In our surroundings, we come across many solids that are combinations of two or more basic solids – a cone with a hemisphere for an ice-cream, a cylindrical tank with hemispherical ends, a cubical block with a hemispherical top, and so on. In this chapter we first recall the surface area and volume formulas for the basic solids, and then learn how to find the surface area and volume of solids formed by combining two or more of these basic solids.

Surface Area and Volume Formulas of Basic Solids

Cuboid

For a cuboid of length l, breadth b and height h:

• Volume = l × b × h
• Total Surface Area = 2(lb + bh + hl)
• Lateral (Curved) Surface Area = 2h(l + b)
• Length of diagonal = √(l² + b² + h²)

Cube

For a cube of side (edge) a:

• Volume = a³
• Total Surface Area = 6a²
• Lateral Surface Area = 4a²
• Length of diagonal = √3 × a

Right Circular Cylinder

For a cylinder of radius r and height h:

• Curved Surface Area (CSA) = 2πrh
• Total Surface Area (TSA) = 2πrh + 2πr² = 2πr(h + r)
• Volume = πr²h

Right Circular Cone

For a cone of radius r, height h and slant height l, where l² = r² + h²:

• Slant height, l = √(r² + h²)
• Curved Surface Area (CSA) = πrl
• Total Surface Area (TSA) = πrl + πr² = πr(l + r)
• Volume = ⅓ πr²h  (one-third the volume of a cylinder with the same base and height)

Sphere

For a sphere of radius r:

• Surface Area = 4πr²
• Volume = 4/3 πr³

Hemisphere

For a hemisphere of radius r (exactly half of a sphere):

• Curved Surface Area = 2πr²
• Total Surface Area = 2πr² + πr² = 3πr²  (curved surface + the flat circular base)
• Volume = 2/3 πr³  (half the volume of a sphere of the same radius)

Surface Area of a Combination of Solids

When two basic solids are joined together to form a new solid, the surface area of the resulting solid is not simply the sum of the surface areas of the two solids – the part of each solid’s surface that gets ‘hidden’ where the two solids are joined must be left out.

TSA of combination = Sum of curved surface areas of all the individual parts that remain exposed

Example 1

Question: Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take π = 22/7)

Solution: Radius of the hemisphere (= radius of the cone) = 3.5/2 = 1.75 cm

Total height of the top = 5 cm, so height of the conical part = 5 − 1.75 = 3.25 cm

Slant height of the cone, l = √(r² + h²) = √(1.75² + 3.25²) = √(3.0625 + 10.5625) = √13.625 ≈ 3.7 cm

Area to be coloured = CSA of hemisphere + CSA of cone = 2πr² + πrl

= 2 × 22/7 × 1.75 × 1.75 + 22/7 × 1.75 × 3.7

= 19.25 + 20.35 ≈ 39.6 cm²

Example 2

Question: A wooden toy rocket is in the shape of a cube of edge 5 cm, with a hemisphere of diameter 4.2 cm fixed on the top. Find the total surface area of the block. (Take π = 22/7)

Solution: Total surface area of the cube = 6 × (edge)² = 6 × 5 × 5 = 150 cm²

Radius of the hemisphere, r = 4.2/2 = 2.1 cm

Note: the circular base of the hemisphere, where it sits on the cube, is not part of the exposed surface, so we subtract the base area of the hemisphere from the cube’s surface area and add the curved surface area of the hemisphere.

Surface area of the block = TSA of cube − base area of hemisphere + CSA of hemisphere

= 150 − πr² + 2πr² = 150 + πr²

= 150 + 22/7 × 2.1 × 2.1

= 150 + 13.86 = 163.86 cm²

Volume of a Combination of Solids

Unlike surface area, when we find the volume of a solid formed by combining two basic solids, no part is ‘hidden’ or lost. The volume of the combined solid is simply the sum (or difference) of the volumes of its parts.

Volume of combination = Sum (or difference) of the volumes of the individual solids

Example 3

Question: A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take π = 3.14)

Solution: Radius of the hemisphere and the cone, r = 4/2 = 2 cm; height of the cone, h = 2 cm

Volume of the toy = Volume of hemisphere + Volume of cone = 2/3 πr³ + ⅓ πr²h

= [⅔ × 3.14 × 2³ + ⅓ × 3.14 × 2² × 2] cm³

= [16.75 + 8.37] ≈ 25.12 cm³

Now, the circumscribing cylinder has radius = 2 cm and height = height of cone + radius of hemisphere = 2 + 2 = 4 cm.

Volume of the cylinder = πr²h = 3.14 × 2² × 4 = 50.24 cm³

Difference of volumes = 50.24 − 25.12 = 25.12 cm³

Example 4

Question: A juice seller was serving his customers using glasses. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of the glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Take π = 3.14)

Solution: Internal radius of the glass, r = 5/2 = 2.5 cm, height h = 10 cm

Apparent capacity of the glass = πr²h = 3.14 × 2.5 × 2.5 × 10 = 196.25 cm³

Volume of the raised hemispherical portion = 2/3 πr³ = 2/3 × 3.14 × 2.5 × 2.5 × 2.5 = 32.71 cm³

Actual capacity of the glass = Apparent capacity − Volume of hemisphere = 196.25 − 32.71 = 163.54 cm³

Example 5

Question: Shanta runs an industry in a shed which is in the shape of a cuboid, surmounted by a half cylinder. If the base of the shed is of dimensions 7 m × 15 m, and the height of the cuboidal part is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies 300 m³ and the 20 workers each occupy 0.08 m³ space on an average, then how much air is in the shed? (Take π = 22/7)

Solution: Cuboidal part: length = 15 m, breadth = 7 m, height = 8 m

Half-cylinder part: diameter = 7 m (radius = 3.5 m), length (= height of shed’s arch) = 15 m

Volume of air in the shed (with no machinery or workers) = Volume of cuboid + ½ Volume of cylinder

= (15 × 7 × 8) + ½ × 22/7 × 3.5 × 3.5 × 15

= 840 + 288.75 = 1128.75 m³

Space occupied by machinery = 300 m³

Space occupied by 20 workers = 20 × 0.08 = 1.6 m³

Volume of air in the shed with machinery and workers present = 1128.75 − (300 + 1.6) = 827.15 m³

Conversion of a Solid from One Shape to Another

When a solid is recast (melted and reshaped) into another solid of a different shape, its shape changes but its volume remains the same. This idea is used to solve problems where a solid is converted from one form into another (for example, melting a metallic cone and recasting it as a sphere).

Example 6

Question: A cone of height 24 cm and base radius 6 cm is made up of modelling clay. A child reshapes it into a sphere. Find the radius of the sphere.

Solution: Volume of the cone = ⅓ × π × 6 × 6 × 24 cm³

Let the radius of the sphere be r. Its volume = 4/3 πr³

Since the volume of clay does not change when it is reshaped:

4/3 πr³ = ⅓ × π × 6 × 6 × 24

r³ = (6 × 6 × 24) / 4 = 3 × 3 × 24 = 216 = 6³

∴ r = 6 cm

Key Points to Remember

• Cuboid: Volume = lbh; TSA = 2(lb + bh + hl). Cube: Volume = a³; TSA = 6a².
• Cylinder (radius r, height h): CSA = 2πrh; TSA = 2πr(h + r); Volume = πr²h.
• Cone (radius r, height h, slant height l = √(r² + h²)): CSA = πrl; TSA = πr(l + r); Volume = ⅓ πr²h.
• Sphere (radius r): Surface Area = 4πr²; Volume = 4/3 πr³.
• Hemisphere (radius r): CSA = 2πr²; TSA = 3πr²; Volume = 2/3 πr³.
• When solids are joined together, the surface area of the combination is the sum of the exposed curved/flat surfaces only – the overlapping (hidden) parts are excluded.
• When solids are joined together, the volume of the combination is simply the sum (or difference) of the individual volumes – no volume is lost.
• When a solid is melted and recast into another shape, its volume remains constant even though its shape changes.

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