Surface Areas and Volumes – Class 10 Maths Chapter 12 Notes & NCERT PDF

Introduction

In Class IX you learnt the surface areas and volumes of the basic solids — cuboid, cone, cylinder and sphere. But real objects are rarely just one solid. An oil tanker on a truck is a cylinder with two hemispheres at its ends; a test tube is a cylinder with a hemisphere; a playing top (lattu) is a cone surmounted by a hemisphere. This chapter teaches you to handle such combinations of solids: first their surface areas (12.2), then their volumes (12.3).

The key idea, in the book's words: whenever we come across a new problem, we first try to see if we can break it down into smaller problems we have earlier solved.

Key Concepts

1. Surface area of a combination (Section 12.2)

When two solids are joined, the faces that are stuck together disappear from the outside. So you add only the surfaces you can actually see (or paint).

The total surface area of the combined solid is not the sum of the total surface areas of its parts.

Sometimes you must subtract a hidden face and add a new curved one. For a cube with a hemisphere on top: surface area = TSA of cube − base area of hemisphere + CSA of hemisphere = 6a² + πr².

2. Volume of a combination (Section 12.3)

Volume behaves differently: no volume disappears when solids are joined, so the volume of the new solid is simply the sum of the volumes of its parts (and a difference when something is scooped out or hollowed).

Methods Step by Step

Worked Examples (from the chapter)

Example 1 — Rasheed's top. Height 5 cm, diameter 3.5 cm, so r = 1.75 cm. Cone height = 5 − 1.75 = 3.25 cm, slant height l ≈ 3.7 cm.

TSA = 2πr² + πrl = (22/7) × 1.75 × (3.5 + 3.7) ≈ 39.6 cm².

Example 2 — Decorative block. Cube of edge 5 cm with a hemisphere of diameter 4.2 cm (r = 2.1 cm).

Area = 150 − πr² + 2πr² = 150 + πr² = 150 + 13.86 = 163.86 cm².

Example 3 — Toy rocket (π = 3.14). Cone: r = 2.5 cm, h = 6 cm, l = 6.5 cm; cylinder: r = 1.5 cm, h = 20 cm.

Orange part = πrl + πr² − π(1.5)² = 3.14 × 20.25 = 63.585 cm²; yellow part = π × 1.5 × (2 × 20 + 1.5) = 195.465 cm². A ring of the cone's base is painted because the cone's base is wider than the cylinder.

Example 4 — Bird-bath. Cylinder (h = 145 cm, r = 30 cm) with a hemispherical depression: TSA = 2πrh + 2πr² = 2πr(h + r) = 33000 cm² = 3.3 m².

Example 5 — Shanta's shed. Cuboid 15 m × 7 m × 8 m plus a half cylinder (diameter 7 m, length 15 m):

Volume = 840 + ½ × (22/7) × 3.5 × 3.5 × 15 = 1128.75 m³. Removing machinery (300 m³) and 20 workers (20 × 0.08 = 1.6 m³) leaves 827.15 m³ of air.

Example 6 — Juice glass (π = 3.14). Apparent capacity = πr²h = 196.25 cm³; the raised hemisphere takes ⅔πr³ ≈ 32.71 cm³, so actual capacity ≈ 163.54 cm³.

Example 7 — Toy in a cylinder. Hemisphere + cone with r = 2 cm, cone height 2 cm: volume = ⅔π(8) + ⅓π(4)(2) = 25.12 cm³. The circumscribing cylinder (r = 2, h = 4) holds 50.24 cm³, so the difference is 25.12 cm³.

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Formula Sheet

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