Surface Areas and Volumes Class 10: Notes and Questions

This guide covers Surface Areas and Volumes Class 10, chapter 12 of NCERT Class 10 Mathematics. You will learn how to break a combined solid into cones, cylinders, hemispheres and cuboids, which faces to count for surface area, and how to add or subtract volumes. Worked examples, important questions and common mistakes are all included.

Combined solids around us

Most real objects are combinations of basic solids:

• An oil tanker: a cylinder with a hemisphere at each end.
• A test tube: a cylinder with a hemisphere.
• A medicine capsule: a cylinder with a hemisphere on each end.
• A playing top (lattu): a cone with a hemisphere.

The method is always the same: split the object into basic solids, then use what you know about each part.

Volume formulas you need

Curved surface areas used in this chapter: cylinder 2πrh, cone πrl (l is the slant height, l = √(r² + h²)), hemisphere 2πr².

Use π = 22/7 unless the question says 3.14. Convert all lengths to the same unit first, and check whether you are given a radius or a diameter.

Surface area of combined solids

When two solids are joined, the faces that touch are hidden inside and are not part of the outer surface.

Rule: add every exposed curved and flat surface. Leave out the shared faces.

• Cone on a hemisphere (a toy or top): the flat bases are hidden, so TSA = CSA of cone + CSA of hemisphere.
• Cone on a cylinder: TSA = CSA of cone + CSA of cylinder + area of the cylinder's bottom base.
• Cylinder with a hemisphere on top: exposed area = 2πrh + 2πr² + πr² = 2πrh + 3πr². The last πr² is the bottom; the joined circle is inside.

Example: the playing top. A top is a cone with a hemisphere on it. Total height 5 cm, diameter 3.5 cm.

1. r = 3.5/2 = 1.75 cm. The hemisphere's height equals its radius, 1.75 cm.
2. Cone height h = 5 − 1.75 = 3.25 cm.
3. Slant height l = √(1.75² + 3.25²) ≈ 3.7 cm.
4. CSA of hemisphere = 2 × (22/7) × 1.75 × 1.75 = 19.25 cm².
5. CSA of cone = (22/7) × 1.75 × 3.7 = 20.35 cm².
6. Total area ≈ 19.25 + 20.35 = 39.6 cm².

A small solid on a bigger face

When a small solid sits on a face of a larger one, part of that face gets covered.

Example: hemisphere on a cube. Cube side a, hemisphere radius r. Start with the cube's 6a², subtract the covered circle πr², and add the hemisphere's curved surface 2πr². The net change is +πr², so

TSA = 6a² + πr²

For a = 5 cm and a hemisphere of diameter 4.2 cm (r = 2.1 cm): TSA = 150 + (22/7) × 2.1 × 2.1 = 150 + 13.86 = 163.86 cm².

If the hemisphere is scooped out of the face instead, the answer is the same: you lose πr² of flat face and gain 2πr² of inner curved surface.

Volume of combined solids

Volume simply adds up. Nothing is hidden, because volume is the space the solid fills.

Example: cylinder with a hemisphere on top. r = 3 cm, cylinder height 6 cm. Cylinder = π × 9 × 6 = 54π cm³. Hemisphere = (2/3)π × 27 = 18π cm³. Total = 72π cm³. The total height is 9 cm, but do not use 9 as the cylinder's height.

Example: factory shed. A cuboid 15 m × 7 m × 8 m with a half-cylinder roof of radius 3.5 m and length 15 m. Cuboid = 15 × 7 × 8 = 840 m³. Half-cylinder = (1/2) × (22/7) × 3.5 × 3.5 × 15 = 288.75 m³. Total = 1128.75 m³.

When volume is taken away

Sometimes part of the space is blocked. A juice glass may be a cylinder with a raised hemispherical bottom.

• Apparent capacity: what the outer shape seems to hold (the full cylinder).
• Actual capacity: the cylinder minus the hemisphere pushing in.

Example: r = 2.5 cm, h = 10 cm, π = 3.14. Apparent = 3.14 × 2.5 × 2.5 × 10 = 196.25 cm³. Hemisphere = (2/3) × 3.14 × 2.5³ ≈ 32.71 cm³. Actual ≈ 196.25 − 32.71 = 163.54 cm³.

Remember this

• Sketch the object and name each basic solid.
• Surface area: count only what is on the outside.
• Volume: add the parts, or subtract hollow parts.
• Hemisphere height = its radius.
• Cone slant height l = √(r² + h²).
• Cube with hemisphere on (or scooped from) a face: 6a² + πr².

Important questions with answers

1. Which solids make up a medicine capsule? A cylinder with a hemisphere attached at each end.

2. A cylinder is topped by a cone. Which surfaces make up the total surface area? CSA of the cone, CSA of the cylinder and the cylinder's bottom base. The joined circle is not included.

3. In the playing top, why is the cone's height 3.25 cm and not 5 cm? The 5 cm includes the hemisphere, whose height is its radius, 1.75 cm. So the cone's height is 5 − 1.75 = 3.25 cm.

4. Find the total surface area of a solid made of a cylinder of radius 7 cm and height 10 cm with a hemisphere of the same radius on top. 2πrh + 3πr² = 2 × (22/7) × 7 × 10 + 3 × (22/7) × 49 = 440 + 462 = 902 cm².

5. A toy is a cone of radius 3 cm and height 4 cm on a hemisphere of radius 3 cm. Find its volume in terms of π. Cone = (1/3)π × 9 × 4 = 12π cm³. Hemisphere = (2/3)π × 27 = 18π cm³. Total = 30π cm³.

6. Two identical cones are joined base to base. What is the total volume? (1/3)πr²h + (1/3)πr²h = (2/3)πr²h.

7. The shed holds 1128.75 m³. 20 workers each take up 0.08 m³ and machinery takes up 300 m³. How much air is left? Occupied = 300 + 20 × 0.08 = 301.6 m³. Air = 1128.75 − 301.6 = 827.15 m³.

8. If the radius of a sphere is doubled, how many times does its volume become? Volume depends on r³, so it becomes 2³ = 8 times.

Common mistakes to avoid

• Adding the full TSA of every part, which counts hidden faces.
• Forgetting the exposed bottom base when a cone sits on a cylinder.
• Using total height in place of one part's height.
• Mixing up diameter and radius.
• Mixing units, such as cm with m, in one calculation.
• Using vertical height where the slant height is needed for the cone's CSA.

Sketch every combined solid and label its parts before you calculate, and study this chapter with Joy for more practice.