Surface Areas and Volumes
CBSE Class 10 · Mathematics · Notes, formulas and practice questions
Revise how to find surface areas and volumes of solids made by joining cones, cylinders, hemispheres and spheres, and solve melting, recasting and liquid-transfer problems. This Class 10 chapter applies the Class 9 volume formulas to combined shapes and volume-conservation questions.
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This chapter builds on the Class 9 formulae for volumes and surface areas of a cuboid, cylinder, cone, sphere and hemisphere. What is new is what happens when two such solids are joined, or when one solid is melted or emptied to make another. The central skill is knowing whether a quantity is preserved: volume is simply added in a combined solid, and volume is unchanged in a recasting or pouring problem. Surface area, on the other hand, is never simply added in full, because the surfaces that meet and glue together cease to be exposed.
Consider a cone standing on a hemisphere with the same radius. The volume of the toy is the sum of the cone's volume and the hemisphere's volume. For total surface area, the cone's base and the hemisphere's flat face are touching each other and are therefore not visible, so only the curved surface of the cone and the curved dome of the hemisphere are counted. The same reasoning applies to a cylinder with a hemisphere at one or both ends. To get such areas right, first draw or visualise the object, label which surfaces face the outside, and then sum exactly those areas.
In conversion problems a solid is melted and recast into a different shape, or a liquid is poured from one container into another. The volume of material or liquid does not change, so you can equate the volume of the original object to the volume of the new object. For example, a melted sphere cast into a cylinder gives (4/3)πr³ = πR²h. This single equation lets you find a missing radius or height. In water-displacement problems, a submerged solid occupies its own volume inside the vessel, so the rise in liquid level equals the solid's volume divided by the vessel's base area.
Many mistakes come from unit conversions and from misreading the geometry of the joint. Always work in the same unit throughout, or convert at the end. Also confirm whether the solid is closed or open: if a cylinder is the body of a vessel, its inner curved surface may be all that is relevant, while a closed container has two end circles. With careful application of the basic formulae and a clear picture of the exposed surfaces, this chapter is mostly about setting up the right sum or equation.
Key terms
- Combined solid
- A solid made by joining two or more standard solids so their plane faces coincide exactly. The common face becomes internal and is not part of the visible surface. Volume adds as the sum of the parts.
- Exposed surface area
- The surface area that you actually see on the outside of a combined solid. If a face of one part is glued to a face of another, both faces disappear from the count, so the total is the sum of only the remaining visible areas.
- Volume of a combined solid
- The sum of the volumes of its separate parts. Since the parts share no empty space at their joint, you simply add the individual volumes without subtracting anything.
- Melting and recasting
- Changing one solid shape into another while conserving volume. The equality of original and new volumes gives an equation that determines an unknown radius, height, or length.
- Liquid displacement
- When a solid is completely submerged in liquid, it pushes aside a volume of liquid equal to its own volume. In a cylindrical vessel, this volume equals the base area of the vessel times the rise in liquid level.
- Total height of a cone-on-hemisphere
- The height measured from the flat base of the hemisphere to the cone's tip. Since the hemisphere's height equals its radius r, the cone's height is the total height minus r. This relationship is needed to find slant height and volume.
- Capsule length
- For a cylinder with hemispherical ends, the total outer length is the cylindrical part's length plus 2r, because each hemispherical cap extends a distance r along the axis. Therefore the cylinder length equals total length minus 2r.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| Volume of a cylinder | V = πr²h | r is the radius of the circular base and h is the height or length of the cylinder. |
| Volume of a cone | V = (1/3)πr²h | r is the base radius and h is the perpendicular height of the cone. |
| Volume of a sphere | V = (4/3)πr³ | r is the radius of the sphere. |
| Volume of a hemisphere | V = (2/3)πr³ | r is the radius of the hemisphere. |
| Curved surface area of a cone | A = πrl | r is the base radius and l is the slant height, with l = √(r² + h²) for a cone of vertical height h. |
| Curved surface area of a cylinder | A = 2πrh | r is the base radius and h is the height of the cylinder. |
| Surface area of a sphere | A = 4πr² | r is the radius of the sphere. |
| Total surface area of a solid hemisphere | A = 3πr² | r is the radius; this is the sum of the curved dome (2πr²) and the flat circular base (πr²). |
Practice questions with answers
1. A toy is a cone mounted on a hemisphere of the same radius. The radius is 3.5 cm and the total height of the toy is 15.5 cm. Take π = 22/7. Find the total surface area of the toy.
Since the hemisphere radius is 3.5 cm, its height along the axis is 3.5 cm, so the cone's height is 15.5 − 3.5 = 12 cm. The cone's slant height is √(3.5² + 12²) = √156.25 = 12.5 cm. Curved area of the cone = (22/7) × 3.5 × 12.5 = 137.5 cm², and curved area of the hemisphere = 2 × (22/7) × 3.5² = 77 cm². Total surface area = 137.5 + 77 = 214.5 cm².
2. A solid metallic sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder.
Volume of the sphere is (4/3)π × 6³ = 288π cm³. The cylinder's volume is π × 4² × h = 16πh. Equating the two volumes gives 16πh = 288π, so h = 18 cm.
3. A solid sphere of radius 3 cm is dropped into a cylindrical vessel of base radius 6 cm partly filled with water. Find the rise in the water level.
The sphere displaces water equal to its volume, (4/3)π × 3³ = 36π cm³. This water forms a cylinder over a base area of π × 6² = 36π cm². So the rise in height is 36π / (36π) = 1 cm.
4. A medicine capsule is a cylinder with two hemispherical ends. The whole length of the capsule is 14 mm and its diameter is 5 mm. Take π = 22/7. Find its surface area.
Radius is 2.5 mm, so the two caps add 5 mm; the cylindrical part is 14 − 5 = 9 mm long. Surface area = curved area of cylinder + curved area of both hemispheres = 2πrh + 4πr² = 2π × 2.5 × 9 + 4π × 2.5² = 45π + 25π = 70π mm². Using π = 22/7, this equals 70 × 22/7 = 220 mm².
5. A solid consists of a cone mounted on a hemisphere of radius 3 cm. If the total height of the solid is 11 cm, find its volume in terms of π.
The cone's height is 11 − 3 = 8 cm. Cone volume = (1/3)π × 3² × 8 = 24π cm³. Hemisphere volume = (2/3)π × 3³ = 18π cm³. Total volume = 24π + 18π = 42π cm³.
6. A right circular cone of height 24 cm and base radius 6 cm is melted and recast into a sphere. Find the radius of the sphere.
Cone volume = (1/3)π × 6² × 24 = 288π cm³. Let the sphere's radius be R. Then (4/3)πR³ = 288π, so R³ = 216 and R = 6 cm.
7. A solid sphere of radius 3 cm is melted and drawn into a cylindrical wire of radius 0.5 mm. Find the length of the wire in metres.
Convert radius of wire: 0.5 mm = 0.05 cm. Sphere volume = (4/3)π × 3³ = 36π cm³. If the wire length is L cm, its volume is π × 0.05² × L = 0.0025πL. Equate: 0.0025πL = 36π, so L = 14400 cm = 144 m.
8. A hemispherical bowl of internal diameter 36 cm is full of a liquid. It is poured into cylindrical bottles, each of radius 3 cm and height 6 cm. How many bottles are needed?
Internal radius of the bowl is 18 cm. Volume of liquid = (2/3)π × 18³ = 3888π cm³. Each bottle holds π × 3² × 6 = 54π cm³. Number of bottles = 3888π / 54π = 72.
9. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the total surface area of the solid. Use π = 22/7.
The hemisphere must fit entirely on one square face, so its diameter cannot exceed the side of the cube. Greatest diameter = 7 cm, radius r = 3.5 cm. Cube's full surface area is 6 × 7² = 294 cm². Subtracting the covered circle leaves 294 − πr²; adding the hemisphere's curved surface gives 294 − πr² + 2πr² = 294 + πr². With π = 22/7, πr² = 38.5 cm², so total surface area = 294 + 38.5 = 332.5 cm².
10. A cylindrical bucket full of sand has height 28 cm and base radius 10 cm. The sand is emptied to form a conical heap of height 21 cm. Find the radius and slant height of the heap.
Volume of sand = π × 10² × 28 = 2800π cm³. For the conical heap, let radius be R; then (1/3)πR² × 21 = 7πR². Equating gives 7πR² = 2800π, so R² = 400 and R = 20 cm. Slant height = √(20² + 21²) = √841 = 29 cm.
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