Class 10 Maths Chapter 12: Surface Areas and Volumes

Surface Areas and Volumes is Chapter 12 of NCERT Class 10 Maths. In Class 9 you found surface areas and volumes of single solids; here you apply the same formulas to objects made by joining two or more solids, such as toys, capsules, tents and pen stands. Watch the three video lessons below, then use the formula list and solved examples to master combination of solids.

What you will learn

Formulas for cuboid, cube, cylinder, cone, sphere and hemisphere in one place
Finding the surface area of a combination of solids by adding only the visible surfaces
Finding the volume of a combination of solids by adding (or subtracting) volumes
Handling solids with a hollow or scooped-out part
Real-life objects like a toy, capsule, tent and pen stand

Video Lessons

Surface Areas and Volumes - Part 1

Surface Areas and Volumes - Part 2

Surface Areas and Volumes - Part 3

Key Concepts and Formulas

Formulas for Single Solids

Keep this list on your fingertips. Here r is the radius, h the height, l the slant height and a the side of a cube.

Cuboid:     TSA = 2(lb + bh + hl), V = lbh
Cube:       TSA = 6a², V = a³
Cylinder:   CSA = 2πrh
            TSA = 2πr(r + h), V = πr²h
Cone:       l = √(r² + h²)
            CSA = πrl, TSA = πr(l + r)
            V = ⅓πr²h
Sphere:     SA = 4πr², V = (4/3)πr³
Hemisphere: CSA = 2πr², TSA = 3πr²
            V = (2/3)πr³

Surface Area of a Combination of Solids

When two solids are joined, the faces that touch each other get hidden. So the total surface area of the new solid is not the sum of the total surface areas of the parts.

Combined SA = sum of exposed surfaces only
Cone on hemisphere: SA = πrl + 2πr²
Hemisphere on cube face: SA = 6a² − πr² + 2πr²
                         SA = 6a² + πr²

Use curved surface areas of the parts wherever a flat face is joined to another solid.

Volume of a Combination of Solids

Volume is the space occupied, and no space is lost when solids are joined. This makes volume simpler than surface area.

Combined V = sum of volumes of the parts
Hollowed solid: V = outer solid − removed part
Cylinder (height H) + cone (height h):
V = πr²(H + h/3)

If a part is removed (a cavity, a hole, a scooped-out hemisphere), subtract its volume.

Problem-Solving Strategy

Draw the figure and label each part with its own radius and height.
Find any missing height or slant height first (for example, total height minus the other part).
Decide which surfaces are visible before writing the surface area.
Use π = 22/7 when the radius is a multiple of 7, otherwise use 3.14 or leave the answer in terms of π.

Solved Examples

Example 1: A toy is a cone of radius 7 cm mounted on a hemisphere of the same radius, with total height 31 cm. Find its total surface area.

Cone height = 31 − 7 = 24 cm
l = √(7² + 24²) = √625 = 25 cm
Cone CSA = (22/7) × 7 × 25 = 550 cm²
Hemisphere CSA = 2 × (22/7) × 49 = 308 cm²
SA = 550 + 308
∴ Surface area = 858 cm²

Class 10 Maths diagram of a toy made of a cone on a hemisphere with radius 7 cm — Add only the curved surfaces of cone and hemisphere

Example 2: A capsule is a cylinder with two hemispherical ends. Its total length is 20 mm and diameter 6 mm. Find its surface area (π = 3.14).

r = 3 mm
Cylinder length = 20 − 6 = 14 mm
SA = 2πrh + 4πr²
= 2π × 3 × 14 + 4π × 9
= 84π + 36π = 120π
= 120 × 3.14
∴ Surface area = 376.8 mm²

Diagram of a capsule as a cylinder with two hemispherical ends showing its surface area, Class 10 Maths Chapter 12 — Cylinder plus two hemispheres gives 120π mm²

Example 3: A solid is a cylinder of radius 7 cm and height 10 cm with a cone of the same radius and height 6 cm on top. Find its volume.

V = πr²H + ⅓πr²h
= πr²(10 + 6/3)
= (22/7) × 49 × 12
= 154 × 12
∴ Volume = 1848 cm³

Example 4: A hemisphere of diameter 7 cm is fixed on top of a cube of side 10 cm. Find the total surface area of the solid.

r = 3.5 cm
SA = 6a² + πr²
= 6 × 100 + (22/7) × 3.5 × 3.5
= 600 + 38.5
∴ Surface area = 638.5 cm²

Example 5: A cone of radius 3 cm and height 9 cm is hollowed out of a solid cylinder of the same radius and height. Find the remaining volume (π = 3.14).

V = πr²h − ⅓πr²h = ⅔πr²h
= ⅔ × π × 9 × 9
= 54π
= 54 × 3.14
∴ Remaining volume = 169.56 cm³

Example 6: A tent is a cylinder of radius 6 m and height 3 m with a conical top of height 8 m. Find the area of canvas needed (π = 3.14).

l = √(6² + 8²) = √100 = 10 m
Canvas = 2πrh + πrl
= 2π × 6 × 3 + π × 6 × 10
= 36π + 60π = 96π
= 96 × 3.14
∴ Canvas = 301.44 m²

Common Mistakes to Avoid

Adding total surface areas of the parts. The joined faces are hidden, so use curved surface areas where parts meet.
Using the vertical height of a cone in the CSA formula. CSA needs the slant height l, so find l = √(r² + h²) first.
Taking the total height of the object as the height of one part. Subtract the other part's height (or radius for a hemisphere).
Using diameter as radius, or mixing cm and m. Convert all lengths to the same unit before you substitute.

Quick Revision

Cone: l = √(r² + h²), CSA = πrl, V = ⅓πr²h
Hemisphere: CSA = 2πr², TSA = 3πr², V = ⅔πr³
Sphere: SA = 4πr², V = (4/3)πr³
Cylinder: CSA = 2πrh, V = πr²h