3D Shapes: Volume and Surface Area

For Edexcel GCSE Mathematics, you need to calculate volumes and surface areas of prisms, cylinders, cones, spheres, pyramids and composite solids. Higher tier students must also handle frustums. This lesson covers all the required formulae and indicates which ones appear on the Edexcel formula sheet.

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Key Vocabulary

Term	Meaning
Prism	A 3D shape with a uniform cross-section (same shape all the way through)
Cylinder	A prism with a circular cross-section
Pyramid	A 3D shape with a polygon base and triangular faces meeting at an apex
Cone	A pyramid with a circular base
Sphere	A perfectly round 3D shape where every point on the surface is the same distance from the centre
Frustum	The solid that remains when the top of a cone or pyramid is cut off by a plane parallel to the base [H]

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Formulae from the Edexcel Formula Sheet

The following are given on the Edexcel formula sheet:

• Volume of a prism = area of cross-section x length
• Curved surface area of a cone = $\pi r l$πrl (where l is the slant height)
• Volume of a cone = $\frac{1}{3}\pi r^2 h$31​πr2h
• Surface area of a sphere = $4\pi r^2$4πr2
• Volume of a sphere = 4/3 x pi x r³

Formulae You Must Memorise

• Volume of a cuboid = l x w x h
• Surface area of a cuboid = 2(lw + lh + wh)
• Volume of a cylinder = $\pi r^2 h$πr2h
• Curved surface area of a cylinder = $2\pi r h$2πrh
• Total surface area of a cylinder = $2\pi r h$2πrh + $2\pi r^2$2πr2
• Volume of a pyramid = 1/3 x base area x height

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Prisms

Volume of a prism = area of cross-section x length

Worked Example 1

A triangular prism has a cross-section that is a right-angled triangle with base 6 cm and height 8 cm. The prism is 15 cm long.

Area of cross-section = 1/2 x 6 x 8 = 24 cm² Volume = 24 x 15 = 360 cm³

Worked Example 2

A hexagonal prism has a cross-sectional area of 42 cm² and length 10 cm.

Volume = 42 x 10 = 420 cm³

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Cylinders

Volume = $\pi r^2 h$πr2h Curved surface area = $2\pi r h$2πrh Total surface area = $2\pi r h$2πrh + $2\pi r^2$2πr2

Worked Example 3

A cylinder has radius 4 cm and height 10 cm. Find (a) the volume and (b) the total surface area. Give answers to 1 d.p.

(a) Volume = pi x 4² x 10 = 160pi = 502.7 cm³

(b) Curved SA = 2 x pi x 4 x 10 = 80pi Two circles = 2 x pi x 4² = 32pi Total SA = 80pi + 32pi = 112pi = 351.9 cm²

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Cones

Volume = $\frac{1}{3}\pi r^2 h$31​πr2h (given on formula sheet) Curved surface area = $\pi r l$πrl (given on formula sheet, l = slant height) Total surface area = $\pi r l$πrl + $\pi r^2$πr2

Worked Example 4

A cone has radius 3 cm and perpendicular height 4 cm. Find (a) the slant height, (b) the volume and (c) the total surface area.

(a) By Pythagoras: l = $\sqrt{3^{2} + 4^{2}}$32+42​ = $\sqrt{9 + 16}$9+16​ = $\sqrt{25}$25​ = 5 cm

(b) Volume = 1/3 x pi x 3² x 4 = 1/3 x 36pi = 12pi = 37.7 cm³ (1 d.p.)

(c) Curved SA = pi x 3 x 5 = 15pi Base = pi x 3² = 9pi Total SA = 15pi + 9pi = 24pi = 75.4 cm² (1 d.p.)

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Spheres

Volume = 4/3 x pi x r³ (given on formula sheet) Surface area = $4\pi r^2$4πr2 (given on formula sheet)

Worked Example 5

A sphere has radius 6 cm. Find (a) the volume and (b) the surface area. Give answers to 1 d.p.

(a) Volume = 4/3 x pi x 6³ = 4/3 x 216pi = 288pi = 904.8 cm³

(b) Surface area = 4 x pi x 6² = 144pi = 452.4 cm²

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Pyramids

Volume = 1/3 x base area x perpendicular height

Worked Example 6

A square-based pyramid has a base with side length 8 cm and a perpendicular height of 9 cm.

Volume = 1/3 x (8 x 8) x 9 = 1/3 x 576 = 192 cm³

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Frustums [H]

A frustum is formed when the top of a cone (or pyramid) is removed by a cut parallel to the base.

Volume of frustum = volume of large cone - volume of small cone removed

Worked Example 7 [H]

A cone has radius 10 cm and height 24 cm. A smaller cone with radius 4 cm is cut from the top. Find the height of the smaller cone and the volume of the frustum.

The cones are similar. The ratio of radii is 4 : 10 = 2 : 5. Height of small cone = (2/5) x 24 = 9.6 cm Height of frustum = 24 - 9.6 = 14.4 cm

Volume of large cone = 1/3 x pi x 10² x 24 = 800pi Volume of small cone = 1/3 x pi x 4² x 9.6 = 51.2pi Volume of frustum = 800pi - 51.2pi = 748.8pi = 2352.1 cm³ (1 d.p.)

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Composite Solids

Break the solid into standard shapes, find each volume separately, then add (or subtract).

Worked Example 8

A solid is made of a cylinder (radius 5 cm, height 12 cm) with a hemisphere (radius 5 cm) on top.

Volume of cylinder = pi x 5² x 12 = 300pi Volume of hemisphere = 1/2 x 4/3 x pi x 5³ = 1/2 x 500pi/3 = 250pi/3 Total = 300pi + 250pi/3 = 900pi/3 + 250pi/3 = 1150pi/3 = 1204.3 cm³ (1 d.p.)

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Unit Conversions

Conversion	Linear	Area	Volume
cm to mm	x 10	x 100	x 1000
m to cm	x 100	x 10,000	x 1,000,000
km to m	x 1000	x 1,000,000	x 1,000,000,000

Worked Example 9

Convert 5.2 m³ to cm³.

5.2 x 1,000,000 = 5,200,000 cm³

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Common Mistakes and Misconceptions