Volume of a Pyramid

Find the volume of a pyramid as one third of base area times height, for any shape of base, work back to a missing height, and find the height from a sloping edge.

In the exam for Cambridge O Level Maths 4024, Cambridge IGCSE Maths 0580, Edexcel Further Pure Maths 4PM1 · 10 real past-paper questions below

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A Third of the Prism

A pyramid has a flat base and sloping sides that meet at a point on top. It holds one third of the prism with the same base and height.

Formula

V = \frac{1}{3} \times base area \times h

$h$ is the perpendicular height: from the top point straight down to the base, at a right angle.

Worked example

1
$square base 6 cm, h = 10 cm$ — The question. h is the straight-up height, so it can go straight into the formula.
2
$base area = 6 \times 6 = 36 cm^{2}$ — Find the base area first, because the formula needs the area of the base, not its side. A square's area is side × side.
3
$V = \frac{1}{3} \times 36 \times 10$ — Put the base area and the height into the formula.
4
$= \frac{360}{3} = 120 cm^{3}$ — 36 × 10 = 360 is the box with the same base. The pyramid is a third of it. Volume is in cm³ because three lengths multiply.

Common mistake

Forgetting the third. $36 \times 10 = 360$ is the whole box, three times too big.

Key idea

The Great Pyramid of Giza was built about 146 m tall on a square base about 230 m wide. So $V \approx \frac{1}{3} \times 23 0^{2} \times 146$ , about 2.6 million m³: enough to fill roughly a thousand Olympic swimming pools.

Any Base: Find Its Area First

The base can be a rectangle, a triangle or any flat shape. Find the base area, then use the same formula.

The base is a right-angled triangle with sides 6 cm and 8 cm. The height is 9 cm.

Worked example

1
$base area = \frac{1}{2} \times 6 \times 8$ — The base is a right-angled triangle, so its area is half × base × height. The two short sides meet at the right angle, so use them.
2
$= 24 cm^{2}$ — 6 × 8 = 48, and half of 48 is 24.
3
$V = \frac{1}{3} \times 24 \times 9$ — Now the pyramid formula, with base area 24 and height 9.
4
$= \frac{216}{3} = 72 cm^{3}$ — 24 × 9 = 216, then take a third of it.

Common mistake

Using only one half. The base triangle has its own half; the pyramid still needs the third as well.

Key idea

A rectangle base 5 cm by 8 cm with height 12 cm: $\frac{1}{3} \times 40 \times 12 = 160 cm^{3}$ .

Working Back: Find the Height

The volume is given and one length is missing. Put in what you know, then undo the third by multiplying by 3.

Volume 200 cm³, base 10 cm by 6 cm. Find the height.

Worked example

1
$base area = 10 \times 6 = 60 cm^{2}$ — Find what you can first: the rectangle base.
2
$\frac{1}{3} \times 60 \times h = 200$ — Put everything you know into the formula. The height h is the only unknown.
3
$60 \times h = 600$ — Multiply both sides by 3. This undoes the third.
4
$h = \frac{600}{60} = 10 cm$ — Divide both sides by 60 to leave h on its own.

Common mistake

Dividing 200 by 60 and forgetting the 3. That gives 3.33 cm, a third of the real height.

Key idea

Missing the side of a square base instead? Same steps, then a square root. V = 150 cm³ and h = 18 cm: base area = 3 × 150 ÷ 18 = 25 cm², so the side is $25 = 5 cm$ .

Surface Area of a Pyramid

Some questions ask for the total surface area: every face added up. A pyramid's faces are its base plus the triangles up the sides.

Formula

surface area = base area + area of all triangle faces

Each triangle face is $\frac{1}{2} \times base edge \times l$ . Here $l$ is the slant height: the height of the face, measured up its middle. It is not the vertical height of the pyramid.

Worked example

1
$square base 10 cm, h = 12 cm$ — The question gives the vertical height. A triangle face needs its own height, the slant height l.
2
$l^{2} = 1 2^{2} + 5^{2}$ — The vertical height, half the base side and the slant height make a right-angled triangle. The slant height is the longest side.
3
$l^{2} = 144 + 25 = 169$ — Square each number and add.
4
$l = 13 cm$ — Square root. Keep the positive answer: a length cannot be negative.
5
$one face = \frac{1}{2} \times 10 \times 13 = 65 cm^{2}$ — Each face is a triangle: half × its base (10) × its height, the slant height.
6
$4 \times 65 = 260 cm^{2}$ — A square base has 4 equal triangle faces.
7
$100 + 260 = 360 cm^{2}$ — Add the square base, 10 × 10 = 100. Area is in cm², not cm³.

Common mistake

Using the vertical height 12 for the triangles. That gives 340 cm², too small: each face leans, so it is taller than the pyramid.

Height From a Sloping Edge (Extended/Higher)

Sometimes you get a sloping edge, not the height. The height, the edge and half the diagonal of the base make a right-angled triangle.

Square base 8 cm. Each sloping edge is 9 cm.

Worked example

1
$square base 8 cm, sloping edge 9 cm$ — The question. The top is above the centre of the square, but no height is given.
2
$centre to corner^{2} = 4^{2} + 4^{2} = 32$ — From the centre, go 4 cm across and 4 cm along to reach a corner (half of 8 each way). Pythagoras gives the distance squared.
3
$h^{2} + 32 = 9^{2}$ — The height, the centre-to-corner line and the sloping edge make a right-angled triangle. The edge is the longest side, so it sits alone.
4
$h^{2} = 81 - 32 = 49$ — Take 32 from both sides.
5
$h = 7 cm$ — Square root. Keep only the positive answer: a height cannot be negative.
6
$V = \frac{1}{3} \times 64 \times 7$ — Now the usual formula. The base area is 8 × 8 = 64.
7
$= \frac{448}{3} \approx 149 cm^{3}$ — 64 × 7 = 448, and 448 ÷ 3 = 149.33…, so 149 cm³ to 3 s.f.

Key idea

Given the slant height up the middle of a face instead, use half the base side. Base 6, face slant 5: $h = 5^{2} - 3^{2} = 4$ .

Key idea

If the top is directly above a corner of the base, one edge stands straight up and is the height. Use a sloping edge and the base side it rises from: edge 13, side 5 gives $h = 1 3^{2} - 5^{2} = 12$ .

A Base With No Height (Extended/Higher)

Sometimes the base is a triangle with no height marked, like an equilateral triangle. Find its area with the sine formula, then use the pyramid formula as usual.

Formula

area of a triangle = \frac{1}{2} ab sin C

$a$ and $b$ are two sides, and $C$ is the angle between them. In an equilateral triangle every angle is 60°.

Worked example

1
$equilateral base, side 10 cm, h = 12 cm$ — The question. The triangle has no height marked, so half × base × height will not work straight away.
2
$base area = \frac{1}{2} \times 10 \times 10 \times sin 6 0^{\circ}$ — Two sides are 10, and the angle between them is 60°, because all three angles of an equilateral triangle are equal.
3
$= 43.30 \dots cm^{2}$ — sin 60° = 0.8660…, and 50 × 0.8660… = 43.30…. Keep all the digits in your calculator.
4
$V = \frac{1}{3} \times 43.30 \dots \times 12$ — Now the usual pyramid formula.
5
$= 173.2 \dots \approx 173 cm^{3}$ — 43.30… × 12 = 519.6…, and a third of it is 173.2…, so 173 cm³ to 3 s.f.

Common mistake

Rounding the base to 43 too early gives 172 cm³, not 173. Also check your calculator is in degrees.

Key idea

A regular polygon base: join the centre to every corner to make n equal triangles. The angle at the centre is 360° ÷ n, so 72° for a pentagon. Find one triangle with $\frac{1}{2} ab sin C$ , then multiply by n.

Which Method? Read the Question

Find the base, then the height.

You see	You do
Base and height given	base area, × height, ÷ 3
Triangle base	half base × height for the base first
Volume given, height missing	× 3, then ÷ base area
A sloping edge, no height	Pythagoras first
Triangle base, no height marked	½ab sin C for the base (Extended)
Mass or density asked	volume first, then density = mass ÷ volume
Faces and surface area	base + every triangle face, using the slant height
Top cut off	Frustum of a Cone
Same shape all the way up	Volume of a Prism

Key idea

Always check you used the perpendicular height, not a sloping length.

Key idea

Write the sum, like $\frac{1}{3} \times 36 \times 10$ , before the answer. There is a mark for that line even if your final answer slips.

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