Surface Area of a Pyramid

Find the surface area of a pyramid by unfolding it into a base and triangles, using the slant height of each face, found with Pythagoras when it is not given.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Edexcel Further Pure Maths 4PM1 · 4 real past-paper questions below

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Unfold the Pyramid

Surface area is the total area of every outside face. Open a pyramid out flat and you get its net: one base with a triangle on each side.

Pyramid	Pieces in the net
Square base	1 square, 4 equal triangles
Rectangle base	1 rectangle, 2 pairs of triangles
Triangle base	4 triangles

Formula

SA = base + all the triangles

Rule

Find each piece's area, then add. The answer is in square units, like $cm^{2}$ .

Slant Height Given

The 2-mark question: you get the net, or the height of each triangle face. That height, measured up the middle of a face, is the slant height $l$ .

Square base 6 cm. Each triangle face has slant height 7 cm.

Worked example

1
$6 \times 6 = 36$ — the base, a square
2
$\frac{1}{2} \times 6 \times 7 = 21$ — one triangle: half base × slant height
3
$4 \times 21 = 84$ — 4 equal triangles
4
$36 + 84 = 120 cm^{2}$ — add the base

Common mistake

Leaving out the base. Total surface area means every face. Only an open pyramid, like a tent with no floor, has no base.

Vertical Height Given: Find the Slant

The 5-mark question: you get the pyramid's vertical height, from the top straight down to the centre of the base. A face needs its slant height, so find it first.

The vertical height, half the base side and the slant height make a right-angled triangle. The slant height is the longest side. Square base 8 cm, vertical height 3 cm.

Worked example

1
$8 \div 2 = 4$ — half the base: centre to the middle of an edge
2
$l^{2} = 3^{2} + 4^{2} = 25$ — Pythagoras
3
$l = 5$ — the slant height
4
$\frac{1}{2} \times 8 \times 5 = 20$ — one triangle
5
$64 + 4 \times 20 = 144 cm^{2}$ — base plus 4 triangles

Common mistake

Using the vertical height 3 in the triangles. That gives 112 cm², too small. Each face leans, so it is taller than the pyramid.

Key idea

When the slant height is a long decimal, keep it in your calculator. Round only the final answer.

Rectangle Base: Two Slant Heights (Extended/Higher)

The 6-mark question: the base is a rectangle, not a square. The faces on the long sides lean at a different slope from the faces on the short sides.

Rectangle base 18 cm by 10 cm, vertical height 12 cm.

Worked example

1
$1 2^{2} + 5^{2} = 13$ — faces on the 18 cm sides use half of 10
2
$2 \times \frac{1}{2} \times 18 \times 13 = 234$ — the two long triangles
3
$1 2^{2} + 9^{2} = 15$ — faces on the 10 cm sides use half of 18
4
$2 \times \frac{1}{2} \times 10 \times 15 = 150$ — the two short triangles
5
$180 + 234 + 150 = 564 cm^{2}$ — base 18 × 10, plus all four

Common mistake

Using one slant height for all four faces. Each pair of faces gets its own right-angled triangle.

From a Sloping Edge (Extended/Higher)

Sometimes you get a sloping edge, from a base corner to the top, not the slant height. The edge, half the base side and the slant height make a right-angled triangle inside one face. Here the edge is the longest side.

Square base 10 cm. Each sloping edge is 13 cm.

Worked example

1
$l^{2} = 1 3^{2} - 5^{2} = 144$ — edge is longest, so subtract half the base
2
$l = 12$ — the slant height
3
$\frac{1}{2} \times 10 \times 12 = 60$ — one triangle
4
$100 + 4 \times 60 = 340 cm^{2}$ — base plus 4 triangles

Common mistake

Using the sloping edge 13 as the face height. That gives 360 cm², too big. The edge runs to a corner, not up the middle.

Working Back From the Total (Extended/Higher)

The question can give the total surface area and ask for a length. Undo the steps in reverse.

Square base 8 cm. Total surface area 144 cm². Find the slant height, then the vertical height.

Worked example

1
$144 - 64 = 80$ — take off the base
2
$80 \div 4 = 20$ — one triangle
3
$\frac{1}{2} \times 8 \times l = 20$ — half base × slant height
4
$4 l = 20, l = 5$ — divide by 4
5
$h = 5^{2} - 4^{2} = 3$ — Pythagoras, slant is longest

Common mistake

Dividing 144 by 4 straight away. Take off the base first: it is not one of the triangles.

Which Method? Read the Question

Find the slant height first. Then it is base plus triangles.

You see	You do
Net, or slant height given	base + triangles
Vertical height given	Pythagoras with half the base
Rectangle base	two slant heights, one per pair
Sloping edge given	edge² − half base², for the slant
Total given, length missing	take off the base, ÷ the triangles
Volume, or height from an edge	Volume of a Pyramid
Angle with the base	Trigonometry in 3D: Pyramids and Cones
Round base	Volume and Surface Area of a Cone
Top cut off	Frustum of a Cone
Prism or cuboid	Surface Area of a Cuboid and Cylinder

Key idea

Write the area of each face on its own line. Method marks are given for correct working, even if the total slips.

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