Surface Area of Pyramids
The surface area of a pyramid is the area of its base plus the areas of its triangular faces. Each triangle needs its slant height, the height of the face, which is not the same as the height of the pyramid. In NSW Year 10 Advanced maths Pythagoras' theorem links the perpendicular height, slant height and sloping edges (NESA outcome MA5‑ARE‑P‑01).
You will find surface areas of square, rectangular and triangular pyramids, in exact form and rounded. You will find slant heights from sloping edges or from the perpendicular height, work with a rectangular base where two slant heights are different, and cost a gift box, a church spire and a glass skylight. Harder questions derive a general formula, solve quadratics for the base edge, and find the area of a crystal made of two pyramids. It is part of Year 10 Advanced Measurement.
Every question with a fully worked solution.
- Surface Area of Pyramids - Video - Surface Area of a Pyramid Watch
Theory
The surface area of a pyramid is the area of the base plus the area of every triangular face. Each triangle uses the slant height, which you usually find with Pythagoras' theorem. This Year 10 Advanced maths page covers square, rectangular and triangular pyramids, open bases, exact surd answers and working backwards.
A right pyramid has a polygon base and triangular faces that meet at the apex, which sits directly above the centre of the base.
The perpendicular height \(h\) is the straight-up distance from the apex to the centre of the base. The slant height \(s\) is the height of a triangular face, measured from the apex to the midpoint of a base edge. A sloping edge \(e\) runs from the apex to a corner of the base. These three lengths are all different: \(e\) is the longest and \(h\) the shortest.
The surface area is the base plus all the triangular faces. A square pyramid has four identical triangles. A rectangular pyramid has two pairs, each with its own slant height.
Leave out the base if the solid has no base (a tent with no floor, a skylight), and leave out any face that is hidden where two solids are joined.
For a square pyramid with base edge \(b\) and slant height \(s\):
Finding the slant height or the height with Pythagoras' theorem:
| You know | Right-angled triangle | Gives |
|---|---|---|
| height \(h\), base edge \(b\) | \(h\), \(\dfrac{b}{2}\), \(s\) (inside the pyramid) | \(s^2 = h^2 + \left(\dfrac{b}{2}\right)^2\) |
| sloping edge \(e\), base edge \(b\) | \(e\), \(\dfrac{b}{2}\), \(s\) (on a face) | \(s^2 = e^2 - \left(\dfrac{b}{2}\right)^2\) |
| sloping edge \(e\), base | \(e\), half-diagonal \(OC\), \(h\) | \(h^2 = e^2 - OC^2\) |
How to find the surface area of a pyramid
- List the faces: the base (if there is one) and each triangle. Group the identical triangles.
- Find each slant height you need with Pythagoras' theorem. Draw the right-angled triangle separately and label it.
- Calculate each area: base, then \(\dfrac{1}{2} \times \text{base edge} \times \text{slant height}\) for each triangle.
- Add them, keeping exact surds or full calculator values until the end.
Other questions you will meet
- Height from a sloping edge: find half the diagonal of the base first, then \(h^2 = e^2 - OC^2\).
- Triangular pyramids: find the height of each different triangle. Ticks show which edges are equal; a face that stands at right angles to the base is a right-angled triangle.
- General formula: with base edge \(2a\) and height \(h\), the slant height is \(\sqrt{a^2 + h^2}\); substitute it and \(b = 2a\) into \(\text{SA} = b^2 + 2bs\), then simplify.
- Unknown base edge: substitute into \(\text{SA} = b^2 + 2bs\), rearrange to a quadratic equal to zero, factorise and keep the positive solution. To check a given value, substitute it and show both sides are equal.
- Joined pyramids (a crystal): count only the outside faces.
- Costs: add any extra percentage for joins, then multiply by the price per square metre. Edge frames are priced by length, so they need the sloping edge.
The triangles need the slant height. It is the hypotenuse of a right-angled triangle with the height and half the base edge, \(15\) cm.
| \(s^2\) | \(=\) | \(8^2 + 15^2\) |
| \(\) | \(=\) | \(289\) |
| \(s\) | \(=\) | \(17\text{ cm}\) |
| \(\text{SA}\) | \(=\) | \(30^2 + 4 \times \dfrac{1}{2} \times 30 \times 17\) |
| \(\) | \(=\) | \(900 + 1020\) |
| \(\) | \(=\) | \(\mathbf{1920}\text{ cm}^2\) |
The faces are not all the same. A face on an \(18\) m edge leans in by half of \(10\); a face on a \(10\) m edge leans in by half of \(18\).
| \(s_1^2\) | \(=\) | \(12^2 + 5^2 = 169\) |
| \(s_1\) | \(=\) | \(13\text{ m} \quad (\text{on the 18 m edges})\) |
| \(s_2^2\) | \(=\) | \(12^2 + 9^2 = 225\) |
| \(s_2\) | \(=\) | \(15\text{ m} \quad (\text{on the 10 m edges})\) |
Add the base and two pairs of matching triangles:
| \(\text{SA}\) | \(=\) | \(18 \times 10\) \(+\ 2 \times \dfrac{1}{2} \times 18 \times 13\) \(+\ 2 \times \dfrac{1}{2} \times 10 \times 15\) |
| \(\) | \(=\) | \(180 + 234 + 150\) |
| \(\) | \(=\) | \(\mathbf{564}\text{ m}^2\) |
(i) the area of canvas
(ii) the height of the tent.
(i) A sloping edge, half a base edge and the slant height make a right-angled triangle on one face:
| \(s^2\) | \(=\) | \(2.6^2 - 1.2^2 = 5.32\) |
| \(s\) | \(=\) | \(2.3065\ldots\) |
| \(A\) | \(=\) | \(4 \times \dfrac{1}{2} \times 2.4 \times 2.3065\ldots\) |
| \(\) | \(=\) | \(11.0712\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{11.07}\text{ m}^2\) |
No floor, so there is no base in the total.
(ii) The height stands above the centre, so use half the diagonal of the base, \(OC\):
| \(OC^2\) | \(=\) | \(1.2^2 + 1.2^2 = 2.88\) |
| \(h^2\) | \(=\) | \(2.6^2 - 2.88 = 3.88\) |
| \(h\) | \(=\) | \(1.9697\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1.97}\text{ m}\) |
(i) Show that one face has area \(\dfrac{\sqrt{3}}{4}x^2\).
(ii) Find \(x\).
(i) The height of an equilateral triangle meets the base at its midpoint, \(\dfrac{x}{2}\) from each corner:
| \(\text{height}^2\) | \(=\) | \(x^2 - \left(\dfrac{x}{2}\right)^2 = \dfrac{3x^2}{4}\) |
| \(\text{height}\) | \(=\) | \(\dfrac{\sqrt{3}}{2}x\) |
| \(\text{area}\) | \(=\) | \(\dfrac{1}{2} \times x \times \dfrac{\sqrt{3}}{2}x = \dfrac{\sqrt{3}}{4}x^2\) |
(ii) Four faces:
| \(4 \times \dfrac{\sqrt{3}}{4}x^2\) | \(=\) | \(100\sqrt{3}\) |
| \(\sqrt{3}\,x^2\) | \(=\) | \(100\sqrt{3}\) |
| \(x^2\) | \(=\) | \(100\) |
| \(x\) | \(=\) | \(\mathbf{10} \quad (x \gt 0)\) |
Common pitfalls
Frequently asked questions
How do you find the surface area of a square pyramid?
Add the area of the square base to the area of the four identical triangles. Each triangle has area \(\dfrac{1}{2} \times b \times s\), so \(\text{SA} = b^2 + 2bs\).
What is the difference between slant height and perpendicular height?
The perpendicular height goes straight down from the apex to the centre of the base. The slant height runs down the middle of a triangular face to the midpoint of a base edge, so it is longer.
How do you find the slant height of a pyramid?
Use Pythagoras' theorem. From the height: \(s^2 = h^2 + \left(\dfrac{b}{2}\right)^2\). From a sloping edge \(e\): \(s^2 = e^2 - \left(\dfrac{b}{2}\right)^2\).
Why does a rectangular pyramid have two slant heights?
Its base edges are different lengths, so the faces lean in by different amounts. The face on a long edge uses half the short edge in Pythagoras' theorem, and the face on a short edge uses half the long edge.
Do you include the base in the surface area of a pyramid?
Yes for a closed pyramid. Leave it out if the pyramid has no base, such as a tent with no floor or a glass skylight, or if the base is hidden because it is joined to another solid.