Surface Area of Pyramids and Cones
Theory
The surface area of a pyramid is its base plus its triangular faces, and the curved surface of a cone is \(\pi r l\). This Year 10A maths page shows how to find slant heights and heights with Pythagoras' theorem (from a height or from a sloping edge), handle square, rectangular and triangular pyramids, roll a sector into a cone, and work backwards to a radius or height.
A right pyramid has a polygon base and triangular faces that meet at the apex, directly above the centre of the base. A right cone has a circular base and an apex above its centre.
The perpendicular height \(h\) goes straight down from the apex to the centre of the base. The slant height runs down the outside surface: for a pyramid it is the height of a triangular face (\(s\)); for a cone it runs from the apex to the edge of the base (\(l\)).
The surface area of a pyramid is the area of the base plus the area of every triangular face. Each face has base = a base edge and height = the slant height. A rectangular base gives two different slant heights.
The curved surface of a cone has area \(\pi r l\). A solid cone adds its circular base, \(\pi r^{2}\).
Three lengths are easy to mix up in a pyramid: the perpendicular height \(h\), the slant height \(s\) (to the middle of a base edge) and the sloping edge \(e\) (to a corner). The sloping edge is the longest and the height the shortest.
| Solid | Surface area |
|---|---|
| Square pyramid, base edge \(b\), slant height \(s\) | \(b^{2} + 4 \times \dfrac{1}{2}bs = b^{2} + 2bs\) |
| Any right pyramid | base area \(+\) area of the triangular faces |
| Cone, curved surface only | \(\pi r l\) |
| Solid cone | \(\pi r l + \pi r^{2}\) |
Finding a length with Pythagoras' theorem:
| You know | Right-angled triangle | Gives |
|---|---|---|
| height \(h\), base edge \(b\) | \(h\), \(\dfrac{b}{2}\), \(s\) | \(s^2 = h^2 + \left(\dfrac{b}{2}\right)^2\) |
| 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 \(OC\), \(h\) | \(h^2 = e^2 - OC^2\) |
| cone, \(r\) and \(h\) | \(r\), \(h\), \(l\) | \(l^2 = r^2 + h^2\) |
How to find the surface area of a pyramid or cone
- List the surfaces. Is there a base? A tent with no floor, a roof or a party hat has none. Group identical triangles.
- Find each slant height. Draw the right-angled triangle on its own: height and half a base edge (pyramid), height and radius (cone), or sloping edge and half a base edge (on a face).
- Calculate each part: the base, \(\dfrac{1}{2} \times \text{edge} \times s\) for each triangle, or \(\pi r l\) and \(\pi r^{2}\).
- Add, keeping surds or full calculator values, then round once and give square units.
Other questions you will meet
- Rectangular base: two slant heights. The faces on the long edges lean in by half the short edge, and the other way round.
- Triangular pyramids: find the height of each different face. Ticks show equal edges; a face standing at right angles to the base is a right-angled triangle.
- Height from a sloping edge: find half the diagonal of the base first, then \(h^2 = e^2 - OC^2\).
- Pyramid backwards: subtract the base from the total, divide by the number of identical faces, solve \(\dfrac{1}{2} \times \text{edge} \times s = \text{face area}\), then use Pythagoras to go from \(s\) to \(h\).
- Cone backwards: for the slant height, \(\pi r l = A\) is linear. For the radius from the total surface area, \(\pi r l + \pi r^2 = \text{SA}\) is a quadratic: divide by \(\pi\), make it equal zero, factorise (or use the quadratic formula) and keep the positive root.
- Costs: find the area, then multiply by the price per square unit. Convert \(\text{cm}^2\) to \(\text{m}^2\) by dividing by \(10\,000\).
(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, \(OC\):
| \(OC^2\) | \(=\) | \(1.2^2 + 1.2^2 = 2.88\) |
| \(h^2\) | \(=\) | \(2.6^2 - 2.88 = 3.88\) |
| \(h\) | \(=\) | \(1.9697\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{1.97}\text{ m}\) |
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 radius of the cone
(ii) its exact perpendicular height
(iii) the exact curved surface area.
(i) The arc of the sector becomes the circumference of the base:
| \(\text{arc}\) | \(=\) | \(\dfrac{240}{360} \times 2\pi \times 15 = 20\pi\) |
| \(2\pi r\) | \(=\) | \(20\pi\) |
| \(r\) | \(=\) | \(\mathbf{10}\text{ cm}\) |
(ii) The sector's radius is the slant height, \(l = 15\):
| \(h^2\) | \(=\) | \(15^2 - 10^2 = 125\) |
| \(h\) | \(=\) | \(\sqrt{25 \times 5} = \mathbf{5\sqrt{5}}\text{ cm}\) |
(iii) The curved surface is the sector itself:
| \(A\) | \(=\) | \(\pi rl = \pi \times 10 \times 15\) |
| \(\) | \(=\) | \(\mathbf{150\pi}\text{ cm}^2\) |
Check: the sector's area is \(\dfrac{240}{360} \times \pi \times 15^2 = 150\pi\) ✓
Substitute \(l = 13\) and divide every term by \(\pi\). The radius is squared in the base, so this is a quadratic:
| \(\pi r(13) + \pi r^2\) | \(=\) | \(90\pi\) |
| \(r^2 + 13r - 90\) | \(=\) | \(0\) |
| \((r + 18)(r - 5)\) | \(=\) | \(0\) |
| \(r\) | \(=\) | \(-18 \text{ or } r = 5\) |
A radius is positive, so \(r = \mathbf{5}\) cm. The slant height is the hypotenuse, so subtract:
| \(h^2\) | \(=\) | \(13^2 - 5^2\) |
| \(\) | \(=\) | \(169 - 25 = 144\) |
| \(h\) | \(=\) | \(\mathbf{12}\text{ cm}\) |
Check: \(\pi(5)(13) + \pi(5)^2 = 65\pi + 25\pi = 90\pi\).
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 four triangular faces. Each triangle has area \(\dfrac{1}{2} \times \text{base edge} \times \text{slant height}\), so \(\text{SA} = b^2 + 2bs\).
What is the formula for the surface area of a cone?
The curved surface is \(\pi r l\), where \(l\) is the slant height. For a solid cone add the base: \(\text{SA} = \pi r l + \pi r^{2}\).
What is the difference between slant height, sloping edge and height?
The perpendicular height goes straight down inside the solid to the centre of the base. The slant height runs down the middle of a face to the midpoint of a base edge. The sloping edge runs from the apex to a corner and is the longest of the three.
How do you make a cone from a sector?
Join the two straight edges. The sector's radius becomes the slant height of the cone, and its arc becomes the circumference of the base, so set the arc length equal to \(2\pi r\) to find the cone's radius.
How do you find the radius of a cone from its surface area?
Substitute the slant height into \(\pi r l + \pi r^2 = \text{SA}\), divide by \(\pi\) and solve the quadratic. Keep the positive solution.