Surface Area of a Pyramid

Open a pyramid out into its net, find the area of each triangular face from its slant height, and find the slant height from the height with Pythagoras' theorem.

Before you start

The area of a triangle is $\frac{1}{2} \times$ base $\times$ height.
In a pyramid the height runs from the apex straight down to the base; the slant height runs down the middle of a triangular face, from the apex to the middle of a side of the base (Volume of a Pyramid).
In a right-angled triangle with shorter sides $a$ and $b$ and hypotenuse $c$ , $c^{2} = a^{2} + b^{2}$ (Pythagoras' theorem).

The Net of a Pyramid

Cut a square pyramid along its four sloping edges and fold the faces down flat: you get its net, the square base with a triangle on each side. The total surface area of the pyramid is the area of its net: the base plus all the triangular faces.

Each triangular face has a side of the base as its base, and the slant height as its height, because the slant height meets that side at right angles in the middle. So one face has area $\frac{1}{2} \times$ side $\times$ slant height.

Total surface area of a pyramid

Total surface area $=$ area of the base $+$ area of all the triangular faces

Area of one triangular face $= \frac{1}{2} \times \text{side of the base} \times \text{slant height}$

For a square pyramid of side $a$ and slant height $l$ : total surface area $= a^{2} + 4 \times \frac{1}{2} \times a \times l = a^{2} + 2al\text{.}$

Example 1: A square pyramid

The square pyramid above has a base of side $8$ cm and a slant height of $10$ cm. Find its total surface area.

Solution

Base: $8 \times 8 = 64 \text{ cm}^{2}$

One triangular face: $\frac{1}{2} \times 8 \times 10 = 40 \text{ cm}^{2}$

Four faces: $4 \times 40 = 160 \text{ cm}^{2}$

Total: $64 + 160 = 224$

The total surface area is $224 \text{ cm}^{2}$ .

Example 2: A rectangular pyramid

A pyramid stands on a rectangle $18$ cm by $10$ cm. The slant height of each face on an $18$ cm side is $13$ cm, and of each face on a $10$ cm side is $15$ cm. Find its total surface area.

Solution

A rectangular pyramid has two pairs of equal faces with different slant heights. Find each pair separately.

Base: $18 \times 10 = 180 \text{ cm}^{2}$

Two faces on the $18$ cm sides: $2 \times \frac{1}{2} \times 18 \times 13 = 234 \text{ cm}^{2}$

Two faces on the $10$ cm sides: $2 \times \frac{1}{2} \times 10 \times 15 = 150 \text{ cm}^{2}$

Total: $180 + 234 + 150 = 564$

The total surface area is $564 \text{ cm}^{2}$ .

The Slant Height from the Height

Often the height is given, not the slant height. Inside the pyramid, the height $VO$ , the line $OM$ from the centre of the base to the middle of a side, and the slant height $VM$ make a right-angled triangle $VOM$ , with the right angle at $O$ .

In a square pyramid of side $a$ , $OM$ is half a side: $OM = \frac{a}{2}$ . The slant height is the hypotenuse, so it is always longer than the height.

Slant height of a square pyramid

$(\text{slant height})^{2} = (\text{height})^{2} + \left(\frac{\text{side}}{2}\right)^{2}$

Find the slant height first, then the areas of the faces. Keep its full value in the calculator if it is not a whole number.

Example 3: From the height

A square pyramid has a base of side $42$ cm and a height of $20$ cm. Find (a) its slant height, (b) its total surface area.

Solution

(a) In $\triangle VOM$ , $VO = 20$ cm and $OM = 42 \div 2 = 21$ cm, with the right angle at $O$ .

$VM^{2} = 20^{2} + 21^{2} = 400 + 441 = 841\text{,}$ so $VM = \sqrt{841} = 29$ cm.

(b) Base: $42 \times 42 = 1764 \text{ cm}^{2}\text{.}$ Four faces: $4 \times \frac{1}{2} \times 42 \times 29 = 2436 \text{ cm}^{2}\text{.}$

Total: $1764 + 2436 = 4200 \text{ cm}^{2}$

(a) $29$ cm (b) $4200 \text{ cm}^{2}$

Example 4: A slant height that is not whole

A square pyramid has a base of side $6$ cm and a height of $5$ cm. Find its total surface area, correct to 3 significant figures.

Solution

$OM = 6 \div 2 = 3$ cm. $VM^{2} = 5^{2} + 3^{2} = 25 + 9 = 34\text{,}$ so $VM = \sqrt{34} = 5.830\ldots$ cm.

Four faces: $4 \times \frac{1}{2} \times 6 \times \sqrt{34} = 12\sqrt{34} = 69.971\ldots \text{ cm}^{2}$

Total: $36 + 69.971\ldots = 105.97\ldots = 106$ (3 s.f.)

Had the slant height been rounded to $5.8$ first, the total would come out as $36 + 69.6 = 105.6$ , which rounds to $106$ here but can give a wrong last figure in other questions. Keep $\sqrt{34}$ in the calculator.

The total surface area is $106 \text{ cm}^{2}$ .

Example 5: The height from the slant height

A square pyramid has a base of side $16$ cm and a slant height of $17$ cm. Find (a) its height, (b) its volume.

Solution

(a) In $\triangle VOM$ the slant height is the hypotenuse: $VM = 17$ cm and $OM = 16 \div 2 = 8$ cm.

$VO^{2} = 17^{2} - 8^{2} = 289 - 64 = 225\text{,}$ so $VO = 15$ cm.

(b) $V = \frac{1}{3} \times 16 \times 16 \times 15 = 1280$

(a) $15$ cm (b) $1280 \text{ cm}^{3}$

Tents and Roofs: a Pyramid without its Base

A tent with no floor, or the roof of a shelter, is a pyramid without its base. Only the triangular faces are covered, so leave the base out of the total. Read each question for what is really there.

Example 6: A tent

A tent with no floor is a square pyramid. Its base has sides of $2.4$ m and its height is $1.6$ m. Find the area of canvas used to make it.

Solution

Slant height: $OM = 2.4 \div 2 = 1.2$ m, so $(\text{slant height})^{2} = 1.6^{2} + 1.2^{2} = 2.56 + 1.44 = 4\text{,}$ and the slant height is $2$ m.

Canvas: four triangles only, no floor. $4 \times \frac{1}{2} \times 2.4 \times 2 = 9.6$

The tent uses $9.6 \text{ m}^{2}$ of canvas.

Example 7: Tiling a roof

The roof of a garden pavilion is a square pyramid with a base of side $6$ m and a height of $2$ m. Roof tiles cost $25 per square metre. Find the cost of tiling the roof, to the nearest dollar.

Solution

$OM = 3$ m. $(\text{slant height})^{2} = 2^{2} + 3^{2} = 13\text{,}$ so the slant height is $\sqrt{13} = 3.605\ldots$ m.

Roof area: $4 \times \frac{1}{2} \times 6 \times \sqrt{13} = 12\sqrt{13} = 43.266\ldots \text{ m}^{2}$ (no base: the roof has no floor).

Cost: $43.266\ldots \times 25 = 1081.66\ldots$

Tiling the roof costs about $1082.

Common Mistakes

Using the height for a face. The height of a triangular face is the slant height. For a square pyramid of side $6$ cm and height $4$ cm, the faces need the slant height $5$ cm, not $4$ cm.
Using the whole side in Pythagoras' theorem. The short side of $\triangle VOM$ is half the side of the base.
Forgetting the base, or counting it when it is not there (a tent with no floor, a roof).
One slant height for a rectangular pyramid. The faces on the long sides and on the short sides have different slant heights.
Rounding the slant height before multiplying. Keep $\sqrt{34}$ in the calculator.

Practice Questions

Try each question before you open its solution. Give answers correct to 3 significant figures where they are not exact.

Question 1

A square pyramid has a base of side $5$ cm and a slant height of $8$ cm. Find its total surface area.

ShowHide solution

Base: $25 \text{ cm}^{2}$ . Four faces: $4 \times \frac{1}{2} \times 5 \times 8 = 80 \text{ cm}^{2}\text{.}$

Total: $25 + 80 = 105$

$105 \text{ cm}^{2}$

Question 2

The net of a square pyramid is drawn below. Find the total surface area of the pyramid.

ShowHide solution

Base: $7 \times 7 = 49 \text{ cm}^{2}$ . Each triangle: $\frac{1}{2} \times 7 \times 9 = 31.5 \text{ cm}^{2}\text{,}$ and four make $126 \text{ cm}^{2}$ .

Total: $49 + 126 = 175$

$175 \text{ cm}^{2}$

Question 3

A square pyramid has a base of side $18$ cm and a height of $12$ cm. Find (a) its slant height, (b) its total surface area.

ShowHide solution

(a) $OM = 9$ cm. $(\text{slant height})^{2} = 12^{2} + 9^{2} = 144 + 81 = 225\text{,}$ so the slant height is $15$ cm.

(b) $324 + 4 \times \frac{1}{2} \times 18 \times 15 = 324 + 540 = 864$

(a) $15$ cm (b) $864 \text{ cm}^{2}$

Question 4

A square pyramid has a base of side $9$ cm and a height of $6$ cm. Find its total surface area.

ShowHide solution

$OM = 4.5$ cm. $(\text{slant height})^{2} = 6^{2} + 4.5^{2} = 36 + 20.25 = 56.25\text{,}$ so the slant height is $7.5$ cm.

$81 + 4 \times \frac{1}{2} \times 9 \times 7.5 = 81 + 135 = 216$

$216 \text{ cm}^{2}$

Question 5

A pyramid on a rectangular base $16$ cm by $12$ cm has a height of $8$ cm, with its apex above the centre of the base. Find (a) the slant heights of its two kinds of face, (b) its total surface area.

ShowHide solution

(a) A face on a $16$ cm side: its slant height runs to the middle of that side, $12 \div 2 = 6$ cm from the centre. $\sqrt{8^{2} + 6^{2}} = \sqrt{100} = 10$ cm.

A face on a $12$ cm side: the middle of that side is $16 \div 2 = 8$ cm from the centre. $\sqrt{8^{2} + 8^{2}} = \sqrt{128} = 11.31\ldots$ cm.

(b) Base: $16 \times 12 = 192 \text{ cm}^{2}$ . Faces: $2 \times \frac{1}{2} \times 16 \times 10 = 160 \text{ cm}^{2}$ and $2 \times \frac{1}{2} \times 12 \times \sqrt{128} = 135.76\ldots \text{ cm}^{2}\text{.}$

Total: $192 + 160 + 135.76\ldots = 487.76\ldots = 488$ (3 s.f.)

(a) $10$ cm and $11.3$ cm (b) $488 \text{ cm}^{2}$

Question 6

A square pyramid has a base of side $14$ cm and a slant height of $25$ cm. Find (a) its height, (b) its volume.

ShowHide solution

(a) $OM = 7$ cm; the slant height is the hypotenuse. $\text{height}^{2} = 25^{2} - 7^{2} = 625 - 49 = 576\text{,}$ so the height is $24$ cm.

(b) $\frac{1}{3} \times 196 \times 24 = 1568$

(a) $24$ cm (b) $1568 \text{ cm}^{3}$

Question 7

A triangular pyramid has a base that is an equilateral triangle of side $6$ cm, and its three sloping faces are equal triangles, each with a slant height of $8$ cm. The height of the base triangle is $\sqrt{27}$ cm. Find the total surface area of the pyramid.

ShowHide solution

Base: $\frac{1}{2} \times 6 \times \sqrt{27} = 15.588\ldots \text{ cm}^{2}$

Three faces: $3 \times \frac{1}{2} \times 6 \times 8 = 72 \text{ cm}^{2}$

Total: $15.588\ldots + 72 = 87.588\ldots = 87.6$ (3 s.f.)

$87.6 \text{ cm}^{2}$

Question 8

A camping tent with no floor is a square pyramid with a base of side $3$ m and a height of $2$ m. Find the area of material needed for the tent.

ShowHide solution

$OM = 1.5$ m. $(\text{slant height})^{2} = 2^{2} + 1.5^{2} = 4 + 2.25 = 6.25\text{,}$ so the slant height is $2.5$ m.

Four faces, no floor: $4 \times \frac{1}{2} \times 3 \times 2.5 = 15$

$15 \text{ m}^{2}$

Question 9

A pyramid-roofed shelter in a park has a roof that is a square pyramid with a base of side $4$ m and a height of $1.5$ m. One litre of paint covers $8 \text{ m}^{2}$ . How many litres are needed to paint the outside of the roof once?

ShowHide solution

$OM = 2$ m. $(\text{slant height})^{2} = 1.5^{2} + 2^{2} = 2.25 + 4 = 6.25\text{,}$ so the slant height is $2.5$ m.

Roof area (four faces, no base): $4 \times \frac{1}{2} \times 4 \times 2.5 = 20 \text{ m}^{2}$

Paint: $20 \div 8 = 2.5$ litres

$2.5$ litres

Question 10

For a square pyramid with a base of side $24$ cm and a height of $5$ cm, Ali writes the total surface area as $576 + 4 \times \frac{1}{2} \times 24 \times 5 = 816 \text{ cm}^{2}\text{.}$ Explain his mistake and find the correct answer.

ShowHide solution

Ali used the height, $5$ cm, as the height of each face. A face needs the slant height.

$OM = 12$ cm, so the slant height is $\sqrt{5^{2} + 12^{2}} = \sqrt{169} = 13$ cm.

$576 + 4 \times \frac{1}{2} \times 24 \times 13 = 576 + 624 = 1200$

He used the height instead of the slant height. The total surface area is $1200 \text{ cm}^{2}$ .

Question 11

A square pyramid has a base of side $20$ cm and a total surface area of $1440 \text{ cm}^{2}$ . Find (a) its slant height, (b) its height.

ShowHide solution

(a) The base is $20 \times 20 = 400 \text{ cm}^{2}$ . The four faces: $1440 - 400 = 1040 \text{ cm}^{2}\text{,}$ so one face is $260 \text{ cm}^{2}$ .

$\frac{1}{2} \times 20 \times l = 260$ , so $10l = 260$ and $l = 26$ .

(b) $OM = 10$ cm. $\text{height}^{2} = 26^{2} - 10^{2} = 676 - 100 = 576\text{,}$ so the height is $24$ cm.

(a) $26$ cm (b) $24$ cm

Question 12

The Great Pyramid of Giza had a square base of side about $230$ m and a height of about $147$ m. Find the area of its four sloping faces, correct to 3 significant figures.

ShowHide solution

$OM = 230 \div 2 = 115$ m. $(\text{slant height})^{2} = 147^{2} + 115^{2} = 21\,609 + 13\,225 = 34\,834$

Slant height $= \sqrt{34\,834} = 186.63\ldots$ m

Four faces: $4 \times \frac{1}{2} \times 230 \times 186.63\ldots = 460 \times 186.63\ldots = 85\,853.7\ldots = 85\,900$ (3 s.f.)

About $85\,900 \text{ m}^{2}$

Key Points

The net of a pyramid is its base with a triangle on each side; the total surface area is the area of the net.
Each triangular face has area $\frac{1}{2} \times$ side of the base $\times$ slant height.
Square pyramid of side $a$ and slant height $l$ : total surface area $= a^{2} + 2al$ .
The slant height is the hypotenuse of $\triangle VOM$ : $(\text{slant height})^{2} = (\text{height})^{2} + \left(\frac{\text{side}}{2}\right)^{2}\text{.}$
A tent or a roof has no base: count only the faces that are there.