Volume of a Pyramid

Find the volume of a pyramid on a square, rectangular or triangular base with $V = \frac{1}{3} \times$ base area $\times$ height, and work back from a volume to a height or a side.

Before you start

The volume of a prism or a cuboid is base area $\times$ height: a cuboid $10$ cm by $10$ cm by $12$ cm holds $10 \times 10 \times 12 = 1200 \text{ cm}^{3}\text{.}$
The area of a triangle is $\frac{1}{2} \times$ base $\times$ height, the height measured at right angles to the base.
Volume is measured in cubic units ( $\text{cm}^{3}$ , $\text{m}^{3}$ ), and $1000 \text{ cm}^{3} = 1$ litre.

The Parts of a Pyramid

A pyramid has a flat base, which is a polygon, and triangular faces that rise from the sides of the base and meet at one point, the apex. A pyramid is named after its base: a square pyramid stands on a square, a rectangular pyramid on a rectangle and a triangular pyramid on a triangle.

The height $h$ of a pyramid (its perpendicular height) is the distance from the apex straight down to the base, along the line that meets the base at right angles. In most pyramids of this lesson the apex is above the centre of the base (the middle of a square or a rectangle), so the height runs from the apex $V$ to the centre $O$ of the base. It is inside the pyramid, so it is drawn dashed.

The slant height is different. It is the height of one of the triangular faces, measured along the face from the apex to the middle of a side of the base, such as $VM$ below. It is always longer than the height, because it is the sloping side of the right-angled triangle $VOM$ . The volume uses the perpendicular height; the slant height is for the area of the faces (Surface Area of a Pyramid).

Height or slant height?

If a length is drawn down the middle of a face, it is a slant height. If it is drawn through the inside of the pyramid, at right angles to the base, it is the height. Read a question carefully: "height" on its own means the perpendicular height.

The Volume of a Pyramid

Volume of a pyramid

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

where $h$ is the perpendicular height of the pyramid. The base area and the height must be in the same unit of length, and the volume is then in cubic units.

Why a third? Take a cube and a corner of its top, $G$ . Join $G$ to the four corners of the bottom face $ABCD$ : this cuts out a pyramid whose base is the bottom of the cube and whose height is the edge of the cube. Two more pyramids of exactly the same shape and size, one on the front face and one on the left face, both with their apex at $G$ , fill the rest of the cube. So three equal pyramids make the cube, and each holds a third of it: $\frac{1}{3} \times$ base area $\times$ height.

You can also see it with water or sand: a hollow pyramid filled and emptied into a prism with the same base and the same height fills it after exactly three pourings. Neither the cube nor the sand is a proof for every pyramid; both show where the third comes from.

The pyramid in the cube has its apex above a corner of its base, not above the middle, and the formula still holds. Wherever the apex is, the volume is $\frac{1}{3} \times$ base area $\times$ height, as long as $h$ is the perpendicular height.

Example 1: A square pyramid

A square pyramid has a base of side $6$ cm and a height of $8$ cm. Find its volume.

Solution

Base area $= 6 \times 6 = 36 \text{ cm}^{2}$

$V = \frac{1}{3} \times 36 \times 8 = 96$

The volume is $96 \text{ cm}^{3}$ .

Example 2: A rectangular pyramid

A pyramid stands on a rectangular base $10$ cm long and $7$ cm wide, and its height is $12$ cm. Find its volume.

Solution

Base area $= 10 \times 7 = 70 \text{ cm}^{2}$

$V = \frac{1}{3} \times 70 \times 12 = 280$

The volume is $280 \text{ cm}^{3}$ .

Example 3: A triangular pyramid

The base of a pyramid is a right-angled triangle $ABC$ with $AB = 8$ cm, $AC = 6$ cm and $\angle BAC = 90^{\circ}$ . The height of the pyramid is $9$ cm. Find its volume.

Solution

The base is a triangle. With $AB$ as its base, its height is $AC$ , since the two are at right angles.

Base area $= \frac{1}{2} \times 8 \times 6 = 24 \text{ cm}^{2}$

$V = \frac{1}{3} \times 24 \times 9 = 72$

Two different heights are used here: the height of the triangle ( $6$ cm, for the base area) and the height of the pyramid ( $9$ cm, for the volume). Keep them apart.

The volume is $72 \text{ cm}^{3}$ .

Example 4: Height, not slant height

A square pyramid has a base of side $10$ cm. Its slant height is $13$ cm and its perpendicular height is $12$ cm. Find its volume.

Solution

The volume needs the perpendicular height, $12$ cm. The $13$ cm runs down a face and is not used.

Base area $= 10 \times 10 = 100 \text{ cm}^{2}$

$V = \frac{1}{3} \times 100 \times 12 = 400$

The volume is $400 \text{ cm}^{3}$ . (With $13$ cm by mistake you would get $433 \text{ cm}^{3}$ , which is wrong.)

Finding a Height or a Side from the Volume

If the volume is known, put it and the other numbers into the formula and solve the equation for the length you want (Changing the Subject of a Formula). Multiplying both sides by $3$ first clears the third: $3V = \text{base area} \times h\text{.}$

Example 5: The height

A pyramid on a rectangular base $9$ cm by $6$ cm has a volume of $252 \text{ cm}^{3}$ . Find its height.

Solution

Base area $= 9 \times 6 = 54 \text{ cm}^{2}$

$252 = \frac{1}{3} \times 54 \times h$

$252 = 18h$

$h = 252 \div 18 = 14$

Check: $\frac{1}{3} \times 54 \times 14 = 252\text{.}$

The height is $14$ cm.

Example 6: The side of the base

A square pyramid has a volume of $400 \text{ cm}^{3}$ and a height of $9$ cm. Find the length of a side of its base, correct to 3 significant figures.

Solution

Let the side be $x$ cm, so the base area is $x^{2} \text{ cm}^{2}$ .

$400 = \frac{1}{3} \times x^{2} \times 9$

$400 = 3x^{2}$

$x^{2} = \frac{400}{3} = 133.33\ldots$

$x = \sqrt{133.33\ldots} = 11.547\ldots = 11.5$ (3 s.f.)

Keep $\frac{400}{3}$ in the calculator and take its square root; round only the final answer.

Each side of the base is $11.5$ cm.

Pyramids in the Real World

Example 7: The Great Pyramid

When it was built, the Great Pyramid of Giza in Egypt had a square base of side about $230$ m and a height of about $147$ m. Find its volume, correct to 3 significant figures.

Solution

Base area $= 230 \times 230 = 52\,900 \text{ m}^{2}$

$V = \frac{1}{3} \times 52\,900 \times 147 = 2\,592\,100$

$= 2\,590\,000$ (3 s.f.)

Its volume was about $2\,590\,000 \text{ m}^{3}$ .

Example 8: Litres in a pyramid

A closed water tank is a square pyramid standing on its base, with a base of side $60$ cm and a height of $50$ cm. How many litres of water does it hold when full?

Solution

Work in centimetres, then change to litres.

Base area $= 60 \times 60 = 3600 \text{ cm}^{2}$

$V = \frac{1}{3} \times 3600 \times 50 = 60\,000 \text{ cm}^{3}$

$1000 \text{ cm}^{3} = 1$ litre, so $60\,000 \text{ cm}^{3} = 60\,000 \div 1000 = 60$ litres.

It holds $60$ litres of water.

Common Mistakes

Forgetting the third. Base area $\times$ height is the volume of the prism with the same base and height, three times too big.
Using the slant height. A length drawn down a face is a slant height. The volume needs the perpendicular height, from the apex straight down to the base.
Using a side for the base area. For a square pyramid of side $6$ cm and height $8$ cm, $\frac{1}{3} \times 6 \times 8 = 16$ is wrong: the base area is $6 \times 6 = 36 \text{ cm}^{2}$ .
Mixing up the two heights of a triangular pyramid: one is the height of the base triangle, the other the height of the pyramid.
Rounding in the middle. Keep $x^{2} = 133.33\ldots$ in the calculator, and round only the last answer.

Practice Questions

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

Question 1

Find the volume of a square pyramid with a base of side $5$ cm and a height of $9$ cm.

ShowHide solution

Base area $= 5 \times 5 = 25 \text{ cm}^{2}$

$V = \frac{1}{3} \times 25 \times 9 = 75$

$75 \text{ cm}^{3}$

Question 2

A rectangular pyramid has a base $8$ m by $6$ m and a height of $5$ m. Find its volume.

ShowHide solution

Base area $= 8 \times 6 = 48 \text{ m}^{2}$

$V = \frac{1}{3} \times 48 \times 5 = 80$

$80 \text{ m}^{3}$

Question 3

The base of a pyramid is a triangle with a base of $10$ cm and a height of $6$ cm. The pyramid is $14$ cm high. Find its volume.

ShowHide solution

Base area $= \frac{1}{2} \times 10 \times 6 = 30 \text{ cm}^{2}$

$V = \frac{1}{3} \times 30 \times 14 = 140$

$140 \text{ cm}^{3}$

Question 4

A pyramid has a base area of $45 \text{ cm}^{2}$ and a volume of $210 \text{ cm}^{3}$ . Find its height.

ShowHide solution

$210 = \frac{1}{3} \times 45 \times h$

$210 = 15h$ , so $h = 14$ .

$14$ cm

Question 5

The square pyramid below has a base of side $12$ cm, a slant height of $10$ cm and a perpendicular height of $8$ cm. Find its volume.

ShowHide solution

Use the perpendicular height, $8$ cm; the slant height is not needed.

Base area $= 12 \times 12 = 144 \text{ cm}^{2}$

$V = \frac{1}{3} \times 144 \times 8 = 384$

$384 \text{ cm}^{3}$

Question 6

A square pyramid has a volume of $147 \text{ cm}^{3}$ and a height of $9$ cm. Find the length of a side of its base.

ShowHide solution

Let the side be $x$ cm.

$147 = \frac{1}{3} \times x^{2} \times 9 = 3x^{2}$

$x^{2} = 49$ , so $x = 7$ .

$7$ cm

Question 7

A pyramid on a rectangular base $15$ cm by $8$ cm has a volume of $1000 \text{ cm}^{3}$ . Find its height.

ShowHide solution

Base area $= 15 \times 8 = 120 \text{ cm}^{2}$

$1000 = \frac{1}{3} \times 120 \times h = 40h$

$h = 1000 \div 40 = 25$

$25$ cm

Question 8

A glass paperweight is a square pyramid with a base of side $6.5$ cm and a height of $5.2$ cm. Find its volume.

ShowHide solution

Base area $= 6.5 \times 6.5 = 42.25 \text{ cm}^{2}$

$V = \frac{1}{3} \times 42.25 \times 5.2 = 73.233\ldots = 73.2$ (3 s.f.)

$73.2 \text{ cm}^{3}$

Question 9

The roof of a bus shelter is a square pyramid with a base of side $3.6$ m and a height of $1.2$ m. Find the volume of the space inside the roof.

ShowHide solution

Base area $= 3.6 \times 3.6 = 12.96 \text{ m}^{2}$

$V = \frac{1}{3} \times 12.96 \times 1.2 = 5.184 = 5.18$ (3 s.f.)

$5.18 \text{ m}^{3}$

Question 10

A cuboid box has a square base of side $9$ cm and a height of $10$ cm. A hollow square pyramid has the same base and the same height. (a) Find the volume of each. (b) How many pyramids full of sand are needed to fill the box?

ShowHide solution

(a) Box: $9 \times 9 \times 10 = 810 \text{ cm}^{3}\text{.}$ Pyramid: $\frac{1}{3} \times 81 \times 10 = 270 \text{ cm}^{3}\text{.}$

(b) $810 \div 270 = 3$

(a) $810 \text{ cm}^{3}$ and $270 \text{ cm}^{3}$ (b) $3$

Question 11

Mei Ling makes agar-agar in jelly moulds, each a square pyramid (standing on its point when filled) with a base of side $8$ cm and a height of $9$ cm. How many moulds can she fill completely with $1$ litre of agar-agar?

ShowHide solution

One mould: $V = \frac{1}{3} \times 8 \times 8 \times 9 = 192 \text{ cm}^{3}\text{.}$

$1$ litre $= 1000 \text{ cm}^{3}$ , and $1000 \div 192 = 5.208\ldots$

Only whole moulds count as filled, so round down.

$5$ moulds

Question 12

A square pyramid with a height of $12$ cm has the same volume as a cube of edge $6$ cm. Find the length of a side of the pyramid's base.

ShowHide solution

The cube: $6^{3} = 216 \text{ cm}^{3}$ . Let the side of the pyramid's base be $x$ cm.

$\frac{1}{3} \times x^{2} \times 12 = 216\text{,}$ so $4x^{2} = 216$ and $x^{2} = 54$ .

$x = \sqrt{54} = 7.348\ldots = 7.35$ (3 s.f.)

$7.35$ cm

Key Points

A pyramid has a polygon for its base and triangular faces meeting at the apex; it is named after its base.
Volume of a pyramid $= \frac{1}{3} \times$ base area $\times$ height.
The height is the perpendicular height, from the apex straight down to the base; the slant height runs down a face and is not used for the volume.
Three equal pyramids fill a cube: a pyramid holds a third of the prism with the same base and height.
To find a height or a side, put the volume into the formula and solve; round only at the end.