Math Core

A prism keeps the same cross section from bottom to top. A pyramid or cone starts with the same base but shrinks to a single point. That tapering removes exactly two thirds of the volume, which gives one of the most elegant formulas in geometry.

Pyramids and cones

A pyramid has a polygon base and triangular faces that meet at a point called the apex (or vertex). A cone is the same idea with a circular base. The height $h$ is the perpendicular distance from the apex to the plane of the base.

A square pyramid. The height h drops straight down to the center of the base; the slant height ℓ runs down the middle of a face.

Pyramids and cones also have a slant height $\ell$ . For a pyramid, it's the height of a triangular face. For a cone, it's the distance from the apex to the edge of the base. The slant height matters for surface area, which comes later in this unit, but it is not the height used for volume.

Why one third?

Here's a way to see the $\frac{1}{3}$ . Take a cube with edge $s$ . Choose one corner of the cube, and connect it to every vertex of the three faces that don't touch that corner. This cuts the cube into three pyramids. Each has one face of the cube as its base and an edge of the cube as its height, so all three are congruent. Each one is exactly one third of the cube:

V_{\text{pyramid}} = \tfrac{1}{3}s^3 = \tfrac{1}{3}(s^2)(s) = \tfrac{1}{3}Bh

The same one-third relationship holds for every pyramid and every cone. Cavalieri's principle carries it from one special case to all the rest: two pyramids with equal base areas and equal heights have equal cross sections at every level, so they have equal volumes, whether the apex sits over the center of the base or leans far to one side.

Volume of a pyramid or cone

V = \frac{1}{3}Bh

where $B$ is the area of the base and $h$ is the perpendicular height. For a cone, $B = \pi r^2$ , so

V = \frac{1}{3}\pi r^2 h

A pyramid or cone holds one third as much as a prism or cylinder with the same base and height.

Worked example: A rectangular pyramid

A pyramid has a rectangular base that is 9 m by 4 m and a height of 10 m. Find its volume.

V = \tfrac{1}{3}Bh = \tfrac{1}{3}(9 \cdot 4)(10) = \tfrac{1}{3}(360) = 120 \text{ m}^3

When you're given the slant height

The height, the slant height and a segment in the base form a right triangle. For a square pyramid, that base segment runs from the center of the base to the midpoint of a side, so its length is half the base edge. For a cone, it's the radius.

A cone with radius r, height h and slant height ℓ. The height, radius and slant height form a right triangle.

Worked example: A square pyramid from its slant height

A square pyramid has base edges of 16 ft and a slant height of 17 ft. Find its volume.

The right triangle has hypotenuse $\ell = 17$ and one leg equal to half the base edge, $8$ :

h = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \text{ ft}

V = \tfrac{1}{3}(16^2)(15) = \tfrac{1}{3}(256)(15) = 1280 \text{ ft}^3

Common mistake

Don't use the slant height in the volume formula. The height must meet the base at a right angle. In the last example, $\tfrac{1}{3}(256)(17) \approx 1450.7$ is wrong. If a problem gives you $\ell$ , your first step is almost always the Pythagorean theorem.

Worked example: A cone from its slant height

A cone has a radius of 7 cm and a slant height of 25 cm. Find its volume in terms of $\pi$ .

h = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 \text{ cm}

V = \tfrac{1}{3}\pi(7)^2(24) = \tfrac{1}{3}(49)(24)\pi = 392\pi \text{ cm}^3

Frustums

Slice the top off a cone or pyramid with a plane parallel to the base, and the bottom piece is called a frustum. Lampshades, buckets and flowerpots are frustums. You don't need a new formula: find the volume of the whole cone and subtract the small cone that was removed.

A cone 12 in tall with radius 6 in, cut halfway up. The top cone has radius 3 in and height 6 in.

Worked example: A bucket

The figure shows a cone 12 in tall with radius 6 in, cut halfway up. Find the volume of the frustum below the cut.

The small cone on top is 6 in tall. Because it is similar to the whole cone with scale factor $\tfrac{6}{12} = \tfrac{1}{2}$ , its radius is 3 in.

\begin{aligned} V &= \tfrac{1}{3}\pi(6)^2(12) - \tfrac{1}{3}\pi(3)^2(6) \\ &= 144\pi - 18\pi \\ &= 126\pi \text{ in}^3 \end{aligned}

Tip

The small cone was only $\tfrac{18}{144} = \tfrac{1}{8}$ of the whole cone, even though it was half as tall. Scaling every length by $\tfrac{1}{2}$ scales volume by $\left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}$ . Most of a cone's volume sits near its base.

Practice

Practice 1

A square pyramid has base edges of 9 cm and a height of 10 cm. Find its volume in cubic centimeters.