Surface Area of Pyramids, Cones, and Spheres

By the end of this lesson, you will be able to find the lateral and total surface areas of regular pyramids, right cones, and spheres using the slant height and standard formulas.

The Big Idea

Pyramids and cones taper to a point; spheres have no flat faces. Each requires its own surface area formula. The key new quantity for pyramids and cones is the slant height — the distance along the slanted face from apex to base edge — which differs from the height of the solid.

Pyramids

A regular pyramid has a regular polygon base and an apex directly above the center.

Slant height $\ell$: the distance from the apex to the midpoint of a base edge (the height of one triangular face).

$$LA = \frac{1}{2} P \ell$$

$$SA = \frac{1}{2} P \ell + B$$

where $P$ = perimeter of base, $B$ = area of base.

Finding slant height: if height $h$ and apothem $a$ are given: $\ell = \sqrt{h^2 + a^2}$

Cones

A right cone has a circular base with apex directly above the center.

Slant height $\ell$: the distance from apex to any point on the circle's edge: $\ell = \sqrt{h^2 + r^2}$

$$LA = \pi r \ell$$

$$SA = \pi r \ell + \pi r^2 = \pi r(\ell + r)$$

Derivation: unroll the lateral surface into a sector of radius $\ell$ and arc length $2\pi r$.

Spheres

A sphere has no flat faces.

$$SA = 4\pi r^2$$

This equals the area of four great circles — a beautiful result.

Example 1 — Square Pyramid

Regular square pyramid: base side = 6, height = 4. Find lateral area and total surface area.

Apothem of square base: $a = 3$ (half the side)

Slant height: $\ell = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$

$P = 24$, $B = 36$

$LA = \frac{1}{2}(24)(5) = 60$

$SA = 60 + 36 = 96 \text{ sq units}$

Example 2 — Cone

Right cone: radius = 5, height = 12. Find total surface area.

Slant height: $\ell = \sqrt{144 + 25} = \sqrt{169} = 13$

$LA = \pi(5)(13) = 65\pi$

$SA = 65\pi + 25\pi = 90\pi \approx 282.7 \text{ sq units}$