How to Find Volume and Surface Area of a Pyramid

Here you will learn to find the surface area and volume of a pyramid. A pyramid is a geometric figure that has a polygon as its base and as many side faces as the base has sides. The side faces are all triangles whose tops meet in a point.

Surface Area

Rule

SurfaceAreaofaPyramid

Imagine that the pyramid is unfolded. The surface area is found by adding the area of the base and the areas of all the side faces.

Example 1

Find the surface area of a pyramid where the base is a square with sides of $\text{}5\text{}\phantom{\rule{0.17em}{0ex}}\text{m}$. The height of the side faces of the pyramid is $\text{}6\text{}\phantom{\rule{0.17em}{0ex}}\text{m}$.

First you need to find the area of the base surface:

 ${A}_{b}=5\phantom{\rule{0.17em}{0ex}}\text{m}\cdot 5\phantom{\rule{0.17em}{0ex}}\text{m}=25\phantom{\rule{0.17em}{0ex}}{\text{m}}^{2}$

Then you can find the area of the sides. Each side surface is a triangle with a base equal to $5$ m and a height of $6$ m:

 ${A}_{s}=\frac{5\cdot 6}{2}=15\phantom{\rule{0.17em}{0ex}}{\text{m}}^{2}$

Since the base is a square, that is, a quadrilateral, you have four such triangles. The total surface area then becomes

 $A={A}_{b}+4\cdot {A}_{s}=25\phantom{\rule{0.17em}{0ex}}{\text{m}}^{2}+4\cdot 15\phantom{\rule{0.17em}{0ex}}{\text{m}}^{2}=85\phantom{\rule{0.17em}{0ex}}{\text{m}}^{2}.$

Volume

Formula

VolumeofaPyramid

 $V=\frac{B\cdot h}{3},$

where $B$ is the area of the base, and $h$ is the height of the pyramid.

Example 2

What is the volume of a pyramid that has a triangular base, where the triangle has a base of $\text{}4\text{}\phantom{\rule{0.17em}{0ex}}\text{m}$ and a height of $\text{}3\text{}\phantom{\rule{0.17em}{0ex}}\text{m}$? The height of the pyramid is $\text{}5\text{}\phantom{\rule{0.17em}{0ex}}\text{m}$.

First you need to calculate the area of the base: $\begin{array}{llll}\hfill B& =\frac{4\cdot 3}{2}=6\phantom{\rule{0.17em}{0ex}}{\text{m}}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill V& =\frac{6\cdot 5}{3}=10\phantom{\rule{0.17em}{0ex}}{\text{m}}^{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$