# What Can the Pythagorean Theorem be Used For?

Now you’ll learn to apply the Pythagorean theorem. Here are some examples:

Example 1

You have a triangle with a hypotenuse of length $\text{}5\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$ and a leg of length $\text{}3\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. Find the length of the other leg.

Here, you want to know the length of the unknown leg. So, you insert the known information into the equation and solve for the unknown. It doesn’t matter which leg in the formula you use and which one you want to figure out. $\begin{array}{llll}\hfill {3}^{2}+{k}^{2}& ={5}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {k}^{2}& ={5}^{2}-{3}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {k}^{2}& =25-9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \sqrt{{k}^{2}}& =\sqrt{16}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& =4\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The other leg is $4$ cm.

Example 2

A triangle has the side lengths of $\text{}3\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$, $\text{}5\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$ and $\text{}7\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. Is this a right triangle?

Here, you have to check if the left-hand side of the Pythagorean theorem is equal to the right-hand side of the Pythagorean theorem. You can write it out like this: $\begin{array}{llll}\hfill {3}^{2}+{5}^{2}& \stackrel{?}{=}{7}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 9+25& \stackrel{?}{=}49\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 34& \ne 49\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The left-hand side is not equal to the right-hand side, so this triangle is not a right triangle.

Example 3

What are the lengths of the sides of an isosceles right triangle with a leg equal to $\text{}4\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$?

Since this is an isosceles triangle, you know that two sides, the legs, are of equal length. Therefore both legs are $4$ cm. Then only the hypotenuse remains unknown. You put the known lengths into the formula and solve for $h$: $\begin{array}{llll}\hfill {4}^{2}+{4}^{2}& ={h}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 16+16& ={h}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \sqrt{32}& =\sqrt{{h}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 5.66& \approx h\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The legs of the triangle are $4$ cm long. It follows that the hypotenuse is about $5.66$ cm.

Remember that you actually get two answers when you take the square root. You choose not to include the negative answer in this case, since it does not make sense to talk about negative lengths.

Example 4

You have an isosceles right triangle with a hypotenuse of length $\text{}6\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. What is the length of the legs?

Since this is an isosceles triangle, the two legs are equal in length. You set the legs as the unknowns and solve the equation. You then get: $\begin{array}{llll}\hfill {k}^{2}+{k}^{2}& ={6}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2{k}^{2}& =36\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {k}^{2}& =\frac{36}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \sqrt{{k}^{2}}& =\sqrt{18}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& \approx 4.24\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Again, it’s an actual length, and so you choose to only include the positive answer. Thus, each leg measures about $4.24$ cm.