Chapter 8.2, Problem 29WE

To determine

To find: x

Answer to Problem 29WE

  $x=20$

Explanation of Solution

Given:

The figure:

Label the given figure as shown below:

Let $\overline{BE}=a$

Now, consider the right triangle $△BAE$ .

Now, $\overline{BE}$ is the hypotenuse and has length a and the lengths of the legs are $AE=12$ and $AB=9$ .

So, by theorem 8-2,

  $\begin{array}{c}B{E}^2=A{B}^2+A{E}^2\\ a^2=9^2+{12}^2\\ a^2=81+144\\ a^2=225\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{a^2}=\sqrt{225}\\ a=15\end{array}$

Now, consider the right triangle $△BED$ .

Now, $\overline{BD}$ is the hypotenuse and has length 25 and the lengths of the legs are $BE=15$ and $ED=x$ .

So, by Pythagoras theorem,

  $\begin{array}{c}B{D}^2=B{E}^2+E{D}^2\\ {25}^2={15}^2+x^2\\ 625=225+x^2\end{array}$

  $\begin{array}{c}625-225=x^2\\ 400=x^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{400}\\ x=20\end{array}$