Chapter 8.2, Problem 27WE

To determine

To find: x

Answer to Problem 27WE

  $x=17$

Explanation of Solution

Given:

The figure:

Label the given figure as shown below:

  $△ACB$

Now, consider the right triangle.

Now, $\overline{AB}$

  $BC=6$

is the hypotenuse and has length 10 and the lengths of the legs are and

  $AC=a$ .

So, by Pythagoras theorem,

  $\begin{array}{c}A{B}^2=B{C}^2+A{C}^2\\ {10}^2=6^2+a^2\\ 100=36+a^2\end{array}$

  $\begin{array}{c}100-36=a^2\\ 64=a^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{a^2}=\sqrt{64}\\ a=8\end{array}$

Now, consider the right triangle $△ACD$ .

Now, $\overline{AD}$ is the hypotenuse and has length x and the lengths of the legs are $AC=8$ and $CD=15$ .

So, by theorem 8-2,

  $\begin{array}{c}A{D}^2=A{C}^2+C{D}^2\\ x^2=8^2+{15}^2\\ x^2=64+225\\ x^2=289\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{289}\\ x=17\end{array}$