Chapter 8.2, Problem 30WE

To determine

To find: x

Answer to Problem 30WE

  $x=3\sqrt{29}$

Explanation of Solution

Given:

The figure:

Label the given figure as shown below:

Let $AD=a$ , $CD=b$

Consider the right triangle . Now, is the hypotenuse and has length 10 and the lengths of the legs are $BD=8$ and $CD=v$ .

So, by theorem 8-2,

  $\begin{array}{c}B{C}^2=B{D}^2+C{D}^2\\ {10}^2=8^2+b^2\\ 100=64+b^2\end{array}$

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

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{b^2}=\sqrt{36}\\ b=6\end{array}$

Now consider the right triangle $△BDA$ .

Now, $\overline{BA}$ is the hypotenuse and has length 17 and the lengths of the legs are $BD=8$ and $AD=a$ .

So, by theorem 8-2,

  $\begin{array}{c}B{A}^2=B{D}^2+A{D}^2\\ {17}^2=8^2+a^2\\ 289=64+a^2\end{array}$

  $\begin{array}{c}289-64=a^2\\ 225=a^2\end{array}$

Now, take square root of both sides,

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

Now, consider the right triangle $△ADC$ .

Now, $\overline{AC}$ is the hypotenuse and has length x and the lengths of the legs are $CD=6$ and $AD=15$ .

So, by Pythagoras theorem,

  $\begin{array}{c}A{C}^2=C{D}^2+A{D}^2\\ x^2=6^2+{15}^2\\ x^2=36+225\\ x^2=261\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{261}\\ x=3\sqrt{29}\end{array}$