Chapter 8.2, Problem 37WE

To determine

To find: h

Answer to Problem 37WE

  $h=12$

Explanation of Solution

Given:

The figure:

Consider the given below:

m:math display='block'>PQ=x

Let, then

  $QR=21-x$

Consider the right triangle $△SQP$ .

Now, $\overline{SP}$ is the hypotenuse and has length 20 and the lengths of the legs are $SQ=h$ and $PQ=x$ .

So, by theorem 8-2,

  $\begin{array}{c}S{P}^2=S{Q}^2+P{Q}^2\\ {20}^2=h^2+x^2\\ 400=h^2+x^2\end{array}$

That is,

$h^2+x^2=400$ ….. (1)

Consider the right triangle $△SQR$ .

Now, $\overline{SR}$ is the hypotenuse and has length 13 and the lengths of the legs are $SQ=h$ and $QR=21-x$ .

So, by theorem 8-2,

  $\begin{array}{c}S{R}^2=S{Q}^2+Q{R}^2\\ {13}^2=h^2+{\left(21-x\right)}^2\\ 169=h^2+\left({21}^2-2\times 21\times x+x^2\right)\end{array}$

  $\begin{array}{c}169=h^2+441-42x+x^2\\ 0=h^2+x^2-42x+441-169\\ 0=h^2+x^2-42x+272\end{array}$

That is,

$h^2+x^2-42x+272=0$ ….. (2)

Now substituting $h^2+x^2=400$ from equation (1) in the equation (2),

  $\begin{array}{c}400-42x+272=0\\ 672-42x=0\\ -42x=-672\end{array}$

  $\begin{array}{c}x=\frac{-672}{-42}\\ x=16\end{array}$

Substituting $x=16$ in equation (1),

  $\begin{array}{c}{h}^2+{16}^2=400\\ h^2+256=400\\ h^2=400-256\\ h^2=144\end{array}$

Taking square root of both sides,

  $\begin{array}{c}\sqrt{h^2}=\sqrt{144}\\ h=12\end{array}$