Chapter 8.2, Problem 38WE

To determine

To find: h

Answer to Problem 38WE

  $h=24$

Explanation of Solution

Given:

The figure:

Consider the given below:

Let $TU=x$ , then

  $SU=11+x$

Consider the right triangle $△PUT$ .

Now, $\overline{PT}$ is the hypotenuse and has length 25 and the lengths of the legs are $PU=h$ and $TU=x$ .

So, by theorem 8-2,

  $\begin{array}{c}P{T}^2=P{U}^2+T{U}^2\\ {25}^2=h^2+x^2\\ 625=h^2+x^2\end{array}$

That is,

$h^2+x^2=625$ ….. (1)

Consider the right triangle $△PUS$ .

Now, $\overline{PS}$ is the hypotenuse and has length 30 and the lengths of the legs are $PU=h$ and $SU=11+x$ .

So, by theorem 8-2,

  $\begin{array}{c}P{S}^2=P{U}^2+S{U}^2\\ {30}^2=h^2+{\left(11+x\right)}^2\\ 900=h^2+\left({11}^2+2\times 11\times x+x^2\right)\end{array}$

  $\begin{array}{c}900=h^2+121+22x+x^2\\ 0=h^2+x^2+22x+121-900\\ 0=h^2+x^2+22x-779\end{array}$

That is,

$h^2+x^2+22x-779=0$ ….. (2)

Now substituting $h^2+x^2=625$ from equation (1) in the equation (2),

  $\begin{array}{c}625+22x-779=0\\ 22x-154=0\\ 22x=154\end{array}$

  $\begin{array}{c}x=\frac{154}{22}\\ x=7\end{array}$

Substituting $x=7$ in equation (1),

  $\begin{array}{c}{h}^2+7^2=625\\ h^2+49=625\\ h^2=625-49\\ h^2=576\end{array}$

Taking square root of both sides,

  $\begin{array}{c}\sqrt{h^2}=\sqrt{576}\\ h=24\end{array}$