Chapter 8.2, Problem 24WE

To determine

To find: x

Answer to Problem 24WE

  $x=28$

Explanation of Solution

Given:

The figure:

Concept Used:

Theorem 3-7: perpendicular lines on same line are parallel to each other.

Theorem 5-5: A quadrilateral is parallelogram if pair of opposite sides is equal.

Theorem 5-1: Parallelogram’s opposite sides are equal.

Theorem 8-2: In right triangle, square of hypogenous is equal to the square of base and perpendicular.

Label the given figure as shown below:

The quadrilateral $ADEF$ is an isosceles trapezium.

It can be observed that the sides $\overline{BF}$ and $\overline{CE}$ perpendicular to the line $\overline{AD}$

So, by Theorem 3-7, it can be said that,

  $\overline{BF}\parallel \overline{CE}$

Now for the quadrilateral $BCEF$ , the sides $\overline{BF}$ and $\overline{CE}$ are both congruent and parallel. So, by theorem 5-5, $BCEF$ is a parallelogram.

By theorem 5-1, opposite sides of a parallelogram are congruent. So,

  $BC=EF=10$

Here,

  $\begin{array}{c}AB=\frac{AD-BC}{2}\\ =\frac{x-10}{2}\end{array}$

Consider the right triangle $△ABF$ .

Now, $\overline{AF}$ is the hypotenuse and has length 15 and the lengths of the legs are $AB=\frac{x-10}{2}$ and $BF=12$ .

So, by theorem 8-2,

  $\begin{array}{c}A{F}^2=A{B}^2+B{F}^2\\ {15}^2={\left(\frac{x-10}{2}\right)}^2+{12}^2\\ 225={\left(\frac{x-10}{2}\right)}^2+144\end{array}$

  $\begin{array}{c}{\left(\frac{x-10}{2}\right)}^2=225-144\\ {\left(\frac{x-10}{2}\right)}^2=81\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{{\left(\frac{x-10}{2}\right)}^2}=\sqrt{81}\\ \frac{x-10}{2}=9\\ x-10=9\times 2\end{array}$

  $\begin{array}{c}x-10=18\\ x=18+10\\ x=28\end{array}$