Chapter 8.2, Problem 23WE

To determine

To find: x

Answer to Problem 23WE

  $x=12$

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 $ABFC$ is an isosceles trapezium.

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

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

  $\overline{AD}\parallel \overline{BE}$

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

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

  $DE=AB=11$

Here,

  $\begin{array}{c}CD=\frac{CF-DE}{2}\\ =\frac{21-11}{2}\\ =\frac{10}{2}\\ =5\end{array}$

Consider the right triangle $△ACD$ .

Now, $\overline{AC}$ is the hypotenuse and has length 13 and the lengths of the legs are $AD=x$ and $CD=5$ .

So, by theorem 8-2,

  $\begin{array}{c}A{C}^2=A{D}^2+C{D}^2\\ {13}^2=x^2+5^2\\ 169=x^2+25\end{array}$

  $\begin{array}{c}{x}^2=169-25\\ x^2=144\end{array}$

Now, take square root of both sides,

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