Chapter 5.5, Problem 20PSB

To determine

Tocalculate: the length of the diagonal AC.

Answer to Problem 20PSB

  $AC=16$

Explanation of Solution

Given information:

  $ABCD$ is an isosceles trapezoid with upper base $\overline{AD}$ .

Lengths of the Sides, $BE=x+7,\text{}CE=y-3,AE=x+5,BD=y+4$

Formula used:

Properties of isosceles trapezoid $ABCD$ :

• $AD\parallel BC$
• Diagonals $\left(AC\text{ and }BD\right)$ are congruent to each other.
• Diagonals of an isosceles trapezoid are divided into proportional parts.
• The lower base angles are congruent and the Upper base angles are congruent implies that $\Delta AED\text{ and }\Delta BED$ are isosceles triangles.

Calculation:

From isosceles triangle $\Delta AED$ , we have

  $\begin{array}{l}\Rightarrow AE=ED\\ \Rightarrow AE=BD-BE\\ \Rightarrow x+5=\left(y+4\right)-\left(x+7\right)\\ \Rightarrow x+5=y+4-x-7\\ \Rightarrow x+x-y=4-7-5\\ \Rightarrow 2x-y=-8\dots \left(i\right)\end{array}$

  

From isosceles triangle $\Delta BED$ , we have

  $\begin{array}{l}\text{From }\Delta BEC\\ \Rightarrow BE=EC\\ \Rightarrow x+7=y-3\\ \Rightarrow x-y=-7-3\\ \Rightarrow x-y=-10\dots \left(ii\right)\end{array}$

  

Solving $\left(i\right)\text{ \& }\left(ii\right)$ , we get

  $\begin{array}{l}2x-y=-8\\ x-y=-10\\ -++\\ \_\_\_\_\_\_\_\_\_\_\\ x=2\end{array}$

On substituting $x=2$ in $\left(i\right)$ ,

  $\begin{array}{l}2\left(2\right)-y=-8\\ \Rightarrow 4-y=-8\\ \Rightarrow y=4+8\\ \Rightarrow y=12\end{array}$

Now,

  $\begin{array}{c}AC=AE+EC\\ =\left(x+5\right)+\left(y-3\right)\\ =\left(2+5\right)+\left(12-3\right)\\ =7+9\\ =16\end{array}$

Hence, the length of the side of an isosceles trapezoid $\left(AC\right)=16\text{ units}$