Chapter 11.3, Problem 31WE

To determine

To Find: The area of the trapezoid.

Answer to Problem 31WE

  $A=156sq.unit$

Explanation of Solution

Given:

  

  $ABCD$ is an isosceles trapezoid.

  

Here

  $\begin{array}{l}\angle A=\angle B=\angle E=\angle F\\ DE=CF\\ AE=FB\\ and\\ EF\parallel DC\\ EF=DC=8\\ now\\ AB=AE+EF+FB\\ AE+FB=10\\ AE=FB=5\end{array}$

Now draw the tangents from the point $AandB$ .

Then

Using tangents to the circle from a point are congruent.

So

  $\begin{array}{l}AQ=AN\\ BM=BN\\ and\\ AQ=BN\\ therefore\\ AQ=BM=AN=BN=9\end{array}$

  $ADandBC$ are congruent

So we can write

  $DQ=CM=DP=CP=4$

Now according to segment addition postulate

  $\begin{array}{l}AD=AQ+DQ\\ AD=9+4\\ AD=13\end{array}$

Now using Pythagoras theorem

  $\begin{array}{l}{13}^2=D{F}^2+5^2\\ DF=12\end{array}$

Area of trapezoid

  $\begin{array}{l}A=\frac{1}{2}h(b_1+b_2)\\ A=\frac{1}{2}(12)(8+18)\\ A=156sq.unit\end{array}$

Hence, the area of the trapezoid is $A=156sq.unit$ .