Chapter 11.3, Problem 13WE

To determine

To Find: The area of trapezoid.

Answer to Problem 13WE

  $A=\frac{27\sqrt{3}}{4}sq.unit$

Explanation of Solution

Given:

  

  

Using trigonometry ratios in triangle $AED$ to find the height

  $\begin{array}{l}\cos 60=\frac{AE}{3}\\ \frac{1}{2}=\frac{AE}{3}\\ 2AE=3\\ AE=1.5\end{array}$

The triangle $AED$ and $BFC$ are congruent.

Thus

Base

  $AE=BF=1.5$

Now

Finding height.

  $\begin{array}{l}\sin 60=\frac{h}{3}\\ \frac{\sqrt{3}}{2}=\frac{h}{3}\\ h=\frac{3\sqrt{3}}{2}\end{array}$

i.e.

  $\begin{array}{l}{b}_1=6\\ b_2=3\\ h=\frac{3\sqrt{3}}{2}\end{array}$

Area of Trapezoid.

  $\begin{array}{l}A=\frac{1}{2}h(b_1+b_2)\\ A=\frac{3\sqrt{3}}{4}(6+3)\\ A=\frac{27\sqrt{3}}{4}sq.unit\end{array}$

Hence, area of the trapezoid is $A=\frac{27\sqrt{3}}{4}sq.unit$ .