Chapter 11.3, Problem 12WE

To determine

To Find: The area of trapezoid.

Answer to Problem 12WE

  $A=21\sqrt{3}sq.unit$

Explanation of Solution

Given:

  

  

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

  $\begin{array}{l}\sin 60=\frac{AB}{BE}\\ \sin 60=\frac{h}{6}\\ \frac{\sqrt{3}}{2}=\frac{h}{6}\\ 6\sqrt{3}=2h\\ h=3\sqrt{3}\end{array}$

Now finding the base of triangle $AEB$

I.e.

  $\begin{array}{l}\cos 60=\frac{AE}{6}\\ \frac{1}{2}=\frac{AE}{6}\\ 2AE=6\\ AE=3\end{array}$

The triangle $AEB$ and $CFD$ have equal base and height.

Thus

Base

  $\begin{array}{l}EF=10-3-3=4\\ EF=BC=4units\end{array}$

Now

  $\begin{array}{l}{b}_1=4\\ b_2=10\\ h=3\sqrt{3}\end{array}$

Area of Trapezoid.

  $\begin{array}{l}A=\frac{1}{2}h(b_1+b_2)\\ A=\frac{3\sqrt{3}}{2}(4+10)\\ A=\frac{42\sqrt{3}}{2}\\ A=21\sqrt{3}sq.unit\end{array}$

Hence, area of the trapezoid is $A=21\sqrt{3}sq.unit$ .