Chapter 11.3, Problem 29WE

To determine

To Find: The area of the trapezoid.

Answer to Problem 29WE

  $A=12.5(7-\sqrt{3})sq.unit$

Explanation of Solution

Given:

  

  

Using trigonometric ratios in triangles $WAZ$

  $\begin{array}{l}\sin 30=\frac{h}{10}\\ \frac{1}{2}=\frac{h}{10}\\ h=5\\ \cos 30=\frac{WA}{10}\\ \frac{\sqrt{3}}{2}=\frac{WA}{10}\\ WA=5\sqrt{3}\end{array}$

Now in triangle $XBY$

  $\begin{array}{l}\tan 45=\frac{h}{BX}\\ 1=\frac{5}{BX}\\ BX=5\end{array}$

Now

  $\begin{array}{l}AB=WX-(WA+BX)\\ AB=20(5\sqrt{3}+5)\\ AB=ZY=15-5\sqrt{3}\end{array}$

Now

Area of trapezoid

  $\begin{array}{l}A=\frac{1}{2}h(b_1+b_2)\\ A=\frac{1}{2}(5)(15-5\sqrt{3}+20)\\ A=12.5(7-\sqrt{3})sq.unit\end{array}$

Hence, $A=12.5(7-\sqrt{3})sq.unit$