Chapter 11.4, Problem 10ST1

To determine

To Find: The area of the polygon.

Answer to Problem 10ST1

  $A=49sq.unit$

Explanation of Solution

Given:

  

  

Using Pythagoras theorem in triangle $ABC$

  $\begin{array}{l}{h}^2={\left(5\sqrt{2}\right)}^2+{\left(5\sqrt{2}\right)}^2\\ h^2=50+50\\ h^2=100\\ h=10\end{array}$

Now using Pythagoras in triangle $BCD$

  $\begin{array}{l}{10}^2=x^2+6^2\\ 100=x^2+36\\ x^2=64\\ x=8\end{array}$

Area of quadrilateral = area of triangle $ABC$ + area of triangle $BCD$

Thus

  $\begin{array}{l}A=\frac{1}{2}\left(5\sqrt{2}\right)\left(5\sqrt{2}\right)+\frac{1}{2}(6)(8)\\ A=\frac{1}{2}(50+48)\\ A=\frac{98}{2}\\ A=49sq.unit\end{array}$

Hence, area is $A=49sq.unit$ .