Chapter 11.4, Problem 20PPS

To determine

To find: The area of each figure.

Answer to Problem 20PPS

The area of figure is $128.15{\text{m}}^2$ .

Explanation of Solution

Given:

The given figure is:

  

Calculation:

To find the area of the figure, add the area of the rectangle and the area of the triangle and then subtract area of semi-circle from that sum.

Area of rectangle with dimension $13\text{m}\times 10\text{m}$ is calculated as:

  $\begin{array}{c}\text{Area}=\text{length×breadth}\\ =13\text{×}10\\ =130\end{array}$

Area of the semi-circle with radius $5\text{m}$ is calculated as:

  $\begin{array}{c}\text{Area}=\frac{1}{2}\pi r^2\\ \approx \frac{1}{2}\left(3.14{\left(5\right)}^2\right)\\ \approx 39.25\end{array}$

The height of the isosceles triangle is calculated by using Pythagorean Theorem is as:

  $\begin{array}{c}h=\sqrt{9^2-5^2}\\ \approx 7.48\end{array}$

Area of isosceles triangle with base $5\text{m}$ is calculated as:

  $\begin{array}{c}Area=\frac{1}{2}\text{×base×height}\\ \approx \frac{1}{2}\text{×10×7}\text{.48}\\ \approx 37.4\end{array}$

Area of shaded region is calculated as:

  $\begin{array}{c}\text{Area of figure}=\text{Area of retangle}+\text{Area of triangle}-\left(\text{Area of semi-circle}\right)\\ \approx 130+37.4-39.25\\ \approx 128.15\end{array}$

Therefore the area of figure is $128.15{\text{m}}^2$ .