Chapter 12.2, Problem 41SR

To determine

The area of the figure:

  

Answer to Problem 41SR

The area of the figure is $228m{m}^2$ .

Explanation of Solution

Given:

  

  $a=b=d=16mm$

Concept Used:

The area of the right angle triangle is given by the following expression:

  $A=\frac{1}{2}\left(a\times b\right)$

Where,

Sides of right angle triangle: $a,b$

The area of semi-circle is given by the following expression:

  $A=\pi \frac{r^2}{2}$

Where,

Radius of semi-circle: $r$

Calculation:

The area of the right angle triangle is given by the following expression:

  $A=\frac{1}{2}\left(a\times b\right)$

Put the given values:

  $\begin{array}{l}A=\frac{1}{2}\left(a\times b\right)\\ A=\frac{1}{2}\left(16\times 16\right)\\ A=128m{m}^2\end{array}$

The area of the semi-circle is given by the following expression:

  $A_2=\pi \frac{r^2}{2}$

Put the given values:

  $\begin{array}{l}{A}_2=\pi \frac{r^2}{2}\\ A_2=\pi \frac{{\left(\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}{2}\\ A_2=\pi \frac{{\left(\raisebox{1ex}{$16$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}{2}\\ A_2=\pi \frac{64}{2}\\ A_2=100.53m{m}^2\\ A_2\approx 100m{m}^2\end{array}$

So, the total area of the figure:

  $\begin{array}{l}A=A_1+A_2\\ A=128m{m}^2+100m{m}^2\\ A=228m{m}^2\end{array}$

Conclusion:

Hence, the total area of figure is $228m{m}^2$