Chapter 12.1, Problem 34SR

To determine

The area of the figure:

  

Answer to Problem 34SR

The area of the figure is $18.24i{n}^2$ .

Explanation of Solution

Given:

  

  $a=b=d=3.2in$

Concept Used:

The area of sqaure is given by the following expression:

  $A=a\times b$

Where,

Sides of sqaure: $a,b$

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 first semi-circle is given by the following expression:

  $A_1=\pi \frac{r_1{}^2}{2}$

  

Put the given values:

  $\begin{array}{l}{A}_1=\pi \frac{r_1{}^2}{2}\\ A_1=\pi \frac{{\left(\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}{2}\\ A_1=\pi \frac{{\left(\raisebox{1ex}{$3.2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}{2}\\ A_1=\pi \frac{2.56}{2}\\ A_1=4.0212i{n}^2\\ A_1\approx 4i{n}^2\end{array}$

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

  $A_2=\pi \frac{r_2{}^2}{2}$

Put the given values:

  $\begin{array}{l}{A}_2=\pi \frac{r_2{}^2}{2}\\ A_2=\pi \frac{{\left(\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}{2}\\ A_2=\pi \frac{{\left(\raisebox{1ex}{$3.2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}{2}\\ A_2=\pi \frac{2.56}{2}\\ A_2=4.0212i{n}^2\\ A_2\approx 4i{n}^2\end{array}$

The area of sqaure is given by the following expression:

  $A=a\times b$

Put the given values:

  $\begin{array}{l}{A}_3=a\times b\\ A_3=3.2\times 3.2\\ A_3=10.24i{n}^2\end{array}$

So, the total area of the figure:

  $\begin{array}{l}A=A_1+A_2+A_3\\ A=4i{n}^2+4i{n}^2+10.24i{n}^2\\ A=18.24i{n}^2\end{array}$

Conclusion:

Hence, the total area of figure is $18.24i{n}^2$