Chapter 0.8, Problem 23E

To determine

To calculate: The area of the figure provided below,

  

Answer to Problem 23E

The area of the figure is ${\text{4 cm}}^2$ .

Explanation of Solution

Given information:

The figure is provided below,

  

Formula used:

Area $\left(A\right)$ of rectangle is product of its length $\left(l\right)$ and breadth $\left(b\right)$ . It is mathematically denoted by $A=l\times b$ .

Area $\left(A\right)$ of a semi-circle with radius r is given by $A=\frac{\pi r^2}{2}$ .

Calculation:

Consider the provided figure,

  

The above figure has a rectangle with length as 2.9 cm and width as 1.2 cm with semi-circle on one width with radius 0.6 cm.

Area of the given figure is the sum of area of semi-circle and area of rectangle.

Recall that area $\left(A\right)$ of a semi-circle with radius r is given by $A=\frac{\pi r^2}{2}$

Apply it,

  $\begin{array}{c}A=\frac{\pi r^2}{2}\\ =\frac{3.14\times {0.6}^2}{2}\\ =\frac{3.14\times 0.36}{2}\\ =0.5652\end{array}$

Now, recall that the area $\left(A\right)$ of rectangle is product of its length $\left(l\right)$ and breadth $\left(b\right)$ . It is mathematically denoted by $A=l\times b$ .

  $\begin{array}{c}A=l\times b\\ =2.9\times 1.2\\ =3.48\end{array}$

Now, area of the given figure is calculated as,

  $\begin{array}{c}A=0.5652+3.48\\ =4.0452\end{array}$

Rounding off to the nearest tenth, we get,

  $A=4{\text{cm}}^{\text{2}}$

Thus, the area of the figure is ${\text{4 cm}}^2$ .