Chapter 11.4, Problem 7CYU

(a)

To determine

To find: The area of the court.

Answer to Problem 7CYU

The area of the blue region is $371{\text{ft}}^2$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the dimension of the given rectangle is $b=10\text{ft}$ and $h=12\text{ft}$ . Also, the width of the semicircle is $d=12\text{ft}$ .

Then, the area of the blue region is,

  $\begin{array}{c}A=bh-\frac{1}{2}\pi r^2\\ =19\left(12\right)-\frac{1}{2}\pi {\left(\frac{12}{2}\right)}^2\\ =228-18\pi \end{array}$

The area of the circle with the blue region is,

  $\begin{array}{c}A=\pi r^2\\ =\pi {\left(\frac{6}{2}\right)}^2\\ =9\pi \end{array}$

Then, the area of the blue region is,

  $\begin{array}{c}A=2\left(228-18\pi \right)+9\pi \\ =371{\text{ft}}^2\end{array}$

(b)

To determine

To find: The area of the court in red.

Answer to Problem 7CYU

The area of the red region is $331{\text{ft}}^2$ .

Explanation of Solution

Calculation:

Each of the one red circle is of the diameter $12\text{ft}$ .

Then, the area is,

  $\begin{array}{c}A=\pi r^2\\ =\pi {\left(\frac{12}{2}\right)}^2\\ =36\pi \end{array}$

One of the red region is he circle with one of the circle excluded.

Then, the area of the circle is,

  $\begin{array}{c}A=\pi {\left(\frac{12}{2}\right)}^2-\pi {\left(\frac{6}{2}\right)}^2\\ =36\pi -9\pi \\ =27\pi \end{array}$

Thus, the area of the red region is,

  $\begin{array}{c}A=2\left(36\pi \right)+27\pi \\ =99\pi \\ =331{\text{ft}}^2\end{array}$