Chapter 13, Problem 7RP

(a)

To determine

To find: The area of the shaded region.

Answer to Problem 7RP

The area of the shaded region is $50$ .

Explanation of Solution

Given:

The area is shown in Figure 1

  

Figure 1

Calculation:

The formula to determine the slope of the line is of the form,

  $m=\frac{y_2-y_1}{x_2-x_1}$

Then, from the Figure 1 the slope of the points $\left(4,5\right),\left(14,5\right)$ is,

  $\begin{array}{c}m=\frac{5-5}{14-4}\\ =0\end{array}$

The slope of the points $\left(0,0\right),\left(10,0\right)$ is of the form,

  $\begin{array}{c}m=\frac{0-0}{10-0}\\ =0\end{array}$

Thus, the lines between the points $\left(4,5\right),\left(14,5\right)$ and $\left(0,0\right),\left(10,0\right)$ is parallel to each other.

Similarly the line between the points $\left(4,5\right),\left(0,0\right)$ and $\left(14,5\right),\left(10,0\right)$ are also parallel as the slopes are equal.

  $\begin{array}{l}\frac{5-0}{4-0}=\frac{5-0}{14-10}\\ \frac{5}{4}=\frac{5}{4}\end{array}$

Thus, the shaded region is the parallelogram weight height 5 and base 10, then the area is,

  $\begin{array}{c}A=5\cdot 10\\ =50\end{array}$

(b)

To determine

To find: The area of the shaded region.

Answer to Problem 7RP

The area of the shaded region is $27$ .

Explanation of Solution

Given:

The area is shown in Figure 2

  

Figure 2

Calculation:

The Figure 2 shows that the figure is the triangle and the area of the triangle is of the form,

  $A=\frac{bh}{2}$

The height of the triangle is $9-3=6$ and the base is $5-\left(-4\right)=9$ then,

  $\begin{array}{c}A=\frac{9\cdot 6}{2}\\ =27\end{array}$

(c)

To determine

To find: The area of the shaded region.

Answer to Problem 7RP

The area of the shaded region is $18\pi$ .

Explanation of Solution

Given:

The area is shown in Figure 3

  

Figure 3

Calculation:

The Figure 3 shows that the shaded region is the semi-circle with the centre at the origin and the radius is of .

The area of the semi-circle is of the form,

  $\frac{1}{2}\pi r^2$

Then,

  $\frac{1}{2}\pi {\left(6\right)}^2=18\pi$