Chapter 9.2, Problem 54PS

Solution

a.

To calculate: The sectors' total regions, double regions, and treble regions.

The sector's total area is approximately $2.19 \pi {\text{in}}^2$ .The double region covers an area of around $0.24 \pi i{n}^2.$ The triple region has a surface area of around $0.14 \pi {\text{in}}^2.$

Given information: Considering that there are $20$ sectors, the size of one sector.

Calculation:

Value of r

  $\begin{array}{c}r=6\frac{5}{8}\\ r=\frac{53}{8}\end{array}$

Value of $r_1$ :

  $\begin{array}{c}{r}_1=r-\frac{3}{8}\\ r_1=\frac{53}{8}-\frac{3}{8}\\ r_1=\frac{50}{8}\end{array}$

Value of $r_2$:

  $\begin{array}{c}{r}_2=r_1-2\frac{1}{8}\\ =r_1-\frac{17}{8}\\ =\frac{50}{8}-\frac{17}{8}\\ r_2=\frac{33}{8}\end{array}$

Value of $r_3$

  $\begin{array}{c}{r}_3=r_2-\frac{3}{8}\\ =\frac{33}{8}-\frac{3}{8}\\ r_3=\frac{30}{8}\end{array}$

The territory of a single sector is divided into $20$ sectors. $A_s$ is $\frac{1}{20}$ of the radius-bound circle's surface area $r=\frac{53}{8}$ .

  $\begin{array}{c}{A}_s=\frac{1}{20}\cdot r^2\pi \\ A_s=\frac{1}{20}{\left(\frac{53}{8}\right)}^2\pi \\ A_s\approx 2.19\pi {\text{in}}^2\end{array}$

The area of double region $A_{{}_{dr}}$ is $\frac{1}{20}$ of the circle's surface area with radius $r=\frac{53}{8}$ minus of the area circle with radius is $r_1=\frac{50}{8}$ .

  $\begin{array}{c}{A}_{dr}=\frac{1}{20}\cdot \left(r^2\pi -r_1^2\pi \right)\\ A_{dr}=\frac{1}{20}\cdot \left(r^2-r_1^2\right)\pi \\ A_{dr}=\frac{1}{20}\cdot \left({\left(\frac{53}{8}\right)}^2-{\left(\frac{50}{8}\right)}^2\right)\pi \\ A_{dr}\approx 0.24\pi {\text{in}}^2\end{array}$

The area of the triple region $A_{tr}\text{}$ is $\frac{1}{20}$ of the area circle with radius $r_2=\frac{33}{8}$ minus the area of circle radius is $r_3=\frac{30}{8}$ .

  $\begin{array}{c}{A}_{tr}=\frac{1}{20}\cdot \left(r_2^2\pi -r_3^2\pi \right)\\ A_{tr}=\frac{1}{20}\cdot \left(r_2^2-r_3^2\right)\pi \\ A_{tr}=\frac{1}{20}\cdot \left({\left(\frac{33}{8}\right)}^2-{\left(\frac{30}{8}\right)}^2\right)\pi \\ A_{tr}\approx 0.14\pi {\mathrm{in}}^2\end{array}$

Conclusion:

The sector's total area is approximately $2.19 \pi {\text{in}}^2$ .The double region covers an area of around $0.24 \pi i{n}^2.$ The triple region has a surface area of around $0.14 \pi {\text{in}}^2.$

b.

To calculate: The likelihood that a geographical mass will be in a double or triple region.

The chance of a double area $P_2\approx 0.1096=10.96\%$ the probability of triple region is $P_3\approx 0.0639=6.39\%$ .

Given information: A dart is thrown at random and lands somewhere inside the sector, with a chance that it will most likely land in the double or triple region.

Calculation:

The ratio between the area of the region $A_{dr}\text{}$ and the area of the complete sector $A_s\text{}$ will represent the likelihood that something will land in the double region.

  $P_2=\frac{A_{dr}}{A_s}$

The ratio is now: Since the area of the double region is approximately $0.24in{}^2$ and the area of the complete sector is approximately $\text{2}{\text{.19 in }}^{\text{2}}$ .

  $\begin{array}{c}{P}_2\approx \frac{0.24\pi i{n}^2}{2.19\pi i{n}^2}\\ P_2\approx 0.1096\\ P_2=10.96\%\end{array}$

The ratio between the area of the sector as a whole and the area of the triple region will represent the likelihood that something will land there.

  $P_3=\frac{A_{tr}}{A_s}$

Given that the total sector covers an area of around $2.19 \pi i{n}^2$ and the triple region's approximate area is $0.14 \pi i{n}^2$ , the ratio.

  $\begin{array}{c}{P}_3\approx \frac{0.14\pi {\text{in}}^2}{2.19\pi {\text{in}}^2}\\ P_3\approx 0.0639\\ P_3=6.39\%\end{array}$

Conclusion:

The chance of a double area $P_2\approx 0.1096=10.96\%$ the probability of triple region is $P_3\approx 0.0639=6.39\%$ .