Chapter 11.6, Problem 5PSA

a.

To determine

To calculate: The area of shaded sector ABC with radius as $2$ .

Answer to Problem 5PSA

The area of shaded sector ABC is $\pi$ .

Explanation of Solution

Given:

Circle with shaded region ABCand radius as 2.

Measure of arcAC is ${90}^{\circ }$ .

Formula Used:

Area of Sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$ r = radius of circle.

Calculation:

  

Area of Sector ABC $=(\frac{90^o}{360^o})\pi {(2)}^2$

  $\begin{array}{l}=\frac{1}{4}*4\pi \\ =\pi \end{array}$

b.

To determine

To calculate: The area of shaded sector PQR with diameter as $16$ .

Answer to Problem 5PSA

The area of shaded sectorPQR is $32\pi$ .

Explanation of Solution

Given:

Circle with shaded region PQR and diameter as 16.

Measure of arcPQ is ${180}^{\circ }$ .

Formula Used:

Area of Sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$ r = radius of circle.

Calculation:

  

Area of sector PQR $=(\frac{180^o}{360^o})\pi {(8)}^2$

  $\begin{array}{l}=\frac{1}{2}*64\pi \\ =32\pi \end{array}$

c.

To determine

To find: The area of shaded sector PQR with radius as 4 and measure of arcPR as ${45}^{\circ }$ .

Answer to Problem 5PSA

The area of shaded sectorPQRis $2\pi$ .

Explanation of Solution

Given:

Circle with shaded region PQR and radius as 4.

Measure of arcPR is ${45}^{\circ }$ .

Formula Used:

Area of Sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$ r = radius of a circle.

Calculation:

  

Area of sector PQR $=(\frac{45^o}{360^o})\pi {(4)}^2$

  $\begin{array}{l}=\frac{1}{8}*16\pi \\ =2\pi \end{array}$

d.

To determine

To calculate: The area of shaded sector ABC with radius as 9.

Answer to Problem 5PSA

The area of shaded sector ABCis $27\pi$ .

Explanation of Solution

Given:

Circle with shaded region ABC and radius as 9.

Measure of arcAC is ${120}^{\circ }$ .

Formula Used:

Area of Sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$ r = radius of a circle.

Calculation:

  

Area of sector ABC $=(\frac{120^o}{360^o})\pi {(9)}^2$

  $\begin{array}{l}=\frac{1}{3}*81\pi \\ =27\pi \end{array}$

e.

To determine

To find: The area of shaded sector XYZ with measure of arc XZ as ${270}^{\circ }$ .

Answer to Problem 5PSA

The area of shaded sector XYZ is $75\pi$ .

Explanation of Solution

Given:

Circle with shaded region XYZ and radius as 10.

Measure of arcXZ is ${270}^{\circ }$ .

Formula Used:

Area of Sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$ r = radius of a circle.

Calculation:

  

Area of sector XYZ $=(\frac{270^o}{360^o})\pi {(10)}^2$

  $\begin{array}{l}=\frac{3}{4}*100\pi \\ =75\pi \end{array}$

f.

To determine

To find: The area of shaded sector LMN with radius as $6$ .

Answer to Problem 5PSA

The area of shaded sector LMN is $9\pi$ .

Explanation of Solution

Given:

Circle with shaded region LMN and radius as 6.

Measure of arcLN is ${90}^{\circ }$ .

Formula Used:

Area of Sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$ r = radius of a circle.

Calculation:

Area of sector LMN $=(\frac{90^o}{360^o})\pi {(6)}^2$

  $\begin{array}{l}=\frac{1}{4}*36\pi \\ =9\pi \end{array}$