Chapter 11.6, Problem 11PSB

a.

To determine

To find: The area of segment AB with radius as $8$ .

Answer to Problem 11PSB

The area ofsegment ABis $18.265{\text{ units}}^2$ .

Explanation of Solution

Given:

Circle with segment AB and radius as 8.

Formula Used:

  $\begin{array}{l}Are{a}_{(segment)}=Are{a}_{(sector)}-Are{a}_{(triangle)}\\ Are{a}_{(segment)}=\left(\frac{measureofarc}{360^o}\right)\pi r^2-\frac{1}{2}*b \ast h\end{array}$

r = radius of a circle.

b = Base of a triangle.

h = Height of a triangle.

Calculation:

  

  $△ABC$ is $45^{o}45^{o}90^o△$

  $Base=8\sqrt{2},Height=4\sqrt{2}$

  $\begin{array}{l}Are{a}_{(segment)}=Are{a}_{(sector)}-Are{a}_{(triangle)}\\ $ Are{a}_{(segment)}=(\frac{{90}^{\circ }}{360^o})\pi {(8)}^2-(\frac{1}{2}(8\sqrt{2} )(4\sqrt{2}))$Are{a}_{(segment)}=\left(\frac{1}{4}\pi (64)\right)-\left(\frac{1}{2}(32)(2)\right)\\ Are{a}_{(segment)}=16\pi -32\\ Are{a}_{(segment)}=18.265{\text{ units}}^2\end{array}$

b.

To determine

To calculate: The area of segment PQ with measure of arc as ${60}^{\circ }$ .

Answer to Problem 11PSB

The area of segmentPQ is $5.797{\text{ units}}^2$ .

Explanation of Solution

Given:

Circle with segment PQ and measure of arc as ${60}^{\circ }$ .

Formula Used:

  $\begin{array}{l}Are{a}_{(segment)}=Are{a}_{(sector)}-Are{a}_{(triangle)}\\ Are{a}_{(segment)}=\left(\frac{measureofarc}{360^o}\right)\pi r^2-\frac{1}{2}*b \ast h\end{array}$

r = radius of a circle.

b = Base of a triangle.

h = Height of a triangle.

Calculation:

  

ΔPQRis an equilateral triangle.

So, $b=8,h=4\sqrt{3}$

  $\begin{array}{l}Are{a}_{(segment)}=Are{a}_{(sector)}-Are{a}_{(triangle)}\\ $ Are{a}_{(segment)}=(\frac{{60}^{\circ }}{360^o})\pi {(8)}^2-(\frac{1}{2}(8)(4\sqrt{3}))$Are{a}_{(segment)}=(\frac{1}{6})\pi (64)-(\frac{1}{2}\times 18\sqrt{3})\\ Are{a}_{(segment)}=\frac{32}{3}\pi -16\sqrt{3}\\ Are{a}_{(segment)}=5.797{\text{ units}}^2\end{array}$

c.

To determine

To find: The area of segment YZ with radius as 6.

Answer to Problem 11PSB

The area of segment YZ is $22.11{\text{ units}}^2$ .

Explanation of Solution

Given:

Circle with radius as 6 and measure of arc YZas ${120}^{\circ }$ .

Formula Used:

  $\begin{array}{l}Are{a}_{(segment)}=Are{a}_{(sector)}-Are{a}_{(triangle)}\\ Are{a}_{(segment)}=\left(\frac{measureofarc}{360^o}\right)\pi r^2-\frac{1}{2}*b \ast h\end{array}$

r = radius of a circle.

b = Base of a triangle.

h = Height of a triangle.

Calculation:

  

ΔABC is $30^o{60}^{o}90^o△$

  $Base=3\sqrt{3},Height=3$

  $\begin{array}{l}Are{a}_{(segment)}=Are{a}_{(sector)}-Are{a}_{(triangle)}\\ $ Are{a}_{(segment)}=(\frac{{120}^{\circ }}{360^o})\pi {(6)}^2-(\frac{1}{2}(6\sqrt{3} )(3))$Are{a}_{(segment)}=(\frac{1}{3})\pi (36)-(\frac{1}{2}\times 18\sqrt{3})\\ Are{a}_{(segment)}=12\pi -9\sqrt{3}\\ Are{a}_{(segment)}=22.11{\text{ units}}^2\end{array}$