Chapter 11.6, Problem 20PSC

a.

To determine

To calculate: The area of shaded region which contains concentric circles.

Answer to Problem 20PSC

The area of shaded region is $60\pi +36\sqrt{3}$ .

Explanation of Solution

Given information:

A radius of inner circle is 6 and outer circle is 12.

Formula used:

Area of a circle: $A=\pi r^2$

r = radius of circle.

Area of triangle: $A=\frac{1}{2}\ast b\ast h$

  $\begin{array}{l}b=baseoftriangle\\ h=heightoftriangle\end{array}$

Area of sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$

r = radius of circle.

Calculation:

  

  $\begin{array}{l}{A}_{Shaded}=A_{Largecircle}-\left(A_{Smallercircle}+A_{Segment}\right)\\ A_{Segment}=A_{Sector}-A_{\Delta AOC}\\ $ A_{Largercircle}=\pi r^2=\pi {(12)}^2=144\pi $$ A_{Smallercircle}=\pi r^2=\pi {(6)}^2=36\pi $\end{array}$

  $\Delta ABOis30°{60}^{\circ }{90}^{\circ }\Delta .$

  $\begin{array}{l}\therefore AB=6\sqrt{3}andAC=12\sqrt{3}\\ \angle AOC={120}^{\circ }\end{array}$

  $\begin{array}{l}{A}_{segment}=\left(\frac{measureofarc}{360^o}\right)\pi r^2-\frac{1}{2}\ast b\ast h\\ $ =(\frac{120^o}{360^o})\pi {(12)}^2-\frac{1}{2}*12\sqrt{3}*6$=\frac{144\pi }{3}-36\sqrt{3}\\ =48\pi -36\sqrt{3}\end{array}$

  $\begin{array}{l}{A}_{Shaded}=A_{Largecircle}-\left(A_{Smallercircle}+A_{Segment}\right)\\ A_{Shaded}=144\pi -84\pi +36\sqrt{3}\\ A_{Shaded}=60\pi +36\sqrt{3}\end{array}$

b.

To determine

To find: The area of shaded region which contains semicircles.

Answer to Problem 20PSC

The area of shaded region is $27\pi$ .

Explanation of Solution

Given information:

Two semicircles with radius 3 and 6.

Formula used:

Area of a circle: $A=\pi r^2$

r = radius of circle.

  $A_{ANNULUS}=\pi (R^2-r^2)$

R = radius of outer circle.

r = radius of inner circle.

Calculation:

  

  $\begin{array}{l}{A}_{Semicircle}=\frac{1}{2}\pi r^2\\ A_{Shadedregion}=2[\frac{1}{2}\pi (R^2-r^2)]\\ A_{Shadedregion}=\pi (6^2-3^2)\\ A_{Shadedregion}=\pi (36-9)\\ A_{Shadedregion}=27\pi \end{array}$

The shaded figure becomes circle with radius $3\sqrt{3}$ .

c.

To determine

To find: The area of shaded region which contains triangle.

Answer to Problem 20PSC

The area of shaded part is $36\sqrt{3}+18\pi$ .

Explanation of Solution

Given information:

An equilateral triangle with sides as 12.The measure of arc is ${60}^{\circ }$ .

Formula used:

Area of equilateral triangle: $A=\frac{s^2}{4}\sqrt{3}$

s = side of equilateral triangle.

Area of sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$

r = radius of circle.

Calculation:

  

The shaded area is formed by subtracting two sectors from the triangle and then adding five circles in the circle.

  $\begin{array}{l}{A}_{Sector}=\left(\frac{measureofarc}{360^o}\right)\pi r^2\\ $ =(\frac{60^o}{360^o})\pi {(6)}^2$=\frac{1}{6}36\pi \\ =6\pi \end{array}$

  $\begin{array}{l}{A}_{Triangle}=\frac{s^2}{4}\sqrt{3}\\ =\frac{{(12)}^2}{4}\sqrt{3}\\ =\frac{144}{4}\sqrt{3}\\ =36\sqrt{3}\\ A_{Triangle}-2A_{Sector}=36\sqrt{3}-2(6\pi )=36\sqrt{3}-12\pi \end{array}$

  $\begin{array}{l}{A}_{Shaded}=A_{Triangle}-2A_{Sector}+5A_{Sectors}\\ A_{Shaded}=36\sqrt{3}-12\pi +5(6\pi )\\ A_{Shaded}=36\sqrt{3}+18\pi \end{array}$