Chapter 11.6, Problem 18PSC

a.

To determine

To calculate: The area of shaded part which contains sectors and triangle.

Answer to Problem 18PSC

The area of shaded part is $48\sqrt{3}-16\pi$ .

Explanation of Solution

Given information:

A triangle with side as 12 and measure of arc of $120^o$ .

Formula used:

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 a circle.

Calculation:

  

A tangent line is perpendicular to the radius drawn to the point of contact. Therefore angle A and angle B are right angles.

  $\angle CDA\cong \angle BDA=60^o$

  $\begin{array}{l}\Delta ABDand\Delta BODare30°{60}^{\circ }{90}^{\circ }\Delta \text{'}s.\\ AB=12,BD=4\sqrt{3.}\end{array}$

  $\begin{array}{l}{A}_{sector}=\left(\frac{measureofarc}{360^o}\right)\pi r^2\\ $ =(\frac{{120}^{\circ }}{360^o})\pi {(4\sqrt{3})}^2$=\frac{1}{3}\ast 48\pi \\ =16\pi \end{array}$

  $A=\frac{1}{2}\ast b\ast h$

  $\begin{array}{l}=\frac{1}{2}\ast 12\ast 4\sqrt{3}\\ =24\sqrt{3}\\ AreaofABDC=2(Areaof△ABD)\\ AreaofABDC=2\ast 24\sqrt{3}=48\sqrt{3}\end{array}$

  $\begin{array}{l}{A}_{shaded}=A_{ABDC}-A_{\sec tor}\\ A_{shaded}=48\sqrt{3}-16\pi \end{array}$

b.

To determine

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

Answer to Problem 18PSC

The area of shaded part is $18\sqrt{3}+24\pi$ .

Explanation of Solution

Given information:

An equilateral triangle with each side12.

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 a circle.

  $A_{shaded}=2A_{triangle}+2A_{\sec tor}$

Calculation:

  

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

  $A=\frac{{(6)}^2}{4}\sqrt{3}=9\sqrt{3}$

  $\begin{array}{l}{A}_{sector}=\left(\frac{measureofarc}{360^o}\right)\pi r^2\\ $ =(\frac{{120}^{\circ }}{360^o})\pi {(6)}^2$=\frac{1}{3}\ast 36\pi \\ =12\pi \end{array}$

Shaded region is sum of two sector and two equilateral triangles.

  $\begin{array}{l}{A}_{shaded}=2A_{triangle}+2A_{sector}\\ A_{shaded}=(2\ast 9\sqrt{3})+(2\ast 12\pi )\\ A_{shaded}=18\sqrt{3}+24\pi \end{array}$