Chapter 11, Problem 13RP

a.

To determine

To calculate:Thearea of shaded regionwhich contains sector ABC .

Answer to Problem 13RP

Thearea of shaded region is $9\pi$ .

Explanation of Solution

Given information:

  $\begin{array}{l}m\stackrel{⌢}{AC}={90}^{\circ }\\ r=6.\end{array}$

Formula used:

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

r = radius of circle.

Calculation:

  

  $\begin{array}{l}{A}_{sector}=\left(\frac{measureofarc}{360^o}\right)\pi r^2\\ A_{sector}=\left(\frac{{90}^{\circ }}{360^o}\right)\pi {\left(6\right)}^2\\ A_{sector}=\left(\frac{1}{4}\right)36\pi \\ A_{sector}=9\pi \end{array}$

b.

To determine

To find: The area of shaded figure of an annulus.

Answer to Problem 13RP

The area of shaded figure of an annulus is $16\pi$ .

Explanation of Solution

Given information:

In an annulus, inner circle is of radius 3 and outer circle is of radius 5.

Formula used:

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

r = radius of a circle.

Calculation:

  

  $\begin{array}{l}{A}_{outercircle}=\pi r^2\\ $ =\pi {(5)}^2$=25\pi \\ A_{innercircle}=\pi r^2\\ $ =\pi {(3)}^2$=9\pi \\ A_{shadedarea}=A_{outercircle}-A_{innercircle}\\ A_{shadedarea}=25\pi -9\pi \\ A_{shadedarea}=16\pi \end{array}$

c.

To determine

To calculate: The area of shaded part which contains circle and square.

Answer to Problem 13RP

The area of shaded part is $21.46$ .

Explanation of Solution

Given information:

A square with side as 10 and circle with diameter as 10.

Formula used:

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

r = radius of a circle.

Area of a square: $A=s^2$

s = side

  $A_{shadedpart}=A_{square}-A_{circle}$

Calculation:

  

  $\begin{array}{l}d=10\\ r=\frac{d}{2}=\frac{10}{2}=5\end{array}$

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

  $A=\pi {(5)}^2=25\pi$

Area of a square: $A=s^2$

  $A={(10)}^2=100$

  $\begin{array}{l}{A}_{shadedpart}=A_{square}-A_{circle}\\ A_{shadedpart}=100-25\pi \\ A_{shadedpart}=21.46\end{array}$