Chapter 11.6, Problem 12PSB

a.

To determine

To find:Thearea of shaded figure of an annulus.

Answer to Problem 12PSB

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$=9\pi \\ A_{shadedarea}=A_{outercircle}-A_{innercircle}\\ A_{shadedarea}=25\pi -9\pi \\ A_{shadedarea}=16\pi \end{array}$

b.

To determine

To calculate: The formula for area of shaded figure in an annulus.

Answer to Problem 12PSB

The area of shaded figure in an annulus is $(R^2\text{- }{r}^2)\pi$ .

Explanation of Solution

Given information:

In an annulus, inner circle is of radius $r$ and outer circle is of radius $R$ .

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 {(R)}^2$A_{innercircle}=\pi r^2\\ =\pi r^2\\ A_{shadedarea}=A_{outercircle}-A_{innercircle}\\ $ A_{shadedarea}=\pi {(R)}^2-\pi {(r)}^2$A_{shadedarea}=\pi (R^2-r^2)\end{array}$