Chapter 11.6, Problem 13PSB

a.

To determine

To find:Thearea of shaded regionfor different values of x .

Answer to Problem 13PSB

The area of shaded region for x as 6 is $18\pi$ , x as 10 is $50\pi$ andx as 7 is $24.5\pi$ .

Explanation of Solution

Given information:

If x is 6, inner circles are of radius 3 and outer circle is of radius 6.

If x is 10, inner circles are of radius 5 and outer circle is of radius 10.

If x is 7, inner circles are of radius 3.5 and outer circle is of radius 7.

Formula used:

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

r = radius of a circle.

  $A_{shadedarea}=A_{outercircle}-2A_{innercircle}$

Calculation:

  

If x =6,

  $\begin{array}{l}{A}_{outercircle}=\pi r^2\\ $ =\pi {(6)}^2$=36\pi \\ A_{innercircle}=\pi r^2\\ $ =\pi {(3)}^2$=9\pi \\ A_{shadedarea}=A_{outercircle}-2A_{innercircle}\\ A_{shadedarea}=36\pi -18\pi \\ A_{shadedarea}=18\pi \end{array}$

If x =10,

  $\begin{array}{l}{A}_{outercircle}=\pi r^2\\ $ =\pi {(10)}^2$=100\pi \\ A_{innercircle}=\pi r^2\\ $ =\pi {(5)}^2$=25\pi \\ A_{shadedarea}=A_{outercircle}-2A_{innercircle}\\ A_{shadedarea}=100\pi -50\pi \\ A_{shadedarea}=50\pi \end{array}$

If x =7,

  $\begin{array}{l}{A}_{outercircle}=\pi r^2\\ $ =\pi {(7)}^2$=49\pi \\ A_{innercircle}=\pi r^2\\ $ =\pi {(3.5)}^2$=12.25\pi \\ A_{shadedarea}=A_{outercircle}-2A_{innercircle}\\ A_{shadedarea}=49\pi -24.5\pi \\ A_{shadedarea}=24.5\pi \end{array}$

b.

To determine

Tofind: An observation on shaded region area.

Answer to Problem 13PSB

The shaded area is half the entire area.

Explanation of Solution

Given information:

If x is 6, inner circlesare of radius 3 and outer circle is of radius 6.

If x is 10, inner circles are of radius 5 and outer circle is of radius 10.

If x is 7, inner circles are of radius 3.5 and outer circle is of radius 7.

Formula used:

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

r = radius of a circle.

  $A_{shadedarea}=A_{outercircle}-2A_{innercircle}$

Calculation:

  

If x =6,

  $\begin{array}{l}{A}_{outercircle}=\pi r^2\\ $ =\pi {(6)}^2$=36\pi \\ A_{innercircle}=\pi r^2\\ $ =\pi {(3)}^2$=9\pi \\ A_{shadedarea}=A_{outercircle}-2A_{innercircle}\\ A_{shadedarea}=36\pi -18\pi \\ A_{shadedarea}=18\pi \end{array}$

If x =10,

  $\begin{array}{l}{A}_{outercircle}=\pi r^2\\ $ =\pi {(10)}^2$=100\pi \\ A_{innercircle}=\pi r^2\\ $ =\pi {(5)}^2$=25\pi & A_{shadedarea}=A_{outercircle}-2A_{innercircle}\\ A_{shadedarea}=100\pi -50\pi \\ A_{shadedarea}=50\pi \end{array}$

If x =7,

  $\begin{array}{l}{A}_{outercircle}=\pi r^2\\ $ =\pi {(7)}^2$=49\pi \\ A_{innercircle}=\pi r^2\\ $ =\pi {(3.5)}^2$=12.25\pi \\ A_{shadedarea}=A_{outercircle}-2A_{innercircle}\\ A_{shadedarea}=49\pi -24.5\pi \\ A_{shadedarea}=24.5\pi \end{array}$

If x is 6, area of entire circle is $36\pi$ and shaded area is $18\pi$ .

If x is 10, area of entire circle is $100\pi$ and shaded area is $50\pi$ .

If x is 6, area of entire circle is $49\pi$ and shaded area is $24.5\pi$ .