Chapter 11, Problem 41RP

To determine

To calculate:The area of shaded region which contains three semicircles and a circle.

Answer to Problem 41RP

Thearea of shaded regionis $11.25\pi$ .

Explanation of Solution

Given information:

A circle with radius 3.

Formula used:

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

r = radius of circle.

Calculation:

  

Solving for x by using a right triangle.

  $\begin{array}{l}{x}^2+{\left(2x-6+3\right)}^2={\left(x+3\right)}^2\\ x^2+{\left(2x-3\right)}^2=x^2+6x+9\\ x^2+4x^2-12x+9=x^2+6x+9\\ 4x^2-18x=0\\ 2x-9=0\\ x=\frac{9}{2}=4.5\end{array}$

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

Area of a semicircle:

r $=9$

  $\begin{array}{l}A=\frac{1}{2}\pi r^2\\ A=\frac{1}{2}\pi {\left(9\right)}^2=\frac{81\pi }{2}\end{array}$

Area of a medium circle:

r $=4.5$

  $\begin{array}{l}A=\pi r^2\\ A=\pi {\left(4.5\right)}^2=\frac{81\pi }{4}\end{array}$

Area of a smallcircle:

r $=3$

  $\begin{array}{l}A=\pi r^2\\ A=\pi {\left(3\right)}^2=9\pi \end{array}$

Shaded Area =Area of a semicircle -Area of a medium circle -Area of a small circle

Shaded Area $=\frac{81\pi }{2}-\frac{81\pi }{4}-9\pi$

Shaded Area $=\frac{162}{4}\pi -\frac{81}{4}\pi -\frac{36}{4}\pi$

Shaded Area $=\frac{45}{4}\pi =11.25\pi$