Chapter 9.1, Problem 69HP

To determine

Find the larger circle exceeds the area of the smaller circles.

Answer to Problem 69HP

The larger circle exceeds the area of the smaller circles by $4\pi m^2$

Explanation of Solution

Given information: Given diameters $d=6,4,2$

Calculation:

Area of the circle $A=\pi r^2$

Plug the value $d=6\to 2r=6\to r=3$

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

Plug the value $\begin{array}{l}d=4\\ 2r=4\\ r=2\end{array}$

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

Plug the value $d=2\to 2r=2\to r=1$

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

Find the combine area of the circles with a diameter of 2 and 4.

  $4\pi +\pi =5\pi$

Subtract the area of the smaller circles from the bigger circle to find the difference.

  $9\pi -5\pi =4\pi$

Thus, the larger circle exceeds the area of the smaller circles by $4\pi m^2$