Chapter 11.6, Problem 25WE

To determine

To Find: The area of region common to both circles.

Answer to Problem 25WE

  $A=24\pi -18\sqrt{3}c{m}^2$

Explanation of Solution

Given: Two circles of radii $6cm$ and their centers are $6cm$ apart.

  

Here

  $OX=OP=3$

Now taking sector $OAPB$

  $\begin{array}{l}\sin OBX=\frac{3}{6}\\ \sin OBX=0.5\\ m\angle OBX=30\\ and\\ \cos 30=\frac{XB}{6}\\ XB=3\sqrt{3}\\ AX=XB=3\sqrt{3}\\ and\\ AX=XB=\frac{1}{2}AB\\ AB=2AX\\ AB=(2)3\sqrt{3}\\ AB=6\sqrt{3}\end{array}$

Now in triangle $AOB$

  $\begin{array}{l}\angle AOB+\angle OBX+\angle OAX=180\\ \angle AOB+60=180\\ \angle AOB=120\\ so\\ m\stackrel{⌢}{AB}=120\\ let\\ \stackrel{⌢}{AB}=x\end{array}$

Now area of sector.

  $\begin{array}{l}{A}_{AOB}=\frac{x}{360}\pi r^2\\ $ A_{AOB}=\frac{120}{360}\pi {(6)}^2$A_{AOB}=12\pi \end{array}$

Now area of shaded region in $OAPB$ is

  $A=12\pi -9\sqrt{3}c{m}^2$

Similarly in the right circle

Shaded region is also $A=12\pi -9\sqrt{3}c{m}^2$

Thus

Common shaded area

  $\begin{array}{l}A=\left(12\pi -9\sqrt{3}\right)+\left(12\pi -9\sqrt{3}\right)c{m}^2\\ A=24\pi -18\sqrt{3}c{m}^2\end{array}$

Hence, $A=24\pi -18\sqrt{3}c{m}^2$