Chapter 11.6, Problem 29WE

To determine

To determine the area of shaded region.

Answer to Problem 29WE

Area is 25.52.

  $\begin{array}{l}\angle CXY={35}^{\circ }\\ \angle BYX={145}^{\circ }\end{array}$

ABYX is a quadrilateral.

Explanation of Solution

Given information :

  

Radius of the two circles are 6 and 2.

Draw a line parallel to AB.

  

Now we know,

  $\begin{array}{l}AB=2\sqrt{6\times 2}\\ AB=4\sqrt{3}=CY\end{array}$

Area of shaded region is area of ABXY minus area of two arc.

Now we know that,

  $\begin{array}{l}\sin \angle CXY=\frac{4}{4\sqrt{3}}\\ \angle CXY={35}^{\circ }\\ \angle BYX={145}^{\circ }\end{array}$

Area of shaded region $=2\left(4\sqrt{3}\right)+4\left(4\sqrt{3}\right)-\left(\frac{{35}^{\circ }}{{360}^{\circ }}\right)\left(3.14\right){\left(6\right)}^2-\left(\frac{{145}^{\circ }}{{360}^{\circ }}\right)\left(3.14\right){\left(2\right)}^2$

Hence area is 25.52.