Chapter 11, Problem 23SGR

To determine

To find: The area covered by decoration part.

Answer to Problem 23SGR

The area covered by decoration part is $59.01{\mathrm{inch}}^2$ .

Explanation of Solution

Given Information: Given diameter is $26incha$ and decoration cover $\frac{1}{9}$ of the circle.

Given diameter is $26incha$ and decoration cover $\frac{1}{9}$ of the circle.

Central angle $=\frac{{360}^{\circ }}{9}={40}^{\circ }$ .

Radius $=\frac{26}{2}=13\mathrm{inch}$ .

Area of each sector:

  $\begin{array}{l}=\frac{\theta }{{360}^{\circ }}\pi r^2\\ =\frac{{40}^{\circ }\times 22\times 13\times 13}{{360}^{\circ }\times 7}\\ =59.01{\mathrm{inch}}^2\end{array}$

Thus, the area covered by decoration part is $59.01{\mathrm{inch}}^2$ .