Chapter 15, Problem 14CR

To determine

To Find: The area of sector formed by the hands of a clock.

Answer to Problem 14CR

  $18.84i{n}^2$

Explanation of Solution

Given:

  

  $\begin{array}{l}\text{Clock is at }2\text{'}O\text{ Clock}\\ \text{Diameter }=12in\end{array}$

Formula Used:

  $\text{Area of sector}=A=\frac{x}{y}\pi r^2$

Calculation:

  $\begin{array}{l}\text{Exterior angle}=180-160={20}^o\\ \text{⇒}{S}_e=360\\ \text{⇒}\frac{360}{20}=18\\ \text{⇒}d=\frac{18\left(18-3\right)}{2}=135\end{array}$

Conclusion:

Thus the area of sector is $18.84i{n}^2$ .