## Trigonometry (11th Edition) Clone

The area of the region that is cleaned is $75.4~in^2$
We can convert the angle to radians: $\theta = (95^{\circ})(\frac{\pi~rad}{180^{\circ}}) = 1.658~rad$ We can find the total area covered by the 10-inch arm. Let $A$ be the total area of the sector. Then the ratio of the angle $\theta$ to $2\pi$ is equal to the ratio of the sector area to the area of the whole circle. $\frac{\theta}{2\pi} = \frac{A}{\pi ~r^2}$ $A = \frac{\theta ~r^2}{2}$ $A = \frac{(1.658~rad)(10~in)^2}{2}$ $A = 82.9~in^2$ We can find the area of the smaller area $A_2$ that is not cleaned, and subtract this area from the total area $A$. $A_2 = \frac{\theta ~r^2}{2}$ $A_2 = \frac{(1.658~rad)(3~in)^2}{2}$ $A_2 = 7.461~in^2$ We can find the area $A_c$ of the region that is cleaned by the wiper: $A_c = A - A_2$ $A_c = 82.9~in^2 - 7.461~in^2$ $A_c = 75.4~in^2$ The area of the region that is cleaned is $75.4~in^2$