Elementary Geometry for College Students (7th Edition)

A = $\frac{100 \pi}{9} - 25 \frac{\sqrt{3}}{3}$ $P = =10 + \frac{20\pi}{3\sqrt3}$
We first find the radius: $\frac{r}{5} = \frac{2}{\sqrt3} \\ r = \frac{10}{\sqrt3}$ We know that 1/3 of the circumference plus AB is the perimeter. Thus: $P = 10 + \frac{2\pi(10)}{3\sqrt3} =10 + \frac{20\pi}{3\sqrt3}$ We find the area. To do so, we first must find the distance between the center of the circle and the line segment (the height of the triangle): $h = \sqrt{( \frac{10}{\sqrt3})^2 - 5^2} = \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}}$ Thus, we subtract the area of the triangle from 1/3 of the area of the circle: $A = \pi(1/3) (\frac{10}{\sqrt3})^2 - 1/2(10)(\frac{5}{\sqrt{3}}) = \frac{100 \pi}{9} - 25 \frac{\sqrt{3}}{3}$