#### Answer

Perimeter =( 16 + $\frac{8}{3}$$\pi$) in
Area = $\frac{32}{3} \pi in^{2}$

#### Work Step by Step

Perimeter of sector = 2*radius + l(AB)
In a circle whose circumference is c, the length l of an arc whose degree measure is m is given by l =$\frac{m}{360}$ * c
Therefore for given circle c = 2$\pi$ r = 2$\pi$ * 8 = 16$\pi$ in
l =$\frac{60}{360}$ * 16$\pi$ in
=$\frac{1}{6}$ * 16$\pi$ in = $\frac{8}{3}$$\pi$ in
Perimeter of sector = 2* radius + l
= 2* 8 + $\frac{8}{3}$$\pi$ in
= 16 + $\frac{8}{3}$$\pi$ in
In a circle of radius length r , the Area A of the sector whose arc has degree measure m is
A = $\frac{m}{360}$ * $\pi r^{2}$
= $\frac{60}{360}$ * $\pi 8^{2}$
= $\frac{1}{6}$ * 64$\pi $ = $\frac{32}{3} \pi in^{2}$
Therefore the area of the sector = $\frac{32}{3} \pi in^{2}$