Answer
Perimeter = (12 + 4$\pi$) in.
Area = (24$\pi$ - 36$\sqrt 3$)$in^{2}$
Work Step by Step
Find Perimeter of the segment
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
l(AB) = $\frac{60}{360}$ * 2$\pi$r
= $\frac{1}{6}$ * 2$\pi$ * 12
= 4$\pi$ in.
As we know $\triangle$ ABC is an equilateral triangle so AB = 12
Perimeter of segment = AB + l(AB)
= (12 + 4$\pi$) in.
Area of sector = Area of Segment + Area of triangle
Area of segment = Area of sector - Area of triangle
In a circle of radius length r , the Area A of the sector whose arc has degree measure m is
A = $\frac{60}{360}$ * $\pi r^{2}$
= $\frac{60}{360}$ * $\pi 12^{2}$
= $\frac{1}{6} * \pi * 144$
= 24$\pi in^{2}$
Area of Triangle : $12^{2}\frac{\sqrt 3}{4}$
= 36$\sqrt 3 in^{2}$
Area of segment = Area of sector - Area of triangle
= 24$\pi in^{2}$ - 36$\sqrt 3 in^{2}$
=(24$\pi$ - 36$\sqrt 3$)$in^{2}$
