In a circular table cover of radius 32 cm, a design (shade) is formed leaving an equilateral traingle ABC in the middle as shown in the adjacent figure.Find the area of the shaded region.

Explanation

Since ABC is an equilateral triangle.
`:. Angle A = 60^(@)`
`rArr angle BOC = 2xx angle BAC = 2 xx 60^(@) = 120^(@)`
(degree measure of an arc is twice the angle subtended by it in alternate segment)
Draw `OM bot BC`
So, we can prove
`Delta OMB ~= Delta OMC` (R.H.S. congruency)
`:. angle BOM = angle COM = 60^(@)` (c.p.c.t)
In right `Delta OMB`, we have
`sin 60^(@) = (BM)/(OB)`
`rArr (sqrt3)/(2) = (BM)/(32) rArr BM = 16 sqrt3 cm`
`:. BC = 2 xx 16 sqrt3 = 32 sqrt3 cm`
`:.` Area of shaded region = Area of circle `- ar (Delta ABC)`
`= pi r^(2) - (("side")^(2) sqrt3)/(4) = pi xx (32)^(2) - (32 sqrt3)^(2) xx (sqrt3)/(4)`
`= (32)^(2) ((22)/(7) - (3 sqrt3)/(4)) = 1024 ((22)/(7) - (3 sqrt3)/(4)) cm^(2)`