Question

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.

Answer 1

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)`