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

We use the concept of areas of circle and equilateral triangle in the problem.
Mark O as center of the circle. Join BO and CO.
Since we know that equal chords of a circle subtend equal angles at the center and all sides of an equilateral triangle are equal,
Each side of `triangle ABC` will subtend equal angles at the center.
`angleBOC = (360^@)/3 = 120^@`
Consider `triangleBOC`. Drop a perpendicular from `OM` to `BC`
We know perpendicular from the center of circle to a chord bisects it.
`BM = MC`
`OB = OC` (radii)
`OM = OM` (common)
`triangleOBM cong triangleOCM` (by SSS congruency)
`angleBOM = angleCOM` (by CPCT)
`2angleBOM = angleBOC = 120^@`
`angleBOM = (120^@)/2 = 60^@`
`sin 60^@ = (BM)/(BO) = sqrt3/2`
`BM = (sqrt3)/2 × BO = sqrt3/2 × 32 = 16sqrt3`
` BC = 2BM = 32sqrt3`
Using the formula of area of equilateral triangle = `sqrt3/4 (side)^2`
We can find the area of `triangleABC` since a side `BC` of `triangleABC` is known.
Visually from the figure, it’s clear that
Area of the design = Area of circle - Area of `triangleABC` = `pir^2 - (sqrt3)/4 (BC)^2`
This can be solved with ease as both the radius of the circle and BC are known.
Radius of circle (r) = `32 cm`
From figure, we observe area of design = Area of circle - Area of `triangleABC`
= `pir^2 - (sqrt3)/4 (BC)^2`
= `22/7 × (32)^2 - sqrt3/4 × (32sqrt3)^2`
= `22/7 × 1024 - (sqrt3)/4 × 1024 × 3`
= `22528/7 - 768sqrt3`
Area of design = `(22528/7 - 768sqrt3) cm²`