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Question
The area of circle, inscribed in equilateral triangle is 154 cms2. Find the perimeter of
triangle.
Solution
Let circle inscribed in equilateral triangle
Be with centre O and radius ‘r’
Area of circle = 𝜋r2
ut given that area = 154 cm2.
𝜋r2 = 154
`22/7xxr^2 = 154`
𝑟2 = 7 × 7
r = 7cms
Radius of circle = 7cms
From fig. at point M, BC side is tangent at point M, BM ⊥ OM. In equilateral triangle, the perpendicular from vertex divides the side into two halves
BM = `1/2 BC = 1/2 (side =x) = x/2`
ΔBMO is right triangle, by Pythagoras theorem
`OB^2= BM^2+MO^2`
`OB=sqrt(r^2+(x^2/4 ))=sqrt(49+x^2/4)`OD=r
Altitude BD`=sqrt(3)/2(side)=sqrt(3)/2x=OB+OD`
BD – OD = OB
⇒`sqrt(3)/2x-r=sqrt(49+x^2/4`
⇒`sqrt(3)/2x-7=sqrt(49+x^2/4`
⇒`(sqrt(3)/2x-7)^2=(sqrt(x^2/4+49))^2`
⇒`3/4x^2-7sqrt(3x)+49=x^2/4+49`
⇒`x/2=7sqrt(3)⇒x=14sqrt(3)cm`
Perimeter =`3x=3xx14sqrt(3)`
`=42sqrt(3)cms`
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