Answer 1

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`