Question

the area of a circle inscribed in an equilateral triangle is 154. find the perimeter of the triangle
(pi=22/7   root3= 1.73) 

Answer 1

Area of the circle = 154 sq cm
⇒$\pi r^2 = 154$
⇒$\frac{22}{7} r^2 = 154$
 ⇒ $r^2 = 49$
⇒r = 7 cm

Let the side of the triangle = a cm
So, s = $\frac{3a}{2}$

But the radius of the incircle, $r = \frac{\Delta}{s}$  Where Δ = Area of the triangle and s = semi-perimeter

$7 = \frac{ \frac{ \sqrt{3} }{4} a^2 }{ \frac{3a}{2} }$

$7 = \frac{ \sqrt{3}a^2 }{2*3a}$

$7 = \frac{ \sqrt{3}a}{6}$

$a = \frac{42}{ \sqrt{3} }$

$a = \frac{42 \sqrt{3}}{3 } = 14 \sqrt{3}$

Perimeter of triangle, $3a = 3*14 \sqrt{3} = 42 \sqrt{3}$

Perimeter of triangle = 42(1.73) = 72.66 cm

Answer 2

The area of the circle=154 cm²
therefore, πr²=154 cm²
or,22r²/7=154 cm²
or,r²=154 cm²*7/22=49 cm²
or,r=7 cm.
Let the side of the triangle be x cm.
Therefore, semi perimeter=3x/2
Radius of the circle=Area/Semi Perimeter of Δ.
or,7=(√3/4*x²)/(3x/2)
or,7=(√3*x)(3*2)
or,7=(√3*x)/6
or,7*6=√3*x
or,42=√3*x
or,42/√3=x
or,x=42/√3
Perimeter=42/√3*3=42√3=42*1.73=72.66 cm.