Question

Area of circle inscribed in an equilateral triangle is 154 cm square. Find the perimeter of triangle.

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

Answer:

72.66

Step-by-step explanation:

Given area of inscribed circle = 154 sq cm

Let the radius of the incircle be r.

⇒ Area of this circle = πr2 = 154

(22/7) × r2 = 154

⇒ r2 = 154 × (7/22) = 49

∴ r = 7 cm

Recall that incentre of a circle is the point of intersection of the angular bisectors.

Given ABC is an equilateral triangle and AD = h be the altitude.

Hence these bisectors are also the altitudes and medians whose point of intersection divides the medians in the ratio 2 : 1

∠ADB = 90° and OD = (1/3) AD

That is r = (h/3)

Þ h = 3r = 3 × 7 = 21 cm

Let each side of the triangle be a, then

Altitude of an equilateral triangle is (√3/2) times its side

That is h = (√3a/2)

∴ a = 14√3 cm

We know that perimeter of an equilateral triangle = 3a

= 3 × 14 √3 = 42√3

= 42 × 1.73 = 72.66 cm