Question

A circle is inscribed in an equilateral triangle ABC is side 12 cm, touching its sides (Fig). Find the radius of the inscribed circle and the area of the shaded part.

Answer 1

Given that ABC is an equilateral triangle of side 12 cm.
Construction:Join O and A, O and B, and O and C.

P, Q, R are the points on BC, CA and AB respectively then,
OP⊥BC
OQ⊥AC
OR⊥AB
Assume the radius of the circle as r cm.

Area of ∆AOB + Area of ∆BOC + Area of ∆AOC = Area of ∆ABC
⇒( 1/2× AB × OR) + (1/2× BC × OP) + (1/2× AC × OQ) = √3/4× $(side{)}^2$.
⇒ (1/2× 12 × r) + (1/2× 12 × r) + (1/2× 12 × r) = √3/4× $(12{)}^2$
⇒ 3 × 1/2× 12 × r = √3/4× 12 × 12
⇒ r = 2√3 = 2 × 1.73 = 3.46
Hence, the radius of the inscribed circle is 3.46 cm.
Area of the shaded region = Area of ∆ABC − Area of the inscribed circle
= [√3/4 ×$(12{)}^2$ − π($2\surd 3{)}^2$]
= [36√3 − 12π]
= [36 × 1.73 − 12 × 3.14]
= [62.28 − 37.68]
= 24.6 $c{m}^2$
∴ The area of the shaded region is 24.6 $c{m}^2$