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√3)2]
= [36√3 − 12π]
= [36 × 1.73 − 12 × 3.14]
= [62.28 − 37.68]
= 24.6 cm2
∴ The area of the shaded region is 24.6 cm2
