In Fig. 3, a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of the inscribed circle and the area of the shaded region. [Use π = 3.14 and √3 = 1.73]
Construction: Join OA, OB, and OC. And draw OE ⊥ BC, OD ⊥ AB, OF ⊥ AC.
We know that the area of equilateral triangle = (√3 a2)/4
And the area of a triangle = bh/2.
Given that, side of ΔABC = 12 cm.
Let the radius of the circle be r cm.
Ar(ΔABC) = ar(ΔAOB) + ar(ΔBOC) + ar(ΔAOC)
⇒ (√3 (12)2)/4 = (AB)r / 2 + (BC)r / 2 + (AC)r / 2
⇒ 36√3 = r(AB + BC + AC) / 2
⇒ 72√3 = r(12 + 12 + 12)
⇒ 72√3 = 36r
⇒ r = 2√3 cm
⇒ r = 2(1.73) = 3.46 cm
We know that the area of circle = πr2
Area of shaded region = area of ΔABC – area of circle
= (√3 (12)2)/4 - 3.14(3.46)2
= 36√3 - 3.14(2√3)2
= 36√3 - 3.14(12)2
= 62.28 – 37.68
= 24.6 cm2
Ans. The radius of inscribed circle is 3.46 cm and the area of shaded region is 24.6 cm2.