Question

In the given figure, 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 $\sqrt{3}=1.73and\pi =3.14$]

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,

$OP\perp BC$

$OQ\perp AC$

$OR\perp AB$

Assume the radius of the circle as r cm.

Area of ∆AOB + Area of ∆BOC + Area of ∆AOC = Area of ∆ABC

⇒$(\frac{1}{2}\times AB\times OR)+(\frac{1}{2}\times BC\times OP)+(\frac{1}{2}\times AC\times OQ)=\frac{\sqrt{3}}{4}\times a^2$

⇒ $(\frac{1}{2}\times 12\times r)+(\frac{1}{2}\times 12\times r)+(\frac{1}{2}\times 12\times r)=\frac{\sqrt{3}}{4}\times (12{)}^2$

⇒ $3\times \frac{1}{2}\times 12\times r=\frac{\sqrt{3}}{4}\times 12\times 12$

⇒ $r=2\sqrt{3}=2\times 1.73=3.46$

Hence, the radius of the inscribed circle is 3.46 cm.

Area of the shaded region = Area of ∆ABC − Area of the inscribed circle

$=[\frac{\sqrt{3}}{4}\times (12{)}^2-\pi \left(2\sqrt{3}{)}^2\right]$

$=[36\sqrt{3}-12\pi ]$

$=[36\times 1.73-12\times 3.14]$

$=[62.28-37.68]=24.6cm2$

∴ The area of the shaded region is $24.6c{m}^2$