Question

A circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region.
[Use $\pi =3.14and\sqrt{3}=1.73$]

Answer 1

It is given that ABC is an equilateral triangle of side 12 cm.

Construction:
Join OA, OB and OC

Draw

$OP\perp BC$

$OQ\perp AC$

$OR\perp AB$

Let the radius of the circle be r cm.

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

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

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

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

⇒ r = $2\sqrt{3}$=2 × 1.73 = 3.46

Therefore the radius of the inscribed circle is 3.46 cm.

Now, area of the shaded region
= Area of ∆ABC – Area of the inscribed circle

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

= $[36\sqrt{3}-12\pi ]c{m}^2$

= [36 ×1.73 – 12 × 3.14] $c{m}^2$

= [62.28 – 37.68] $c{m}^2$

= 24.6 $c{m}^2$

Therefore, the area of the shaded region is 24.6 $c{m}^2$.