Question

In an equilateral triangle of side $24cm$, a circle is inscribed touching its sides. Find the area of the remaining portion of the triangle [Take $\sqrt{3}=1.732$]

• A. $68.21c{m}^2$
• B. $98.55c{m}^2$
• C. $112.67c{m}^2$
• D. $154.12c{m}^2$

Answer 1

The correct option is B $98.55c{m}^2$

Let ABC be an equilateral triangle of side $24cm$, and let AD be perpendicular from A on BC. Since the triangle is equilateral, so AD bisects BC

$\therefore$ BC=CD$=12$cm

The center of the inscribed circle will coincide with the centroid of $△$ABC.

$\therefore$ OD=$\frac{AD}{3}$

In $△$ABD, we have

$A{B}^2=A{D}^2+B{D}^2$ [Using Phythagoras Theorem]

$\Rightarrow {24}^2=A{D}^2+{12}^2$

$\Rightarrow AD=\sqrt{{24}^2-{12}^2}$

$=\sqrt{(24-12)(24+12)}$

$=\sqrt{36\times 12}$

$=12\sqrt{3}$cm.

$OD=\frac{1}{3}AD=(\frac{1}{3}\times 12\sqrt{3})cm=4\sqrt{3}$cm

Area of the incircle$=\pi (OD{)}^2$

$=\left[\frac{22}{7}\times (4\sqrt{3}{)}^2\right]c{m}^2$

$=\left[\frac{22}{7}\times 48\right]c{m}^2$

$=150.85c{m}^2$

Area of the triangle ABC=$\frac{\sqrt{3}}{4}(side{)}^2$

$=\frac{\sqrt{3}}{4}(24{)}^4=249.4c{m}^2$

$\therefore$ Area of the remaining portion of the triangle

$=(249.4-150.85)c{m}^2=98.55c{m}^2$