Question

In the given figure, $ABC$ is an equilateral triangle whose side is $2\sqrt{3}cm$. A circle is drawn which passes through the midpoints $D,E$ and $F$ of its sides. The area of the shaded region is

• A. $\frac{1}{4}\left(\begin{array}{c}4\pi -3\sqrt{3}\end{array}\right)c{m}^2$
• B. $\frac{1}{4}\left(\begin{array}{c}2\pi -\sqrt{3}\end{array}\right)c{m}^2$
• C. $\frac{1}{4}\left(\begin{array}{c}\pi -3\sqrt{3}\end{array}\right)c{m}^2$
• D. $\frac{1}{4}\left(\begin{array}{c}3\pi -\sqrt{3}\end{array}\right)c{m}^2$

Answer 1

The correct option is A $\frac{1}{4}\left(\begin{array}{c}4\pi -3\sqrt{3}\end{array}\right)c{m}^2$

$\because △ABC$ is an Equilateral triangle,the radius of the incircle is given by formula:

$r=\frac{s}{2\sqrt{3}}$; where, $s=$side of an equilateral triangle

$\therefore r=\frac{2\sqrt{3}}{2\sqrt{3}}=1$

Area of Incircle $A_c=\pi r^2$

$A_c=\pi c{m}^2$

By midpoint theorem,

sides $DF=FE=DE=\frac{1}{2}(BC=AB=BC)$

Area of triangle DEF is also an equilateral triangle

$A_t=\frac{\sqrt{3}}{4}{s}^2$ ; where $s=$ side of a triangle

$\therefore A_t=\frac{\sqrt{3}}{4}(\frac{2\sqrt{3}}{2}{)}^2$

$\therefore A_t=\frac{3\sqrt{3}}{4}$

Area of shaded region $=$ Area of incircle $-$ Area of Triangle DEF

$\therefore A_s=A_c-A_t$

$\therefore A_s=\pi -\frac{3\sqrt{3}}{4}$

$\therefore A_s=\frac{1}{4}(4\pi -3\sqrt{3})c{m}^2$

Hence, option $A$ is correct.