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. $\frac{1}{4}(4\pi -3\sqrt{3})c{m}^2$
• B. $\frac{1}{4}(2\pi -\sqrt{3})c{m}^2$
• C. $\frac{1}{4}(\pi -3\sqrt{3})c{m}^2$
• D. $\frac{1}{4}(3\pi -\sqrt{3})c{m}^2$

Answer 1

Answer: (A)

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

$\because DF||BC$
$\therefore$ From mid point theorem
$DF=\frac{1}{2}BC=\frac{1}{2}\times 2\sqrt{3}=\sqrt{3}$
$\Delta$DEF is also equilateral triangle
$\Delta =r.s$
$\frac{\sqrt{3}}{4}(2\sqrt{3}{)}^2=r(3)\times \sqrt{3}$
$\Rightarrow \frac{\sqrt{3}}{4}\times \frac{4\times 3}{3\sqrt{3}}=r$
$\therefore =1$
$\therefore$ Area of the shaded portion
$=\pi r^2-\frac{\sqrt{3}}{4}(\sqrt{3}{)}^2=\pi (1{)}^2-\frac{3\sqrt{3}}{4}$
$=\frac{4\pi -3\sqrt{3}}{4}=\frac{1}{4}(4\pi -3\sqrt{3})c{m}^2$