$\triangle$ABC is an equilateral triangle of side 14 cm. A semi circle on BC as diameter is drawn to meet AB at D, and AC at E. Find the area of the shaded region.
$49\left(\dfrac{\pi}{3}-\sqrt{3}\right)cm^{2}$
$49\left(\dfrac{\pi}{3}-\dfrac{\sqrt3}{2}\right)cm^{2}$
$49cm^{2}$
None
Explanation:
O is the centre of the circle and the mid-point of BC.DO is parallel to AC.
So, $\angle$DOB=$60^{o}$
Area of $\triangle$BDO=$\dfrac{49\sqrt{3}}{4}$
Area of sector OBD=$\dfrac{49\pi}{6}$
Hence area of the shaded region
=$2\left(\dfrac{49\pi}{6}-\dfrac{49\sqrt3}{4}\right)cm^{2}$
=$49\left(\dfrac{\pi}{3}-\dfrac{\sqrt3}{2}\right)cm^{2}$
