Question

In the given fig. $\theta$ ABC is an equilateral triangle inscribed in a circle of radius 4 cm and centre O. Show that the area of the shaded region is $\frac{4}{3}(4\pi -3\sqrt{3})c{m}^2$. [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

In $\Delta OBD$ Let BD = a, OB = 4cm, $sin{60}^{\circ }=\frac{BD}{OB}$

$a=\frac{4\sqrt{3}}{2}=2\sqrt{3}cm$

$BC=2a=4\sqrt{3}$

$OD=4cos{60}^{\circ }=2cm$

$\therefore$ Area of shaded region = Area of sector OBPC - Area of $\Delta OBC$

$\frac{120}{360}\times \pi \times 4^2-0.5\times 4\sqrt{3}\times 2$

$\frac{4}{3}[4\pi -3\sqrt{3}]c{m}^2$