Question

In the figure, three circles of radius 2 cm touch one another externally .These circles are circumscribed by a circle of radius R cm . Find the value of R and the area of the shaded region in the term of $\pi and\sqrt{3}$

Answer 1

Given: $\Delta ABC$ is an equilateral triangle of side $4cm.$

In $\Delta BDO$ we have ,

cos$\angle OBD=\frac{BD}{OB}$

$\Rightarrow cos{30}^{\circ }=\frac{2}{OB}[\because \angle OBD={30}^{\circ }]$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{2}{OB}$

$\Rightarrow OB=\frac{4}{\sqrt{3}}$

Produce OB such that it meets the larger circle at P. OP is the radius of larger circle.
$\therefore$ $OP=OB+BP$

$\Rightarrow R=(\frac{4}{\sqrt{3}}+2)cm$

Area of the shaded region = Area of the larger circle of radius R - 3 $\times$ Area of smaller circle of radius 2 cm + 3(Area of a sector of angle ${60}^{\circ }$ in a circle of radius 2 cm) - [Area of $\Delta ABC$ - 3(Area of sector of angle ${60}^{\circ }$ in a circle of radius 2cm)]

$\Rightarrow$ Area of the shaded region = Area of the larger circle of radius R - 3 $\times$ Area of smaller circle of radius 2 cm + 6 $\times$ Area of a sector angle ${60}^{\circ }$ in a circle of radius of 2 - Area of $\Delta ABC$

$=[\pi {(\frac{4}{\sqrt{3}}+2)}^2-3\times \pi \times 2^2+6\times (\frac{60}{360}\times \pi \times 2^2)-\frac{\sqrt{3}}{4}\times 4^2]c{m}^2$

$=[\pi (\frac{16}{3}+4+\frac{16}{\sqrt{3}})-12\pi +4\pi -4\sqrt{3}]c{m}^2$

$=[\pi (\frac{4}{3}+\frac{16}{\sqrt{3}})-4\sqrt{3}]c{m}^2$

$=\left[\frac{4\pi }{3}\right(4\sqrt{3}+1)-4\sqrt{3}]c{m}^2$