Question

In the given figure, an equilateral triangle has been inscribed in a circle of radius 4 cm. Find the area of the shaded region.

Answer 1

Draw $OD\perp BC$.
Because $∆ABC$ is equilateral, $\angle A=\angle B=\angle C=60°$.
Thus, we have:

$$\begin{gathered} \angle OBD=30° \\ \Rightarrow \frac{OD}{OB}=\sin 30° \\ \Rightarrow \frac{OD}{OB}=\frac{1}{2} \\ \Rightarrow OD=\left(\frac{1}{2}\times 4\right)\mathrm{cm}\left[\because OB=\text{radius}\right] \\ \Rightarrow OD=2\mathrm{cm} \end{gathered}$$

$$\begin{gathered} {\mathrm{BD}}^2=\left({\mathrm{OB}}^2-{\mathrm{OD}}^2\right)\left[\mathrm{B}y\mathrm{Pythagoras}\text{'}\mathrm{theorem}\right] \\ \Rightarrow {\mathrm{BD}}^2=\left(4^2-2^2\right){\text{cm}}^2 \\ \Rightarrow {\mathrm{BD}}^2=\left(16-4\right){\text{cm}}^2 \\ \Rightarrow {\mathrm{BD}}^2=12{\text{cm}}^2 \\ \Rightarrow \mathrm{BD}=2\sqrt{3}\text{cm} \end{gathered}$$

Also,
$$\begin{gathered} \mathrm{BC}=2\times \mathrm{BD} \\ =\left(2\times 2\sqrt{3}\right)\text{cm} \\ =4\sqrt{3}\text{cm} \end{gathered}$$
∴ Area of the shaded region = (Area of the circle) $-$ (Area of $∆ABC$)
$$\begin{gathered} =\left|\left(3.14\times 4\times 4\right)-\left(\frac{\sqrt{3}}{4}\times 4\sqrt{3}\times 4\sqrt{3}\right)\right|{\text{cm}}^2 \\ =\left|50.24-\left(12\times 1.73\right)\right|{\text{cm}}^2 \\ =\left(50.24-20.76\right){\text{cm}}^2 \\ =29.48{\text{cm}}^2 \end{gathered}$$