Question

Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex of an equilateral triangle of side 12 cm as centre and a sector of circle of radius 6 cm with centre B is made. [Use $\sqrt{3}=1.73,\mathrm{\pi }=3.14$]
[CBSE 2014]

Answer 1

In equilateral traingle all the angles are of 60°
∴ ∠ABO = ∠AOB = 60°
Area of the shaded region = (Area of triangle AOB − Area of sector having central angle 60°) + Area of sector having central angle (360° − 60°)
$$\begin{gathered} =\frac{\sqrt{3}}{4}{\left(\mathrm{AB}\right)}^2-\frac{60°}{360°}\mathrm{\pi }{\left(6\right)}^2+\frac{300°}{360°}\mathrm{\pi }{\left(6\right)}^2 \\ =\frac{1.73}{4}{\left(12\right)}^2-\frac{1}{6}\times 3.14{\left(6\right)}^2+\frac{5}{6}\times 3.14{\left(6\right)}^2 \\ =62.28-18.84+94.2 \\ =137.64{\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of shaded region is 137.64 cm²