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 $\sqrt{3}=1.73, \pi=3.14]$
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°)
$=\frac{\sqrt{3}}{4}(\mathrm{AB})^{2}-\frac{60^{\circ}}{360^{\circ}} \pi(6)^{2}+\frac{300^{\circ}}{360^{\circ}} \pi(6)^{2}$
$=\frac{1.73}{4}(12)^{2}-\frac{1}{6} \times 3.14(6)^{2}+\frac{5}{6} \times 3.14(6)^{2}$
$=62.28-18.84+94.2$
$=137.64 \mathrm{~cm}^{2}$
Hence, the area of shaded region is 137.64 cm2
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