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# Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

"

Given:

A circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

To do:

We have to find the area of the shaded region.

Solution:

Area of the equilateral triangle $OAB=\frac{\sqrt{3}}{4}(\text { side })^{2}$

$=\frac{\sqrt{3}}{4}(12)^{2}$

$=\frac{\sqrt{3} \times 12 \times 12}{4}$

$=36 \sqrt{3} \mathrm{~cm}^{2}$

$\mathrm{AOB}=60^{\circ}$ (Equilateral triangle)

Exterior angle of $\angle \mathrm{AOB}=360^{\circ}-60^{\circ}$

$=300^{\circ}$

Radius $=6 \mathrm{~cm}$

Sector angle $=300^{\circ}$

Area of the sector $=\frac{\pi r^{2} \theta}{360^{\circ}}$

$=\frac{22}{7} \times \frac{6 \times 6 \times 300^{\circ}}{360^{\circ}}$

$=\frac{22 \times 30}{7}$

$=\frac{660}{7} \mathrm{~cm}^{2}$

Total area of the shaded part $=$ Area of the triangle $+$ Area of the sector

$=36 \sqrt{3}+\frac{660}{7}$

$=(\frac{660}{7}+36 \sqrt{3}) \mathrm{cm}^{2}$

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