How do you find the area of the intersection of circles?

Question

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Thank you for your help.

Answer 1

A sector that subtends an angle of $60^{\circ}$ has an area equal to $\frac{60^{\circ}}{360^{\circ}} = \frac{1}{6}$ of the area of the whole circle

Area of the sector $PMQ$ = Area of sector $PNQ$ = $\frac{1}{6} \times \pi \times 7^2 = \frac{49}{6}\pi$

Note that triangles $PMQ$ and $PNQ$ are equilateral

Area of triangle $PMQ$ = Area of triangle $PNQ$ = $\frac{\sqrt{3}}{4} \times 7^2 = \frac{49\sqrt{3}}{4}$

Note that the shaded portion is equal to:

$($Sector $PMQ$ minus Triangle $PMQ$$)$ $+$ $($Sector $PNQ$ minus Triangle $PNQ$$)$

Area of shaded portion = $\left(\frac{49}{6}\pi - \frac{49\sqrt{3}}{4}\right) + \left(\frac{49}{6}\pi - \frac{49\sqrt{3}}{4}\right) = 49\left(\frac{\pi}{3}- \frac{\sqrt{3}}{2}\right)$