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Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60° (Use $\pi$ = 3.14).

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Area of sector – Area of triangle

Radius = 12 cm

Angle = 60°

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

$\\=\frac{3.14 \times 12 \times 12}{360}\times 60\\\\ =75.36 cm^{2}$

DAOB is isosceles triangles

Let $\angle OAB=OBA=X$
$\\OA=OB=12 cm\\ \angle AOB=60^{0}$

$\angle OAB +\angle OBA+\angle AOB=180^{0}$ { $\because$ Sum of all interior angles of a triangle is 180°}

x + x +60=180

2x =120

x=60

Here all three angles are 60° $\therefore$ given triangle is an equilateral triangle.

Area of $\triangle AOB=\frac{\sqrt{3}}{4}(side)^{2}$ { $\because$ Area of equilateral triangle $=\frac{\sqrt{3}}{4}(side)^{2}$ }

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

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

Area of segment = Area of sector OBCA – Area of DAOB

= $\left (75.36-36\sqrt{3} \right )cm^{2}$

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