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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


Here all the three angles are 60°        \thereforegiven 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)


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

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


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