# In a circle of ra

Given: Radius of the circle = 21 cm and angle subtended by the arc = 60°

(i) We know that the length of the arc

Length of BDC

(ii) We know that the area of the minor sector

Area of ABDC

Area of ABDC = 231cm2

(iii) Area of the segment BDC = area of sector ABDC – area of triangle ABC

In ∆ABC,

A = 60°, AB = AC = 21 cm {radius of the circle}

ABC = ACB {angles opposite to equal sides are equal}

By the angle sum property of the triangle,

ABC + ACB + A = 180°

2ABC = 180° - 60°

ABC = 60°

Hence, ∆ABC is an equilateral triangle.

Area of a equilateral trianglewhere a is the side of it.

Area of ∆ABC

Area of ∆ABC = 190.95cm2

Area of the segment BDC = area of sector ABDC – area of triangle ABC

Area of the segment BDC = 231 – 190.95

Area of the segment BDC = 40.05cm2

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