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°

⇒ 2∠ABC = 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