A chord of a circle of radius 3.5 cm subtends an angle of 90° at the centre. Find the area of the minor segment of the circle.

Given: Radius of the circle = 3.5 cm and angle subtended by the arc = 90°

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

We know that the area of the minor sector

⇒ Area of ABDC

⇒ Area of ABDC = 9.625cm²

In ∆ABC,

∠A = 90°, AB = AC = 3.5 cm {radius of the circle}

∵ ∆ABC is a right angled triangle.

Area of ∆ABC

⇒ Area of ∆ABC

⇒ Area of ∆ABC = 6.125cm²

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

⇒ Area of the segment BDC = 9.625 – 6.125

⇒ Area of the segment BDC = 3.5cm²