A chord of a circle of radius 14 cm makes a right angle at the centre.

1.	A chord of a circle of radius 14 cm makes a right angle at the centre. Find the areas of the minor and major segments of the circle.
Answer» Radius of the circle = 14 cm Angle subtend at center = 90° By Pythagoras theorem = AB² = OA² + OB² = 14² +14² AB = \(14\sqrt{2}\) Area of sector OAB = \(\frac{90}{360}\timesπr^2\) = \(\frac{1}{4}πr^2\) = \(\frac{1}{4}\times\frac{22}7\times14\times14\) = 154 cm² Area of triangle AOB = \(\frac{1}2\times14\times14\) = 98 cm² So area of minor segment – OACB =area of sector – area of triangle = 154 – 98 = 56 cm² Area of major segment = area of circle - area of minor segment = \(\frac{22}7\times14\times14\) - 56 = 44 ×14 – 56 = 560 cm²

A chord of a circle of radius 14 cm makes a right angle at the centre. Find the areas of the minor and major segments of the circle.

Answer»

Radius of the circle = 14 cm

Angle subtend at center = 90°

By Pythagoras theorem = AB² = OA² + OB²

= 14² +14²

AB = \(14\sqrt{2}\)

Area of sector OAB = \(\frac{90}{360}\timesπr^2\)

= \(\frac{1}{4}πr^2\)

= \(\frac{1}{4}\times\frac{22}7\times14\times14\) = 154 cm²

Area of triangle AOB = \(\frac{1}2\times14\times14\) = 98 cm²

So area of minor segment – OACB =area of sector – area of triangle

= 154 – 98 = 56 cm²

Area of major segment = area of circle - area of minor segment

= \(\frac{22}7\times14\times14\) - 56

= 44 ×14 – 56 = 560 cm²