Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre.

#### Solution

Let the radius of the circle be r.

OA = OB = r cm

We have,

AB = 5 cm

Central angle of AOBA = 90°

In ΔAOB, AB^{2} = OA^{2} + OB^{2}

5^{2} = r^{2} + r^{2}

r = `5/sqrt(2)` cm

Also, AD = DB = `5/2` cm

Now, In ΔADO,

OA^{2} = OD^{2} + AD^{2}

= `(5/sqrt(2))^2 - (5/2)^2`

= `5/2` cm

Area of isosceles ΔAOB = `1/2` × base × height

= `1/2 x 5 xx 5/2`

= `25/4` cm^{2}

Area of sector AOBA = `(pir^2)/360 xx theta`

= `(pi(5/sqrt(2))^2)/360 xx 90`

= `(25pi)/8`

Area of minor segment = Area of sector AOBA - Area of isosceles ΔAOB

= `((25pi)/8 - 25/4)`

Now, Area of circle = `pir^2`

= `pi(5/sqrt(2))^2`

= `(25pi)/2`

Area of major segment = Area of circle – Area of minor segment

= `(25pi)/2 - ((25pi)/8 - 25/4)`

= `((75pi)/8 + 25/4)`

Now, Difference = Area of major segment Δ Area of minor segment

= `((75pi)/8 + 25/4) - ((25pi)/8 - 25/4)`

= `(25pi)/4 + 25/2 cm^2`