Question

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.

Answer 1

Let the radius of the circle be r.

Given that, length of chord of a circle, AB = 5 cm

And central angle of the sector AOBA (θ) = 90°

Now, in ΔAOB,

(AB)² = (OA)² + (OB)²   ...[By Pythagoras theorem]

(5)² = r² + r²

⇒ 2r² = 25

∴ r = `5/sqrt(2) "cm"`

Now, in ΔAOB we draw a perpendicular line OD, which meets at D on AB and divides chord AB into two equal parts.

So, AD = DB

= `"AB"/2`

= `5/2 "cm"`   ...[∵ The perpendicular drawn from the centre to the chord of a circle divides the chord into two equal parts]

By Pythagoras theorem, in ΔADO,

OA² = OD² + AD²

⇒ OD² = OA² – AD² 

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

= `25/2 - 25/4`

= `(50 - 25)/4`

= `25/4`

⇒ OD = `5/2 "cm"`

∴ Area of an isosceles ΔAOB

= `1/2 xx "AB" xx "OD"`

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

= `25/4 "cm"^2` 

Now, area of sector AOBA

= `(pi"r"^2)/360^circ xx θ`

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

= `(pi xx 25)/(2 xx 4)`

= `(25pi)/8 "cm"^2`

∴ Area of minor segment

= Area of sector AOBA – Area of an isosceles ΔAOB

= `((25pi)/8 - 25/4) "cm"^2`

Now, area of the circle

= πr²

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

= `(25pi)/2 "cm"^2`

∴ Area of major segment

= Area of circle – Area of minor segment

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

= `(25pi)/8 (4 - 1) + 25/4`

= `((75pi)/8 + 25/4) "cm"^2`

∴ Difference of the areas of two segments of a circle

= Area of major segment – Area of minor segment

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

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

= `(75pi - 25pi)/8 + 50/4`

= `(50pi)/8 + 50/4`

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

Hence, the required difference of the areas of two segments is `((25pi)/4 + 25/2) "cm"^2`.