Question

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. [Use π = 3.14.]

A chord of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding minor segment. [Use π = 3.14]

Answer 1

Let AB be the chord of a circle subtending an angle of 90° at the centre O of the circle.

Area of ​​sector = `theta/(360^circ) x pi^2`

`= (90^circ)/(360^circ) xx 314 /100 xx 10xx10` cm²

= `1/4 xx 314` cm²

`= 157/2` cm²

= 78.5 cm² 

Corresponding minor segment = ΔAOB

= 78.5 cm² − `[1/2 xx10 xx10]`

= 78.5 cm² − 50 cm²

= 28.5 cm² 

Answer 2

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

`=theta/360^circpi("OA")^2 - 1/2xx` OA× OB `

`= 90^circ/360^circxx3.14(10)^2 - 1/2xx10xx10`

= 78.5 − 50

= 28.5 cm²

Hence, the area of minor segment is 28.5 cm²