Q.
A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding :
(i) minor segment
(ii) major sector. (Use π = 3.14)
see full answer
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Detailed Solution
We know that, the formula for the area of the sector of a circle
Area of the sector =
Area of the segment = Area of the sector - Area of the triangle

Radius = r = 10 cm, Angle = θ = 90°
In the given figure,
AB is the chord that forms a right angle at the center.
Thus we get the formulas as follows,
Area of minor segment APB = Area of sector OAPB - Area of right triangle AOB
Area of major segment AQB = - Area of minor segment APB
Area of the right triangle ΔAOB = 1/2 OA OB
(i) Area of minor segment APB = Area of sector OAPB - Area of right ΔAOB
= θ/360° × r2 - 1/2 × OA × OB
= 90°/360° × r2 - 1/2 r r
= 1/4 r2 -1/2 r2
= r2 (1/4 - 1/2)
= r2 (3.14 - 2)/4
= (r2 1.14)/4
= (10 10 1.14)/4
= 28.5 cm2
(ii) Area of major sector AOBQ = r2 - Area of minor sector OAPB
= r2 - θ/360° r2
= r2 (1 - 90°/360°)
= 3.14 102 3/4
= 235.5 cm2

