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)

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Detailed Solution

We know that, the formula for the area of the sector of a circle

Area of the sector = θ3600 × πr2

Area of the segment  = Area of the sector - Area of the triangle

 

Chord of circle of radius 10 cm subtends a right angle at centre. Find the  area of the corresponding : (i) minor segment (ii) major sector. (Use pi =  3.14)

 

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 = πr2 - 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° × πr2 - 1/2 × OA × OB

= 90°/360° × πr2 - 1/2 × r × r

= 1/4 πr2 -1/2 r2

= r2 (1/4 π - 1/2)

= r(3.14 - 2)/4

= (r2 × 1.14)/4

= (10 × 10 × 1.14)/4 

= 28.5 cm2

(ii) Area of major sector AOBQ = πr2 - Area of minor sector OAPB

πr2 - θ/360° × πr2

πr2 (1 - 90°/360°)

= 3.14 ×10× 3/4

= 235.5 cm2

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