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Question

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


Solution


Let r be the radius of the circle and AB be the chord, which subtend angle of 90 at centre O. AB = 5 cm in the right angled triangle OAB, using Pythagoras theorem:
OA2+OB2=AB2
r2+r2=52
2r2=25
r=52cm
Therefore area of the circle =
πr2=227×252=39.2857 cm2
The area of the minor segment (shaded area) = area of the sector OAB - area of the triangle OAB area of the sector OAB =
πr2×90360=39.2857×14=9.8214 cm2
Area of triangle OAB =
12×OA×OB=12r2=12×252=254=6.25 cm2

The area of minor segment = 9.8214 - 6.25 = 3.5714 sq cm ……(1)
The area of the major segment = area of the circle - area of the minor segment
= 39.2857 -  3.5714
= 35.7143 sq cm     ….(2)
Difference of the areas of two segments = 35.7143 - 3.5714
= 32.1429 cm2


 

All India Test Series

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