Q. 144.2( 10 Votes )

The radius of a circle with centre O is 7 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of minor and major segments.

Answer :


Consider the above figure,


From here we can conclude that the portion or the segment below the chord AB is the minor segment and the segment above AB is major segment.


Also we know,


Area of minor segment = Area of sector – Area of ∆AOB eqn1


Now, Area of sector


Where R = radius of the circle and θ = central angle of the sector


Given, R = 7 cm and θ = 90°


Putting these values in the equation 2, we get






Area of sector = 38.5 cm2 eqn3


Area of AOB = 1/2 × base × height



As triangle is isosceles therefore height and base both are 7 cm.


Area of AOB = 1/2×7×7


= 24.5 cm2 eqn4


Putting values of equation 2 and 4 in equation 1 we get


Area of minor segment = 38.5 – 24.5


Area of minor segment = 14 cm2


Area of major segment = πR2 – Area of minor segment eqn5


Put the value of R, and Area of minor segment in equation 5


= π(72) – 14


= 49π - 14



= (22×7) - 14


= 154 - 14


= 140 cm2


Area of minor segment is 14 cm2 and of major segment is 140 cm2.


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