A chord of a circle of the radius 10cm subtents of the right angle of the center find the area of the corresponding :i)minor segment ,i)major segment
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SOLUTION :
radius r=10cm
angle =90
area of sector A=90/360pir^2
A=3.14x10x10/4
A=25x3.14
A=78.5sq.cm
Let the angle subtended and radius form an arc AOB,then
area of AOB=rxrsin90/2
AOB=10x10x1/2
ar. of AOB =50sq.cm
area of minor segment =78.5-50
=28.5aq.cm
Then area of circle =pir^2
=3.14x10x10
=314 Sq.cm
area of major segment =area of circle -area of minor segment
=314-28.5
=285.5sq.cm
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