Question

A Chord of a circle of radius 5cm substends an angle of 60 degree at the centre . Find the area of corresponding major and minor segment.

Answer 1

we know that
ar of minor segment =
$\alpha \div 360 \times \pi r { }^{2} + 1 \div 2 \times r {}^{2} \sin( \alpha )$
where α denotes centre angle,
ATQ
α=60 (minor segment)
using formula
ar of minor segment = 37cm^2
and ar major segment = 42cm^2

Answer 2

Area of minor sector
=(60/360)×(22/7)×5×5
=13.09cm.sq.
Area of circle
=(22/7)×5×5
=78.57cm.sq
Area of major sector
=area of circle -area of minor sector
=78.57-13.09
=65.48cm.sq
Area of major sector=65.48cm.sq
Area of minor sector=13.09cm.sq
Thank you