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Question
Find the areas of both the segments of a circle of radius 42 cm with central angle 120°.
Solution
Area of the minor sector`=120/360xxpixx42xx42`
`=1/3xxpixx42xx42xx42`
`= pixx14xx42`
= 18488 cm2
Area of the triangle `= 1/2 "R"^2sin theta`
Here, R is the measure of the equal sides of the isosceles triangle and θ is the angle enclosed by the equal sides.
Thus, we have;
`1/2xx42xx42xxsin(120^circ)`
= 762.93 cm2
Area of the minor segment = Area of the sector - Area of the triangle
= 1848 - 762.93 = 1085.07 cn
Area of the major segment = Area of the circle - Area of the minor segment
=(π × 42 × 42)- 1085.07
= 5544 - 1085.07
=4458.93 cm2
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