Question

Find the areas of both the segments of a circle of radius 42 cm with central angle 120°. [Given, sin 120° = `sqrt(3)/2` and `sqrt(3) = 1.73`.]

Answer 1

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