Question

A chord AB of a circle of radius 14cm makes an angle of 60° at the center of the circle. Find the area of minor segment of the circle...?
{ please answer it..., It's my humble request. }​

Answer 1

GIVEN:

radius of the circle = 14cm

angle of the sector = 60°

RTP:

area of the minor segment

SOLUTION:

in ∆AOB

_________

in ∆AOC

OA = 14cm

AOC = 30°

sin30° = opp/hyp = AC/OA

1/2 = AC/14

AC = 14/2

AC = 7cm

AC = CB

CB = 7cm

base = AC + CB = AB = 14cm

cos30° = adj/hyp = OC/AO

√3/2 = OC/14

OC = 14√3/2

OC = 7√3cm = height

area of ∆AOB = 1/2 × b × h

1/2 × 14 × 7√3

7 × 7√3

49√3cm^2

area of the sector = x/360° × πr^2

60°/360° × 22/7 × 14 × 14

1/6 × 22 × 28

11 × 28/3

308/3cm^2

area of the minor segment = area of sector - area of ∆AOB

308/3 - 49√3

308 - 147/3

161/3

__________

53.6cm^2

__________

therefore the area of minor segment = 53.6cm^2

i think this is the answer.................I hope this helps u