(Q9) A chord of a circle of radius 12 cm. subtends an angle of 120° at the center.Find the area of the corresponding minor segment of the circle (use π =3.14 and √3 = 1.732)

PQO12cm120o30o
360
 ×3.14×12×12 = .72

Drop a perpendicular from 'O' to the chord PQ.

△OPM = △OQM [OP = OQ ∠p = ∠Q angles opp. to equal sides OP & OQ; ∠OMP = ∠OMQ by A.A.S]

△OPQ = △OPM+△OQM = 2.△OPM

Area of △OPM =
1
2
 ×PM×OM
But, cos30° =
PM
OP

In △OPQ
√3
2
 =
PM
12

∠POQ = 120°

∠OPQ=∠OQP

=
180° - °
2
=
°
2
 = °
PM =
12×√
2
 = √3

= 18×1.732 = 31.176cm

△OPQ = 2×31.176 = 62.352cm2

Area of the minor segment

PQ = (Area of the sector) - (Area of the △OPQ)

= .72 - .352 = .368 cm2