Question

a chord of circle of radius 12cm subtends an angle of 120° at the centre find the area of the corresponding minor segment of circle​

Answer 1

Radius (r) = 12 cm

θ = 120°

Area of segment APB = Area of sector OAPB - Area of ΔOAB

$Area of sector OAPB = \frac{\theta}{360}*\pi r^2\\\\=\frac{120}{360}*3.14*(12)^2\\ \\=\frac{1}{3}*3.14 *(12)^2\\\\=1*3.14*4*12\\\\= 150.72cm^2\\\\Area of \triangle AOB=\frac{1}{2}*base*height$

Draw OM ⊥ AB.

∵ ∠OMB = ∠OMA = 90°

In ΔOMA and ΔOMB ;

∠OMA =∠OMB { Both 90° }

OA = OB { Both radius }

OM = OM { Common }

∴ ΔOMA ≅ ΔOMB { RHS congruency }

Hence,

∠AOM = ∠BOM (CPCT)

∴ ∠AOM = ∠BOM = 1/2 × 120 = 60°

Also, since ΔOMB ≅ ΔOMA

∴ BM = AM

Also, BM = AM = 1/2 AB  ___(1)

In ΔOMA ;

sin60 = AM/AO

√3/2 = AM / 12

AM = 6√3

In ΔOMA ;

cos60 = OM / AO

1/2 = OM/12

OM = 6

From (1) ;

AB = 2 AM

AB = 2 × 6√3 = 12√3

Now area of Δ AOB = 1/2 × 12√3 × 6

= 62.28 cm²

Area of segment APB = 150.72 - 62.28 = 88.44 cm²

Answer 2

Answer:

hey mate your answer is 88.24