Question

a chord of a circle of radius 12cm. subtend an angle of 120°at the center. find the area of the corresponding minor segment of the circle ​

Answer 1

ANSWER

Given that:-

Radius of circle (r)=12cm

θ=120°

To find:-

Area of segment APB=?

Solution:-

Area of sector OAPB(A1 )= 360°/θ ×πr^2

⇒A1 = 360/120 ×3.14×12^2 =150.72cm^2

Let M be the point on AB such that AB⊥OM

∴∠OMA=∠OMB=90°

Now in △OMA and △OMB,

∠OMA=∠OMB[ each 90°]

OA=OB[∵OA=OB=r]

OM=OM[ common ]

By R.H.S. congruency,

△OMA≅△OMB

Now by CPCT,

∠AOM=∠BOM

AM=BM

Therefore,

∠AOM=∠BOM=1/2∠AOB=60°

AM=BM=1/2AB.....(1)

Now in right angled △AOM

sin60= OA/AM

[∵sinθ= hypotenuse/

Perpndicular ]

⇒sin60= 12AM

⇒AM=6 √3 cm

cos60= OA/OM

[∵cosθ=

hypotenuse

base ]

⇒cos60= 12OM

⇒OM=6cm

Now, from eqn

(1), we have

AB=2AM=12 3cm

Now, area of △AOB(A^2 ) will be-

A^2= 1/2×AB×OM= 1/2×12√ 3×6=36√3 cm^2

=62.28cm

∴ Area of segment APB=A^1 −A^2

=150.72−62.28=88.44cm^2

Hence the aea of the corresponding segment of the circle is 88.44cm

2

Answer 2

Answer:

the above answer is correct