Question

A chord PQ of length 12 cm subtends an angle of 120° at the centre of a circle. Find the area of the minor segment cut off by the chord PQ.

Answer 1

Answer:

The area of the minor segment is 4 {(4π - 3√3)} cm².

Step-by-step explanation:

Given :

Chord of a circle, PQ = 12 cm  

∠POQ = 120°, Then, ∠POL = ∠QOL = 60°

PL = PQ/2

PL = 12/2 = 6 cm

In ∆PLO,

sin θ = Perpendicular /Hypotenuse  = PL/OP

sin 60° = 6/OP  

√3/2 = 6/OP

[sin 60°  = √3/2]

√3 OP = 6×2  

OP = 12/√3

OP = 12 √3 /√3 ×√3

[On rationalising the denominator]

OP = 12√3/3 = 4√3 cm

Radius of the circle, r (OA) = 4√3 cm

Area of the minor segment ,A = {πθ/360 - sin θ /2 cos θ/2 }r²

A = {120°π/360° - sin 120°/2 cos 120°/2 }r²

A = {π/3  - sin 60°cos 60°} (4√3)²

A = {π/3  - √3/2 × 1/2} × 48

A = {π/3  - √3/4 } × 48

A = {(4π - 3√3)/12} × 48

A = {(4π - 3√3)} × 4

Area of the minor segment = 4 {(4π - 3√3)} cm²

Hence, the area of the minor segment is 4 {(4π - 3√3)} cm².

HOPE THIS ANSWER WILL HELP YOU….

Here are more questions of the same chapter :

In the following figure, shows a sector of a circle, centre O, containing an angle  θ°. Prove that:

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AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm and divides the circle into two segments. Find the area of the minor segment.

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Answer 2

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