Question

If AB is a chord of length 5√3 cm of a circle with centre O and radius 5 cm, then area of sector OAB is
(a)$\frac{3\pi}{8}cm^{2}$
(b)$\frac{8\pi}{3}cm^{2}$
(c)25πcm²
(d)$\frac{25\pi}{3}cm^{2}$

Answer 1

Answer:

Area of sector of a circle formed by chord AB is 25π/3 cm² .

Among the given options option (d) 25π/3 cm² is the correct answer.

Step-by-step explanation:

Given :

Radius of circle,(OA,OB),r = 5 cm

Chord of a circle, AB = 5√3 cm

Let ∠AOB = 2θ, Then, ∠AOM = ∠BOM = θ

In ∆AOM,

sin θ = Perpendicular /Hypotenuse  = AM/OA

sin θ = (5√3/2)/5

sin θ = (5√3/2) × 1/5  

sin θ =  √3/2

sin θ = sin 60°

[sin 60°  = √/3/2]

θ = 60°  

∠AOB = 2θ

∠AOB = 2 × 60° = 120°

∠AOB = 120°  

 

Area of the sector of a circle, AOB = (θ/360) × πr²

= (120°/360°) × π × 5²

= 1/3 × π × 25

= 25π/3 cm²

Area of sector of a circle, = 25π/3 cm²

Hence, Area of sector of a circle formed by chord AB is 25π/3 cm² .

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Answer 2

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