Multiple choice

What is the area of the circle in which a chord of length 2a makes an angle θ at its centre?

  1. πa2Cot2(θ/2)

  2. 2πa2(1+Cot2(θ/2))

  3. πa2(1+Cot2(θ/2))

  4. 4πa2(1+Cot2(θ/2))

  5. (πa2/4)(1+Cot2(θ/2))
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C Correct answer
Explanation

According to the question, a chord of length 2a makes an angle θ at the centre. So, in the triangle, side opposite to angle θ is 2a. r (r = radius of the circle) In that triangle, applying the Cosine Law, Cos θ = (r2 + r2 - (2a)2)/(2r2) Cos θ = (2r2 - 4a2)/(2r2) Solving, we get r2 = (2a2)/(1-Cosθ)------------(1) Now, Cos θ can be written as Cos 2(θ/2) (Cos 2x = (1 - tan2x)/(1 + tan2x)) Cos 2(θ/2) = (1 - tan2(θ/2))/(1 + tan2(θ/2)) Cos θ = (1 - tan2(θ/2))/(1 + tan2(θ/2)) 1 - Cos θ = 2tan2(θ/2)/(1 + tan2(θ/2)) It can be written as:1 - Cos θ = 2/(1 + Cot2(θ/2)) Divinding numerator and denominator by tan2(θ/2) and substituting the value of 1 - Cos θ in (1), we get r2 = a2(1 + Cot2(θ/2)) Now, area of circle is πr2. So, area = πa2(1+Cot2(θ/2)) (Correct Answer)

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