Question:
If two parallel chords of a circle, having diameter 4 units, lie on the opposite sides of the centre and subtend angles \(\cos^{-1}\left(\frac{1}{7}\right)\) and \(\sec^{-1}(7)\) at the centre respectively, then the distance between these chords, is :
If two parallel chords of a circle, having diameter 4 units, lie on the opposite sides of the centre and subtend angles \(\cos^{-1}\left(\frac{1}{7}\right)\) and \(\sec^{-1}(7)\) at the centre respectively, then the distance between these chords, is :
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Both angles are equal because \(\sec^{-1}7 = \cos^{-1}\frac17\); each chord lies at \(2\cos(\theta/2) = \frac{4}{\sqrt7}\) from the centre, so the separation is twice that.
Updated On: Nov 11, 2026
- \(\frac{4}{\sqrt{7}}\)
- \(\frac{8}{\sqrt{7}}\)
- \(\frac{8}{7}\)
- \(\frac{16}{7}\)
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The Correct Option is B
Solution and Explanation
Radius \(r = 2\). A chord subtending angle \(\theta\) at the centre is at distance \(r\cos\frac{\theta}{2}\) from the centre.
\(\sec^{-1}(7) = \cos^{-1}\left(\frac17\right)\), so both chords subtend the same angle \(\theta\) with \(\cos\theta = \frac17\).
\(\cos^2\frac{\theta}{2} = \frac{1 + 1/7}{2} = \frac47\), so \(\cos\frac{\theta}{2} = \frac{2}{\sqrt7}\).
Distance of each chord from the centre \(= 2 \cdot \frac{2}{\sqrt7} = \frac{4}{\sqrt7}\).
The chords are on opposite sides of the centre, so the distance between them \(= \frac{4}{\sqrt7} + \frac{4}{\sqrt7} = \frac{8}{\sqrt7}\).
Answer: (2) \(\frac{8}{\sqrt{7}}\)
\(\sec^{-1}(7) = \cos^{-1}\left(\frac17\right)\), so both chords subtend the same angle \(\theta\) with \(\cos\theta = \frac17\).
\(\cos^2\frac{\theta}{2} = \frac{1 + 1/7}{2} = \frac47\), so \(\cos\frac{\theta}{2} = \frac{2}{\sqrt7}\).
Distance of each chord from the centre \(= 2 \cdot \frac{2}{\sqrt7} = \frac{4}{\sqrt7}\).
The chords are on opposite sides of the centre, so the distance between them \(= \frac{4}{\sqrt7} + \frac{4}{\sqrt7} = \frac{8}{\sqrt7}\).
Answer: (2) \(\frac{8}{\sqrt{7}}\)
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