If two parallel chords of a circle, having diameter 4 units, lie on...

Given condition,

$∴ \cos 2\theta = \dfrac{1}{7}$ $⇒ 2\cos^{2}\theta - 1 = \dfrac{1}{7}$

$⇒ 2\cos^{2} \theta = \dfrac{8}{7}$

$⇒ cos\theta = \dfrac{2}{\sqrt{7}}$

Then, $C{P}_{1} = rcos\theta = \dfrac{4}{\sqrt{7}}$

Since, $\cos^{- 1}\dfrac{1}{7} = \sec^{- 1}7$, hence both the chords are at equal distance from center of the circle.

Hence, the distance between the chords $= 2 \times \dfrac{4}{\sqrt{7}} = \dfrac{8}{\sqrt{7}} .$