If two parallel chords of a circle,having dia | vedclass

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(B) The diameter of the circle is $4 \, 	ext{units}$,so the radius $r = 2 \, 	ext{units}$.Let the angles subtended by the chords at the centre be $2

Vedclass · Accepted Answer

$\frac{8}{\sqrt{7}}$

Vedclass · Answer

(B) The diameter of the circle is $4 \, 	ext{units}$,so the radius $r = 2 \, 	ext{units}$.Let the angles subtended by the chords at the centre be $2	heta$ and $2\phi$.Given $2	heta = \cos^{-1}(1/7) \Rightarrow \cos(2	heta) = 1/7$.Using the formula $\cos(2	heta) = 2\cos^2	heta - 1$,we have $2\cos^2	heta - 1 = 1/7$ $\Rightarrow 2\cos^2	heta = 8/7$ $\Rightarrow \cos^2	heta = 4/7$ $\Rightarrow \cos	heta = 2/\sqrt{7}$.The distance of the first chord from the centre is $d_1 = r \cos	heta = 2 	imes (2/\sqrt{7}) = 4/\sqrt{7}$.Given $2\phi = \sec^{-1}(7)$ $\Rightarrow \sec(2\phi) = 7$ $\Rightarrow \cos(2\phi) = 1/7$.Using the formula $\cos(2\phi) = 2\cos^2\phi - 1$,we have $2\cos^2\phi - 1 = 1/7$ $\Rightarrow 2\cos^2\phi = 8/7$ $\Rightarrow \cos^2\phi = 4/7$ $\Rightarrow \cos\phi = 2/\sqrt{7}$.The distance of the second chord from the centre is $d_2 = r \cos\phi = 2 	imes (2/\sqrt{7}) = 4/\sqrt{7}$.Since the chords lie on opposite sides of the centre,the total distance between them is $d_1 + d_2 = 4/\sqrt{7} + 4/\sqrt{7} = 8/\sqrt{7}$.

If two parallel chords of a circle,having diameter $4 \, \text{units}$,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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