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Question
A chord of a length 16.8 cm is at a distance of 11.2 cm from the centre of a circle . Find the length of the chord of the same circle which is at a distance of 8.4 cm from the centre.
Solution
AF = FB = 8.4 cm
And DE = EC ----(1)
(Perpendicular from centre to a chord bisects the chord)
In right Δ ODA,
By Pythagoras theorem, OA2 = OF2 + AF2
= (11.2)2 + (8.4)2
= 125.44 + 70.56
OA2 = 196
OA = 14 cm
OA = OD = 14cm (radii of same circle)
Similarly, In Δ DEO
OD2 = OE2 + DE2
DE2 = 142 + 8.42
= 196 - 70.56
DE2 = 125.44
DE = 11.2 cm
∴ length of chord DC = 2DE = 2(11.2) = 22.4 cm
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