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Question
A chord of length 6 cm is at a distance of 7.2 cm from the centre of a circle. Another chord of the same circle is of length 14.4 cm. Find its distance from the centre.
Solution
AF = FB = 3cm
CE = ED = 7.2cm
(Perpendicular from centre to a chord bisects the chord)
In right Δ AFO, By Pythagoras theorem,
OA2 =OF2+ AF2
OA2 = (7.2)2 + (3)2
OA2 = 51.84 + 9
OA2 = 60.84
OA = 7.8cm
OA = OC = 7.8cm (radii of same circle)
Similarly, In right Δ OFC,
OC2 = OE2 + EC2
OE2 = (7 .8)2 - (7.2)2
= 60.84 - 51.84
OE2 = 9
OE = 3cm
Distance from centre of chord CD with length 14.4cm is 3cm.
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