AB and CD are two parallel chords of a circle such that AB = 24 cm and CD = 10 cm. If the
radius of the circle is 13 cm. find the distance between the two chords.
Join OA and OC.
Since the perpendicular from the centre of the circle to a chord bisects the chord.
Therefore, N and M are the mid-points of AB and CD respectively.
`AN = NB = 1/2 AB = 1/2 xx 24 = 12` cm and
`CM = MD = 1/2 CD = 1/2 xx 10` = 5 cm
In right-angled triangles ANO and CMO, we have
OA2 = ON2 + AN2 and OC2 = OM2 + CM2
⇒ 132 = ON2 + 122 and 132 = OM2 + 52
⇒ ON2 = 132 - 122 and OM2 = 132 - 52
⇒ ON2 = 169 - 144 and OM2 = 169 - 25
⇒ ON2 = 25 and OM2 = 144
⇒ ON = 5 and OM = 12
Now, NM = ON + OM = 5 + 12 = 17cm
Hence, the distance between the two chords is 17 cm.
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