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.

#### Solution

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.

Consequently

`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

OA^{2} = ON^{2} + AN^{2} and OC^{2} = OM^{2} + CM^{2}

⇒ 13^{2} = ON^{2} + 12^{2} and 13^{2} = OM^{2} + 5^{2}

⇒ ON^{2} = 13^{2} - 12^{2} and OM^{2} = 13^{2} - 5^{2}

⇒ ON^{2} = 169 - 144 and OM^{2} = 169 - 25

⇒ ON^{2} = 25 and OM^{2} = 144

⇒ ON = 5 and OM = 12

Now, NM = ON + OM = 5 + 12 = 17cm

Hence, the distance between the two chords is 17 cm.