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Question
In the adjoining figure, AB and CD are two parallel chords and ‘O’ is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of lengths 24 cm and 18 cm, respectively.

Sum
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Solution
Given:
AB = 24 cm, CD = 18 cm, O is the centre, radius = 15 cm.
OM ⟂ AB and ON ⟂ CD perpendiculars from centre to chords.
Step-wise calculation:
1. Half-lengths:
AM = MB = 12 cm
CN = ND = 9 cm
2. In right ΔOMB:
`OM = sqrt(OB^2 - MB^2)`
= `sqrt(15^2 - 12^2)`
= `sqrt(225 - 144)`
= `sqrt(81)`
= 9 cm
3. In right ΔOND:
`ON = sqrt(OD^2 - ND^2)`
= `sqrt(15^2 - 9^2)`
= `sqrt(225 - 81)`
= `sqrt(144)`
= 12 cm
4. The chords lie on opposite sides of O as in the figure.
So, MN = OM + ON ...(If they were on the same side, MN would be |ON – OM| = 3 cm)
= 9 + 12
= 21 cm
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