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 length 24 cm and 18 cm respectively.

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Solution

In the figure, chords AB ∥ CD
O is the centre of the circle
Radius of the Circle = 15 cm
Length of AB = 24 cm and CD = 18 cm
Join OA and OC

AB = 24 cm and OM ⊥ AB
AM = MB= $\frac{24}{2}$ =12 cm
Similarly ON ⊥ CD
CN = ND = $\frac{18}{2}$ = 9 cm
Now, In right ∆ CMO
OA² = AM² + OM²
(15)² = (12)² + OM²
225 = 144 + OM²
OM²= 225 – 144 = 81 = (9)²
OM = 9 cm

Similarly, In right ∆ CNO
OC² = CN² + ON²
(15)² = (9)² + ON²
225 = 81 + ON²
ON²= 225 – 81 = 144 = (12)²
ON = 12 cm
Now MN = OM + ON = 9 + 12 = 21 cm

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