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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 - Mathematics

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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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Chapter 14: Circles - Exercise 14A [Page 274]

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Nootan Mathematics [English] Class 9 ICSE
Chapter 14 Circles
Exercise 14A | Q 5. | Page 274
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