Question

Two parallel chords are drawn on the same side of the centre of a circle of radius 20 . It is found that they subtend 60° and 120° angles at the centre , then what is the perpendicular distance between the chords

Answer 1

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Answer 2

The perpendicular distance between the chords is 7.32

1) As in the diagram,

AB and DE are two parallel chords.

CM and CN are two perpendiculars drawn to AB and DE respectively.

2) Given,

angle ACB = 60°

angle DCE = 120°

3) As we know that the perpendicular to any chord in the circle divides the central angle in to two equal angles,

Therefore,

angle ACM = angle MCB = 30°

angle DCN = angle NCE = 60°

3) In triangle ACM,

AC = 20, angle ACM = 30°

Cos 30° = (adjacent side)/(hypotenuse)

Cos 30° = CM/AC

0.866 = CM/20

CM = 20×0.866

CM = 17.32

4) Similarly,

In triangle DCN,

angle DCN =60°, DC = 20

Cos 60° = (adjacent side)/(hypotenuse)

Cos 60° = CN/DC

0.5 = CN/20

CN = 20×0.5

CN = 10

4) The perpendicular distance between the cords is,

CM - CN = 17.32 - 10 = 7.32

5) Therefore the perpendicular distance between the two chords is 7.32