A chord of length 42 cm is drawn in a circle having diameter 58 cm. What is the minimum distance of other parallel chord of length 40 cm in the same circle from 42 cm long chord?

1 cm

We know that,

H² = B² + P²

We have,

The diameter of the circle = 58 cm

The length of one chord = 42 cm

The length of another chord = 40 cm

AB and PQ are two chords, and O is the center of the circle.

M is the midpoint of AB and N is the midpoint of PQ

OB = OQ = 29 cm [radius of the circle]

AB = 42 cm and OB = 29

Then,

AM = MB = \(\frac{42}{2}\) = 21

In ΔMOB

(OB)² = (OM)² + (MB)²

= (29)² = (OM)² + (21)²

= 841 = (OM)² + 441

= OM = 20 cm

Now,

PQ = 40 cm and OB = 29

NQ = PN = \(\frac{40}{2}\) = 20

In ΔONQ

(OQ)² = (ON)² + (NQ)²  = (29)² = (ON)² + (20)²

= 841 = (ON)² + 400

= ON = 21 cm

So according to the question. = 21 – 20 = 1