A chord of length 40 cm is drawn in a circle having diameter 50 cm. What is the minimum distance of other parallel chord of length 30 cm in the same circle from 40 cm long chord?

Given:

A chord of length 40 cm is drawn in a circle having a diameter of 50 cm.

Length of another chord = 30 cm

Concept used:

1. The perpendicular from the centre of a circle to a chord bisects the chord.

2. Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“.

3. Diameter = Radius × 2

Calculation:

Let C be the centre of the circle and AB and CD be the shorter and longer chord respectively.

Since the minimum distance is required, the chords must be on the same side of the centre of the circle.

AB = 30 and CD = 40

OF is the perpendicular distance between O and AB. 

Since AB and CD are parallel, OF is also perpendicular to CD at E.

Radius of the circle = 50/2 = 25 cm

OC = OD = OA = OB = 25 cm = Radii

​According to the concept,

AF = FB = 30/2 = 15 cm

CE = ED = 40/2 = 20 cm

ΔOEC is a right-angle triangle at E.

So, OE = \(\sqrt {25^2 - 20^2}\) = 15 cm

Likewise, OF = \(\sqrt {25^2 - 15^2}\) = 20 cm

Distance between AB and CD, EF = 20 - 15 = 5 cm

∴ The minimum distance is 5 cm.