The lengths of two parallel chords of a circle are $6$ cm and $8$ cm. The smaller chord is at a distance of $4$ cm from the centre,t hen the radius of the circle is?

Draw a perpendicular from the center O to the 8 cm chord, forming a right triangle where the base is 4 cm and the hypotenuse is the radius r. Because a perpendicular from the center bisects a chord, the base is half of 8 cm, which is 4 cm. Applying the Pythagorean theorem to the smaller chord which is at a distance of 4 cm from the center gives r^2 = 4^2 + 3^2. Therefore, r^2 = 25 and r = 5 cm.

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