Question

AD is a diameter of a circle and AB is a chord. If AD=34cm, AB=30cm, the distance of AB from the centre of the circle is :

Answer 1

The distance is 8 cm

Explanation:

The chord and the diameter form one leg and the hypotenuse of a right triangle. The other leg, BD, has length ...

BD² +AB² = AD²

BD² = AD² -AB² = 34² -30² = 256

BD = √256 = 16

The segment from the center of the circle to the midpoint of the chord is the midline of triangle ABD, so is half the length of BD.

distance from AB to the center = 16/2 = 8 . . . cm