In circle A, radius AB and chord CD have lengths of 12 cm. Find the distance from A to CD.

Answer 1

Draw a diagram.

Draw a radius so that it meets chord CD at a right angle. Let's say that the point of intersection is P. Then P bisects the chord. So, CP = 6.

Draw radius AC. Since any two radii are congruent, AC = 12 and AC is the hypotenuse of right triangle APC.

AP is the distance from A to CD.

Using the Pythagorean Theorem, (AP)² + 6² = 12².

So, AP = √108 = 6√3

Answer 2

AB=12= CD

AC = AD = 12

AC, AD and CD are 3 sides of an equilateral triangle, all = 12 cm

use one side as the base and calculate the altitude of the triangle

the altitude and half the base are 2 sides of a right triangle with hypotenuse = 12

and half the base = 6

12^2 = 6^2 + a^2

a^2 = 144-36 = 108

a = sqr108 = 6sqr3 = about 10.4 cm = the distance from the chord to the circle's center