A circle has a radius of $15.$ Let $\overline{AB}$ be a chord of the circle, such that $AB = 5$. What is the distance between the chord and the center of the circle?
Ket the center of the circle = O
Draw a line from the center perpendicular to the chord.. call this OM...this will bisect the chord
OA = 15
AM = AB / 2 = 5/2 = 2.5
OM = what we are looking for
Triangle AOM is right with AMO = 90°
So....by the Pythagorean Theorem
OM = sqrt [ OA^2 -AM^2] = sqrt [ 15^2 - 2.5^2 ] = sqrt (875 / 4) = (5/2)sqrt (35) ≈ 14.79