In circle T, the length of the diameter AB is 22 cm. AB bisects CD at W. Find the distance of chord AC from the center if CD = 16 and BW = 4?

Answer 1

AC^2 = 18^2 + 8^2 = 388

AC= sqr388 = about 19.7

AC/2 = 9.85,

AC/2 and AO are two sides of a right triangle with AC/2 and d as the legs with AO as the hypotenuse

distance from AC to circle's center = d

d^2 =AO^2 - (AC/2)^2 (22/2)^2 - (19.7/2)^2 = 23.9775

d = sqr23.9775 = about 4.9 cm from CD to the circle's center