The segments AB and CD are the chords of a circle. Find the distance from the center of the circle to the chord CD, if AB = 12

The segments AB and CD are the chords of a circle. Find the distance from the center of the circle to the chord CD, if AB = 12, CD = 16, and the distance from the center of the circle to the chord AB is 8

Draw the segments OB and OC, as shown in the figure.
The distance from a point to a line is the length of the perpendicular drawn to the line. Therefore, OE is perpendicular to AB, and OF is perpendicular to CD. Points E and F divide their chords in half (by chord property)
It turns out that the triangles OEB and OCF are rectangular, EB = AB / 2 and CF = CD / 2.
By the Pythagorean theorem:
OB ^ 2 = OE ^ 2 + EB ^ 2
OB ^ 2 = 8 ^ 2 + (12/2) ^ 2
OB ^ 2 = 64 + 36 = 100
OB = 10
OB = OC = 10 (since OB and OC are the radii of a circle)
By the Pythagorean theorem:
OC2 = CF2 + FO2
OC ^ 2 = (CD / 2) ^ 2 + FO ^ 2
10 ^ 2 = (16/2) ^ 2 + FO ^ 2
100 = 64 + FO2
Fo ^ 2 = 36
Fo = 6
Answer: the distance from the center of the circle to the CD chord is 6

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