A circle has a radius of 15. Let $\overline{AB}$ be a chord of the circle, such that AB = 4. What is the distance between the chord and the center of the circle?
Let's set the center of the circle as O.
Draw a line from O perpendicular to the chord. Let this line be OM.
\(OA = 15 \\ AM = AB / 2 = 4/2 = 2\)
Triangle AOM is right, with angle AMO being 90 degrees.
By using the pythagorean theorem, we have
\(OM = \sqrt{ OA^2 -AM^2} = \sqrt{ 15^2 - 2^2 } = \sqrt{221}\)
So sqrt221 is our answer.
Thanks! :)
Let's set the center of the circle as O.
Draw a line from O perpendicular to the chord. Let this line be OM.
\(OA = 15 \\ AM = AB / 2 = 4/2 = 2\)
Triangle AOM is right, with angle AMO being 90 degrees.
By using the pythagorean theorem, we have
\(OM = \sqrt{ OA^2 -AM^2} = \sqrt{ 15^2 - 2^2 } = \sqrt{221}\)
So sqrt221 is our answer.
Thanks! :)