A and B are two points on a circle with center O, and C lies outside the circle, on ray AB. Given AB = 24, BC = 30, and OA = 16, find OC.
+179 OC=\(\sqrt{1876}\)
Proof
The proof is difficult to explain concisely and without a diagram, but basically, draw AB, O, and the perpendicular from O to the midpoint of AB. The intersection will be labeled D. Thus, AD, DO, and OA will form a right triangle, hypotenuse of 16 and one leg of AD of 12. The other leg (DO) is \(\sqrt{112}\) through the Pythagorean theorem. Now, consider the right triangle DCO. DC= 30+12 = 42, and DO =\(\sqrt{112}\). From the Pythagorean theorem, OC = \(\sqrt{112+42^2} \) or \(\sqrt{1876}\).
Have a great day!
+876 "The proof is difficult to explain concisely and without a diagram."
This reminds me of the quote "I have discovered a truly remarkable proof of this theorem which this margin is too small to contain"
+129852 Here is what EnchantedLava is saying
AB is a chord on the circle.....call the mid point D.....then AD = DB = 24/2 = 12
And we can form right triangle AOC with AO = 16 = the hypotenuse, and leg AD = 12
And OD will form the other leg such that AO^2 - AD^2 = OD^2
16^2 - 12^2 = 112 = OD^2
And we can form another right triangle DOC such that OD and DC are legs.....and DC = ( DB + BC) = 12 + 30 = 42
And OC will form the hypotenuse of this triangle....such that
OC = sqrt ( OD^2 + DC^2) = sqrt (112 + 42^2) = sqrt (1876)
See the image here :
