Question 388780: an isosceles triangle, whose legs are each 13, is inscribed in a circle. if the altitude to the base of the triangle is 5, find the radius of the circle.
Found 3 solutions by scott8148, Edwin McCravy, robertb: Answer by scott8148(6628) (Show Source): Answer by Edwin McCravy(20056) (Show Source): Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website! Consider the isosceles triangle ABC, with A the vertex angle.Then AB = AC = 13. If we let D be the midpoint of segment BC, then segment AD is the altitude of triangle ABC from A to segment BC, and AD = 5, as given. This altitude AD must lie on a diameter, due to symmetry. Let AE be the diameter that contains the altitude AD. Then triangle ACE (in that order) is a right triangle, with C as the right angle.By similarity of the right triangle ACE with the right triangle ADC, we must have, by proportionality of corresponding sides, that , or d = 169/5, or d = 33.8, where d = diameter of the circle. Thus r = 16.9.

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