SOLUTION: 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.

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): You can put this solution on YOUR website!
the altitude is on a diameter of the circle

5:13::13:d

5d = 169 ___ 10r = 169 ___ r = 16.9
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!

The other tutor's solution is incorrect. Draw in the radii to the vertices: Extend the vertical radius down to the base of the given triangle, which we label D, and we let x be the measure of OD dividing the base into two equal parts, each measuring 2.5: From right triangle BOD From right triangle BAD Ignore the negative: Substitute for in Add these two equations: 2r = 13.24728445 r = 6.62363632224 Edwin

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.

Hope this helps!
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