SOLUTION: Please show me how to find the altitude of a triangle that is 13 on each side and 10 on the bottom. 13/_\13 10

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I'm going to do the best i can to explain this.
The triangle is iscosceles. Therefore AB=AC (A is the top vertex, B & C are the bottom vertex.) The height (in this case altitude) of the triangle is perpendicular to to base. So the height bisects the line BC at point D. So BD=CD. Now you can use the pathagorean theorm.
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BD = height
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AB = 13 -- this is going to be c
AC = 13
BC = 10
BD & CD = 5 -- this is going to be a
BD = ? -- this is going to be b
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a^2 + b^2 = c^2
5^2 + b^2 = 13^2
25 + b^2 = 169 -----subtract 25 from each side
b^2 = 144 ----------take the square root of each side
b = 12
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The height (altitude) of the triangle is 12