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In any isosceles triangle $ABC$ with $AB=AC$, the altitude $AD$ bisects the base $BC$ so that $BD=DC$. As shown in $\triangle ABC$, $AB=AC=25$ and $BC=14$. Determine the length of the altitude $AD$.

 Jun 11, 2018
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By Pythagoras's Theorem:

 

25^2 - (14/2)^2 = AD^2

AD^2 = 576

AD = sqrt(576)

AD = 24

 Jun 12, 2018

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