+0  
 
+1
127
1
avatar+198 

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
 #1
avatar
0

By Pythagoras's Theorem:

 

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

AD^2 = 576

AD = sqrt(576)

AD = 24

 Jun 12, 2018

26 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.