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In any isosceles triangle $ABC$ with $AB=AC$, the altitude $AD$ bisects the base $BC$ so that $BD=DC$.

 

Determine the area of \($\triangle \)ABC$.

 

 

 Apr 2, 2020
 #1
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altitude  of  triangle  =   sqrt  [25^2  - 7^2  ]  =  sqrt  [ 625  - 49 ]  =  sqrt  [ 576] =  24

 

Area  of triangle =  (1/2) (14) (24) =  7 * 24   =   168 units^2

 

 

cool cool cool

 Apr 2, 2020

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