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In triangle ABC, medians AD and CE intersect at P , PE=1.5, PD=2 and .DE= 2.5 What is the area of AEDC?

 Jun 21, 2022
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Area of triangle DPE  can be found as

 

Semiperimemter, S,   =   [2.5 + 2 + 1.5 ]  / 2  =  3

Area  =  sqrt [  3 ( 3 - 2) (3 - 2.5) ( 3 -1.5) ] =   1.5  

 

 

By the property  of  medians

 

PC =  2 * PE  =  2 * 1.5  =  3

PA =  2 *  PD =  2 * 2  =  4

 

And triangles  DPE and APC are similar  with each side of APC twice as long as that of DPC

 

So scale factor of triangle   APC   to  triangle DPE   =   2

Area of triangle  APC =  area of DPE * scale factor^2  =   1.5 * 2^2  =  6

 

And by another median property areas of triangle  AEP  plus triangle  CDP   = area of triangle APC

 

So....area of  AEDC  =   1.5 + 6  + 6   =     13.5

 

 

cool cool cool  

 Jun 22, 2022

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