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We have triangle  ABC$ where $AB = AC$ and $AD$ is an altitude. Meanwhile, $E$ is a point on $AC$ such that $AB \parallel DE.$ If $BC = 12$ and the area of $\triangle ABC$ is $180,$ what is the area of $ABDE$?

 Aug 16, 2018
 #1
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Triangle  ABC  is isosceles with  AB  = AC

AD  is an altitude which is also  a median to BC...so  DC  = DB  = 6

And  DE is parallel to AB

 

So  angle EDC  = angle ABC

And angle ACB  = angle ECD

So...by AA congruency, ΔABC  ~ Δ  EDC

 

Since the area  of  Δ  ABC = 180....we can  find the altitude AD thusly

 

180  = (1/2) (AC) (AD)

180  = (1/2) (12) (AD)

180 = 6  (AD)   divide each side by 6

30  = AD

 

Now  Δ ABC  and Δ EDC  are similar.... and BC  = 2DC

So......the altitude of Δ ABC  is also twice that of  Δ EDC...so  the altitude of Δ  EDC   = 15

And the area of Δ EDC  = (1/2)DC( altitude of EDC)  = (1/2)(6)(15)  = .45

 

So...the area  of ABDE  = area of  Δ ABC  -  area  of Δ  EDC  =   180  - 45  =  135  units^2

 

 

cool cool cool

 Aug 16, 2018
edited by CPhill  Aug 16, 2018

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