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 $ABDE$?

newuseeer
Aug 16, 2018

#1**+1 **

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

CPhill
Aug 16, 2018