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We have triangle \(\triangle ABC\) 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 \(ABDE\)?

Guest Aug 25, 2018
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Since  AB  =AC....then triangle  ABC  is isosceles

And...in an isosceles triangle the altitude bisects the base

So...let BC  be the base and we have that

So...DC = 1/2 BC

And area of  triangle ABC  = (1/2) BC (AD)

180  = (1/2) (12) (AD)

180  = 6 AD

30  = AD

 

And  since  AB and DE are parallel...then

angle BAC  = angle DEC

angle ACB  =angle ECB

So...by AA congruency

 ΔABC  ~ ΔEDC

And DC = 1/2 *BC  = 1/2 *  12   = 6

 

So....the altitude of Δ EDC  = 1/2 AD =  1/2 * 30  = 15

 

So...the area of ΔEDC = (1/2)(6)(15)  = 45

 

So...the are of ABDE  = area of  ΔABC  - area of ΔEDC  =

 

180  - 45   =

 

135  units^2

 

 

cool cool cool

CPhill  Aug 25, 2018

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