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

#1**+2 **

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

CPhill Aug 25, 2018