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 \(ABDE\)?
+128474 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
