+1348 In triangle $ABC$, we have that $AB = AC,$ and $\overline{AD}$ is an altitude. Also, $E$ is a point on $\overline{AC}$ such that $\overline{BE} \perp \overline{AC}.$ If $BC = 12$ and the area of triangle $ABC$ is $180,$ what is the area of $ABDE$?
+129895 A
E
B D C
12
Since AB = AC, the triangle is isoceles
AB bisects BC so that DC = 6
And
[ABC ] = (1/2) (BC) (AD)
180 = (1/2) ( BC) (AD)
180 = (1/2) (12) (AD)
180 = 6 * AD
180 / 6 = AD = 30
And angle ADC = angle BEC = 90°
And angle DCA = angle ECB
So triangle CDA is similar to triangle CEB
So CD / AD = CE / BE
So 6/30 = CE / BE
So 1/5 = CE / BE
5CE = BE
And
BC^2 = CE^2 + BE^2
12^2 = CE^2 + (5CE)^2
144 = CE^2 + 25CE^2
144 = 26CE^2
144/26 = CE^2
72/13 = CE^2
CE = sqrt [ 72/13] = (6 sqrt (2)) / sqrt (13)
tan (angle ACD) = AD / DC = 30/6 = 5/1
sin (ACD) = sin (BCE) = 5 /sqrt (5^2 + 1^2) = 5/sqrt (26)
Area of triangle ECD = (1/2)(DC) (CE) = (1/2)(6)(6sqrt (2) /sqrt (13) ) * (5 /sqrt (26)) =
90sqrt (2) / sqrt (13) / [ sqrt (2) * sqrt (13) ] = 90 / 13
[ ABDE ] = [ ABC ] - [ ECD ] = (180 ) - ( 90/13 ) = 2250 / 13
