Problem:
In the Figure, ∠EAB and ∠ABC are right angles, AB=4,BC=6,AE=8, and AC and BE intersect at D. What is the difference between the areas of △ADE and △BDC?
Answer Choices:
A. 2
B. 4
C. 5
D. 8
E. 9
Solution:
Let x,y, and z be the areas of â–³ADE,â–³BDC, and â–³ABD, respectively. The area of â–³ABE is (1/2)(4)(8)=16=x+z, and the area of â–³BAC is (1/2)(4)(6)=12=y+z. The requested di erence is
x−y=(x+z)−(y+z)=16−12=4​
The problems on this page are the property of the MAA's American Mathematics Competitions
