Expert: Sherman D. - 9/20/2004

Question
How do u solve this problem:
In the figure,<EAB  and <ABC are right angles. AB=4, BC=6, AE=8, and segment AC and segment BE intersect at D. What is the difference between the areas of triangle ADE and triangle BDC?

Please reply ASAP.
Thanks,
Leti

Answer
When you draw this out, it should look like a rectangle, with the right side missing, and the bottom length is shorter than the top length. Now draw 2 diagonals where each diagonal connects with the opposite vertex.

AE is the top
BC is the bottom
AB is the left side
D is where the 2 diagonals meet.

if you have the values in their right places, then

Since <ABC and <EAB are right triangles, you can use the formula A = (ab)/2

<ABC = (6 * 4)/2
<ABC = 24/2
<ABC = 12

<EAB = (8 * 4)/2
<EAB = 32/2
<EAB = 16

Here comes the easy part

If you take the difference of <ABC and <EAB you get the answer of 4. Now take that and subtract that from <ABC and <EAB, and you get

Triangle ADE = 16 - 4 = 12
Triangle BDC = 12 - 4 = 8

So the area of Triangle ADE is 12 and the area of Triangle BDC is 8, either way you put it, the mathematical difference is 4.