2004 AMC 10A Problems/Problem 9
Problem
In the figure, 
and 
are right angles. 
, and 
and 
intersect at 
. What is the difference between the areas of 
and 
?
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
| 2004 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
Imagine drawing a line XY parallel to AB through point D. We can then call the length of XD 4x and the length of YD, 3x. Why? Triangles ADE and CDB are similar based on Vertical angles and AIA Thm. So, 7x = 4. Therefore, x = 4/7.
The problem asks for the difference between ABE and BDC. The area of triangle ABE is (1/2)bh= (1/2)(4)(8) = 16.
From this, we subtract the area of BDC = (1/2)bh = (1/2)(6)(3x) = (9x) = (9)(4/7) = 36/7.
Thus, the difference is 16 - (36/7) = (112-36)/7 = 76/7.
I suppose the answer choices are wrong.
