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Difference between revisions of "2004 AMC 10A Problems/Problem 9"
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131320 AoPS topic]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131320 AoPS topic]
 
{{AMC10 box|year=2004|ab=A|num-b=8|num-a=10}}
 
{{AMC10 box|year=2004|ab=A|num-b=8|num-a=10}}
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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.
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The problem asks for the difference between ABE and BDC. The area of triangle ABE is (1/2)bh=
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(1/2)(4)(8) = 16.
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From this, we subtract the area of BDC = (1/2)bh = (1/2)(6)(3x) = (9x) = (9)(4/7) = 36/7.
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Thus, the difference is 16 - (36/7) = (112-36)/7 = 76/7.
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I suppose the answer choices are wrong.

Revision as of 22:22, 18 February 2008

Problem

In the figure, $\angle EAB$ and $\angle ABC$ are right angles. $AB=4, BC=6, AE=8$, and $AC$ and $BE$ intersect at $D$. What is the difference between the areas of $\triangle ABC$ and $\triangle BDC$?

AMC10 2004A 9.gif

$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 4 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 8 \qquad \mathrm{(E) \ } 9$

Solution

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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.

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