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| − | ==Problem==
| + | #redirect [[2004 AMC 12A Problems/Problem 8]] |
| − | In the figure, <math>\angle EAB</math> and <math>\angle ABC</math> are right angles. <math>AB=4, BC=6, AE=8</math>, and <math>AC</math> and <math>BE</math> intersect at <math>D</math>. What is the difference between the areas of <math>\triangle ABC</math> and <math>\triangle BDC</math>?
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| − | <center>[[Image:AMC10_2004A_9.gif]]</center>
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| − | <math> \mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 4 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 8 \qquad \mathrm{(E) \ } 9 </math>
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| − | ==Solution==
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| − | {{solution}}
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| − | == See also ==
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| − | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131320 AoPS topic]
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| − | {{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.
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