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2004 AMC 10A Problem 9

Problem 9 of 25EasierGeometry

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

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Solution

Let [ABD][ABD] be the area shared by both large triangles. Then [ABE]=[ADE]+[ABD][ABE] = [ADE] + [ABD] and [ABC]=[BDC]+[ABD].[ABC] = [BDC] + [ABD]. Subtracting, [ADE]−[BDC]=[ABE]−[ABC]. \begin{aligned} &[ADE] - [BDC] \\ &= [ABE] - [ABC]. \end{aligned} Since ∠EAB\angle EAB and ∠ABC\angle ABC are right angles, [ABE]=12(4)(8)=16,[ABC]=12(4)(6)=12. \begin{aligned} [ABE] &= \tfrac12(4)(8) = 16, \\ [ABC] &= \tfrac12(4)(6) = 12. \end{aligned} The difference is 16−12=4.16 - 12 = 4. Thus, the correct answer is B.

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Concepts: triangle area · area decomposition

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