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Equal areas again

Source: 1959 AMC 12 Problem 10

August 13, 2013
geometryAMCAMC 12triangle area

Problem Statement

In triangle ABCABC with AB‾=AC‾=3.6\overline{AB}=\overline{AC}=3.6, a point DD is taken on ABAB at a distance 1.21.2 from AA. Point DD is joined to EE in the prolongation of ACAC so that triangle AEDAED is equal in area to ABCABC. Then AE‾\overline{AE} is: <spanclass=′latex−bold′>(A)</span> 4.8<spanclass=′latex−bold′>(B)</span> 5.4<spanclass=′latex−bold′>(C)</span> 7.2<spanclass=′latex−bold′>(D)</span> 10.8<spanclass=′latex−bold′>(E)</span> 12.6 <span class='latex-bold'>(A)</span>\ 4.8 \qquad<span class='latex-bold'>(B)</span>\ 5.4\qquad<span class='latex-bold'>(C)</span>\ 7.2\qquad<span class='latex-bold'>(D)</span>\ 10.8\qquad<span class='latex-bold'>(E)</span>\ 12.6
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