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by sanju09 » Fri Jul 23, 2010 2:08 am
In the figure supplied, AD = 4, AB = 3 and CD = 9. What is the area of triangle AEC?
(A) 18
(B) 13.5
(C) 9
(D) 4.5
(E) 3


[spoiler]Source: majortests.com[/spoiler]
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by outreach » Fri Jul 23, 2010 6:55 am
is C the answer
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by GMATGuruNY » Fri Jul 23, 2010 7:19 am
Please see the attached file.
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by kvcpk » Fri Jul 23, 2010 7:24 am
Complete the rectangle ABFD

Now, Area of triangle BFC = 1/2 * (9+3) * (4)
=24

Area of whole quadrilateral = area of triangle CAD + area of rectangle ABFD
= 18 + 12
=30

Hence area of triangle ABC = 30-24 = 6

CED + ABFD -ABE = Area of triangle BFC =24

CED + 12 - ABE = 24
CED - ABE = 12

Triangles CED and AEB are similar.
Hence ratios of areas = ratio of one of its sides
hence CED/ AEB = 9/3 = 3
2AEB = 12
AEB = 6

Oops.. I came up with a funny answer of 0. Can someone look at this procedure and tell where I am going wrong.
My brain has drained I believe :(
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by sanju09 » Tue Aug 31, 2010 8:57 pm
outreach wrote:is C the answer

Apply similarity of triangles AEB and CED and obtain AE = 1. Now, in triangle AEC, base AE = 1 and height CD = 9; area AEC = ½ × 1 × 9 = [spoiler]4.5.

D[/spoiler]
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