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# GMATPREP #2 - similar triangles

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Senior Manager
Joined: 19 Aug 2006
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GMATPREP #2 - similar triangles [#permalink]  19 Mar 2009, 17:51
I thought this was an interesting problem.
I'll post the OA later.
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Senior Manager
Joined: 08 Jan 2009
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Re: GMATPREP #2 - similar triangles [#permalink]  19 Mar 2009, 19:34
From the figure we can notice that both are similar triangles according to AAA rule.

therefore ( ratio of Areas) = square of corresponding sides
( AL)/ ( AS) = (S/s ) ^2

Given AL = 2AS therefore S^2 = 2*s^2 S = (2)^1/2 s Ans C.
Senior Manager
Joined: 19 Aug 2006
Posts: 250
Followers: 2

Kudos [?]: 7 [0], given: 0

Re: GMATPREP #2 - similar triangles [#permalink]  22 Mar 2009, 12:06
tkarthi4u wrote:
From the figure we can notice that both are similar triangles according to AAA rule.

therefore ( ratio of Areas) = square of corresponding sides
( AL)/ ( AS) = (S/s ) ^2

Given AL = 2AS therefore S^2 = 2*s^2 S = (2)^1/2 s Ans C.

Yes, thank you, the OA is C.
Does anyone know any good links to more similar triangle theory and/or problems?
Re: GMATPREP #2 - similar triangles   [#permalink] 22 Mar 2009, 12:06
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