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Manager
Joined: 01 Dec 2007
Posts: 52

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05 Sep 2008, 04:08
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could not understand following
i want to ask how could in PQR (a+b)^2=a^2+b^2

according to me in triangle PQR
hypotenes =pq =x^2
perpendicuar qr=y^2
base pr=(a+b)^2

please explain in light of above solutions

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Current Student
Joined: 11 May 2008
Posts: 552
Re: OG 11 DS #152 triangle pqr [#permalink]

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05 Sep 2008, 05:59
dont get confused.
if u know abt similar triangles,
each of the two triangles contained in the big right angle triangle are similar to the big right triangle.
so given with each of the two values+ ratio of corresponding sides of the similar triangles, we can get the diameter.
eg,
given a=1.
let the point where Q meets the diameter(where a and b meet ) be S
from similar triangles, QS^2 = PS*SR.(since from similar triangles, QS/PS=SR//QS).
SR= 2^2/4=1. SO DIA = 5.

there is no need to repeat the process using condition 2. as we have got SR=1
and that is what is given in the second condtition also.
so D is the ans

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: OG 11 DS #152 triangle pqr   [#permalink] 05 Sep 2008, 05:59
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