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Manager
Joined: 01 Dec 2007
Posts: 54

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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: 555

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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

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Re: OG 11 DS #152 triangle pqr   [#permalink] 05 Sep 2008, 05:59
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