OG 11 DS #152 triangle PQR

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OG 11 DS #152 triangle PQR

by ifthyder » Fri Sep 05, 2008 2:56 am
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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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by Ian Stewart » Fri Sep 05, 2008 3:58 am
In triangle PQR, the largest of the three triangles in the diagram, the right angle is at Q, and the diameter PR is the hypotenuse.

It's worth knowing the following: If AB is a diameter of a circle, and C is another point on the circle, triangle ABC will have a right angle at C, and AB will be the hypotenuse of ABC.
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