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$Old$ 11-12-2008, 11:13 AM

bearej50 $bearej50 is offline$

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Two chords KV and QR of circle O are perpendicular at oint P, with PQ = 6 and PR = 8. If the radius of circle O is √65, find KP and PV.

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$Old$ 11-13-2008, 09:26 AM

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Let $OM\perp RQ$ and $ON\perp KV$ .
$RQ=14\Rightarrow MQ=7\Rightarrow ON=MP=MQ-PQ=7-6=1$
In the triangle OMQ: $OM=\sqrt{OQ^2-MQ^2}=\sqrt{65-49}=\sqrt{16}=4\Rightarrow PN=4$
In the triangle ONK: $KN=\sqrt{OK^2-ON^2}=\sqrt{65-1}=8$
Then $KP=KN-PN=8-4=4$ and $PV=KV-KP=16-4=12$

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$Old$ 11-13-2008, 10:02 AM

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Red dog your a genius any chance you can help with my problem?
http://www.mathhelpforum.com/math-he...arger-one.html

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$Old$ 11-13-2008, 10:36 AM

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thank you very much red dog!

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