• November 12th 2008, 09:13 AM
bearej50
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.
• November 13th 2008, 07:26 AM
red_dog
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 $PV=KV-KP=16-4=12$
• November 13th 2008, 08:02 AM
lmcph24
Red dog your a genius any chance you can help with my problem?
http://www.mathhelpforum.com/math-he...arger-one.html
• November 13th 2008, 08:36 AM
bearej50
thank you very much red dog!