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In triangle $PQR$, we have $PQ = PR = 13$ and $QR = 10$. Let $M$ be the midpoint of $\overline{PQ}$ and $N$ be on $\overline{QR}$ such that $\overline{PN}$ is an altitude of triangle $PQR$. If $\overline{PN}$ and $\overline{RM}$ intersect at $X$, then what is $PX$?

 Dec 5, 2019
 #1
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By similar triangles, PX = sqrt(13^2 - 10^2)/2 = sqrt(69)/2.

 Dec 5, 2019
 #2
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Drop the perpendicular from M to QR at Y.  Then MY = 7/2, so PX = 3*MY = 3*7/2 = 21/2.

 Dec 5, 2019
 #3
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See  the following image   :

 

 

PQR  is isosceles  with  PQ  = PR.....so  QN  = (1/2) (QR)  = (1/2)(10)  = 5

 

By the Pythagorean Theorem

 

sqrt  ( QP^2  - QN^2)  =  PN

 

sqrt ( 13^2  - 5^2)  =  sqrt  ( 169 - 25)   = sqrt (144)  = 12 =  PN

 

Since  PQR is isosceles with PQ = PR then,  X  is a median......so

 

PX  =  (2/3) (12)   =    8

 

 

cool cool cool

 Dec 5, 2019

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