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The midpoint of PQ is M.  The midpoint of PR is N.  QN and RM intersect at O. If QN is perpendicular to PR, QN = 10, and PR = 16, then find OR.

 

 Apr 17, 2022
 #1
avatar+9459 
+2

Connect QR. Then O is the centroid of \(\triangle PQR\).

 

PR = 16 implies NR = 8.

By property of centroid, QO : ON = 2 : 1. Then QN = 10 implies ON = 10/3.

Then,

\(OR = \sqrt{8^2 + \left(\dfrac{10}3\right)^2} = \dfrac{26}3\)

 Apr 17, 2022
edited by MaxWong  Apr 17, 2022
 #2
avatar+124598 
+2

In triangle PMR  Draw atltitude MS   to  PR

 

Now  triangles  MPS and  QPN are similar

 

So    MP  /  QP   =  PS  / PN       

 

          1/2 =  PS /PN   ⇒   PS   = 4   ⇒  MS = (1/2) QN  =  5

 

And triangles   RSM  and  RNO   are similar

 

So    SR   = 12      MS  = 5     RN  =  8

 

MS / SR   =  ON  / RN

 

5 / 12  =  ON / 8

 

40 /12  =  ON   =   10/3

 

And since   triangle RNO  is right, then

 

OR   = sqrt   [ RN^2  + ON^2  ]   =   sqrt  [  8^2  + (10/3)^2 ]  =

 

sqrt [  64 +  100 / 9  ]  =  sqrt  [ (576 + 100) / 9  ]   =   sqrt  [ 676 ] / 3  =   26  / 3

 

 

cool cool cool

 Apr 17, 2022
 #3
avatar+2455 
+1

Because \(PN = 8 \)\(QN = 10\), and \(\angle N = 90\)\(PQ = \sqrt{164} = 2\sqrt{41}\)

 

We know that \(M\) is the midpoint, so \(PM = \sqrt{41}\).

 

Now, draw \(MA\), so that \(MA \) is perpendicular to \(PR\).

 

Because of similar triangles, \(MA = 5\) and \(AN = 4\)

 

Using the Pythagorean Theorem, we find that \(RM = 13\).

 

Because of similar triangles, \(OR = {2 \over 3} \times 13 = \color{brown}\boxed{26\over3}\) 

 Apr 17, 2022

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