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# help geometry

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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
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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
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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   Apr 17, 2022
#3
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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