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Triangle $MNO$ is an isosceles triangle with $MN = NO = 25\;\text{cm}$. A line segment, drawn from the midpoint of $\overline{MO}$ perpendicular to $\overline{MN}$, intersects $\overline{MN}$ at point $P$ with $NP:PM = 2:1$. What is the length of the altitude drawn from point $N$ to $\overline{MO}$? Express your answer in simplest radical form.

 Sep 10, 2023
 #1
avatar+129439 
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Call the midpoint of MO =  Q

Call the altitude A

 

NP =  (2/3)(25) =  50/3

PM =  25/3

 

QN^2 =  [ A^2 - (50/3)^2 ]

QN^2  = [ MQ^2  - (25/3)^2 ]

 

A^2  - (50/3)^2  = MQ^2 - (25/3)^2

A^2 - 2500/9  = MQ^2 - 625/9

MQ^2 = A^2 - 1875/9

 

MN ^2  =  MQ^2  + A^2

25^2  =   A^2 - 1875/9 + A^2

625 = 2A^2 - 1875/9

5625  = 18A^2 - 1875

18A^2 = 5625 + 1875

18A^2 = 7500

A^2  = (7500 / 18)  = ( 1250/3)

A = sqrt (1250/3)  =  sqrt ( 625 * 2 / 3)  =  25 sqrt (2/3)  =  25sqrt (6) / 3

 

cool cool cool

 Sep 10, 2023

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