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In right triangle $ABC,$ $\angle C = 90^\circ.$  Median $\overline{AM}$ has a length of $\sqrt{5},$ and median $\overline{BN}$ has a length of $\sqrt{5}.$  What is the length of the hypotenuse of the triangle?

 Dec 20, 2023
 #1
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A

 

 N              

 

C         M            B

 

We have the equations

 

AC^2 + CM^2  =  AM^2

NC^2 + BC^2 = BN^2                            CM = (BC/2)       CN  =  (AC/2)

 

AC^2  +(BC/2)^2  = 5     (1)

(AC/2)^2 + BC^2  = 5     (2)              subtract (2)  from (1)

 

(3/4)AC^2  - (3/4)BC^2   =  0

(3/4)AC^2 = (3/4)BC^2

AC^2 = BC^2

AC = BC

 

Therefore using (1)

 

AC^2 + (AC/2)^2 =  5

(5/4)AC^2  = 5

AC^2  = 5(4/5)

AC^2 = 4

AC = 2 =  BC

 

hypotenuse AB = sqrt ( AC^2 + BC^2)  = sqrt (2^2 + 2^2)  =  sqrt (8) = 2sqrt (2)

 

 

cool cool cool

 Dec 21, 2023

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