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In right triangle $ABC,$ $\angle C = 90^\circ$. Median $\overline{AM}$ has a length of 1, and median $\overline{BN}$ has a length of 1. What is the length of the hypotenuse of the triangle?

 May 14, 2024
 #1
avatar+129514 
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Call  AC = x   and  BC =  y

 

Using the Puyhagorean Theroem twice, we have this system

 

x^2  + (y/2)^2  = 1

(x/2)^2 + y^2  = 1           simplify

 

x^2 + y^2/4  = 1  →  4x^2 + y^2  = 4  →   4x^2 + y^2 = 4        (1)

x^2/4 + y^2  = 1 →    x^2/4 + y^2  =1 → -x^2/4 - y^2  = -1       (2)

 

Add (1) and (2)

 

(15/4)x^2  = 3

 

x^2 = 12/15 =  4/5

 

To find y

12/15 + y/2/4  = 1

y^2 / 4 = 3/15

y^2 = 12/15 = 4/5

 

Hypotenuse =  sqrt [ x^2 + y^2 ] =  sqrt [ 2 * 4/5 ] = sqrt (8/5)  = 2sqrt (2/5) =  2sqrt (10) / 5

 

cool cool cool

 May 14, 2024

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