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In right triangle $ABC,$ $\angle C = 90^\circ$. Median $\overline{AM}$ has a length of 2, and median $\overline{BN}$ has a length of 3. What is the length of the hypotenuse of the triangle?

 Jan 16, 2025
 #1
avatar+130317 
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Triangles  AMC  and BNC  are right

CM = 1/2 BC

CN = 1/2 AC

 

So

 

AM^2  - [ (1/2)BC ] ^2  = AC^2

 2^2  - BC^2 / 4 = AC^2

AC^2 + BC^2/4 = 4    →  4AC^2 + BC^2  = 16   ( 1)

 

And

BN^2 - [ (1/2)AC]^2  = BC^2

3^2 - AC^2/4  = BC^2

BC^2 + AC^2 / 4  = 9   →  4BC^2 + AC^2  = 36  →  -16BC^2 - 4AC^2 = -144   (2)

 

Add (1) , (2)

 

-15BC^2  =  -128

BC^2 = 128/15

 

4AC^2 + BC^2  =  16

4AC^2 + 128/15 = 16

4AC^2  = 16 - 128/15

4AC^2  = 112/15

AC ^2 = 112/60  = 28/15

 

AB = the hypotenuse  = sqrt [ BC^2 + AC^2] = sqrt [ 128/15 + 28/15 = sqrt [ 156 / 16] =sqrt [10.4] ≈ 3.225

 

cool cool cool

 Jan 16, 2025

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