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In right triangle ABC, C=90. Median ¯AM has a length of 1, and median ¯BN has a length of 1. What is the length of the hypotenuse of the triangle?

 Jan 23, 2025

Best Answer 

 #2
avatar+130380 
+1

Let x = AC   and  y = BC      and AB is the hypotenuse

CM = y/2

CN = x/2

 

Triangles  AMC  and BNC are both right 

AC^2 + ( BC/2) = AM

(AC/2)^2 + BC^2 = BN

 

So.....by the Pythagorean Theorem we have these two equations

x^2 + (y/2)^2   =1         

(x/2)^2 + y^2  = 1              simplify

 

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

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

 

Add the last two equations and we  get that

3x^2 - 3y^2  = 0

3x^2 = 3y^2

x^2 = y^2

 

Sub this  back into (1) for y^2

4x^2 + x^2  = 4

5x^2  = 4

x^2  = 4/5

and

y^2 = 4/5

 

AC^2 + BC^2  =  AB^2

 

4/5 + 4/5  = AB^2

 

AB =   sqrt [ 4/5 + 4/5 ]

 

AB = sqrt [ 8/5]  =  2sqrt [2/5] = (2/5)sqrt 10

 

cool cool cool

 Jan 23, 2025
 #1
avatar+28 
+1

Using Apollonius's Theorem we know that the median from the right angle is half the hypotenuse so the answer is 2x1 which is 2

 Jan 23, 2025
 #2
avatar+130380 
+1
Best Answer

Let x = AC   and  y = BC      and AB is the hypotenuse

CM = y/2

CN = x/2

 

Triangles  AMC  and BNC are both right 

AC^2 + ( BC/2) = AM

(AC/2)^2 + BC^2 = BN

 

So.....by the Pythagorean Theorem we have these two equations

x^2 + (y/2)^2   =1         

(x/2)^2 + y^2  = 1              simplify

 

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

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

 

Add the last two equations and we  get that

3x^2 - 3y^2  = 0

3x^2 = 3y^2

x^2 = y^2

 

Sub this  back into (1) for y^2

4x^2 + x^2  = 4

5x^2  = 4

x^2  = 4/5

and

y^2 = 4/5

 

AC^2 + BC^2  =  AB^2

 

4/5 + 4/5  = AB^2

 

AB =   sqrt [ 4/5 + 4/5 ]

 

AB = sqrt [ 8/5]  =  2sqrt [2/5] = (2/5)sqrt 10

 

cool cool cool

CPhill Jan 23, 2025

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