+0  
 
+1
2
2
avatar+202 

I'm studying for this test about the Pythagorean Theorem when I came across this question, and ummm, yeah my small brain can't figure it out... :)

 

In a right triangle ABC, length of the medians to the sides AB and BC are \(2\sqrt{61}\)​ and \(\sqrt{601}\)​ respectively. Find the length of its hypotenuse.

 

I don't reall know where to start exactly. My geuss is to set variables x for BD or something like that, but wouldn't that just take a while and be terribly inefficient?

 

Thanks!

 May 7, 2024
 #1
avatar+128796 
+1

Not as bad as you might  think.....

 

CE = 2sqrt 61  = sqrt 244

AD = sqrt 601 

 

Note  that we have this system

AD^2  = (BC/2)^2  + (AB)^2

CE^2  = (BC)^2 + (AB/2)^2         simplify

 

601  = BC^2/4 + AB^2

244 = BC^2 + (AB/2)^2

 

601 = BC^2 /4  + AB^2

244 = BC^2 + AB^2/4                mutiply the top equation through by -4

 

-2404 = -BC^2- 4AB^2

244  = BC^2  + AB^2/4     add these

 

-2160  =  -(15/4)AB^2

 

(-2160) * (-4/15)  =AB^2  = 576

AB = sqrt (576)  = 24

 

To find BC

244 = BC^2 + (24/2)^2

244 = BC^2 + 144

100 = BC^2

10= BC

 

Hypotenuse  =  sqrt [ 24^2 + 10^2 ] =  sqrt [676 ]  =  26

 

cool cool cool

 May 7, 2024
 #2
avatar+202 
+1

Ahhh, I see. I didn't think about using the two equations we make and combining them in the end. 

 

I was more focued on setting a variable, which I see is not needed. 

 

Thanks CPhill, I understand how to solve this problem now!

NotThatSmart  May 7, 2024

2 Online Users

avatar