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!

NotThatSmart May 7, 2024

#1**+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

CPhill May 7, 2024

#2**+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