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Pythagorean Theorem!

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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

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Not as bad as you might  think.....

CE = 2sqrt 61  = sqrt 244

Note  that we have this system

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

May 7, 2024
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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