+98 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!
+128732 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
+98 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!
