+1537 In right triangle $ABC,$ $\angle C = 90^\circ$. Median $\overline{AM}$ has a length of 1, and median $\overline{BN}$ has a length of 1. What is the length of the hypotenuse of the triangle?
+129847 Call AC = x and BC = y
Using the Puyhagorean Theroem twice, we have this system
x^2 + (y/2)^2 = 1
(x/2)^2 + y^2 = 1 simplify
x^2 + y^2/4 = 1 → 4x^2 + y^2 = 4 → 4x^2 + y^2 = 4 (1)
x^2/4 + y^2 = 1 → x^2/4 + y^2 =1 → -x^2/4 - y^2 = -1 (2)
Add (1) and (2)
(15/4)x^2 = 3
x^2 = 12/15 = 4/5
To find y
12/15 + y/2/4 = 1
y^2 / 4 = 3/15
y^2 = 12/15 = 4/5
Hypotenuse = sqrt [ x^2 + y^2 ] = sqrt [ 2 * 4/5 ] = sqrt (8/5) = 2sqrt (2/5) = 2sqrt (10) / 5
