In right triangle $ABC,$ $\angle C = 90^\circ$. Median $\overline{AM}$ has a length of 2, and median $\overline{BN}$ has a length of 3. What is the length of the hypotenuse of the triangle?
Triangles AMC and BNC are right
CM = 1/2 BC
CN = 1/2 AC
So
AM^2 - [ (1/2)BC ] ^2 = AC^2
2^2 - BC^2 / 4 = AC^2
AC^2 + BC^2/4 = 4 → 4AC^2 + BC^2 = 16 ( 1)
And
BN^2 - [ (1/2)AC]^2 = BC^2
3^2 - AC^2/4 = BC^2
BC^2 + AC^2 / 4 = 9 → 4BC^2 + AC^2 = 36 → -16BC^2 - 4AC^2 = -144 (2)
Add (1) , (2)
-15BC^2 = -128
BC^2 = 128/15
4AC^2 + BC^2 = 16
4AC^2 + 128/15 = 16
4AC^2 = 16 - 128/15
4AC^2 = 112/15
AC ^2 = 112/60 = 28/15
AB = the hypotenuse = sqrt [ BC^2 + AC^2] = sqrt [ 128/15 + 28/15 = sqrt [ 156 / 16] =sqrt [10.4] ≈ 3.225