+1439 In right triangle $ABC,$ $\angle C = 90^\circ.$ Median $\overline{AM}$ has a length of $\sqrt{5},$ and median $\overline{BN}$ has a length of $\sqrt{5}.$ What is the length of the hypotenuse of the triangle?
+129881 A
N
C M B
We have the equations
AC^2 + CM^2 = AM^2
NC^2 + BC^2 = BN^2 CM = (BC/2) CN = (AC/2)
AC^2 +(BC/2)^2 = 5 (1)
(AC/2)^2 + BC^2 = 5 (2) subtract (2) from (1)
(3/4)AC^2 - (3/4)BC^2 = 0
(3/4)AC^2 = (3/4)BC^2
AC^2 = BC^2
AC = BC
Therefore using (1)
AC^2 + (AC/2)^2 = 5
(5/4)AC^2 = 5
AC^2 = 5(4/5)
AC^2 = 4
AC = 2 = BC
hypotenuse AB = sqrt ( AC^2 + BC^2) = sqrt (2^2 + 2^2) = sqrt (8) = 2sqrt (2)
