+842 Suppose that there is a right triangle $\triangle ABC$ such that the legs $AB$ and $BC$ have lengths $4\sqrt{2}$ and $5,$ respectively. What is the length of the median $BM$?
+129895 A
B C
AC = sqrt [ (4sqrt 2)^2 + 5^2 ] = sqrt [ (sqrt (32))^2 + 5^2 ] = sqrt (57)
cos (ACB) = 5/sqrt (57)
Law of Cosines
BM^2 = (AC / 2)^2 + BC^2 - 2 ( AC/2 * BC) (5/ sqrt 57)
BM^2 = AC^2 / 4 + BC^2 - (AC*BC) ( 5/sqrt 57)
BM^2 = 57 / 4 + 25 - (sqrt (57) * 5) (5 /sqrt 57)
BM^2 = 57/4 + 25 - 25
BM = sqrt (57) / 2 = AC / 2
