Hypotenuse is √(6² + 8²) = √100 = 10
from mathopenref: median: "A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side."
if we draw a median to the hypotenuse, then we have to draw from the right angle to the hypotenuse.
we use a theorem: "In a right triangle, the median drawn to the hypotenuse has the measure half the hypotenuse. "
the length of the median to the hypotenuse is half the hypotenuse: 5
for altitude:
we draw line on hypotenuse that is perpendicular to hypotenuse to a vertex of the angle opposite to it.
look at my diagram: h is the altitude length
x and y are segments of the original hypotenuse
we have ΔABD with legs of x and h and hypotenuse of 6
ΔABD: x² + h² = 6²
we have ΔBCD with legs of 10-x and h and hypotenuse of 8
ΔBCD: (10-x)² + h² = 8²
this is a system of equations
x² + h² = 6²
(10-x)² + h² = 8²
isolate h² in x² + h² = 6² so that we can substitute:
h² = 6² - x²
h² = 36 - x²
substituting into (10-x)² + h² = 8² we have
(10-x)² + h² = 8²
(10-x)² + (36 - x²) = 8² ...............substitution done
(10-x)² expands to 100 - 20x + x²
100 - 20x + x² + (36 - x²) = 64
+x² and -x² cancel
100 - 20x + 36 = 64
136 - 20x = 64
-20x = -72
x = 3.6
Since h² = 36 - x², then h = √(36 - x²) and
h = √(36 - 3.6²) =4.8
the altitude is 4.8
