How do i solve problem a. and problem b? (using the pythagorean theorem). a. the median to the hypotenuse. b. the altitude to the hypotenuse.

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

The distance from point a to point c across the coordinate plane is aproxamatly 8.94 rounded to the nearest hundredth. you use a^2 + b^2 =c^2