SOLUTION: A small commuter airline flies to three cities whose locations form the vertices of a right triangle. The total flight distance (from city A to city B to city C and back to city A)

Algebra ->  Pythagorean-theorem -> SOLUTION: A small commuter airline flies to three cities whose locations form the vertices of a right triangle. The total flight distance (from city A to city B to city C and back to city A)      Log On


   

Question 420202: A small commuter airline flies to three cities whose locations form the vertices of a right triangle. The total flight distance (from city A to city B to city C and back to city A) is 1400 miles. It is 600 miles between the two cities that are furthest apart (city A to city B). Find the other two distances between cities.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Farthest distance = 600 miles will be the hypotenuse
the balance 800 is the sum of the two legs of the right triangle.
if one side is x
other side = 800-x
..
x^2+(800-x)^2=600^2
x^2+640000-1600x+x^2=360000
2x^2-1600x+280000=0
/2
x^2-800x+140000=0
solve using quadratic equation
b^2-4ac= 120000
one side = 541.4 miles
other side =258.6 miles