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 254766: 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. Find the other two distances between cities.

Thanks
Brit
Found 2 solutions by stanbon, dabanfield:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A small commuter airline flies to three cities whose locations form the vertices of a right triangle.
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The total flight distance (from city A to City B to city C and back to city A) is 1400 miles.
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It is 600 miles between the two cities that are furthest apart. Find the other two distances between cities.
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hypotenuse = 600
Let one leg be "x".
Then 2nd leg = (1400-600-x) = 800-x
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Equation:
x^2 + (800-x)^2 = 600^2
x^2 + 640000-1600x+x^2 = 360000
2x^2 -1600x + 280000 = 0
x^2 - 800x + 140000 = 0
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I graphed and got x = 258.58 miles (one leg)
800-x = 541.42 miles (other leg)
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Cheers,
stan H.

Answer by dabanfield(803) About Me  (Show Source):
You can put this solution on YOUR website!
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. Find the other two distances between cities.
Let's suppose the longest side of the right triangle is AC. Then AC = 600 miles. Since the perimeter of the triangle is 1400, the sum of the other sides, AB and BC is 1400-600 = 800.
Since AB + BC = 800, AB = 800 - BC.
By the Pythagorean Theorem we have:

AC^2 = AB^2 + BC^2 so
600^2 = (800-BC)^2 + BC^2
3600 = 6400 - 1600*BC + BC^2 + BC^2
2*BC^2 - 1600*BC + 2800 = 0
BC^2 - 800*BC + 1400 = 0
Solve this quadratic equation for BC then calculate AC = 800-BC.