SOLUTION: An architect is designung a building with a right triangle footprint. The hypotenuse of the triangle is 80 feet longer than one leg of the triangle and 40 feet longer than the othe

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: An architect is designung a building with a right triangle footprint. The hypotenuse of the triangle is 80 feet longer than one leg of the triangle and 40 feet longer than the othe      Log On


   

Question 722822: An architect is designung a building with a right triangle footprint. The hypotenuse of the triangle is 80 feet longer than one leg of the triangle and 40 feet longer than the other leg. Use Pythagorean Theorem to find the dimensions of the footprint of the building.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x= length of the hypotenuse (in feet)
x-80 = length (in feet) of one leg of the triangle, and
x-40 = length (in feet) of one leg of the triangle
must be positive numbers.

The Pythagorean Theorem says that
%28x-80%29%5E2%2B%28x-40%29%5E2=x%5E2 --> x%5E2-160x%2B6400%2Bx%5E2-80x%2B1600=x%5E2 --> 2x%5E2-240x%2B8000=x%5E2 --> x%5E2-240x%2B8000=0

We can solve that quadratic equation by factoring, or by "completing the square", or by using the quadratic formula.

By factoring:
x%5E2-240x%2B8000=0 --> %28x-200%29%28x-40%29=0 --> x=200 or x=40

Completing the square:
x%5E2-240x%2B8000=0 --> x%5E2-240x=-8000 --> x%5E2-240x%2B120%5E2=120%5E2-8000 --> %28x-120%29%5E2=14400-8000 --> %28x-120%29%5E2=6400 --> %28x-120%29%5E2=80%5E2 --> x=120+%2B-+80 --> x=200 or x=40

Using the quadratic formula:
x%5E2-240x%2B8000=0 --> x+=+%28-%28-240%29+%2B-+sqrt%28%28-240%29%5E2-4%2A1%2A8000%29%29%2F%282%2A1%29+ --> x+=+%28240+%2B-+sqrt%2857600-32000%29%29%2F2 --> x+=+%28240+%2B-+sqrt%2825600%29%29%2F2 --> x+=+%28240+%2B-+160%29%2F2 --> x=120+%2B-+80 --> x=200 or x=40

x=200 yields x-80=120 and x-40=160 for a building with sides measuring highlight%28120%29, highlight%28160%29, and highlight%28200%29 feet.
x=40 is a solution to the quadratic equation, but it is not a solution to the problem, because it yields negative numbers for x-80 and x-40 and those negative numbers cannot be the lengths of the sides of a building.