Question 545262
The hypotenuse of a right triangle is 5 feet long. One leg is 3 feet longer then the other leg. Find the perimeter of the triangle. 
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Let x = length of "other leg"
then
x+3 = length of "one leg"
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applying Pythagorean theorem:
x^2 + (x+3)^2 = 5^2
x^2 + (x+3)(x+3) = 25
x^2 + x^2+6x+9 = 25
2x^2+6x+9 = 25
2x^2+6x-16 = 0
x^2+3x-8 = 0
applying the "quadratic formula" results in:
x = {1.7, -4.7}
Throw out the negative solution leaving:
x = 1.7 feet (other leg)
"one leg":
x+3 = 1.7+3 = 4.7 feet
Perimeter:
4.7 + 1.7 + 5 = 11.4 feet
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Details of quadratic to follow:
*[invoke quadratic "x", 1, 3, -8 ]