Question 193575
One leg of a right triangle is 96 inches. Find the hypotenuse and teh other leg if the length of the hypotenuse exceeds 2 and 1/2 times teh other leg by 4 inches.
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Let x = length of "other leg" of triangle
then 
(5/2)x-4 = hypotenuse
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The applying the Pythagorean theorem:
x^2 + ((5/2)x-4) = 96^2
x^2 + ((5/2)x-4)((5/2)x-4) = 96^2
x^2 + (25/4)x^2-20x +16 = 96^2
(29/4)x^2-20x +16 = 9216
(29/4)x^2-20x -9200 = 0
29x^2 - 80x -36800 = 0
Solving using the quadratic equation we get:
x ={37.03, -34.26}
Throwing away the negative solution we're left with:
x = 37.03 inches (other leg)
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Hypotenuse:
(5/2)x-4 = (5/2)37.03-4 = 88.58 inches (hypotenuse)
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Details of the quadratic equation:
*[invoke quadratic "x", 29, -80, -36800 ]