Question 156850
The hypotenuse of a right triangle is twice as long as one of the legs and 10 inches longer than the other. What are the lengths of the sides of the triangle?
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So far I know that i should use the Pythagorean theorem.
I've drawn the triangle and know:
1st Leg: h/2
2nd Leg: h-10
Hypotenuse: h 
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(h-10)^2 + (h/2)^2 = h^2  <<-- formula looks good!
4(h^2-20h-100+ (h^2/4))=h^2  <<--why didn't you multiply the right side by 4?
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Taking it from the top:
(h-10)^2 + (h/2)^2 = h^2
(h^2-20h+100) + (h^2/4) = h^2
4(h^2-20h+100) + 4(h^2/4) = 4h^2
4(h^2-20h+100) + h^2 = 4h^2
4h^2-80h+400 + h^2 = 4h^2
5h^2-80h+400 = 4h^2
h^2-80h+400 = 0
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Can't factor so you must use the quadratic equation.
Doing so results in:
h = {74.64, 5.36}
If we pick h=5.36, our 2nd leg would be negative -- throw out solution.
Conclusion:
h = 74.64
1st Leg: h/2 = 37.32 inches
2nd Leg: h-10 = 64.64 inches
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Reference:  calculations for the quadratic
*[invoke quadratic "x", 1, -80, 400 ]