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The hypotenuse of a right triangle is $5 m$ if the smaller is doubles and longer is triples the new hypotenuse is $6\sqrt{5} m$. FInd the sides of the triangle.

What I found so far: After coming up the equation in both we will come upon a case where we will find a quadratic equation in one variable where we will have to apply Discriminant Formula or Factorisation formula to come up with the solution.

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So what is the difficulty? Have you found the quadratic equation, but you can't solve it? –  Gerry Myerson Oct 29 '12 at 23:44
    
Nope, I can easily find the solution. But I am having difficulty in coming up with that Quadratic Equation. How can I get that Quadratic Equation? –  Alpha Oct 29 '12 at 23:48

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up vote 3 down vote accepted

1. $$x^2 + y^2=5^2$$ $$x^2 = 25-y^2$$ 2. $$(2x)^2 + (3y)^2 = (6\sqrt{5})^2$$ $$4x^2 + 9y^2 = 180$$ 3. Substitute $x^2 = 25-y^2$ $$4(25-y^2) + 9y^2 = 180$$ $$100-4y^2+9y^2=180$$ $$5y^2=80$$ $$y^2=16$$ $$y=4$$ $$x=3$$ Negative root rejected because side lengths are positive.

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Did you not notice the "homework" tag? Did you really want to leave OP nothing to do at all? –  Gerry Myerson Oct 29 '12 at 23:55

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