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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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    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?2012-10-29

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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.