Two sides of a triangle forming a right angle are 6x^2 and (2x^2-1). If the area of the triangle is 84 square units, then what is the perimeter of the triangle?

Explanation

Area = \frac{1}{2}(6x^2)(2x^2-1) = 84 \implies x^2(2x^2-1) = 28. Let y = x^2, so 2y^2 - y - 28 = 0 \implies (2y+7)(y-4) = 0 \implies x^2 = 4. The sides are 6(4)=24 and 2(4)-1=7. Hypotenuse = \sqrt{24^2+7^2} = 25. Perimeter = 24+7+25 = 56.

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