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The area of a right triangle is numerically equal to its perimeter. Let the smaller sides of the triangle be x and y. The third side is `sqrt(x^2 +y^2)` . The area of the triangle is `(x*y)/2`
This gives : `x + y + sqrt(x^2 + y^2) = (x*y)/2`
=> `x + y = (x*y)/2 - sqrt(x^2 + y^2)`
=> `x^2 + y^2 + 2xy = (x^2*y^2)/4 + (x^2 + y^2) - x*y*sqrt(x^2 + y^2)`
=> `2xy = (x^2*y^2)/4 - x*y*sqrt(x^2 + y^2)`
=> `2 = (xy)/4 - sqrt(x^2 + y^2)`
=> `8 = xy - 4*sqrt(x^2 + y^2)`
There isn't a unique solution for this equation as it has two variables and no restrictions on the values of x and y.
It is not possible to determine a unique result for the dimensions of the triangle.
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