Question 1204789
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Find the perimeter of a right triangle whose hypotenuse is 2 and whose area is 1.
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<pre>
Let x and y be the legs.


Then we have 

    x^2 + y^ 2 = 4     (1)    (Pythagorean equation)

       xy      = 2     (2)    (from the area equation)


Multiply equation (2) by 2 and subtract if from equation (1).  Then you get

    x^2 - 2xy + y^2 = 0,

or

    (x-y)^2 = 0.


It implies x = y.


So, our triangle is isosceles right angled triangle.


Then from equation (2), we have

    x^2 = 2,

which implies x = y = {{{sqrt(2)}}}.


The perimeter then is  x + y + 2 = {{{sqrt(2)}}} + {{{sqrt(2)}}} + 2 = 2 + {{{2*sqrt(2)}}}  units.
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Solved.