Determine hypotenuse of right triangle if its area is 6 cm^2 and perimeter is 12 cm?

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Let the smaller sides of the triangle have a length a and b. The length of the hypotenuse is sqrt( a^2 + b^2)

The area of the triangle is (1/2)*a*b = 6

=> ab = 12

=> a = 12/b

The perimeter is a + b + sqrt (a^2 + b^2) = 12

a + b + sqrt (a^2 + b^2) = 12

replace 12 with ab

=> a + b + sqrt (a^2 + b^2) = ab

=> ab - (a + b) = sqrt (a^2 + b^2)

square the two sides

=> a^2*b^2 + (a +b)^2 - 2ab*(a + b) = a^2 + b^2

=> a^2*b^2 + a^2 + b^2 + 2ab - 2ab*(a + b) = a^2 + b^2

=> a^2*b^2  + 2ab - 2ab*(a + b) = 0

divide all the terms by ab

=> ab + 2 - 2(a + b) = 0

substitute a = 12/b

=> b*12/b + 2 - 2(12/b + b) = 0

=> 14 = 2*(12/b + b)

=> 7 = 12/b + b

=> 7b = 12 + b^2

=> b^2 - 7b + 12 = 0

=> b^2 - 4b - 3b + 12 = 0

=> b(b - 4) - 3(b - 4) = 0

=> (b - 3)(b - 4) = 0

b = 3 or b = 4

a = 4 or a = 3

The hypotenuse is sqrt (a^2 + b^2) = sqrt ( 3^2 + 4^2) = sqrt 25 = 5

The required hypotenuse = 5.

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