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

We'll put the length of the legs of the right triangle as
x and y.

A = x*y/2

6 = x*y/2
=> x*y = 12 (1)

Since the perimeter is the sum of
the lengths of the legs and hypotenuse of right angle triangle, we'll
get

P = x + y + sqrt(x^2 +
y^2)

Hypothenuse = sqrt(x^2 + y^2) (Pythagorean
identity)

12 = x + y + sqrt(x^2 + y^2)
(2)

We'll equate (1) and
(2):

xy = x + y + sqrt(x^2 +
y^2)

We'll subtract both sides  x +
y:

xy - (x+y) = sqrt(x^2 +
y^2)

We'll raise to square both
sides:

[xy - (x+y)]^2 = [sqrt(x^2 +
y^2)]^2

(xy)^2 + x^2 + 2xy + y^2 - 2xy(x+y) = x^2 +
y^2

We'll eliminate x^2 +
y^2:

(xy)^2 + 2xy - 2xy(x+y) =
0

We'll factorize by xy:

xy(xy
+ 2 - 2x - 2y) = 0

We'll put xy =
0

Since xy = 0 => 2 - 2x - 2y =
0

-2(x+y) = -2

x + y =
1

We'll create the quadratic equation whose sum is 1 and
product is 0:

x^2 - x = 12

x^2
- x - 12 = 0

x1 = [1+sqrt(1 +
48)]/2

x1 = (1+7)/2

x1 =
4

x2 = -3

Since a length of a
side cannot be negative, we'll reject x = -3.

We'll put x =
4

6 = 4*y/2

12 =
4*y

y = 3

The lengths of the
legs of the triangle are x = 4 and y = 3.

The hypothenuse
is:

H = sqrt(3^2 + 4^2)

H =
sqrt25

H = 5
cm

The length of hypotenuse is H = 5
cm.