Question 291783
Ok so lets call one leg x
The other leg is three times the length 3x and 2 cm less so 3x-2.
The Area of a triangle is {{{A=(1/2)*b*h}}}
For a right Triangle the base is one leg and the height is the other leg.
We also know that the Area is 48. {{{A=48}}}
So {{{A=48=(1/2)*x(3x-2). x is one side/leg, and 3x-2 is the other side/leg
{{{48=(1/2)*(3x^2-2x)}}} Multiply by 2 to cancel the {{{1/2}}}
{{{96=3x^2-2x}}} Subtract 96 from both sides.
{{{3x^2-3x-96=0}}} Factor out the 3.
{{{3(x^2-x-32)=0}}} Use the qudratic formula to solve/
*[invoke quadratic "x", 1, -1, -32 ]

And since you cannot have a negative length in a triangle, {{{6.17890834580027}}} is the length of the leg we called x. To find the length of the other leg multiply by 3 and then subtract 2 to get the length of the other leg.
{{{3*6.17890834580027-2=16.53672503740081}}}
So the length of the other leg is {{{16.53672503740081}}}