Question 1113878
{{{ Length = 5/2 }}} units 
{{{ Width = 6/5 }}} units

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The Workout (assumes the max rectangle will have 1/2 the area of the 3-4-5 triangle, which I do not prove here, but it has been proven elsewhere):
Area of the 3-4-5 triangle is  {{{ (1/2)*4*3 = 6 }}} sq units.<br>

Area of largest rectangle is 1/2 that of the triangle, or  3 sq units.<br>
If the triangle is drawn with side=3 along the y-axis, side=4 along the x-axis, and the hypotenuse connecting the points (0,3) with (4,0),  then the corner of the rectangle that meets the x-axis is at (2,0) and the corner that meets the y-axis is at (0,3/2).

Thus the length of the rectangle is  {{{ sqrt(2^2 + (3/2)^2) = sqrt(25/4) = 5/2 }}} units.
and the width of the rectangle is  {{{ 3 / ((5/2))  = 6/5 }}} units.<br>

The picture below has blue lines drawn to help visualize the 3 pairs of congruent triangles.  Exactly 3 of the 6 form the rectangle. <br>

{{{ drawing ( 300, 300, 0, 5, 0, 5, grid(0),
                     line(0,3,4,0),
                     green(line(0,1.5,2,0)),
                     green(line(0, 1.5, 0.7, 2.5)),
                     green(line(2, 0,  2.7, 1)),
                    blue(line(0,1.5, 2,1.5)),
                    blue(line(2,1.5, 2,0))
    )
}}}