Question 657795
let l be the length of the rectangle and w be the width of the rectangle.  With the diagonal being 20 feet long, using the Pythagorean Theormem, we get 
{{{sqrt(l^2 + w^2) = 20}}}
{{{l^2 + w^2 = 400}}}

With a perimeter of 52 feet, we have
{{{2*l + 2*w = 52}}}
{{{l + w = 26}}}
{{{l = 26 - w}}}

Substituting, we get
{{{(26-w)^2 + w^2 = 400}}}
{{{(676-52w+w^2) + w^2 = 400}}}
{{{2*w^2 - 52w + 276 = 0}}}

We can solve for w using the quadratic equation with a=2, b=-52, and c=276:

{{{w = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{w = (-(-52) +- sqrt( (-52)^2-4*2*276))/(2*2) }}}
{{{w = (52 +- sqrt(2704-2208)) / 4 = (52 +- sqrt(16*31))/4 = (52 +- 4sqrt(31))/4}}}
{{{w = 13 +- sqrt(31) }}}
{{{ w = 13 + sqrt(31) }}} and {{{ l = 26-w = 26 - (13+sqrt(31)) = 13-sqrt(31) }}}, or
{{{ w = 13 - sqrt(31) }}} and {{{ l = 26-w = 26 - (13-sqrt(31)) = 13+sqrt(31) }}}
Both of these are basically the same solution with the l and w values being switched.  So the area is:
{{{ a = l * w = (13+sqrt(31))*(13-sqrt(31)) = 169+13sqrt(31)-13sqrt(31)-(sqrt(31))^2 = 169 - 31 = 138}}}

The area is 138 square feet.