Question 369221
From the given, {{{b^2 + h^2 = 100}}}, from the Pythagorean theorem.  Also, {{{A = 20 = (1/2)bh}}}, or {{{bh = 40}}}, or {{{b = 40/h}}}.  By direct substitution, {{{1600/h^2 + h^2 = 100}}}, or
{{{h^4-100h^2+1600 = 0}}},
{{{(h^2-80)(h^2-20) = 0}}}.  From this we get
{{{h = sqrt(80) = 4sqrt(5)}}} or {{{h = sqrt(20) = 2sqrt(5)}}} . For each value of h we get {{{b = 2sqrt(5)}}} or {{{b = 2sqrt(5)}}} respectively.  Therefore there is only basically one right triangle formed, one whose leg is {{{2sqrt(5)}}} and the other leg {{{4sqrt(5)}}} .