Question 1030260
Let x be such that {{{0 <= x <= 2a}}}.  Then the adjacent sides of the rectangle would have sides x and 2a - x. (Why?)
The area of the rectangle would be {{{A(x) =  x(2a - x)}}}.

The area under the curve is given by {{{int(x(2a - x), dx, 0,2a) = int((2ax - x^2), dx, 0,2a) = (ax^2 - x^3/3)[0]^(2a) = (4/3)a^3}}} square units.

==> The cdf for the area is given by {{{F(x) = (3/(4a^3))int((2aw - w^2), dw, 0,x) = (3ax^2 - x^3)/(4a^3)}}}.

Now solve for the bounds on x that will give a rectangular area of {{{a^2/2}}}:

{{{A(x) =  x(2a - x) = a^2/2}}}

<==> {{{0 = x^2 - 2ax + a^2/2}}} after simplifying.

==> {{{x = ((2-sqrt(2))/2)a}}}, {{{x = ((2+sqrt(2))/2)a}}}.

==> {{{P(A >= a^2/2) = P(((2-sqrt(2))/2)a <= x <= ((2+sqrt(2))/2)a)}}}

={{{F(((2+sqrt(2))/2)a) - F(((2-sqrt(2))/2)a) = (8+5*sqrt(2))/16 - (8+5*sqrt(2))/16 = (5*sqrt(2))/8}}}.

This is approximately equal to {{{highlight(0.88388)}}} to five significant figures.