Question 10295
Let x = width of the rectangle
y = length of the rectangle
5 = diagonal of the rectangle


By the Theorem of Pythagoras, 
{{{ x^2 + y^2 = 25}}}  By the way, this equation would be the equation of a circle, which is probably why this problem is in your conic section section of the book!!


Solve for y:
{{{y^2 = 25 - x^2}}}
{{{y = sqrt(25 - x^2) }}}


Area = xy = 10
{{{ x*sqrt(25- x^2) = 10}}}


Square both sides:
{{{x^2 * (25-x^2) = 100 }}}


Set the equation equal to zero by taking everything to the right side of the equation, in order to get the x^4 to have a positive coefficient.
{{{25x^2 - x^4 = 100 }}}
{{{0 = x^4 - 25x^2 + 100}}}


It doesn't always happen, but it sure feels good when it does--that math comes out even!  This does indeed factor!!

{{{0 = (x^2 - 20)(x^2 - 5) }}}

{{{x^2 = 20}}}
{{{x = 0+-sqrt(20)}}}
{{{x= 2*sqrt(5) or -2*sqrt(5)}}}


{{{x^2 = 5}}}
{{{x = sqrt(5) or -sqrt(5) }}}


It turns out that if {{{x = sqrt(5)}}}, then {{{y = 2sqrt(5)}}} and if {{{x=2sqrt(5)}}}, then {{{y =sqrt(5)}}}.  So there is actually only one solution.  The rectangle is {{{sqrt(5)}}} by {{{2sqrt(5)}}}.  


R^2 at SCC