Question 101752
Ok first lets sketch out what we know
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From the drawing we can see that the diagonal forms a right triangle.
So we can use the pythagorean formula to solve for x.
The pythagorean formula is {{{a^2+b^2=c^2}}} where c is the hypotenuse of a right triangle. a and b are the other two sides of a right triangle.
So lets start:
{{{x^2+(x-3)^2=15^2}}}
lets start on the right side and go ahead and square 15
{{{x^2+(x-3)^2=225}}}
now on the left side square (x-3) use the foil method to do this 
{{{x^2+x^2-3x-3x+9=225}}}
combine like terms
{{{2x^2-6x+9=225}}}
Ok the pythagorean formula got us this far but now we need to use the quadratic formula to continue solving for x. To do this we first need to set the the equation equal to zero. so...
{{{2x^2-6x+9=225}}}
subtract 225 from both sides
{{{2x^2-6x-216=0}}}
now we can use the quadratic formula to solve for x
*[invoke quadratic "x", 2, -6, -216]

Ok so when solving quadratic equations you get two answers.
the answers we got here are x = -9 and x = 12
They are both correct answers mathematically.
But for our problem we are trying to find the length and width of a rectangle and neither of these can be a negative number.
So we will throw out the x = -9 answer as an 'extraneous solution' 
That leaves just x = 12
Ok back to the rectangle from the diagram we started with we can see that the:
Length = x    and the  width = x - 3
so
Length = 12   and the width = 12 - 3 = 9
Find area of rectangle by Length times width
12 * 9 = 108
So finally the area of the rectangle is {{{108cm^2}}}