Question 198598
I don't know how to a diagram, sorry.  But I do have the solution.

{{{a[1]=21*15}}} and {{{a[2]=216}}} where {{{a[1]}}} is area of the uncut rectangle and {{{a[2]}}} is the smaller one.  {{{x}}} will be the width cut from the initial rectangle.

{{{a[2]=(21-x)(15-x)}}} is the new dimensions of the rectangle.
Solve
{{{a[1]=21*15=315}}}{{{a[2]=216}}}
Therefore
{{{216=(21-x)(15-x)}}}
Solve for x
{{{x^2-36x+315=216}}}{{{x^2-36x+99=0}}}
Quad From
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} when a=1, b=-36, c=99
Therefore
Real solutions:
Root 1: 3
Root 2: 33

Since 33 is a bit to large to take away from 21 or 15, the answer is 3.
3cm needs to be cut from each side.


Anymore questions feel free to ask me.
Anthony Allie
arallie@gmail.com
http://arallie.webs.com/tutoring.htm