Question 198599
I don't know how to a diagram, sorry.  But I do have the solution.

{{{a[1]=20*16}}} and {{{a[2]=.60a[1]}}} 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]=(20-x)(16-x)}}} is the new dimensions of the rectangle.
Solve
{{{a[1]=20*16=320}}}{{{a[2]=.60(320)=192}}}
Therefore
{{{192=(20-x)(16-x)}}}
Solve for x
{{{x^2-36x+320=192}}}{{{x^2-36x+128=0}}}
Quad From
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} when a=1, b=-36, c=128
Therefore
Real solutions:
Root 1: 4
Root 2: 32

Since 32 is a bit to large to take away from 20 or 16, the answer is 4.
4cm 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