Question 140787
Let call the length and width of the rectangle L and W.
1.{{{L^2+W^2=20}}}
The original perimeter is 2L+2W.
2.{{{2(L+2)+2(W+2)=(5/3)(2L+2W)}}}
Simplifying 2 (multiply both sides by 3 to get rid of denominators), 
{{{6(L+2)+6(W+2)=5(2L+2W)}}}
{{{6L+12+6W+12=10L+10W}}}
{{{4L+4W=24}}}
{{{L+W=6}}}
{{{L=6-W}}}
Now plug this value into 1.
1.{{{L^2+W^2=20}}}
{{{(6-W)^2+W^2=20}}}
{{{(36-12W+W^2)+W^2=20}}}
{{{2W^2-12W+36=20}}}
{{{2W^2-12W+16=0}}}
{{{W^2-6W+8=0}}}
{{{(W-4)(W-2)=0}}}
The two solutions are W=4 and W=2.
For W=4, L=6-W=2.
For W=2, L=6-W=4.
The dimensions of the rectangle are 2 cm by 4 cm.