Question 847449
The square, side s.
Perimeter is {{{4s}}} and area is {{{s^2}}}.


The rectangle, sides x and 2x.
Perimeter is {{{2*x+2*2x=6x}}}.


Both perimeters are specified to be equal.
{{{4s=6x}}}
{{{2s=3x}}}.
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Solving for s, {{{s=(3/2)x}}} and this is to be used for finding area for the square and for the rectangle, as expressions.


Area for the square:  {{{((3/2)x)^2=highlight((9/4)x^2)}}}.
Area for the rectangle:   {{{x*2x=highlight(2x^2)}}}.
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RATIO of the areas of the square to the rectangle:  
{{{((9/4)x^2)/(2x^2)}}}
{{{(9/4)/2}}}
{{{highlight(9/8)}}}.