Question 986989
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L=length; W=width=L-2m; A=Area=0.003m^2; P=perimeter
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{{{LW=A}}}
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{{{L(L-2m)=0.003m^2}}}
{{{L^2-2L=0.003}}}
{{{L^2-2L-0.003=0}}}
*[invoke quadratic "L", 1, -2, -0.003 ]
Length=2.0015 meter
Width=2.0015-2=0.0015 meter
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P=2(L+W)=2(2.0015m+0.0015)=2(2.003)=4.006 meters
ANSWER: The perimeter is 4.006 meters.
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CHECK:
{{{LW=A}}}
{{{(2.0015m)(0.0015m)=0.003m^2}}}
{{{0.003m^2=0.003m^2}}}
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I think the problem was entered incorrectly.
A 30cm^2 room would be very small.
If the difference in length and width was in centimeters, 
or the area was in square meters, you would get a different 
result.  Process would be the same.