Question 947304
Let L=length; Let W=Width; Perimeter=P=2(L+W); Diagonal=D=hypotenuse
L and W are two legs of a right triangle with the diagonal as hypotenuse, so:
P=2(L+W)
22=2(L+W) Divide each side by 2
11=L+W Solve for W, subtract L from each side
11-L=W
{{{L^2+W^2=D^2}}}
{{{L^2+W^2=(sqrt(61))^2}}} 
{{{L^2+W^2=61}}} Substitute for W
{{{L^2+(11-L)^2=61}}}
{{{L^2+(121-22L+L^2)=61}}}
{{{2L^2-22L+121=61}}} Subtract 61 from each side
{{{2L^2-22L+60=0}}}
*[invoke quadratic "L", 2, -22, 60 ]
ANSWER:Length is 6 or 5
If L=6 units, then W=11-L=11-6=5 units
If L=5 units, then W=11-L=11-5=6 units
ANSWER: The dimensions are 6 units x 5 units (or 5 units x 6 units)
The area is the same either way: Area =L x W=6 units x 5 units= 30 square units