Question 847137
x and y is width and length of rectangle P.
xy=A for rectangle P
2x+2y=p for rectangle P, using lowercase p for perimeter.


"area of rect P is 12 ft^2";
A=xy=12.  
You could find y as a formula and substitute into the perimeter equation for rectangle P.
y=12/x;
2x+2(12/x)=p
2x+24/x=p

--
--
While being unproductive here, one should know that if the perimeters of P and Q are in a certain ratio, then the side lengths are also in this ratio.  If the two rectangles are similar, then the problem could be easier.


Rectangle P.
x and y as before.
2x+2y=p and xy=A, for P.


2/5 is Perimeter P to perimeter Q.
Rectangle Q.
2(5/2)x+2(5/2)y=perimQ
2(5/2)(x+y)=perimQ
{{{highlight((5/2)(2x+2y)=perimQ)}}}
-
(5/2)x(5/2)y=areaQ
{{{highlight((5/2)^2*xy=areaQ)}}}


Recall that {{{highlight_green(xy=12*ft^2)}}}.
Substituting this value for xy, the area of P, into the "areaQ" equation, we can compute the area of rectangle Q.
-
{{{(5/2)^2*12=areaQ}}}
{{{(25/4)*12}}}
{{{highlight(highlight(areaQ=75))}}}