Arectangle is twice twice as long as it is wide and has the same perimeter as a square whose area is 9 square feet feet larger than that of the rectangle. what are the dimensions of both the rectangle and the square?

If the width of the rectangle is w and the length is l, then l=2w (since the rectangle is twice as long as it is wide) and w*l=w*(2w), as well as 2w+2l being 2w+2(2w)=2w+4w=6w. Since the perimeter (2w+2l) is the same as the perimeter of the square (if one side length of the square is x, then the perimeter is 4x), 6w=4x and by dividing both sides by 4, we get 6w/4=x=3w/2. 
Since the area of the square is x^2, we can plug 3w/2 in for x to get
(3w/2)^2=9w^2/4 as the area of the square. Since the area of the square is 9 feet larger than the rectangle, 9w^2/4-9=2*l=w*(2w)=2w^2. Subtracting 2w^2 from both sides of 2w^2=9w^2/4-9 as well as adding 9 to both sides, we get w^2/4=9 and by multiplying 4 to both sides we get w^2=46 and by square rooting both sides we get w=sqrt(36)=6. Since the length is 2w, 6*2=12=the length. Since the area of the rectangle is l*w=6*12=72, the square's area is 9 more than that and is 72+9=81. Since x^2 is the area of the square which equals 81, 81=x^2 and by square rooting both sides we get x=9

No i refuse to tell you the answer but the estimate is 10