Bordered up

Let the dimensions (in feet) of the rectangle be $x,y$ . Then the dimensions of the outside of the border are $x + 2$ and $y + 2$ . The area of the border is then $(x + 2)(y + 2) - xy$ , which we require to equal the area of the rectangle, i.e.,

$(x + 2)(y + 2) - xy = xy \Longrightarrow (xy + 2x + 2y + 4) - xy = xy \Longrightarrow 2y + 4 = xy - 2x$

$\Longrightarrow 2(y + 2) = x(y - 2) \Longrightarrow x = 2\left(\dfrac{y + 2}{y - 2}\right) = 2\left(1 + \dfrac{4}{y - 2}\right) = 2 + \dfrac{8}{y - 2}$ .

Now for $x$ to be an integer we require that $(y - 2)|8 \Longrightarrow y - 2 = \pm 1, \pm 2, \pm 4, \pm 8$ .

Since $y \lt 10$ must be a positive integer the only valid options are $y - 2 = -1, 1, 2, 4 \Longrightarrow y = 1, 3, 4, 6$ .

This in turn yields solution pairs $(x,y) = (-6,1), (3,10), (4,6), (6,4)$ . Since both of $x$ and $y$ must be positive integers less than $10$ the only possible dimensions are $(4,6), (6,4)$ , which in both cases yield an area for the rectangle of $\boxed{24}$ square feet.