CEMC Gauss (Grade 7), 2023

Since $ABCD$ is a square and
its side lengths are integers, then its area is equal to a perfect
square.

Since the product of the areas of $ABCD$ and $EFGH$ (the rectangle) is equal to 98, then
the area of $ABCD$ is a divisor of
98.

The positive divisors of 98 are 1, 2, 7, 14, 49, and 98.

There are exactly two divisors of 98 that are perfect squares, namely 1
and 49.

Since the area of $ABCD$ is greater
than the area of $EFGH$, then the
area of $ABCD$ is 49, and so the area
of $EFGH$ is 2 (since $49\times2=98$).

Square $ABCD$ has area 49, and so
$AB=BC=CD=DA=7$.

The perimeter of $ABCDEFGH$ is
equal to $$\begin{align*} &AB+BC+CD+DE+EF+FG+GH+HA & \\ = &7+7+7+DE+EF+EH+GH+HA & \text{(since $EH=FG$)}\\ = &21+DE+EH+HA+EF+GH & \text{(reorganizing)}\\ = &21+DA+EF+GH & \text{(since $DE+EH+HA=DA$)}\\ = &21+7+EF+GH & \text{(since $DA=7$)}\\ = &28+2\times GH & \text{(since $EF=GH$)}\end{align*}$$
Since the side lengths are integers and the area of $EFGH$ is 2, then either $GH=1$ (and $FG=2$), or $GH=2$ (and $FG=1$).

If $GH=1$, then the perimeter of
$ABCDEFGH$ is $28+2\times1=30$.

Since 30 is not given as a possible answer, then $GH=2$ and the perimeter is $28+2\times2=32$.