Eight-sided polygon $A B C D E F G H$ has integer side lengths. It can be divided into a rectangle and a square, as shown. The area of the square is greater than the area of the rectangle. The product of the two areas is equal to 98 . Which of the following could be the perimeter of $A B C D E F G H$ ?
(A) 51
(B) 32
(C) 44
(D) 34
(E) 33

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-004.jpg?height=358&width=460&top_left_y=1122&top_left_x=1277)
Since $A B C D$ 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 $A B C D$ and $E F G H$ (the rectangle) is equal to 98 , then the area of $A B C D$ 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 $A B C D$ is greater than the area of $E F G H$, then the area of $A B C D$ is 49 , and so the area of $E F G H$ is 2 (since $49 \times 2=98)$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-007.jpg?height=377&width=464&top_left_y=1053&top_left_x=1557)

Square $A B C D$ has area 49 , and so $A B=B C=C D=D A=7$.

The perimeter of $A B C D E F G H$ is equal to

$$
\begin{aligned}
& A B+B C+C D+D E+E F+F G+G H+H A \\
= & 7+7+7+D E+E F+E H+G H+H A \quad(\text { since } E H=F G) \\
= & 21+D E+E H+H A+E F+G H \quad(\text { reorganizing) } \\
= & 21+D A+E F+G H \quad(\text { since } D E+E H+H A=D A) \\
= & 21+7+E F+G H \quad(\text { since } D A=7) \\
= & 28+2 \times G H \quad(\text { since } E F=G H)
\end{aligned}
$$

Since the side lengths are integers and the area of $E F G H$ is 2, then either $G H=1$ (and $F G=2$ ), or $G H=2$ (and $F G=1$ ).

If $G H=1$, then the perimeter of $A B C D E F G H$ is $28+2 \times 1=30$.

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

ANSWER: (B)