There is no other measuring tool, but it was found that by laying this carpet flat in each of the two store rooms and securing each corner to the room's different measurement points - if the dimensions of the two rooms are 38 feet $\times 55$ feet and 50 feet $\times 55$ feet, find the area of the carpet.
2. Let the side lengths of the carpet be $x, y$.
It is easy to see that
$\triangle A E H \cong \triangle C G F \backsim \triangle B F E \cong \triangle D H G$.
Let $\frac{y}{x}=k$, which is the similarity ratio of the two sets of triangles. Let $A E=a, A H=b$. By similarity, $B E=k b$, $D H=k a$. Therefore,
$$
\begin{array}{l}
a+k b=50, \\
k a+b=55 .
\end{array}
$$

Thus,
$$
\begin{array}{c}
a=\frac{55 k-50}{k^{2}-1}, \\
b=\frac{50 k-55}{k^{2}-1} . \\
x^{2}=\left(\frac{55 k-50}{k^{2}-1}\right)^{2}+\left(-\frac{0 k-55}{k^{2}-1}\right)^{2} \\
=\frac{3025 k^{2}-5500 k+2500+2500 k^{2}}{\left(k^{2}-1\right)} \\
-\frac{5500 k-3025}{\left(k^{2}-1\right)^{2}} .
\end{array}
$$

Similarly, from the other room we get
$$
x^{2}=\left(\frac{55 k-38}{k^{2}-1}\right)^{2}+\left(\frac{38 k-55}{k^{2}-1}\right)^{2} .
$$

Simplifying, we have
$$
\begin{aligned}
x^{2}\left(k^{2}-1\right)^{2} & =3025 k^{2}-4180 k+1444 \\
& +1444 k^{2}-4180 k+3025 \\
& =4469 k^{2}-8360 k+4469
\end{aligned}
$$

Comparing the two equations, we get
$$
\begin{aligned}
& 5525 k^{2}-11000 k+5525 \\
= & 4469 k^{2}-8360 k+4469,
\end{aligned}
$$

which simplifies to
$$
1056 k^{2}-2640 k+1056=0,
$$
$$
2 k^{2}-5 k+2=0 \text {. }
$$
$k=2$ or $\frac{1}{2}$, meaning one side is twice the length of the other. Without loss of generality, let $y=2 x$, then $a+2 b$
$$
\begin{array}{l}
=50,2 a+b=55 . \text { Thus } a=20, b=15 . \\
x^{2}=a^{2}+b^{2}=20^{2}+15^{2}=400+225=625, \\
x=25, y=50 .
\end{array}
$$

The dimensions of the carpet are 25 ft $\times 50$ ft.