3. The carpet owner has a rectangular carpet, the size of which is unknown. Unfortunately, his measuring tape is broken, and he has no other measuring tools. However, he discovers that if he lays the carpet flat in either of his two store rooms, each corner of the carpet exactly meets a different wall. He knows that the length of the carpet in feet is an integer, and the two rooms have the same length (unknown), with one side being 88 feet and the other side being 55 feet. Find the dimensions of the carpet.
3. Let the unknown side of the room be \( q \) chi. From the previous problem, we have
\[
\begin{array}{l}
(k q-50)^{2}+(50 k-q)^{2} \\
=(k q-38)^{2}+(38 k-q)^{2}
\end{array}
\]

Thus,
\[
\begin{aligned}
& k^{2} q^{2}-100 k q+2500+2500 k^{2}-100 k q+q^{2} \\
= & k^{2} q^{2}-76 k q+1444+1444 k^{2}-76 k q+q^{2} .
\end{aligned}
\]
\[
\text{Therefore, } 1056 k^{2}-48 k q+1056=0 \text{, }
\]

Hence,
\[
\begin{array}{l}
22 k^{2}-k q+22=0, \\
k q=22\left(k^{2}+1\right), \\
q=22\left(k+\frac{1}{k}\right) .
\end{array}
\]

Since \( k=\frac{y}{x} \) is a rational number, let \( k=\frac{c}{d} \), where \( c \) and \( d \) are positive integers, and \( (c, d)=1 \). Then
\[
q=22\left(\frac{c}{d}+\frac{d}{c}\right) .
\]

Thus, \( 22\left(c+\frac{d^{2}}{c}\right)=d q \) is an integer. Therefore, \( c \) is a divisor of 22, and similarly, \( d \) is a divisor of 22, so
\[
\text{c, } d \in\{1,2,11,22\} \text{. }
\]

Since \( (c, d)=1 \), we can assume \( c > d \),
\[
k=1,2,11,22, \frac{11}{2} \text{. }
\]

Correspondingly,
\[
q=44,55,244,485,125 .
\]

Similar to the previous problem,
\[
x^{2}\left(k^{2}-1\right)^{2}=\left(k^{2}+1\right)\left(q^{2}-1900\right) \text{. }
\]
\( k=1 \) leads to \( 0=72 \), which is a contradiction. For \( k=2, q=55 \), we get the solution from the previous problem.

If \( k=11 \), then \( 61 \mid\left(k^{2}+1\right), 61^{2} \mid\left(k^{2}+1\right) \). \( 61 \mid q^{2} \), but 61 does not divide 1900. Therefore, \( 61 \mid \) the right side of (3.1), but \( 61^{2} \) does not divide the right side of (3.1). This contradicts the fact that the left side of (3.1) is a perfect square.
If \( k=22 \), similarly, \( 97 \mid \) the right side of (3.1), but \( 97^{2} \) does not divide the right side of (3.1), leading to a contradiction.
If \( k=\frac{11}{2}, k^{2}+1=\frac{125}{4}, k^{2}-1=\frac{117}{4} \),
\( q=125 \), then
\[
\begin{aligned}
x^{2}(117)^{2} & =4(125)\left(125^{2}-1900\right) \\
& =50^{2} \times 5 \times 549 .
\end{aligned}
\]
\( 5^{5} \) divides the right side, but \( 5^{6} \) does not, leading to a contradiction.
Therefore, \( k=2, q=55 \), \( x=25 \) chi, \( y=50 \) chi.
Comment: The second problem is a variation of this problem.