7. (3 points) As shown in the figure, a line segment divides a rectangle with a perimeter of $30 \mathrm{~cm}$ into a square and a smaller rectangle. If the perimeter of the smaller rectangle is $16 \mathrm{~cm}$, then the area of the original rectangle is $\qquad$ $\mathrm{cm}^{2}$.
【Analysis】The perimeter of the large rectangle is reduced to that of the small rectangle by: $30-16=14$ (cm), which is equivalent to reducing two sides of the square, so the side length of the square is: $14 \div 2=7$ (cm), which means the original width of the rectangle is 7 cm; thus, the original length of the rectangle is: $16 \div 2-7+7=8$ (cm), and the area is: $8 \times 7=56 \mathrm{~cm}^{2}$.

【Solution】Solution: According to the analysis,
$30-16=14$ (cm),
Side length of the square: $14 \div 2=7$ (cm),
Original length of the rectangle: $16 \div 2-7+7=8$ (cm),
Area: $8 \times 7=56$ (square cm);
Answer: The original area of the rectangle is $56 \mathrm{~cm}^{2}$.
Therefore, the answer is: 56.