3. Given is a rectangle $A B C D$. If we decrease the length of each side of the first pair of opposite sides of rectangle $A B C D$ by $3 \mathrm{~cm}$, and increase the length of each side of the second pair of opposite sides of that rectangle by $7 \mathrm{~cm}$, then a new rectangle is formed whose area is equal to the area of rectangle $A B C D$. If we increase the length of each side of the first pair of opposite sides of rectangle $A B C D$ by $3 \mathrm{~cm}$, and decrease the length of each side of the second pair of opposite sides of that rectangle by $5 \mathrm{~cm}$, then a new rectangle is formed whose area is again equal to the area of rectangle $A B C D$. What is the area of rectangle $A B C D$?
3. 

![](https://cdn.mathpix.com/cropped/2024_05_30_3cbbf7de75a41eab3027g-03.jpg?height=828&width=405&top_left_y=363&top_left_x=400)

The first image shows a rectangle $A B C D$ with sides $a$ and $b$, and a new rectangle with sides $a-3$ and $b+7$. Their areas are equal. If we subtract the same number (the area of their common part) from both, we will still have equal numbers. Therefore, the areas of the highlighted rectangles in the image are equal. The sides of one rectangle are $3 \mathrm{~cm}$ and $b \mathrm{~cm}$, and the other are $7 \mathrm{~cm}$ and $a-3 \mathrm{~cm}$, so $3 b=7 a-21$.

![](https://cdn.mathpix.com/cropped/2024_05_30_3cbbf7de75a41eab3027g-03.jpg?height=757&width=471&top_left_y=1820&top_left_x=333)

The second image shows the initial rectangle $A B C D$ with sides $a$ and $b$, and a new rectangle with sides $a+3$ and $b-5$. Similarly, it can be concluded that the areas of the highlighted rectangles in the image are equal. The sides of one are $3 \mathrm{~cm}$ and $b-5 \mathrm{~cm}$, and the other are $5 \mathrm{~cm}$ and $a \mathrm{~cm}$. It follows that $3 \cdot(b-5)=5 a$ or $3 b=5 a+15$.

From $3 b=7 a-21$ and $3 b=5 a+15$, we get $7 a-21=5 a+15$, so $a=18 \mathrm{~cm}$.

Then $b=35 \mathrm{~cm}$.

The area of rectangle $A B C D$ is $630 \mathrm{~cm}^{2}$.