Problem 4. In the rectangle $A B C D$, the length $\overline{A B}$ is twice the width $\overline{B C}$. On the side $\overline{A B}$, a point $P$ is chosen such that $P$ is the midpoint of $\overline{A B}$. The segment $\overline{P C}$ divides the rectangle into one quadrilateral and one triangle whose perimeters differ by $20 \mathrm{~cm}$. Determine the area of the rectangle $A B C D$.
Solution. If we denote $\overline{B C}=b$, then $\overline{A P}=\overline{P B}=\overline{A D}=b$. The perimeter of quadrilateral $A P C D$ is $b+b+2 b+\overline{P C}$, and the perimeter of triangle $P B C$ is $b+b+\overline{P C}$. From the condition of the problem, we have

![](https://cdn.mathpix.com/cropped/2024_06_05_a1a5b9bf3a5d296219b1g-05.jpg?height=260&width=450&top_left_y=1336&top_left_x=1015)
that the perimeters differ by 20, so we get $2 b=20$, hence $b=10 \mathrm{~cm}$. For the length of the rectangle $\overline{A B}=a$ we have $a=2 b=20 \mathrm{~cm}$, and for the area $P=a b=10 \cdot 20=200 \mathrm{~cm}^{2}$.