Problem 5. From point $O$ in rectangle $A B C D$, lines are drawn parallel to the sides of the rectangle, dividing it into four smaller rectangles $P O E D, O M C E, A K O P$, and $K B M O$ (see diagram).

1) If the areas of rectangles

![](https://cdn.mathpix.com/cropped/2024_06_05_dc4d34c0a9dd2a3ba572g-04.jpg?height=308&width=387&top_left_y=1730&top_left_x=1075)
POED, OMCE, AKOP, are 2, 4, 6 respectively, determine the area of rectangle $A B C D$.

2) If the perimeters of rectangles POED, OMCE, AKOP, are 6, 10, 10 respectively, calculate the perimeter of rectangle $A B C D$.
Solution.1) Since the area of OMCE is twice the area of $P O E D$, it follows that side $O M$ is twice as large as side $O P$. Similarly, since the area of $A K O P$ is three times the area of $P O E D$, it follows that $O K$ is three times larger than $O E$. We conclude that the area of rectangle BMOK is $2 \cdot 3=6$ times larger than the area of rectangle $P O E D$, i.e., it is $6 \cdot 2=12$. The area of $A B C D$ is the sum of the areas of the four rectangles and is equal to $2+4+6+12=24$.

2) From the fact that the perimeter of $O M C E$ is 4 units larger than the perimeter of $P O E D$, it follows that $O M$ is $4: 2=2$ units larger than $O P$. Similarly, $O K$ is two units larger than $O E$. It follows that the perimeter of $K B M O$ is $2 \cdot 2+2 \cdot 2=8$ units larger than the perimeter of $P O E D$, and is $6+8=14$. To determine the perimeter of rectangle $A B C D$, the perimeters of the four rectangles need to be summed and the result divided by 2, i.e., the perimeter of $A B C D$ is $\frac{6+10+10+14}{2}=20$.

## VI section