5. Mislav cut out two rectangles of different lengths but the same width from paper.
![](https://cdn.mathpix.com/cropped/2024_06_03_47603db0a075197c1d6ag-6.jpg?height=150&width=970&top_left_y=336&top_left_x=565)

He glued them so that they partially overlap as shown in the figure and obtained a new rectangle.

![](https://cdn.mathpix.com/cropped/2024_06_03_47603db0a075197c1d6ag-6.jpg?height=123&width=822&top_left_y=607&top_left_x=614)

The overlapping (shaded) part has an area of $18 \mathrm{~cm}^{2}$, and its width is equal to the width of the original rectangles. The length of the overlapping part is equal to one-sixth of the length of the smaller rectangle and one-eighth of the length of the larger rectangle.

What is the perimeter of the newly obtained rectangle if the width of the larger rectangle is four times smaller than its length?

## Tasks for 10 points:
5. Mislav cut out two rectangles of different lengths but the same width from paper.
![](https://cdn.mathpix.com/cropped/2024_05_30_1ae5e90785d026149547g-16.jpg?height=148&width=970&top_left_y=1942&top_left_x=565)

He glued them together so that they partially overlap as shown in the figure and obtained a new rectangle.

![](https://cdn.mathpix.com/cropped/2024_05_30_1ae5e90785d026149547g-16.jpg?height=137&width=820&top_left_y=2213&top_left_x=618)

The overlapping (shaded) part has an area of $18 \mathrm{~cm}^{2}$, and its width is equal to the width of the original rectangles. The length of the overlapping part is equal to one-sixth of the length of the smaller rectangle and one-eighth of the length of the larger rectangle.

What is the perimeter of the new rectangle obtained if the width of the larger rectangle is four times smaller than its length?

## Solution.

Let the length of the overlapping part be $x$, then the length of the larger rectangle is $8 x$, and the length of the smaller rectangle is $6 x$.

![](https://cdn.mathpix.com/cropped/2024_05_30_1ae5e90785d026149547g-17.jpg?height=263&width=1405&top_left_y=474&top_left_x=317)

From the area of the overlapping part, we calculate the length $x$:

$$
\begin{array}{lr}
x \cdot 2 x=18 & 1 \text { POINT } \\
x \cdot x=9 & \\
x=3 \mathrm{~cm} . & 1 \text { POINT } \\
\text { The perimeter of the desired rectangle } o=2 \cdot 13 x+2 \cdot 2 x=30 x, & 1 \text { POINT } \\
\text { so } o=30 \cdot 3 \mathrm{~cm}=90 \mathrm{~cm} . & 1 \text { POINT } \\
\text {.......................................................................................... }
\end{array}
$$

The perimeter of the desired rectangle $o=2 \cdot 13 x+2 \cdot 2 x=30 x$, ..... 1 POINT
so $o=30 \cdot 3 \mathrm{~cm}=90 \mathrm{~cm}$. ..... 1 POINT

## Note:

If some conclusions (the lengths of the larger and smaller rectangles or the width of the rectangle) are visible from the sketch but are not fully stated in words, they should be scored with ONE or 2 POINTS as in the proposed solution.