## 6. Area

In the figure, a rectangle with side lengths of 48 $\mathrm{mm}$ and $32 \mathrm{~mm}$ is shown. The longer sides of the rectangle are divided into three equal parts by marked points, and the shorter sides into two equal parts.

The perimeter of each yellow rectangle is equal to the length of the longer side of the rectangle. Determine the area of the blue-colored shape, in square millimeters.

![](https://cdn.mathpix.com/cropped/2024_05_30_97e0360945497c9bac66g-4.jpg?height=434&width=625&top_left_y=1025&top_left_x=1224)

Result: $\quad 768$
## Solution, first method.

The longer side of the rectangle is divided into three lengths of $48: 3=16 \mathrm{~mm}$.

The shorter side of the rectangle is divided into two lengths of $32: 2=16 \mathrm{~mm}$.

The yellow rectangles have perimeters of $48 \mathrm{~mm}$ and have two sides of length $16 \mathrm{~mm}$ each.

![](https://cdn.mathpix.com/cropped/2024_05_30_97e0360945497c9bac66g-4.jpg?height=512&width=654&top_left_y=1957&top_left_x=655)

The lengths of the remaining sides of the yellow rectangles are $(48-16-16): 2=16: 2=8 \mathrm{~mm}$.

We will solve the problem further by dividing the rectangle into a square grid of squares with side lengths of $8 \mathrm{~mm}$, as shown in the figure.

The area of each such square is $\mathrm{P}_{1}=8 \cdot 8=64 \mathrm{~mm}^{2}$, and the area of a triangle that is half of such a square is $P_{2}=64: 2=32 \mathrm{~mm}^{2}$.

Notice that the "blue" shape consists of 8 squares with an area of $64 \mathrm{~mm}^{2}$ and 8 halves of such squares, so its area is $P=8 \cdot 64+8 \cdot 32=512+256=768 \mathrm{~mm}^{2}$.

## Solution, second method.

As in the first method, we calculate the lengths of the sides of the yellow rectangles and divide the initial rectangle into a square grid. Let P and R be the midpoints of segments $\overline{N K}$ and $\overline{M L}$, respectively.

The yellow triangles DJI, AEJ, BGF, CHG have the same area as the blue triangles EPJ, IJP, FGR, GHR.

Also, the yellow rectangles HINM and EFLK have the same area as the blue rectangles MNPR and KLRP.

Therefore, the area of the blue shape is equal to the area of the remaining yellow part of the rectangle, i.e., the area of the blue shape is half the area of the initial rectangle.

The answer is $P=48 \cdot 32: 2=768 \mathrm{~mm}^{2}$.

## Solution, third method.

The longer side of the rectangle is divided into three lengths of $48: 3=16 \mathrm{~mm}$. The shorter side of the rectangle is divided into two lengths of $32: 2=16 \mathrm{~mm}$.

The yellow rectangles have perimeters of $48 \mathrm{~mm}$ and have two sides of length $16 \mathrm{~mm}$ each.

The lengths of the remaining sides of the yellow rectangles are $(48-16-16): 2=16: 2=8 \mathrm{~mm}$.

We will solve the problem further by subtracting from the area of the large rectangle ABCD:

- the areas of two yellow rectangles (one side is $8 \mathrm{~mm}$, and the other is $16 \mathrm{~mm}$), and
- the areas of four yellow triangles (these are right triangles of the same shape and area, and two together form a square with a side length of $16 \mathrm{~mm}$)

The area of the shape colored blue, in square millimeters, is

$P=48 \cdot 32-(2 \cdot 8 \cdot 16+2 \cdot 16 \cdot 16)=1536-(256+512)=1536-768=768$