## 9. Diagonal Squares

Vlado covered the diagonal of a large square with a side length of $2020 \mathrm{~cm}$ using a row of smaller squares with a side length of $4 \mathrm{~cm}$ cut from green collage paper. The diagonals of the green squares align with the diagonal of the large square, and the intersection of any two consecutive green squares is a square with a side length of 1 cm. Calculate the area of the shape formed by the green squares. Express the result in square decimeters and round to the nearest whole number.

## Result: $\quad 101$
## Solution, First Method.

Let's calculate how many squares of side length $4 \mathrm{~cm}$ there are in total.

Start by drawing squares starting from the bottom left corner of the large square, which we will label as $A_{0}$. The top right corner of the first drawn square will be labeled as $A_{1}$, the top right corner of the second drawn square as $A_{2}$, and so on.

From vertex $\mathrm{A}_{0}$ to vertex $\mathrm{A}_{1}$, one can move $4 \mathrm{~cm}$ to the right and $4 \mathrm{~cm}$ up. To reach vertex $A_{2}$, one needs to move another $3 \mathrm{~cm}$ to the right and another $3 \mathrm{~cm}$ up, and to reach vertex $A_{3}$, another $3 \mathrm{~cm}$ to the right and another $3 \mathrm{~cm}$ up. Thus, after the first square, each subsequent square moves to a point that is $3 \mathrm{~cm}$ to the right and $3 \mathrm{~cm}$ up relative to the last point on the diagonal of the large square.

Since adding squares moves equally upwards and to the right, it is sufficient to look at the rightward movements. From the starting point, the sum of all movements must equal the length of the side of the large square, which is $2020 \mathrm{~cm}$. For the first square, the movement is $4 \mathrm{~cm}$, and for each subsequent square, it is $3 \mathrm{~cm}$.

Let $n$ be the total number of squares of side length $4 \mathrm{~cm}$ in this sequence.

Then we have $4 \mathrm{~cm} + (n-1) \cdot 3 \mathrm{~cm} = 2020 \mathrm{~cm}$.

This simplifies to $3n + 1 = 2020$, from which we get $n = 673$.

When calculating the area, we observe that the first square is complete, and all subsequent squares are squares of side length $4 \mathrm{~cm}$ with a square of side length $1 \mathrm{~cm}$ cut out. The area is $16 + 672 \cdot 15 = 10096 \mathrm{~cm}^{2} = 100.96 \mathrm{~dm}^{2}$.

Rounded to the nearest whole number, the area is approximately $101 \mathrm{~dm}^{2}$.

## Solution, Second Method.

From the area of the initial square with side length $2020 \mathrm{~cm}$, we will subtract the parts that are not green. We have two congruent parts, each consisting of rectangles of height $3 \mathrm{~cm}$ and lengths $3 \mathrm{~cm}, 6 \mathrm{~cm}, \ldots, 2016 \mathrm{~cm}$.

The sum of their areas in $\mathrm{cm}^{2}$ is

$$
\begin{gathered}
2 \cdot (3 \cdot 3 + 3 \cdot 6 + \cdots + 3 \cdot 2016) = 2 \cdot 3 \cdot 3 \cdot (1 + 2 + \cdots + 672) = 2 \cdot 3 \cdot 3 \cdot \frac{672 \cdot 673}{2} \\
= 2016 \cdot 2019
\end{gathered}
$$

Thus, the final result in $\mathrm{cm}^{2}$ is:

$$
\begin{aligned}
2020 \cdot 2020 - & 2016 \cdot 2019 = 2020 \cdot 2020 - (2020 - 4) \cdot (2020 - 1) \\
& = 2020 \cdot 2020 - 2020 \cdot 2020 + 4 \cdot 2020 + 1 \cdot 2020 - 4 \\
& = 5 \cdot 2020 - 4 \\
& = 10096
\end{aligned}
$$

The result is $100.96 \mathrm{~dm}^{2}$. Rounded to the nearest whole number, the area is approximately $101 \mathrm{~dm}^{2}$.