6. Marko drew a square $A A_{1} C_{1} B_{1}$ and polygons $A_{1} A_{2} C_{2} B_{2} B_{1} C_{1}, A_{2} A_{3} C_{3} B_{3} B_{2} C_{2}$, $A_{3} A_{4} C_{4} B_{4} B_{3} C_{3}, A_{4} A_{5} C_{5} B_{5} B_{4} C_{4}$, where adjacent sides are mutually perpendicular and it holds that: $\left|A_{1} A_{2}\right|=\left|A_{2} A_{3}\right|=\left|A_{3} A_{4}\right|=\left|A_{4} A_{5}\right|=\left|B_{1} B_{2}\right|=\left|B_{2} B_{3}\right|=\left|B_{3} B_{4}\right|=\left|B_{4} B_{5}\right|$ (see figure).

If the area of the polygon $A_{1} A_{2} C_{2} B_{2} B_{1} C_{1}$ is $80 \mathrm{~m}^{2}$, and the area of the polygon $A_{4} A_{5} C_{5} B_{5} B_{4} C_{4}$ is $104 \mathrm{~m}^{2}$, determine the area of the square $A A_{1} C_{1} B_{1}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_47603db0a075197c1d6ag-9.jpg?height=617&width=628&top_left_y=617&top_left_x=728)
6. Marko drew a square $A A_{1} C_{1} B_{1}$ and polygons $A_{1} A_{2} C_{2} B_{2} B_{1} C_{1}, A_{2} A_{3} C_{3} B_{3} B_{2} C_{2}, A_{3} A_{4} C_{4} B_{4} B_{3} C_{3}$, $A_{4} A_{5} C_{5} B_{5} B_{4} C_{4}$, where adjacent sides are mutually perpendicular and it holds that: $\left|A_{1} A_{2}\right|=\left|A_{2} A_{3}\right|=\left|A_{3} A_{4}\right|=\left|A_{4} A_{5}\right|=\left|B_{1} B_{2}\right|=\left|B_{2} B_{3}\right|=\left|B_{3} B_{4}\right|=\left|B_{4} B_{5}\right|$ (see figure).

If the area of the polygon $A_{1} A_{2} C_{2} B_{2} B_{1} C_{1}$ is $80 \mathrm{~m}^{2}$, and the area of the polygon $A_{4} A_{5} C_{5} B_{5} B_{4} C_{4}$ is $104 \mathrm{~m}^{2}$, determine the area of the square $A A_{1} C_{1} B_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_1ae5e90785d026149547g-32.jpg?height=580&width=579&top_left_y=538&top_left_x=767)

## Solution.

Let $x$ be the length of the side of the square, and let the extension of the side be denoted by $y$. The area of the polygon $A_{1} A_{2} C_{2} B_{2} B_{1} C_{1}$ is:

$$
\begin{aligned}
& (x+y)^{2}-x^{2}=80 \\
& 1 \text{ POINT} \\
& x^{2}+2 x y+y^{2}-x^{2}=80 \\
& 2 x y+y^{2}=80
\end{aligned}
$$

The area of the polygon $A_{4} A_{5} C_{5} B_{5} B_{4} C_{4}$ is:
$(x+4 y)^{2}-(x+3 y)^{2}=104$
1 POINT
$x^{2}+8 x y+16 y^{2}-x^{2}-6 x y-9 y^{2}=104$
$2 x y+7 y^{2}=104$

From (1) and (2) it follows:

$2 x y=80-y^{2}$

$80-y^{2}+7 y^{2}=104$

$1 \text{ POINT}$

$6 y^{2}=104-80$

$6 y^{2}=24$

$y^{2}=4 \quad(y>0)$

$y=2$

$1 \text{ POINT}$

$$
\begin{aligned}
& 2 x \cdot 2=80-2^{2} \\
& 4 x=80-4 \\
& 4 x=76 \\
& x=19
\end{aligned}
$$

Note 1: The student can express the areas of the polygons $A_{1} A_{2} C_{2} B_{2} B_{1} C_{1}$ and $A_{4} A_{5} C_{5} B_{5} B_{4} C_{4}$ using the difference of squares:

The area of the polygon $A_{1} A_{2} C_{2} B_{2} B_{1} C_{1}$ is:
$(x+y)^{2}-x^{2}=80$
$1 \text{ POINT}$
$(x+y-x) \cdot(x+y+x)=80$
$1 \text{ POINT}$
$y \cdot(2 x+y)=80$
$2 x y+y^{2}=80$
1 POINT

The area of the polygon $A_{4} A_{5} C_{5} B_{5} B_{4} C_{4}$ is:
$(x+4 y)^{2}-(x+3 y)^{2}=104$
$1 \text{ POINT}$
$(x+4 y-x-3 y)(x+4 y+x+3 y)=104$
$1 \text{ POINT}$
$y(2 x+7 y)=104$
$2 x y+7 y^{2}=104$
1 POINT

The solution continues in the same way as described in the previous solution.

Note 2: The problem can also be solved by dividing the square $A A_{5} C_{5} B_{5}$ into squares with side length $y$ and rectangles with side lengths $x$ and $y$, as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_30_1ae5e90785d026149547g-33.jpg?height=557&width=600&top_left_y=1178&top_left_x=728)

The area of the polygon $A_{1} A_{2} C_{2} B_{2} B_{1} C_{1} A_{1}$ is

$$
x y+y^{2}+y x=80
$$

1 POINT

or

$$
2 x y+y^{2}=80
$$

The area of the polygon $A_{4} A_{5} C_{5} B_{5} B_{4} C_{4}$ is

$$
x y+y^{2}+y^{2}+y^{2}+y^{2}+y^{2}+y^{2}+y^{2}+y x=104, \quad 1 \text { POINT }
$$

or

$$
2 x y+7 y^{2}=104
$$

The solution continues in the same way as described in the previous solution.