## 9. Rhombi

Rhombus $A_{0} B_{0} C_{0} D_{0}$ is located inside rhombus $A B C D$ such that their sides are parallel, and their centers are at the same point. The segments $\overline{A A_{0}}, \overline{B B_{0}}, \overline{C C_{0}}$, and $\overline{D D_{0}}$ are divided into five congruent parts. The resulting points are connected by segments parallel to the sides of the given rhombi, thus forming four new rhombi.

The area of rhombus $A B C D$ is $2250 \mathrm{~cm}^{2}$, and the area of rhombus $A_{0} B_{0} C_{0} D_{0}$ is $810 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_12924d306592e3a90c60g-08.jpg?height=505&width=733&top_left_y=701&top_left_x=1164)

What is the total colored area in the figure?

## Result: $\quad 864$
## First Solution.

Due to symmetry, we will consider a quarter of the desired area, as shown in the figure. We observe 6 right-angled triangles, whose areas we will denote, starting from the smallest, as $P_{0}, P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$.

With the labels as in the figure, we have:

$P_{0}=\frac{1}{2} a \cdot b$,

$P_{1}=\frac{1}{2}(a+x) \cdot(b+y)=\frac{1}{2}(a b+a y+x b+x y)$

$P_{2}=\frac{1}{2}(a+2 x) \cdot(b+2 y)=\frac{1}{2}(a b+2 a y+2 x b+4 x y)$

$P_{3}=\frac{1}{2}(a+3 x) \cdot(b+3 y)=\frac{1}{2}(a b+3 a y+3 x b+9 x y)$

![](https://cdn.mathpix.com/cropped/2024_05_30_12924d306592e3a90c60g-08.jpg?height=451&width=691&top_left_y=1756&top_left_x=1228)

$P_{4}=\cdots=\frac{1}{2}(a b+4 a y+4 x b+16 x y)$

$P_{5}=\cdots=\frac{1}{2}(a b+5 a y+5 x b+25 x y)$

The area of the smallest colored quadrilateral is $P_{1}-P_{0}=\frac{1}{2}(a y+x b+x y)$

The area of the next largest colored quadrilateral is $P_{3}-P_{2}=\frac{1}{2}(a y+x b+5 x y)$

The area of the largest colored quadrilateral is $P_{5}-P_{4}=\frac{1}{2}(a y+x b+9 x y)$

The sum of the areas of these three quadrilaterals is $\frac{1}{2}(3 a y+3 x b+15 x y)=\frac{3}{2}(a y+x b+5 x y)$

Since,

$P_{5}-P_{0}=\frac{1}{2}(a b+5 a y+5 x b+25 x y)-\frac{1}{2} a \cdot b=\frac{1}{2}(5 a y+5 x b+25 x y)$

$=\frac{5}{2}(a y+x b+5 x y)$.

It can be observed that the sum of the areas of the three colored quadrilaterals is actually equal to $\frac{3}{5}\left(P_{5}-P_{0}\right)$.

From the given areas of the largest and smallest rhombuses, we calculate the difference in the areas of the largest and smallest triangles:

$P_{5}-P_{0}=\frac{1}{4}(2250-810)=360 \mathrm{~cm}^{2}$

Therefore, $\frac{3}{5}\left(P_{5}-P_{0}\right)=216 \mathrm{~cm}^{2}$. The total colored area is $4 \cdot 216=864 \mathrm{~cm}^{2}$.