Triangle $A B C$ has side $A C$ measuring $24 \, \text{cm}$ and the height from vertex $B$ is $25 \, \text{cm}$. Side $A B$ is divided into five equal parts, with the division points sequentially from $A$ to $B$ labeled as $K, L, M, N$. Through each of these points, a line parallel to side $A C$ passes. The intersections of these parallel lines with side $B C$ are sequentially from $B$ to $C$ labeled as $O, P, Q, R$.

Calculate the sum of the areas of trapezoids $K L Q R$ and $M N O P$.

(I. Jančigová)
The area of the entire triangle $ABC$ is

$$
\frac{1}{2} \cdot 24 \cdot 25 = 300 \left(\mathrm{~cm}^{2}\right)
$$

Parallel lines to side $AB$ passing through points $O, P, Q, R$ and parallel lines to side $BC$ passing through points $K, L, M, N$ divide the triangle $ABC$ into smaller, congruent triangles. The congruence of these triangles follows from the construction and some basic theorems about the congruence of triangles (SAS, SSS, ASA). The smallest parts in this division are triangles congruent to triangle $NBO$, and there are 25 of them in total.

![](https://cdn.mathpix.com/cropped/2024_04_17_18c503b3c9e047833e05g-3.jpg?height=411&width=914&top_left_y=588&top_left_x=571)

The area of triangle $NBO$ is $\frac{1}{25} \cdot 300 = 12 \left(\mathrm{~cm}^{2}\right)$. The trapezoids $KLRQ$ and $MNOP$ consist of 7 and 3 such triangles, respectively. The sum of the areas of these trapezoids is therefore

$$
(7 + 3) \cdot 12 = 120 \left(\mathrm{~cm}^{2}\right)
$$

Another solution. We complete the triangle $ABC$ into a parallelogram $ABDC$. The four parallel lines, when extended, divide the parallelogram into five congruent parallelograms:

![](https://cdn.mathpix.com/cropped/2024_04_17_18c503b3c9e047833e05g-3.jpg?height=411&width=1240&top_left_y=1442&top_left_x=408)

The area of the entire parallelogram $ABDC$ is

$$
24 \cdot 25 = 600 \left(\mathrm{~cm}^{2}\right)
$$

The sum of the areas of the two gray parallelograms is

$$
\frac{2}{5} \cdot 600 = 240 \left(\mathrm{~cm}^{2}\right)
$$

The triangles $ABC$ and $DCB$ are congruent, including the shaded parts. Therefore, the sum of the areas of the trapezoids $KLRQ$ and $MNOP$ is equal to half the sum of the areas of the gray parallelograms:

$$
\frac{1}{2} \cdot 240 = 120 \left(\mathrm{~cm}^{2}\right)
$$

Another solution. The triangles $ABC, KBR, LBQ, MBP$, and $NBO$ are similar to each other because they share the same angle at vertex $B$ and the opposite sides are parallel. The similarity ratios between the first and the remaining four triangles are $\frac{4}{5}, \frac{3}{5}, \frac{2}{5}$, and $\frac{1}{5}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_04_17_18c503b3c9e047833e05g-4.jpg?height=412&width=899&top_left_y=525&top_left_x=584)

The area of the entire triangle $ABC$ is

$$
\frac{1}{2} \cdot 24 \cdot 25 = 300 \left(\mathrm{~cm}^{2}\right)
$$

The areas of triangles $NBO, MBP, LBQ, KBR$ are, respectively,

$$
\begin{aligned}
& \frac{1}{2} \left(\frac{1}{5} \cdot 24\right) \left(\frac{1}{5} \cdot 25\right) = 12 \left(\mathrm{~cm}^{2}\right), \\
& \frac{1}{2} \left(\frac{2}{5} \cdot 24\right) \left(\frac{2}{5} \cdot 25\right) = 48 \left(\mathrm{~cm}^{2}\right), \\
& \frac{1}{2} \left(\frac{3}{5} \cdot 24\right) \left(\frac{3}{5} \cdot 25\right) = 108 \left(\mathrm{~cm}^{2}\right), \\
& \frac{1}{2} \left(\frac{4}{5} \cdot 24\right) \left(\frac{4}{5} \cdot 25\right) = 192 \left(\mathrm{~cm}^{2}\right).
\end{aligned}
$$

The area of trapezoid $KLRQ$ is the difference between the areas of triangles $KBR$ and $LBQ$, and the area of trapezoid $MNOP$ is the difference between the areas of triangles $MBP$ and $NBO$. The sum of the areas of these two trapezoids is therefore

$$
(192 - 108) + (48 - 12) = 120 \left(\mathrm{~cm}^{2}\right).
$$

Evaluation. 2 points for initial considerations (division, completion, or similarity); 2 points for auxiliary calculations and the result; 2 points for the quality of the commentary.

Note. In all the solutions provided, it is advantageous to work with expressions that represent dependencies on the area of triangle $ABC$ and the values given in the problem, and to substitute the values only at the end. If we denote the area of triangle $ABC$ by $S$, then the calculations (1) and (2) correspond to

$$
\frac{7 + 3}{25} S = \frac{2}{5} S, \quad \frac{1}{2} \cdot \frac{2}{5} \cdot 2 S = \frac{2}{5} S
$$

and the relationship (3) can be expressed as

$$
\left(\left(\frac{4}{5}\right)^{2} - \left(\frac{3}{5}\right)^{2} + \left(\frac{2}{5}\right)^{2} - \left(\frac{1}{5}\right)^{2}\right) S = \frac{16 - 9 + 4 - 1}{25} S = \frac{2}{5} S.
$$