B4 Parallel lines $A B$ and $D E$, separated by $12 \mathrm{~cm}$, are intersected by a line with the intersection point $\mathrm{v}$ at point $C$ such that $|B C|:|C D|=2: 3$.

![](https://cdn.mathpix.com/cropped/2024_06_07_91db77e4bb8bf03ce3c4g-4.jpg?height=871&width=1008&top_left_y=1044&top_left_x=567)

A Write down the measures of the angles $\angle C A B, \angle A B C$ and $\angle B C A$.

B Calculate the distance $|A B|$, if the distance $|D E|=9 \mathrm{~cm}$.

C Calculate the area of triangle $A B C$.

## 10th Mathematics Knowledge Competition for Vocational School Students  National Competition, April 17, 2010

## Solutions to the problems and scoring guide

A contestant who arrives at the solution by any correct method (even if not anticipated by the scoring guide) receives all possible points.

For a correct method, we consider any procedure that

- sensibly takes into account the wording of the problem,
- leads to the solution of the problem,
- is mathematically correct and complete.

$V$ in the table, the correct answers to the multiple-choice questions are recorded. Each correct answer is scored with 2 points, an incorrect answer with $-0.5$ points, and if the question is not answered, 0 points.

| 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| B | E | C | B | D | C |
B4. From the figure, we understand that $\angle C A B=\angle C E D$ and $\angle A B C=\angle E D C$, as these are alternate interior angles. Thus, $\angle B A C=65^{\circ}$ and $\angle C B A=55^{\circ}$. Since the sum of the interior angles in a triangle is $180^{\circ}$, $\angle A C B=180^{\circ}-\left(65^{\circ}+55^{\circ}\right)=60^{\circ}$. Triangles $\triangle A B C$ and $\triangle D E C$ are similar, so it follows that: $\frac{|A B|}{|D E|}=\frac{|B C|}{|C D|}$. From the equation $\frac{|A B|}{9 \mathrm{~cm}}=\frac{2}{3}$, we find that $|A B|=6 \mathrm{~cm}$. The area of triangle $\triangle A B C$ is $S=\frac{|A B| \cdot v_{A B}}{2}=\frac{6 \mathrm{~cm} \cdot 4.8 \mathrm{~cm}}{2}=14.4 \mathrm{~cm}^{2}$. The height $v_{A B}=4.8 \mathrm{~cm}$ is obtained by considering $v_{A B}+v_{D E}=12 \mathrm{~cm}$ and $\frac{v_{A B}}{v_{D E}}=2: 3$.

![](https://cdn.mathpix.com/cropped/2024_06_07_91db77e4bb8bf03ce3c4g-7.jpg?height=874&width=994&top_left_y=811&top_left_x=568)

A Recorded angle measures: $\angle B A C=65^{\circ}, \angle C B A=55^{\circ}, \angle A C B=60^{\circ}$, three times 1 point each 3 points

B Calculated length $|A B|=6 \mathrm{~cm}$ 2 points

C Calculated height of triangle $\triangle A B C, v_{A B}=4.8 \mathrm{~cm}$ 1 point Calculated area of triangle $\triangle A B C, S=14.4 \mathrm{~cm}^{2}$ 1 point