16. In the isosceles triangle \(ABC\), where \(\overline{AB} = \overline{BC} = 5 \, \text{cm}\) and \(\overline{AC} = 6 \, \text{cm}\), the height from vertex \(C\) intersects side \(AB\) at point \(D\). By how many \(\text{cm}^2\) is the area of triangle \(BCD\)?
(A) \(\frac{10}{3}\)
(B) \(\frac{56}{15}\)
(C) \(\frac{17}{5}\)
(D) \(\frac{84}{25}\)
(E) \(\frac{24}{7}\)

## Unione Matematica Italiana PROGETTO OLIMPIADI DI MATEMATICA  Ministry of Education, University and Research

- The test consists of 16 problems. Each question is followed by 5 answers indicated by the letters \(\mathbf{(A), (B), (C), (D), (E)}. \) Only one of these answers is correct, the other 4 are incorrect.
- Each correct answer is worth 5 points, each incorrect answer is worth 0 points. For each unanswered or illegible answer, 1 point will be assigned.
- For each of the problems, you must write the letter corresponding to the answer you believe is correct in the grid provided below. No erasures or corrections are allowed on the grid. The use of any type of calculator or communication device is not permitted.

You have 110 minutes to complete the test. Good luck and have fun!

NAME SURNAME CLASS date of birth: email (optional):
![](https://cdn.mathpix.com/cropped/2024_04_17_94b06074249d24c20b4bg-3.jpg?height=100&width=1230&top_left_y=1218&top_left_x=140)
(16) The correct answer is $(D)$.

According to the Pythagorean theorem, the height relative to side $A C$ measures (in $\mathrm{cm}$) $\sqrt{5^{2}-3^{3}}=4$. Therefore, the area of the triangle is $\frac{4 \cdot 6}{2}=12 \mathrm{~cm}^{2}$. It follows that the measure of height $C D$ (still in $\mathrm{cm}$) is $\frac{2 \cdot 12}{5}=24 / 5$. It follows that $\overline{B D}=\sqrt{5^{2}-(24 / 5)^{2}}=7 / 5$ and thus the area of $B C D$ is $\frac{(7 / 5) \cdot (24 / 5)}{2}=84 / 25$.

Question proposed by Paolo Francini.

As always, the problems proposed have been, in addition to being checked and selected, in many cases extensively modified or reformulated by the competition organizers. Therefore, when indicating the names of the proposers, it is understood that the responsibility for the selection of the questions, as well as for any errors, imprecisions, or questionable formulations, lies exclusively with the organizers themselves and not at all with the proposers. Our thanks go to all of them, and also to those who contributed with questions that were not selected, for the variety, originality, and quality of the proposals presented.

The organizers of the Archimedes Games, Paolo Francini, Andrea Sambusetti