Task B-3.7. What is the maximum value that the area of a triangle can have if its perimeter is $20 \mathrm{~cm}$, and the length of one side is $8 \mathrm{~cm}$?
## Second Solution.

The area of a triangle can be calculated using the formula $P=\frac{1}{2} a b \sin \gamma$. (or $P=\frac{1}{2} b c \sin \alpha$)

Since $a+b+c=20$, then $a+c=12$, or $a=12-c$.

(1 point)

Using the cosine rule, we get $c^{2}=(12-c)^{2}+64-16(12-c) \cos \gamma$ or

$$
\begin{gathered}
\cos \gamma=\frac{26-3 c}{2(12-c)}, \text{ and } \\
\sin \gamma=\sqrt{1-\left(\frac{26-3 c}{2(12-c)}\right)^{2}}=\frac{-5 c^{2}+60 c-100}{2(12-c)}
\end{gathered}
$$

Then the area of the triangle is $P=\frac{1}{2} \cdot 8 \cdot(12-c) \sin \gamma=2 \cdot \sqrt{-5 c^{2}+60 c-100}$. (1 point) Similarly to the first solution, we look for the maximum value of the expression

$$
-5 c^{2}+60 c-100
$$

which is achieved for $c=6 \text{ cm}$.

(2 points)

Then $a=6 \text{ cm}$ and $P=8 \sqrt{5} \text{ cm}^{2}$.

(3 points)

## ZUPANIJSKO NATJECANJE IZ MATEMATIKE

## 4th grade - high school - B variant

## February 15, 2013.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.