2. The lengths of the sides of a certain triangle are three consecutive even natural numbers, and its perimeter is $84 \mathrm{~cm}$. If the length of the height to the second largest side is $v_{2}=24 \mathrm{~cm}$, calculate the lengths of the remaining two heights $v_{1}$ and $v_{3}$. Round the result to two decimal places.
2. If we denote the even number by $n$, then the lengths of the sides of the triangle are:
$n, n+2$ and $n+4$.
2 points

Since the perimeter of the triangle is $84 \mathrm{~cm}$, we have the equation:

$n+n+2+n+4=84 \mathrm{~cm}$.

1 point

Solving this equation, we get:

$3 n+6=84$, or $3 n=78$, from which it follows that $n=26$. 1 point

Thus, the lengths of the sides of the triangle are: $26 \mathrm{~cm}, 28 \mathrm{~cm}$ and $30 \mathrm{~cm}$. 2 points

Given that the length of the height to the second largest side is $v_{2}=24 \mathrm{~cm}$,

the area of the given triangle will be:

$p=(28 \cdot 24): 2=672: 2=336 \mathrm{~cm}^{2}$.

2 points

Since the area of the triangle is also $p=\left(26 \cdot v_{1}\right): 2$, we have:

$336=\left(26 \cdot v_{1}\right): 2$, or $v_{1}=25.84615 \ldots \mathrm{cm} \approx 25.85 \mathrm{~cm} \quad 1$ point

Finally, the area of the triangle is also $p=\left(30 \cdot v_{3}\right): 2$, or $336=\left(30 \cdot v_{3}\right): 2$,

so $v_{3}=22.4 \mathrm{~cm}$

1 point

TOTAL 10 POINTS