1. Determine the smallest possible area of triangle $ABC$, whose altitudes satisfy the inequalities $v_{a} \geqq 3 \mathrm{~cm}, v_{b} \geqq 4 \mathrm{~cm}, v_{c} \geqq 5 \mathrm{~cm}$.
1. Let $a, b, c$ be the lengths of the sides of triangle $ABC$. For its height $v_{b}$, the inequality

$$
c \geqq v_{b},
$$

holds because $v_{b}$ is the length of the shortest segment connecting vertex $B$ to a point on line $AC$. For the area $S$ of triangle $ABC$, we have:

$$
S=\frac{c v_{c}}{2} \geqq \frac{v_{b} v_{c}}{2} \geqq 10 \mathrm{~cm}^{2}
$$

If there exists a triangle $ABC$ satisfying the conditions of the problem with an area of exactly $10 \mathrm{~cm}^{2}$, then in both inequalities $S=\frac{1}{2} c v_{c} \geqq \frac{1}{2} v_{b} v_{c} \geqq 10 \mathrm{~cm}^{2}$, equality must hold. Therefore, $c=v_{b}=4 \mathrm{~cm}$ and simultaneously $v_{c}=5 \mathrm{~cm}$. From the first equality, it follows that such a triangle must be a right triangle with a right angle at vertex $A$. For the length of its leg $AC$, we have $b=v_{c}=5 \mathrm{~cm}$, and the length $a$ of its hypotenuse $BC$ is $\sqrt{41} \mathrm{~cm}$. From the formula $S=\frac{1}{2} a v_{a}$ for its height $v_{a}$, we get

$$
v_{a}=\frac{2 S}{a}=\frac{20}{\sqrt{41}} \mathrm{~cm}>3 \mathrm{~cm}
$$

A right triangle $ABC$ with legs $b=5 \mathrm{~cm}$ and $c=4 \mathrm{~cm}$ thus satisfies the conditions of the problem.

The smallest possible area of triangle $ABC$, whose heights satisfy the conditions of the problem, is $10 \mathrm{~cm}^{2}$.

Note. We get the same good estimate from the inequality $b v_{b} \geqq v_{c} v_{b}$, while other combinations yield weaker estimates.