## 252. Math Puzzle $5 / 86$

A round wooden beam has a mass of $73.5 \mathrm{~kg}$. What would its mass be if it had twice the diameter but only half the length?
For the solution, the volume calculation suffices: $V=\pi \cdot r^{2} \cdot l$

For $r=1$, $V=\pi \cdot 1^{2} \cdot l$

For $r=2$, $V=\pi \cdot 2^{2} \cdot l$

With the diameter doubled, the volume quadruples, and thus the mass as well; it now amounts to $294 \mathrm{~kg}$. In our case, however, the length is halved, as is the volume and the mass. The corresponding beam thus has a mass of $147 \mathrm{~kg}$.