## SUBJECT II

Find the dimensions of a rectangular parallelepiped, knowing that the lengths of the diagonals of its faces are inversely proportional to $\frac{\sqrt{7}}{7}, \frac{\sqrt{11}}{11}$, and $\frac{\sqrt{6}}{6}$, and the length of its diagonal is equal to $\sqrt{12} \text{ cm}$.

Supplement GM no.10/2013
## Solution:

$$
d=\sqrt{a^{2}+b^{2}+c^{2}}=\sqrt{12} \Rightarrow d^{2}=a^{2}+b^{2}+c^{2}=12
$$

The diagonals of the faces have lengths: $\sqrt{a^{2}+b^{2}}, \sqrt{a^{2}+c^{2}}$, and $\sqrt{b^{2}+c^{2}} \Rightarrow$

$$
\begin{aligned}
& \Rightarrow \sqrt{a^{2}+b^{2}} \cdot \frac{\sqrt{7}}{7}=\sqrt{a^{2}+c^{2}} \cdot \frac{\sqrt{11}}{11}=\sqrt{b^{2}+c^{2}} \cdot \frac{\sqrt{6}}{6}=k \\
& \Rightarrow \sqrt{a^{2}+b^{2}}=\sqrt{7} k, \sqrt{a^{2}+c^{2}}=\sqrt{11} k, \sqrt{b^{2}+c^{2}}=\sqrt{6} k
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_908719aa324d964aa379g-3.jpg?height=276&width=306&top_left_y=119&top_left_x=338)

$$
\begin{aligned}
& 2\left(a^{2}+b^{2}+c^{2}\right)=24 k^{2} \Rightarrow a^{2}+b^{2}+c^{2}=12 k^{2} \Rightarrow \text { substituting in (1) } \\
& \Rightarrow k^{2}=1 \Rightarrow k=1
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_908719aa324d964aa379g-3.jpg?height=54&width=52&top_left_y=523&top_left_x=1710)

$$
\begin{aligned}
& \Rightarrow a^{2}+b^{2}=7 \Rightarrow c^{2}=5 \Rightarrow c=\sqrt{5} \text{ cm} \\
& a^{2}+c^{2}=11 \Rightarrow b^{2}=1 \Rightarrow b=1 \text{ cm} \\
& b^{2}+c^{2}=6 \Rightarrow a^{2}=6 \Rightarrow a=\sqrt{6} \text{ cm}
\end{aligned}
$$