2.43. Find the distance between the midpoints of two skew edges of a cube, the total surface area of which is $36 \mathrm{~cm}^{2}$.
2.43. Since the total surface area of the cube $S_{\text {full }}=36 \mathrm{~cm}^{2}$, the area of one face $S=6 \mathrm{~cm}^{2}$ and the edge of the cube

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-478.jpg?height=401&width=407&top_left_y=1457&top_left_x=733)

Fig. 2.37
$A D=a=\sqrt{6}$ cm (Fig. 2.37). The required distance

$$
\begin{gathered}
K M=\sqrt{A K^{2}+A M^{2}} \text {, where } A K=\frac{a}{2}=\frac{\sqrt{6}}{2}(\text { cm }), \\
A M=\sqrt{A D^{2}+D M^{2}}=\sqrt{a^{2}+\frac{a^{2}}{4}}=\frac{a \sqrt{5}}{2}=\frac{\sqrt{30}}{2} \text { (cm). }
\end{gathered}
$$

Finally, we get $K M=\sqrt{\frac{6}{4}+\frac{30}{4}}=3$ (cm).

Answer: 3 cm.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-479.jpg?height=375&width=354&top_left_y=712&top_left_x=52)

Fig. 2.38