[ Distance between skew lines ]

The base of the pyramid is a rectangle with sides 6 and 8. One of the lateral edges is perpendicular to the base plane and equals 6. Find the distance between this edge and the diagonal of the base that it intersects, as well as the lateral surface area of the pyramid.
Let the rectangle $A B C D$ with sides $A B=C D=6$ and $B C=A D=8$ be the base of the pyramid $P A B C D$ with height $P A=6$. Drop the perpendicular $A M$ to the diagonal $B D$. Since $A P$ is perpendicular to the plane $A B C D$, and $A M \perp P A$, it follows that $A M$ is the common perpendicular to the skew lines $P A$ and $B D$. $A M$ is the height of the right triangle $A B D$ drawn from the vertex of the right angle, so $B D \cdot A M=A B \cdot A D$, from which we find that

$$
A M=\frac{A B \cdot A D}{B D}=\frac{6 \cdot 8}{\sqrt{36+64}}=\frac{48}{10}=4.8
$$

Since $A B$ is the orthogonal projection of the slant $P B$ onto the plane $A B C D$ and $A B \perp B C$, by the theorem of three perpendiculars, $P B \perp B C$, so the triangle $B C P$ is right-angled. Similarly, we can prove that the triangle $C D P$ is also right-angled. Let $S$ be the lateral surface area of the pyramid $P A B C D$. Then

$$
B P=\sqrt{A B^{2}+A P^{2}}=6 \sqrt{2}, D P=\sqrt{A D^{2}+A P^{2}}=10
$$

$$
\begin{aligned}
& =\frac{1}{2} \cdot 6 \cdot 6+\frac{1}{2} \cdot 8 \cdot 6+\frac{1}{2} \cdot 8 \cdot 6 \sqrt{2}+\frac{1}{2} \cdot 6 \cdot 10= \\
& =18+24+24 \sqrt{2}+30=72+24 \sqrt{2}
\end{aligned}
$$

## Answer

$$
4.8 ; 72+24 \sqrt{2}
$$