2. Paving the ground with three types of regular polygon tiles, with vertices fitting together perfectly, and fully covering the ground. Let the number of sides of the regular polygon tiles be $x$, $y$, and $z$. Then,
(A) $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$
(B) $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}$
(C) $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$
(D) $\frac{1}{x}+\frac{1}{y}=\frac{2}{z}$
2. B.

Let the number of sides of the tiles be $x, y, z$, and the internal angles of each tile be $\alpha, \beta, \gamma$. Then
$$
\begin{array}{l}
(x-2) 180^{\circ}=\alpha x, (y-2) 180^{\circ}=\beta y, \\
(z-2) 180^{\circ}=\gamma z . \\
\text { Hence } \frac{x-2}{x}+\frac{y-2}{y}+\frac{z-2}{z}=\frac{\alpha+\beta+\gamma}{180^{\circ}}=\frac{360^{\circ}}{180^{\circ}}=2,
\end{array}
$$

i.e., $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}$.