1. If the points $(x, y)$ in a plane region satisfy the inequality
$$
\sqrt{\frac{x^{2}}{25}}+\sqrt{\frac{y^{2}}{16}} \leqslant 1 \text {, }
$$

then the area of the plane region is ( ).
(A) 30
(B) 40
(C) 50
(D) 60
-,1.B.
The original inequality is $\frac{|x|}{5}+\frac{|y|}{4} \leqslant 1$. It represents the boundary and interior of a rhombus with vertices at $(5,0)$, $(0,-4)$, $(-5,0)$, and $(0,4)$. Its area is
$$
S=\frac{1}{2} \times 10 \times 8=40 \text{. }
$$