$19 \cdot 49$ As shown in the figure, in the arrow-shaped polygon, the angles at vertices $A, C, D, E$, and $F$ are all right angles, $B C=F G=5, C D$ $=F E=20, D E=10, A B=A G$. The area of this polygon is closest to
(A) 288 .
(B) 291 .
(C) 294 .
(D) 297 .
(E) 300 .
[Solution] As shown in the right figure, connect $C F$, then $B G=10+5+5=20$. Also, $\triangle A B G$ is an isosceles right triangle, so
$$
\begin{aligned}
S_{\triangle A B G} & =\frac{1}{2} \cdot B G \cdot h=\frac{1}{2} \cdot 20 \cdot 10 \\
& =100,
\end{aligned}
$$

Also, $S_{\square(D E F)}=20 \cdot 10=200$.
Thus, $S_{\bigcirc}=100+200=300$.
Therefore, the answer is $(E)$.