如图，矩形$ABCD$内接在直径为$\overline{FE}$的半圆中。已知$DA = 16$，$FD = AE = 9$，那么$ABCD$的面积是多少？

A. 240

B. 248

C. 256

D. 264

E. 272

1. 2020 AMC 8 官方解析

Let $O$ be the center of the circle.Then $DO = OA = 8$ and radius $OE = OA + AE = 8 + 9 = 17$. Draw radius $OB$ which also has length $17$. Then $\triangle OAB$ is a right triangle with hypotenuse $17$ and base leg $8$. By the Pythagorean Theorem
$AB = \sqrt{172^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15$.
The area of rectangle $ABCD$ is $DA \cdot AB = 16 \cdot 15 = 240$.

设$O$为圆心，则$DO = OA = 8$，半径$OE = OA + AE = 8 + 9 = 17$。连接半径$OB$，其长度也为$17$。则$\triangle OAB$是一个直角三角形，斜边为$17$，底边为$8$。由毕达哥拉斯定理可得
$AB = \sqrt{172^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15$，因此矩形$ABCD$的面积为
$DA \cdot AB = 16 \cdot 15 = 240$。