AMC8 2020

AMC8 2020 · Q18

AMC8 2020 · Q18. It mainly tests Pythagorean theorem, Area & perimeter.

Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$

矩形 $ABCD$ 铭刻在一个以直径 $\overline{FE}$ 为直径的半圆中，如图所示。设 $DA=16$，且 $FD=AE=9$。$ABCD$ 的面积是多少？

(A) 240 240

(B) 248 248

(D) 264 264

(E) 272 272

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$, so $OC = 17$. By symmetry, $O$ is the midpoint of $DA$, so $OD=OA=\frac{16}{2}= 8$. By the Pythagorean theorem in right-angled triangle $ODC$ (or $OBA$), we have that $CD$ (or $AB$) is $\sqrt{17^2-8^2}=15$. Accordingly, the area of $ABCD$ is $16\cdot 15=\boxed{\textbf{(A) }240}$.

设 $O$ 为半圆的圆心。半圆的直径是 $9+16+9=34$，所以 $OC = 17$。由对称性，$O$ 是 $DA$ 的中点，所以 $OD=OA=\frac{16}{2}= 8$。在直角三角形 $ODC$（或 $OBA$）中，由勾股定理，$CD$（或 $AB$）是 $\sqrt{17^2-8^2}=15$。因此，$ABCD$ 的面积是 $16\cdot 15=\boxed{\textbf{(A) }240}$。

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