AMC10 2012 A

AMC10 2012 A · Q21

AMC10 2012 A · Q21. It mainly tests Coordinate geometry, 3D geometry (volume).

Let points A = (0, 0, 0), B = (1, 0, 0), C = (0, 2, 0), and D = (0, 0, 3). Points E, F, G, and H are midpoints of line segments BD, AB, AC, and DC respectively. What is the area of EFGH?

设点 A = (0, 0, 0), B = (1, 0, 0), C = (0, 2, 0), D = (0, 0, 3)。点 E, F, G, H 分别是线段 BD, AB, AC, DC 的中点。EFGH 的面积是多少？

(A) \sqrt{2} \sqrt{2}

(B) \frac{2\sqrt{5}}{3} \frac{2\sqrt{5}}{3}

(D) \sqrt{3} \sqrt{3}

(E) \frac{2\sqrt{7}}{3} \frac{2\sqrt{7}}{3}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The midpoint formula gives $E=\left(\frac12,0,\frac32\right)$, $F=\left(\frac12,0,0\right)$, $G=(0,1,0)$, and $H=\left(0,1,\frac32\right)$. Note that $\overline{EF}=\overline{GH}=\frac32$, $\overline{EF}\perp\overline{EH}$, $\overline{GF}\perp\overline{GH}$, and $$ EH=FG=\sqrt{\left(\frac12\right)^2+1^2}=\frac{\sqrt5}{2}. $$ Therefore $EFGH$ is a rectangle with area $\frac32\cdot\frac{\sqrt5}{2}=\frac{3\sqrt5}{4}$.

答案（C）：中点公式给出 $E=\left(\frac12,0,\frac32\right)$，$F=\left(\frac12,0,0\right)$，$G=(0,1,0)$，以及 $H=\left(0,1,\frac32\right)$。注意 $\overline{EF}=\overline{GH}=\frac32$，$\overline{EF}\perp\overline{EH}$，$\overline{GF}\perp\overline{GH}$，并且 $$ EH=FG=\sqrt{\left(\frac12\right)^2+1^2}=\frac{\sqrt5}{2}. $$ 因此 $EFGH$ 是一个矩形，其面积为 $\frac32\cdot\frac{\sqrt5}{2}=\frac{3\sqrt5}{4}$。

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