AMC10 2011 A

AMC10 2011 A · Q11

AMC10 2011 A · Q11. It mainly tests Triangles (properties), Area & perimeter.

Square $EFGH$ has one vertex on each side of square $ABCD$. Point $E$ is on $AB$ with $AE = 7 \cdot EB$. What is the ratio of the area of $EFGH$ to the area of $ABCD$?

正方形 $EFGH$ 的每个顶点分别位于正方形 $ABCD$ 的一条边上。点 $E$ 在 $AB$ 上，且 $AE = 7 \cdot EB$。$EFGH$ 与 $ABCD$ 的面积比是多少？

(A) $\frac{49}{64}$ $\frac{49}{64}$

(B) $\frac{25}{32}$ $\frac{25}{32}$

(D) $\frac{5\sqrt{2}}{8}$ $\frac{5\sqrt{2}}{8}$

(E) $\frac{\sqrt{14}}{4}$ $\frac{\sqrt{14}}{4}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Without loss of generality, assume that $F$ lies on $BC$ and that $EB = 1$. Then $AE = 7$ and $AB = 8$. Because $EFGH$ is a square, $BF = AE = 7$, so the hypotenuse $EF$ of $\triangle EBF$ has length $\sqrt{1^2 + 7^2} = \sqrt{50}$. The ratio of the area of $EFGH$ to that of $ABCD$ is therefore $\frac{EF^2}{AB^2} = \frac{50}{64} = \frac{25}{32}$.

不失一般性，假设 $F$ 在 $BC$ 上，且 $EB = 1$。则 $AE = 7$，$AB = 8$。因为 $EFGH$ 是正方形，$BF = AE = 7$，所以 $\triangle EBF$ 的斜边 $EF$ 的长度为 $\sqrt{1^2 + 7^2} = \sqrt{50}$。因此 $EFGH$ 与 $ABCD$ 的面积比为 $\frac{EF^2}{AB^2} = \frac{50}{64} = \frac{25}{32}$。

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