AMC10 2014 A

AMC10 2014 A · Q13

AMC10 2014 A · Q13. It mainly tests Triangles (properties), Area & perimeter.

Equilateral $\triangle ABC$ has side length 1, and squares $ABDE$, $BCHI$, and $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$?

等边$\triangle ABC$边长为1，在三角形外部有正方形$ABDE$、$BCHI$和$CAFG$。六边形$DEFGHI$的面积是多少？

(A) $\frac{12 + 3\sqrt{3}}{4}$ $\frac{12 + 3\sqrt{3}}{4}$

(B) $\frac{9}{2}$ $\frac{9}{2}$

(D) $\frac{6 + 3\sqrt{3}}{2}$ $\frac{6 + 3\sqrt{3}}{2}$

(E) 6 6

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The three squares each have area $1$, and $\triangle ABC$ has area $\frac{\sqrt{3}}{4}$. Note that $\angle EAF = 360^\circ - 60^\circ - 2\cdot 90^\circ = 120^\circ$. Thus the altitude from $A$ in isosceles $\triangle EAF$ partitions the triangle into two $30$-$60$-$90$ right triangles, each with hypotenuse $1$. It follows that $\triangle EAF$ has base $EF=\sqrt{3}$ and altitude $\frac{1}{2}$, so its area is $\frac{\sqrt{3}}{4}$. Similarly, triangles $GCH$ and $DBI$ each have area $\frac{\sqrt{3}}{4}$. Therefore the area of hexagon $DEFGHI$ is $3\cdot \frac{\sqrt{3}}{4} + 3\cdot 1 + \frac{\sqrt{3}}{4} = 3+\sqrt{3}$.

答案（C）：三个正方形的面积各为 $1$，且 $\triangle ABC$ 的面积为 $\frac{\sqrt{3}}{4}$。注意到 $\angle EAF = 360^\circ - 60^\circ - 2\cdot 90^\circ = 120^\circ$。因此，在等腰三角形 $\triangle EAF$ 中，从 $A$ 作的高把该三角形分成两个 $30$-$60$-$90$ 的直角三角形，每个的斜边为 $1$。由此可得 $\triangle EAF$ 的底边 $EF=\sqrt{3}$，高为 $\frac{1}{2}$，所以其面积为 $\frac{\sqrt{3}}{4}$。同理，三角形 $GCH$ 和 $DBI$ 的面积也各为 $\frac{\sqrt{3}}{4}$。因此，六边形 $DEFGHI$ 的面积为 $3\cdot \frac{\sqrt{3}}{4} + 3\cdot 1 + \frac{\sqrt{3}}{4} = 3+\sqrt{3}$。

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