Question

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

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

$\dfrac{12+3\sqrt3}4$

$\dfrac92$

$3+\sqrt3$

$\dfrac{6+3\sqrt3}2$

$6$

答案：C

知识点：等边三角形正方形（几何）面积分割

难度评级：1540

解答：

求出各个小块的面积，再把它们相加。

中央等边三角形的面积为 $\dfrac{1^2 \sqrt{3}}{4} = \dfrac{\sqrt{3}}{4}.$

所有正方形的总面积为 $3 \cdot 1^2 = 3.$

另外， $\angle EAF = 360^{\circ} - 60^{\circ} - 2 \cdot 90^{\circ}$ $= 120^{\circ}.$

又因为 $AE=AF=1$ ，且 $\angle EAF=120^\circ$ 。从 $A$ 向底边作高可得 $EF=\sqrt3$ ，高为 $\frac12$ ，所以 $[EAF]=\frac{\sqrt3}{4}$ 。另外两个外侧三角形面积相同。因此它们的总面积为 $\frac{3\sqrt3}{4}$ 。

总面积为 $\dfrac{\sqrt{3}}{4} + \dfrac{3\sqrt{3}}{4} + 3 = 3 + \sqrt{3}.$

所以正确答案是 C。

We can find the areas of all the individual pieces and then add them up together.

The area of the center equilateral triangle is $\dfrac{1^2 \sqrt{3}}{4} = \dfrac{\sqrt{3}}{4}.$

We have that the areas of all the squares is $3 \cdot 1^2 = 3.$

We also have that $\angle EAF = 360^{\circ} - 60^{\circ} - 2 \cdot 90^{\circ}$ $= 120^{\circ}.$

Also, $AE=AF=1$ and $\angle EAF=120^\circ$ . Dropping the altitude from $A$ shows that $EF=\sqrt3$ and the altitude is $\frac12$ , so $[EAF]=\frac{\sqrt3}{4}$ . The other two outer triangles have the same area. Thus their combined area is $\frac{3\sqrt3}{4}$ .

The total area is then $\dfrac{\sqrt{3}}{4} + \dfrac{3\sqrt{3}}{4} + 3 = 3 + \sqrt{3}.$

Thus, C is the correct answer.

2014 AMC 10A 第 13 题

答案：C

难度评级：1540

解答：

其他年份的第 13 题