AMC8 2015

AMC8 2015 · Q21

AMC8 2015 · Q21. It mainly tests Triangles (properties), Area & perimeter.

In the given figure hexagon ABCDEF is equiangular, ABJI and FEHG are squares with areas 18 and 32 respectively, △JBK is equilateral and FE = BC. What is the area of △KBC?

在给定的图形中，六边形ABCDEF是等角的，ABJI和FEHG是面积分别为18和32的正方形，△JBK是等边三角形，且FE = BC。△KBC的面积是多少？

(A) $6\sqrt{2}$ $6\sqrt{2}$

(B) 9 9

(D) $9\sqrt{2}$ $9\sqrt{2}$

(E) 32 32

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The area of the square $ABJI$ is $18$ and $\triangle KJB$ is equilateral, so $KB = JB = \sqrt{18} = 3\sqrt{2}$. The area of the square $FEHG$ is $32$, so $BC = FE = \sqrt{32} = 4\sqrt{2}$. Each interior angle of the hexagon is $120^\circ$, so $\angle KBC = 360^\circ - 60^\circ - 90^\circ - 120^\circ = 90^\circ$ and $\triangle KBC$ is a right triangle. Its area is $\frac{1}{2}\cdot 3\sqrt{2}\cdot 4\sqrt{2} = 12$.

答案（C）：正方形 $ABJI$ 的面积是 $18$，且 $\triangle KJB$ 为等边三角形，因此 $KB = JB = \sqrt{18} = 3\sqrt{2}$。正方形 $FEHG$ 的面积是 $32$，所以 $BC = FE = \sqrt{32} = 4\sqrt{2}$。六边形的每个内角为 $120^\circ$，因此 $\angle KBC = 360^\circ - 60^\circ - 90^\circ - 120^\circ = 90^\circ$，从而 $\triangle KBC$ 是直角三角形。其面积为 $\frac{1}{2}\cdot 3\sqrt{2}\cdot 4\sqrt{2} = 12$。

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