AMC 8 2015 Problem 21

AMC8 2015 年第 21 题

四边形与多边形★★★☆☆

In the given figure hexagon ABCDEF is equiangular, ABJI and FEHG are squares with areas 18 and 32 respectively,

\triangle JBK

is equilateral and FE=BC . What is the area of

\triangle KBC

如图所示，六边形 ABCDEF 是个等角六边形，ABJI 和 FEHG 是面积分别为 18 和 32 的正方形。是等边三角形，且 FE=BC，则的面积是多少？

正确答案 · Correct Answer

解析 · Solution

Solution 1

Clearly, since

\overline{FE}

is a side of a square with area 32,

\overline{FE} = \sqrt{32} = 4 \sqrt{2}

. Now, since

\overline{FE} = \overline{BC}

, we have

\overline{BC} = 4 \sqrt{2}

. We know that

\overline{JB}

is a side of a square with area 18, so

\overline{JB} = \sqrt{18} = 3 \sqrt{2}

. Since

\Delta JBK

is equilateral,

\overline{BK} = \overline{JB} = 3 \sqrt{2}

. Lastly,

\Delta KBC

is a right triangle. We see that

\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}

, so

\Delta KBC

is a right triangle with legs

3 \sqrt{2}

and

4 \sqrt{2}

. Now, its area is

\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} =

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