AMC10 2006 A

AMC10 2006 A · Q7

AMC10 2006 A · Q7. It mainly tests Area & perimeter, Transformations.

The 8 × 18 rectangle ABCD is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$?

8 × 18 的矩形 ABCD 被切成两个全等的六边形，如图所示，使得这两个六边形可以重新排列而不重叠形成一个正方形。$y$ 是多少？

(A) 6 6

(B) 7 7

(D) 9 9

(E) 10 10

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A.) Let $E$ represent the end of the cut on $\overline{DC}$, and let $F$ represent the end of the cut on $\overline{AB}$. For a square to be formed, we must have $DE=y=FB$ and $DE+y+FB=18,$ so $y=6.$ The rectangle that is formed by this cut is indeed a square, since the original rectangle has area $8\cdot 18=144,$ and the rectangle that is formed by this cut has a side of length $2\cdot 6=12=\sqrt{144}.$

（A）设 $E$ 表示在 $\overline{DC}$ 上切割的端点，设 $F$ 表示在 $\overline{AB}$ 上切割的端点。要形成一个正方形，必须满足 $DE=y=FB$ 且 $DE+y+FB=18,$ 因此 $y=6.$ 由这次切割形成的矩形确实是一个正方形，因为原矩形的面积为 $8\cdot 18=144,$ 而切割后形成的矩形有一条边长为 $2\cdot 6=12=\sqrt{144}.$

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