2003 AMC 8 第 25 题

先试着解答 2003 AMC 8 第 25 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2003 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

25.

图中,正方形 WXYZWXYZ 的面积为 25 cm225 \text{ cm}^2。四个较小正方形边长为 11 cm,边要么与大正方形的边平行,要么重合。在 △ABC\triangle ABC 中,AB=ACAB = AC,且当 △ABC\triangle ABC 沿边 BC‾\overline{BC} 折叠时,点 AA 与 OO(正方形 WXYZWXYZ 的中心)重合。△ABC\triangle ABC 的面积是多少平方厘米?
In the figure, the area of square WXYZWXYZ is 25 cm2.25 \text{ cm}^2. The four smaller squares have sides 11 cm long, either parallel to or coinciding with the sides of the large square. In △ABC,\triangle ABC, AB=AC,AB = AC, and when △ABC\triangle ABC is folded over side BC‾,\overline{BC}, point AA coincides with O,O, the center of square WXYZ.WXYZ. What is the area of △ABC,\triangle ABC, in square centimeters?
154\dfrac{15}{4}
214\dfrac{21}{4}
274\dfrac{27}{4}
212\dfrac{21}{2}
272\dfrac{27}{2}
答案:C
知识点:折纸三角形面积
小提示:
设 MM 为 BC‾\overline{BC} 的中点
Let MM be the midpoint of BC‾.\overline{BC}.
大提示:
折叠说明 AA 到 BC‾\overline{BC} 的距离等于 OO 到 BC‾\overline{BC} 的距离
Folding makes the distance from AA to BC‾\overline{BC} equal the distance from OO to BC‾.\overline{BC}.
解答:
正方形 WXYZWXYZ 的边长为 55 cm,因为 25=5\sqrt{25} = 5。
WZ‾\overline{WZ} 到 BC‾\overline{BC} 的距离是 22,因为它等于 22 个单位正方形的边长之和。
最后,AA 到 BC‾\overline{BC} 的距离等于 BC‾\overline{BC} 到 OO 的距离,即 2+52=92 cm。 2 + \dfrac{5}{2} = \dfrac{9}{2} \text{ cm}\text{。}
现在可求 BCBC:WZ−2=5−2=3 cm。 WZ - 2 = 5 - 2 = 3 \text{ cm}\text{。}
因此 △ABC\triangle ABC 的面积为 12⋅3⋅92=274 cm2。 \dfrac{1}{2} \cdot 3 \cdot \dfrac{9}{2} = \dfrac{27}{4} \text{ cm}^2\text{。}
所以正确答案是 C。
The side length of WXYZWXYZ is 55 cm, since 25=5.\sqrt{25} = 5.
We also know that the distance from WZ‾\overline{WZ} to BC‾\overline{BC} is 22 since it is the sum of the side lengths of 22 unit squares.
Finally, the distance from AA to BC‾\overline{BC} is the same as the distance from BC‾\overline{BC} to O,O, which is 2+52=92 cm. 2 + \dfrac{5}{2} = \dfrac{9}{2} \text{ cm}.
Now, we can find BC,BC, which is WZ−2=5−2=3 cm. WZ - 2 = 5 - 2 = 3 \text{ cm}.
Therefore, the area of △ABC\triangle ABC is 12⋅3⋅92=274 cm2. \dfrac{1}{2} \cdot 3 \cdot \dfrac{9}{2} = \dfrac{27}{4} \text{ cm}^2.
Thus, C is the correct answer.
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