AMC 8 2003 Problem 25

AMC8 2003 年第 25 题

三角形★★★★☆
In the figure, the area of square WXYZWXYZ is 25 cm225 \text{ cm}^2 . The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In △ABC\triangle ABC , AB = AC , and when △ABC\triangle ABC is folded over side BC‾\overline{BC} , point AA coincides with OO , the center of square WXYZWXYZ . What is the area of △ABC\triangle ABC , in square centimeters?
在下图中，正方形 WXYZ 的面积为 25 平方厘米。四个小正方形边长为 1 厘米，它们的各边与大正方形的边平行或重合。在 △ABC\triangle ABC 中，AB=AC，若将 △ABC\triangle ABC 沿着边 BC‾\overline{BC} 折叠，则点 A 和正方形的中心 OO 重合。那么 △ABC\triangle ABC 的面积是多少平方厘米？
AMC 8 2003 Problem 25 figure
  1. A.154\displaystyle \frac{15}4
  2. B.214\displaystyle \frac{21}4
  3. C.274\displaystyle \frac{27}4
  4. D.212\displaystyle \frac{21}2
  5. E.272\displaystyle \frac{27}2

正确答案 · Correct Answer

C

解析 · Solution

解法1

常规解法 · 定底与高 大正方形面积 25，边长 5，中心 OO。由图 BCBC 两端各被单位小方缩进 1，故 BC=5-1-1=3。△ABC\triangle ABC 等腰，沿 BCBC 翻折后顶点落到 OO，说明它在 BCBC 上的高 = BCBC 到 OO 的距离 =1+1+52=92=1+1+\tfrac52=\tfrac92。面积 =12⋅3⋅92=274\displaystyle =\tfrac12\cdot3\cdot\tfrac92=\tfrac{27}4。

Solution 1

Elementary · fix base and height The large square has area 25, side 5, center OO. From the figure BC=5-1-1=3 (each end pulled in by a unit square). Folding △ABC\triangle ABC over BCBC sends AA to OO, so the height from AA to BCBC equals the distance from OO to BCBC, namely 1+1+52=921+1+\tfrac52=\tfrac92. The area is 12⋅3⋅92=274\displaystyle \tfrac12\cdot3\cdot\tfrac92=\tfrac{27}4.

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