AMC8 2003

AMC8 2003 · Q25

AMC8 2003 · Q25. It mainly tests Vieta / quadratic relationships (basic), Area & perimeter.

In the figure, the area of square WXYZ is 25 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, AB = AC, and when △ABC is folded over side BC, point A coincides with O, the center of square WXYZ. What is the area of △ABC, in square centimeters? [figure]

图中正方形WXYZ的面积是25 cm$^2$。四个小正方形的边长为1 cm，与大正方形的边平行或重合。在△ABC中，AB = AC，当沿BC折叠△ABC时，点A与O（WXYZ正方形中心）重合。求△ABC的面积（平方厘米）？[figure]

(A) \frac{15}{4} \frac{15}{4}

(B) \frac{21}{4} \frac{21}{4}

(D) \frac{21}{2} \frac{21}{2}

(E) \frac{27}{2} \frac{27}{2}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let M be the midpoint of BC. Since △ABC is isosceles, AM is an altitude to base BC. Because A coincides with O when △ABC is folded along BC, it follows that AM = MO = $\frac{5}{2} + 1 + 1 = \frac{9}{2}$ cm. Also, BC = $5 -1 -1 = 3$ cm, so the area of △ABC is $\frac{1}{2} \cdot BC \cdot AM = \frac{1}{2} \cdot 3 \cdot \frac{9}{2} = \frac{27}{4}$ cm$^2$.

设M为BC中点。由于△ABC是等腰三角形，AM是底BC的高。因为沿BC折叠时A与O重合，故AM = MO = $\frac{5}{2} + 1 + 1 = \frac{9}{2}$ cm。另外，BC = $5 -1 -1 = 3$ cm，故△ABC面积为 $\frac{1}{2} \cdot BC \cdot AM = \frac{1}{2} \cdot 3 \cdot \frac{9}{2} = \frac{27}{4}$ cm$^2$。

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