Question

图中，正方形 $WXYZ$ 的面积为 $25 \text{ cm}^2$ 。四个较小正方形边长为 $1$ cm，边要么与大正方形的边平行，要么重合。在 $\triangle ABC$ 中， $AB = AC$ ，且当 $\triangle ABC$ 沿边 $\overline{BC}$ 折叠时，点 $A$ 与 $O$ （正方形 $WXYZ$ 的中心）重合。 $\triangle ABC$ 的面积是多少平方厘米？

In the figure, the area of square $WXYZ$ is $25 \text{ cm}^2.$ The four smaller squares have sides $1$ cm long, either parallel to or coinciding with the sides of the large square. In $\triangle ABC,$ $AB = AC,$ and when $\triangle ABC$ is folded over side $\overline{BC},$ point $A$ coincides with $O,$ the center of square $WXYZ.$ What is the area of $\triangle ABC,$ in square centimeters?

$\dfrac{15}{4}$

$\dfrac{21}{4}$

$\dfrac{27}{4}$

$\dfrac{21}{2}$

$\dfrac{27}{2}$

答案：C

知识点：折纸三角形面积

难度评级：1680

解答：

正方形 $WXYZ$ 的边长为 $5$ cm，因为 $\sqrt{25} = 5$ 。

$\overline{WZ}$ 到 $\overline{BC}$ 的距离是 $2$ ，因为它等于 $2$ 个单位正方形的边长之和。

最后， $A$ 到 $\overline{BC}$ 的距离等于 $\overline{BC}$ 到 $O$ 的距离，即。 $2 + \dfrac{5}{2} = \dfrac{9}{2} \text{ cm}$

现在可求 $BC$ ， $WZ - 2 = 5 - 2 = 3 \text{ cm}$

因此 $\triangle ABC$ 的面积为 $\dfrac{1}{2} \cdot 3 \cdot \dfrac{9}{2} = \dfrac{27}{4} \text{ cm}^2$

所以正确答案是 C。

We get that the side lengths of $WXYZ$ are $5$ cm, since $\sqrt{25} = 5.$

We also know that the distance from $\overline{WZ}$ to $\overline{BC}$ is $2$ since it is the sum of the side lengths of $2$ unit squares.

Finally, the distance from $A$ to $\overline{BC}$ is the same as the distance from $\overline{BC}$ to $O,$ which is $2 + \dfrac{5}{2} = \dfrac{9}{2} \text{ cm}$

Now, we can find $BC,$ which is $WZ - 2 = 5 - 2 = 3 \text{ cm}$

Therefore, the area of $\triangle ABC$ is $\dfrac{1}{2} \cdot 3 \cdot \dfrac{9}{2} = \dfrac{27}{4} \text{ cm}^2$

Thus, C is the correct answer.

2003 AMC 8 第 25 题

答案：C

难度评级：1680

解答：

其他年份的第 25 题