AMC12 2022 B

AMC12 2022 B · Q13

AMC12 2022 B · Q13. It mainly tests Triangles (properties), Similarity.

The diagram below shows a rectangle with side lengths $4$ and $8$ and a square with side length $5$. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?

下图显示了一个边长为 $4$ 和 $8$ 的矩形和一个边长为 $5$ 的正方形。正方形的三个顶点位于矩形的三个不同边上，如图所示。平方与矩形重叠区域的面积是多少？

(A) 15\dfrac{1}{8} 15\dfrac{1}{8}

(B) 15\dfrac{3}{8} 15\dfrac{3}{8}

(D) 15\dfrac{5}{8} 15\dfrac{5}{8}

(E) 15\dfrac{7}{8} 15\dfrac{7}{8}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let us label the points on the diagram. By doing some angle chasing using the fact that $\angle ACE$ and $\angle CEG$ are right angles, we find that $\angle BAC = \angle DCE = \angle FEG$. Similarly, $\angle ACB = \angle CED = \angle EGF$. Therefore, $\triangle ABC \sim \triangle CDE \sim \triangle EFG$. As we are given a rectangle and a square, $AB = 4$ and $AC = 5$. Therefore, $\triangle ABC$ is a $3$-$4$-$5$ right triangle and $BC = 3$. $CE$ is also $5$. So, using the similar triangles, $CD = 4$ and $DE = 3$. $EF = DF - DE = 4 - 3 = 1$. Using the similar triangles again, $EF$ is $\frac14$ of the corresponding $AB$. So, \begin{align*} [\triangle EFG] &= \left(\frac14\right)^2 \cdot [\triangle ABC] \\ &= \frac{1}{16} \cdot 6 \\ &= \frac38. \end{align*} Finally, we have \begin{align*} [ACEG] &= [ABDF] - [\triangle ABC] - [\triangle CDE] - [\triangle EFG] \\ &= 7 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac38 \\ &= 28 - 6 - 6 - \frac38 \\ &= \boxed{\textbf{(D) }15\dfrac{5}{8}}. \end{align*}

让我们标记图中的点。通过使用 $\angle ACE$ 和 $\angle CEG$ 是直角的事实进行一些角度追踪，我们发现 $\angle BAC = \angle DCE = \angle FEG$。类似地，$\angle ACB = \angle CED = \angle EGF$。因此，$\triangle ABC \sim \triangle CDE \sim \triangle EFG$。给定矩形和正方形，$AB = 4$ 且 $AC = 5$。因此，$\triangle ABC$ 是 $3$-$4$-$5$ 直角三角形，$BC = 3$。 $CE$ 也是 $5$。所以，使用相似三角形，$CD = 4$ 且 $DE = 3$。 $EF = DF - DE = 4 - 3 = 1$。再次使用相似三角形，$EF$ 是对应 $AB$ 的 $\frac14$。所以， \begin{align*} [\triangle EFG] &= \left(\frac14\right)^2 \cdot [\triangle ABC] \\ &= \frac{1}{16} \cdot 6 \\ &= \frac38. \end{align*} 最后，我们有 \begin{align*} [ACEG] &= [ABDF] - [\triangle ABC] - [\triangle CDE] - [\triangle EFG] \\ &= 7 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac38 \\ &= 28 - 6 - 6 - \frac38 \\ &= \boxed{\textbf{(D) }15\dfrac{5}{8}}. \end{align*}

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