AMC10 2024 B

AMC10 2024 B · Q11

AMC10 2024 B · Q11. It mainly tests Triangles (properties), Area & perimeter.

In the figure below $WXYZ$ is a rectangle with $WX=4$ and $WZ=8$. Point $M$ lies $\overline{XY}$, point $A$ lies on $\overline{YZ}$, and $\angle WMA$ is a right angle. The areas of $\triangle WXM$ and $\triangle WAZ$ are equal. What is the area of $\triangle WMA$? Note: On certain tests that took place in China, the problem asked for the area of $\triangle MAY$.

下图中 $WXYZ$ 是一个矩形，$WX=4$，$WZ=8$。点 $M$ 在 $\overline{XY}$ 上，点 $A$ 在 $\overline{YZ}$ 上，且 $\angle WMA$ 是直角。$\triangle WXM$ 和 $\triangle WAZ$ 的面积相等。求 $\triangle WMA$ 的面积。注：某些在中国举行的考试中，该题询问 $\triangle MAY$ 的面积。

(A) 13 13

(B) 14 14

(D) 16 16

(E) 17 \qquad 17 \qquad

Answer

Correct choice: (C)

正确答案：（C）

Solution

We know that $WX = 4$, $WZ = 8$, so $YZ = 4$ and $YX = 8$. Since $\angle WMA = 90^\circ$, triangles $WXM$ and $MYA$ are similar. Therefore, $\frac{WX}{MY} = \frac{XM}{YA}$, which gives $\frac{4}{8 - XM} = \frac{XM}{4 - ZA}$. We also know that the areas of triangles $WXM$ and $WAZ$ are equal, so $WX \cdot XM = WZ \cdot ZA$, which implies $4 \cdot XM = 8 \cdot ZA$. Substituting this into the previous equation, we get $\frac{4}{8 - 2ZA} = \frac{2ZA}{4 - ZA}$, yielding $ZA = 1$ and $XM = 2$. Thus, \[\triangle WMA = 4 \cdot 8 - \frac{4 \cdot 2}{2} - \frac{8 \cdot 1}{2} - \frac{6 \cdot 3}{2} = \boxed{\textbf{(C) }15}\]

我们知道 $WX = 4$，$WZ = 8$，因此 $YZ = 4$，$YX = 8$。由于 $\angle WMA = 90^\circ$，$\triangle WXM$ 和 $\triangle MYA$ 相似。因此，$\frac{WX}{MY} = \frac{XM}{YA}$，即 $\frac{4}{8 - XM} = \frac{XM}{4 - ZA}$。我们还知道 $\triangle WXM$ 和 $\triangle WAZ$ 的面积相等，因此 $WX \cdot XM = WZ \cdot ZA$，即 $4 \cdot XM = 8 \cdot ZA$。将此代入前式，得 $\frac{4}{8 - 2ZA} = \frac{2ZA}{4 - ZA}$，解得 $ZA = 1$，$XM = 2$。于是， \[\triangle WMA = 4 \cdot 8 - \frac{4 \cdot 2}{2} - \frac{8 \cdot 1}{2} - \frac{6 \cdot 3}{2} = \boxed{\textbf{(C) }15}\]

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