AMC10 2004 B

AMC10 2004 B · Q18

AMC10 2004 B · Q18. It mainly tests Triangles (properties), Area & perimeter.

In right triangle $\triangle ACE$, we have $AC = 12$, $CE = 16$, and $EA = 20$. Points $B$, $D$, and $F$ are located on $AC$, $CE$, and $EA$, respectively, so that $AB = 3$, $CD = 4$, and $EF = 5$. What is the ratio of the area of $\triangle BDF$ to that of $\triangle ACE$?

在直角三角形$\triangle ACE$中，$AC = 12$，$CE = 16$，$EA = 20$。点$B$、$D$和$F$分别位于$AC$、$CE$和$EA$上，使得$AB = 3$，$CD = 4$，$EF = 5$。$\triangle BDF$的面积与$\triangle ACE$的面积之比是多少？

(A) \frac{1}{4} \frac{1}{4}

(B) \frac{9}{25} \frac{9}{25}

(D) \frac{11}{25} \frac{11}{25}

(E) \frac{7}{16} \frac{7}{16}

Answer

Correct choice: (E)

正确答案：（E）

Solution

The area of $\triangle ACE$ is $\frac{1}{2}(12)(16) = 96$. Draw $FQ \perp CE$. By similar triangles, $FQ = 3$ and $QE = 4$. The area of trapezoid $BCQF$ is $\frac{1}{2}(3 + 9)(12) = 72$. Since $\triangle BCD$ and $\triangle FDQ$ have areas 18 and 12, respectively, the area of $\triangle BDF$ is $72 - 18 - 12 = 42$. The desired ratio is $\frac{42}{96} = \frac{7}{16}$.

$\triangle ACE$的面积为$\frac{1}{2}(12)(16) = 96$。作$FQ \perp CE$。由相似三角形，$FQ = 3$且$QE = 4$。梯形$BCQF$的面积为$\frac{1}{2}(3 + 9)(12) = 72$。由于$\triangle BCD$和$\triangle FDQ$的面积分别为18和12，$\triangle BDF$的面积为$72 - 18 - 12 = 42$。所需比值为$\frac{42}{96} = \frac{7}{16}$。

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