AMC10 2008 B

AMC10 2008 B · Q6

AMC10 2008 B · Q6. It mainly tests Fractions, Ratios in geometry.

Points $B$ and $C$ lie on $\overline{AD}$. The length of $\overline{AB}$ is 4 times the length of $\overline{BD}$, and the length of $\overline{AC}$ is 9 times the length of $\overline{CD}$. The length of $\overline{BC}$ is what fraction of the length of $\overline{AD}$?

点 $B$ 和 $C$ 在 $\overline{AD}$ 上。 $\overline{AB}$ 的长度是 $\overline{BD}$ 长度的 4 倍，$\overline{AC}$ 的长度是 $\overline{CD}$ 长度的 9 倍。$\overline{BC}$ 的长度是 $\overline{AD}$ 长度的几分之几？

(A) $\frac{1}{36}$ $\frac{1}{36}$

(B) $\frac{1}{13}$ $\frac{1}{13}$

(D) $\frac{5}{36}$ $\frac{5}{36}$

(E) $\frac{1}{5}$ $\frac{1}{5}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Because $AB + BD = AD$ and $AB = 4BD$, it follows that $BD = \frac{1}{5}\cdot AD$. By similar reasoning, $CD = \frac{1}{10}\cdot AD$. Thus \[ BC = BD - CD = \frac{1}{5}\cdot AD - \frac{1}{10}\cdot AD = \frac{1}{10}\cdot AD. \]

答案（C）：因为 $AB + BD = AD$ 且 $AB = 4BD$，可得 $BD = \frac{1}{5}\cdot AD$。用类似的推理，$CD = \frac{1}{10}\cdot AD$。因此 \[ BC = BD - CD = \frac{1}{5}\cdot AD - \frac{1}{10}\cdot AD = \frac{1}{10}\cdot AD。 \]

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