AMC10 2005 B

AMC10 2005 B · Q10

AMC10 2005 B · Q10. It mainly tests Triangles (properties), Pythagorean theorem.

In $\triangle ABC$, we have $AC = BC = 7$ and $AB = 2$. Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD = 8$. What is $BD$?

在$\triangle ABC$中，$AC = BC = 7$且$AB = 2$。设$D$是线$AB$上的一点，使得$B$在$A$和$D$之间，且$CD = 8$。$BD$是多少？

(A) 3 3

(B) $2\sqrt{3}$ $2\sqrt{3}$

(D) 5 5

(E) $4\sqrt{2}$ $4\sqrt{2}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $\overline{CH}$ be an altitude of $\triangle ABC$. Applying the Pythagorean Theorem to $\triangle CHB$ and to $\triangle CHD$ produces $8^2 - (BD + 1)^2 = CH^2 = 7^2 - 1^2 = 48$, so $(BD + 1)^2 = 16$. Thus $BD = 3$.

设$\overline{CH}$为$\triangle ABC$的高。对$\triangle CHB$和$\triangle CHD$应用勾股定理，得$8^2 - (BD + 1)^2 = CH^2 = 7^2 - 1^2 = 48$，所以$(BD + 1)^2 = 16$。因此$BD = 3$。

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