AMC12 2009 B

AMC12 2009 B · Q13

AMC12 2009 B · Q13. It mainly tests Triangles (properties), Pythagorean theorem.

Triangle $ABC$ has $AB = 13$ and $AC = 15$, and the altitude to $\overline{BC}$ has length $12$. What is the sum of the two possible values of $BC$?

三角形 $ABC$ 中 $AB = 13$ 且 $AC = 15$，到 $\overline{BC}$ 的高长为 $12$。$BC$ 的两个可能取值之和是多少？

(A) 15 15

(B) 16 16

(D) 18 18

(E) 19 19

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $D$ be the foot of the altitude to $\overline{BC}$. Then $BD = \sqrt {13^2 - 12^2} = 5$ and $DC = \sqrt {15^2 - 12^2} = 9$. Thus $BC = BD + BC = 5 + 9 = 14$. Otherwise, assume that the triangle is obtuse at angle $B$ then $BC = DC - BD = 9 -5 = 4$. The sum of the two possible values is $14 + 4 = \boxed{18}$. The answer is $\mathrm{(D)}$.

设 $D$ 为到 $\overline{BC}$ 的高的垂足。则 $BD = \sqrt {13^2 - 12^2} = 5$，$DC = \sqrt {15^2 - 12^2} = 9$。因此 $BC = BD + BC = 5 + 9 = 14$。否则，若三角形在角 $B$ 处为钝角，则 $BC = DC - BD = 9 -5 = 4$。两个可能值之和为 $14 + 4 = \boxed{18}$。答案是 $\mathrm{(D)}$。

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