Question

三角形 $ABC$ 满足 $AB = 13$ 、 $AC = 15$ ，且到 $BC$ 的高长为 $12$ 。 $BC$ 的两个可能值之和是多少？

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

$15$

$16$

$17$

$18$

$19$

高的垂足会分割 $BC$ ；在两个直角三角形中使用勾股定理。

The foot of the altitude splits $BC$ ; use the Pythagorean theorem in the two right triangles

两段垂足距离为 $\sqrt{13^2 - 12^2} = 5$ 和 $\sqrt{15^2 - 12^2} = 9$ ； $BC$ 是它们的和或差。

The foot distances are $\sqrt{13^2 - 12^2} = 5$ and $\sqrt{15^2 - 12^2} = 9$ ; $BC$ is their sum or difference

设 $D$ 为从 $A$ 作高所得的垂足。则 $BD = \sqrt{13^2 - 12^2} = 5$ ， $DC = \sqrt{15^2 - 12^2} = 9$ 。

若 $D$ 在 $B$ 与 $C$ 之间，则 $BC = 5 + 9 = 14$ ；若三角形为钝角三角形，则 $BC = 9 - 5 = 4$ 。两值之和为 $14 + 4 = 18$ 。

所以正确答案是 D。

Let $D$ be the foot of the altitude from $A.$ Then $BD = \sqrt{13^2 - 12^2} = 5$ and $DC = \sqrt{15^2 - 12^2} = 9.$

If $D$ lies between $B$ and $C,$ then $BC = 5 + 9 = 14$ ; if the triangle is obtuse, $BC = 9 - 5 = 4.$ The sum is $14 + 4 = 18.$

Thus, the correct answer is D.