AMC10 2002 A

AMC10 2002 A · Q13

AMC10 2002 A · Q13. It mainly tests Triangles (properties), Pythagorean theorem.

The sides of a triangle have lengths of 15, 20, and 25. Find the length of the shortest altitude.

一个三角形的边长为 15、20 和 25。求最短的高的长度。

(A) 6 6

(B) 12 12

(D) 13 13

(E) 15 15

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) First notice that this is a right triangle, so two of the altitudes are the legs, whose lengths are 15 and 20. The third altitude, whose length is $x$, is the one drawn to the hypotenuse. The area of the triangle is $\frac{1}{2}(15)(20)=150$. Using 25 as the base and $x$ as the altitude, we have $\frac{1}{2}(25)x=150$, so $x=\frac{300}{25}=12$. OR Since the three right triangles in the figure are similar, $\frac{x}{15}=\frac{20}{25}$, so $x=\frac{300}{25}=12$.

（B）首先注意这是一个直角三角形，所以其中两条高就是两条直角边，长度分别为 15 和 20。第三条高长度为 $x$，是向斜边作的高。三角形的面积为 $\frac{1}{2}(15)(20)=150$。以 25 为底、$x$ 为高，则有 $\frac{1}{2}(25)x=150$，所以 $x=\frac{300}{25}=12$。或者由于图中的三个直角三角形相似， $\frac{x}{15}=\frac{20}{25}$，所以 $x=\frac{300}{25}=12$。

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