1951 AMC 12 第 49 题

Question

从一个直角三角形的两个锐角顶点所作的中线长分别为

5

和

\sqrt{40}

。斜边长为：

The medians of a right triangle which are drawn from the vertices of the acute angles are

5

and

\sqrt{40}.

The value of the hypotenuse is:

10

2\sqrt{40}

\sqrt{13}

2\sqrt{13}

以上答案均不正确

None of these

设两条直角边为

a,b

，并将直角置于二者的公共端点

Let the legs be

a,b

and place the right angle at their common endpoint

由两条中线的长度可得

a^2+\frac{b^2}{4}=25

和

\frac{a^2}{4}+b^2=40

The two median lengths give

a^2+\frac{b^2}{4}=25

and

\frac{a^2}{4}+b^2=40

设两条直角边为

a,b

。从其相对的锐角顶点所作中线的长度平方满足

a^2+\frac{b^2}{4}=25,\qquad \frac{a^2}{4}+b^2=40\text{。}

解得

a^2=16

和

b^2=36

。因此，斜边

c

满足

c^2=a^2+b^2=52\text{，}

所以

c=2\sqrt{13}

。

因此，正确答案是 D。

Let the legs be

a,b.

The medians from their opposite acute vertices have squared lengths

a^2+\frac{b^2}{4}=25,\qquad \frac{a^2}{4}+b^2=40.

Solving gives

a^2=16

and

b^2=36.

Thus the hypotenuse

c

satisfies

c^2=a^2+b^2=52,

so

c=2\sqrt{13}.

Thus, the correct answer is D.

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