2000 AMC 12 第 21 题

Question

过直角三角形斜边上的一点，作两条分别平行于两条直角边的直线，把三角形分成一个正方形和两个较小的直角三角形。若其中一个小直角三角形的面积是正方形面积的

m

倍，则另一个小直角三角形面积与正方形面积的比是多少？

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is

m

times the area of the square. What is the ratio of the area of the other small right triangle to the area of the square?

\dfrac{1}{2m + 1}

m

1 - m

\dfrac{1}{4m}

\dfrac{1}{8m^2}

答案：D

知识点：相似面积比

难度评级：1970

小提示：

设正方形边长为

1

；一个小三角形沿正方形的一条边有一条长为

r

的直角边，所以它的面积是

\dfrac{r}{2} = m

。

Let the square have side

1

; one small triangle has a leg

r

along the square, so its area is

\dfrac{r}{2} = m

大提示：

两个小三角形相似，所以另一个三角形对应的直角边长为

\dfrac{1}{r}

。

The two small triangles are similar, so the other triangle’s matching leg is

\dfrac{1}{r}

解答：

设正方形边长为

1

。与正方形一边相邻的小三角形有一条长为

r

的垂直边，所以它的面积为

\tfrac12 \cdot 1 \cdot r = m

，从而

r = 2m

。

两个小三角形相似，所以另一个三角形沿正方形的直角边长为

\dfrac1r

，面积为

\frac12 \cdot 1 \cdot \frac1r = \frac{1}{2r} = \frac{1}{4m}\text{。}

因此，正确答案是 D。

Let the square have side

1.

The small triangle sharing one side of the square has a perpendicular leg

r,

so its area is

\tfrac12 \cdot 1 \cdot r = m,

giving

r = 2m.

The two small triangles are similar, so the other triangle’s leg along the square is

\dfrac1r,

and its area is

\frac12 \cdot 1 \cdot \frac1r = \frac{1}{2r} = \frac{1}{4m}.

Thus, the correct answer is D.

2000 AMC 12 第 21 题

答案：D

难度评级：1970

小提示：

大提示：

解答：

其他年份的第 21 题