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

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

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

解答：

其他年份的第 21 题