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

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. The ratio of the area of the other small right triangle to the area of the square is

$\dfrac{1}{2m + 1}$

$m$

$1 - m$

$\dfrac{1}{4m}$

$\dfrac{1}{8m^2}$

答案：D

知识点：相似面积比直角三角形

难度评级：1750

解答：

设正方形边长为 $1$ 一个小三角形的直角边为 $1$ 和 $r$ ，面积为 $\tfrac12 r = m$ ，所以 $r = 2m$ 。

两个小三角形相似，因此另一个小三角形的直角边为 $1$ 和 $\tfrac1r$ ，面积为 $\tfrac12 \cdot \tfrac1r = \tfrac{1}{4m}$ 。

由于正方形面积为 $1$ ，所求比值是 $\dfrac{1}{4m}$ 。

所以正确答案是 D。

Let the square have side $1.$ One small triangle has legs $1$ and $r,$ with area $\tfrac12 r = m,$ so $r = 2m.$

The two small triangles are similar, so the other has legs $1$ and $\tfrac1r,$ with area $\tfrac12 \cdot \tfrac1r = \tfrac{1}{4m}.$

Since the square has area $1,$ the desired ratio is $\dfrac{1}{4m}.$

Thus, the correct answer is D.

2000 AMC 10 第 19 题

答案：D

难度评级：1750

解答：

其他年份的第 19 题