Hello, Vicky!

I couldn't follow your work, but it looks reasonable.

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 with two smaller right triangles.

The area of one of the two small right triangle 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

. . $\displaystyle a)\;\frac{1}{2m+1} \qquad b)\;m\qquad c)\;1-m\qquad d)\;\frac{1}{4m}\qquad e)\;\frac{1}{8m^2}$ Code:

- *
|::*
b |:::::*
|:::A2:::*
- *-----------*
| x |::*
| |:::::*
x | S x |::::::::*
| |:::::A1::::*
| | :::::::::::::*
- *-----------*-----------------*
: x : a :

The side of the square is $\displaystyle x.$

. . Its area is: .$\displaystyle S \:=\:x^2$

The area of $\displaystyle A_1$ is: .$\displaystyle A_1 \:=\:\tfrac{1}{2}ax$

We are told: .$\displaystyle A_1 \:=\:m\cdot S \quad\Rightarrow\quad \tfrac{1}{2}ax \:=\:mx^2 \quad\Rightarrow\quad a \:=\:2mx$ .[1]

The area of $\displaystyle A_2$ is: .$\displaystyle A_2 \:=\:\tfrac{1}{2}xb$ .[2]

From the similar right triangles: .$\displaystyle \frac{b}{x} \:=\:\frac{x}{a} \quad\Rightarrow\quad b \:=\:\frac{x^2}{a}$

Substitute into [2]: .$\displaystyle A_2 \:=\:\tfrac{1}{2}x\left(\frac{x^2}{a}\right) \:=\:\frac{x^3}{2a}$

The ratio of $\displaystyle A_2$ to $\displaystyle S$ is: .$\displaystyle \frac{A_2}{S} \:=\:\frac{\frac{x^3}{2a}}{x^2} \:=\:\frac{x}{2a}$

Substitute [1]: .$\displaystyle \frac{A_2}{S} \;=\;\frac{x}{2(2mx)} \;=\;\frac{1}{4m}$