1. Mar 21, 2008

### kingredg

my teacher gave my class this problem earlier this week
and we need to know it by the end of next week

anyone knows how to solve this

2. Mar 21, 2008

### slider142

First, you can use the obvious Pythagorean theorem to get two equations in 3 unknowns. Get the third equation by using CF and the ratios of the two similar triangles to their larger counterparts.

3. Mar 21, 2008

### kingredg

^^^how can i use the Pythagorean theorem when i only have 1 side length for both

and by the way i'm only a freshman

what's the CF?

4. Mar 22, 2008

### kamerling

if you guess that AE = x. you can calculate AB and DE with pythagoras, and then
you have CF/AF = DE/AE because the triangles AED and AFC are similar. That gets you
AF, and you can find EF in the same way. Finally you must have AE = AF + EF.
I believe you'll get a 4th degree equation in x^2 after two rounds of squaring and collecting terms. I'm going to solve this numerically.

5. Mar 22, 2008

### kingredg

^^^i'll try that

6. Mar 23, 2008

### janhaa

Is AE $$\;\; \approx \;\,26\,\text meters\,?$$

7. Mar 24, 2008

### kamerling

I get the solution of $$\frac {CF}{\sqrt{{AD}^2 - x^2}} + \frac {CF}{\sqrt{{BE}^2 - x^2}} = 1$$

wich is 26.0328775442....

8. Mar 25, 2008

### kingredg

^^^sorry for asking such a dumb question but
how did you find the X to solve it

9. Mar 25, 2008

### kamerling

I just set AE equal to x. this gives you

AF = CF * x / DE using the fact that CF/AF = DE/AE because the triangles AED and AFC are similar.

combined this gives $$AF = \frac {x(CF)}{\sqrt{{AD}^2 - x^2}}$$

you can do the same on the other side to get $$EF = \frac {x (CF)}{\sqrt{{BE}^2 - x^2}}$$

Finally you must have AF + EF = AE = x

Substituting the previous expressions for AF and EF in this and dividing by x gives:

$$\frac {CF}{\sqrt{{AD}^2 - x^2}} + \frac {CF}{\sqrt{{BE}^2 - x^2}} - 1 = 0$$

The easiest way to solve this is to type 10/sqrt(1600-x^2)+10/sqrt(900-x^2)-1