# Thread: Triangle Ratio Problem (I think)

1. ## Triangle Ratio Problem (I think)

It's one of those problems where they give you a right triangle, and then there is an altitude, and it's usually a variable. They also give you the length of the hypotenuse. I'm not very good at describing what I mean, (sorry) but I believe it should be set up as x+9/18=8/x+9

Do you understand what I'm talking about? I'm so sorry for the terrible description. ^^; The link is to the question, it's the third question on the page.
G52900

I've been working on it forever. I believe there is some foil-ing and then un-foil-ing, but I'm really not sure. I just know that I've tried it three different ways, and all of them were wrong. Can someone at least tell me how to set it up? What is it I'm doing wrong?

2. So, if I was to start this off, I'd start it by multiplying both sides by 18, so that the question became x+9=144/x+9, right? That's how I started it out. After that, you'd multiply both sides by x+9, so that the equation became x^2+18x+81, right? What next? I've tried moving it all on to one side, so that the equation became x^2+18x+63, but that doesn't seem to work. I'm just really lost.

Also, I know this is probably useless, but I really need help on this tonight. It's part of a larger program (as you can see on the link) that I need to complete for tomorrow. It was a tough assignment, because if you get stuck, (like I did!) then you can't finish the work. If I don't figure this question, I can't just skip it and move on. I probably won't get credit for the work. I'd really, really appreciate someone helping me! Just give me a hint!

3. ratio equation ...

$\displaystyle \dfrac{18}{x+9} = \dfrac{x+9}{8}$

$\displaystyle (x+9)^2 = 8 \cdot 18$

finish it.

4. x+9 = 12, right? Thanks.

5. ## triangle ratio problem

Hi KaieaAi,
The altitude on the hypothenuse of a right triangle is the geometric mean of the hypthenuse segments (X+ 9)^2 = 18 *8

bjh