# How to find other length of the base on a Trinagle?

• Sep 9th 2012, 10:23 AM
Chaim
How to find other length of the base on a Trinagle?
Attachment 24749
Hi, I was confused on with this triangle.
At first I thought I should've started off with Pythagoras' Therom
I got the square root of (x2+180) = c
After I got c, I didn't know if it mattered.

I tried getting the smaller triangle's hypotenuse, which is the square root of (4 + x2) = 2 + x
After that, I made the 2 hypotenuses equal to equal other
2 + x = square root of (x2 + 180)
4 + x2 = x2 + 180
Yeah.. I got lost here since I will be losing x...
Can someone give me a tip or explain? :) Thanks
• Sep 9th 2012, 10:42 AM
Plato
Re: How to find other length of the base on a Trinagle?
Quote:

Originally Posted by Chaim

$\displaystyle \frac{x+12}{6}=\frac{x}{2}$
• Sep 9th 2012, 10:42 AM
skeeter
Re: How to find other length of the base on a Trinagle?
side lengths of similar triangles are proportional ...

$\displaystyle \frac{6}{12+x} = \frac{2}{x}$