# Thread: Special Right Triangles Help Needed

1. ## Special Right Triangles Help Needed

Well I am new here but I am really stuck on one of my problems that I was assigned to for HW.

It is..

The length of an altitude of an equilateral triangle is 12 feet. Find the length of a side of the triangle.

I kinda get the concept but this one just blows my mind.

2. ## Re: Special Right Triangles Help Needed

Try solving $\displaystyle 12^2 = (2x)^2+x^2$

3. ## Re: Special Right Triangles Help Needed

Hello, GeometryBeginner!

Welcome aboard!

The length of an altitude of an equilateral triangle is 12 feet.
Find the length of a side of the triangle.

Did you make a sketch?

Code:
              *
/|\
/ | \
x /  |  \ x
/   |   \
/    |12  \
/     |     \
/      |      \
* - - - + - - - *
:  x/2  :  x/2  :
See the right triangle?

Pythagorus says: .$\displaystyle \left(\tfrac{x}{2}\right)^2 + 12^2 \:=\:x^2$

Now solve for $\displaystyle x.$

You should get: .$\displaystyle x \,=\,8\sqrt{3}$

4. ## Re: Special Right Triangles Help Needed

A trigonometric approach:

$\displaystyle \sin(60^{\circ})=\frac{12}{x}$

$\displaystyle x=\frac{12}{\sin(60^{\circ})}=\frac{12}{\frac{ \sqrt{3}}{2}}=\frac{24}{\sqrt{3}}=8\sqrt{3}$