# Special Right Triangles Help Needed

• Sep 26th 2012, 05:36 PM
GeometryBeginner
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.
• Sep 26th 2012, 05:56 PM
pickslides
Re: Special Right Triangles Help Needed
Try solving $\displaystyle 12^2 = (2x)^2+x^2$
• Sep 26th 2012, 09:01 PM
Soroban
Re: Special Right Triangles Help Needed
Hello, GeometryBeginner!

Welcome aboard!

Quote:

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: .$\displaystyle x \,=\,8\sqrt{3}$
• Sep 26th 2012, 09:07 PM
MarkFL
Re: Special Right Triangles Help Needed
A trigonometric approach:

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

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