Special Right Triangles Help Needed

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September 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.
September 26th 2012, 05:56 PM
pickslides

Re: Special Right Triangles Help Needed

Try solving $12^2 = (2x)^2+x^2$
September 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: . $\left(\tfrac{x}{2}\right)^2 + 12^2 \:=\:x^2$

Now solve for $x.$

You should get: . $x \,=\,8\sqrt{3}$
September 26th 2012, 09:07 PM
MarkFL

Re: Special Right Triangles Help Needed

A trigonometric approach:

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

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

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