More area (with more radicals)...

[The Preacher]

Poor Preacher. He doesn't get the radicals.

Think of an equilateral triangle in which the length of a side is s, the perimeter (distance around the triangle) is p, and the area is A. Find the missing parts.

I have to draw a line up the middle, that's the altitude, and with that I'll be able to find the area. Here's what I've got so far:

Draw auxiliary line (for altitude).

The auxiliary line is also a leg for two right triangles. Use the Pythagorean Theorem to find the length. One leg (l) = 3, the other side is the hypotenuse (h), and it = 6.

\(\displaystyle l^2\)=9
\(\displaystyle h^2\)=36
36-9=27

Just like in that last problem, this is where I get stuck. I have to find the square root of 27, but I don't know how to get it in radical form.


EDIT: Err... wait. Should I just factor 27 to get the answer? Or is my thinking faulty somewhere?

DOUBLE EDIT: Alright, I got it! Sorry for this (now worthless) post. I'm so glad to have gotten it myself. Thanks guys, you helped me understand the concepts so I could get this question. =]

 

[galactus]

For curiosity, the area of a equilateral triangle can be given by:

\(\displaystyle \frac{sqrt{3}}{4}s^{2}\)

 

[zerodegrees]

hey...doesn't an altitude in an equilateral triangle set up a 30-60-90 triangle situation?

 

[Denis]

Yes...so ?

 

[The Preacher]

For curiosity, the area of a equilateral triangle can be given by:

\(\displaystyle \frac{sqrt{3}}{4}s^{2}\)

I just found that in my lesson. I'm so worthless at math. Lol.

And yeah, zerodegrees, it might have been easier had I been thinking of it that way. I make everythink in geometry more complicated than it has to be.