# is this formula already in use?

## Recommended Posts

hi,

While solving a calculus problem today related to "cross sections" I derived a formula to calculate the area of an equilateral triangle. So I was just curious whether the formula I came up with is already in use or it's....?

The formula:

$A(s)=\frac{s}{2}\sqrt{s^2 - \frac{s^2}{4}}$

What do you guys think?

##### Share on other sites

I think this is pretty impractical. It needs some serious simplification at least.

Look, you have s^2 - 1/4 s^2. Why not just write 3/4 s^2?

Then if you do that, you have the square root of a square.

##### Share on other sites

And then, using the simplifications NeonBlack suggested, you get the result using trigonometry anyway...

I was a little confused about your use of 's' as the variable at first, because I thought you were trying to replicate Heron's formula.

Let s be the semiperimeter, s=0.5*(a+b+c)

a,b,&c are the lengths of each of the three sides.

Heron's forumla for the area of the triangle is then:

Area = [ s*(s-a)*(s-b)*(s-c) ]^(1/2)

I just saw that squareroot and the s and thought... now that isn't right, though if s is the length of the side of the equilateral triangle, then it is fine -- it does warrant simplification though.

##### Share on other sites

yeah that simpifies down to (root 3)/4 s

##### Share on other sites

Are you sure?

It seems to me that an expression for an area should have a term which is second order in length.

If I double the length of the side the area should quadruple rather than double.

##### Share on other sites
Should be [sqrt(3)/4]*s^2.

That is exactly the formula that computerages posted originally (just not reduced to simplest form):

$A(s)=\frac{s}{2}\sqrt{s^2 - \frac{s^2}{4}} = \frac{s^2}{2}\sqrt{1 - \frac1 4} = \frac{s^2}{2}\sqrt{\frac3 4} = s^2\frac{\sqrt3}4$

## Create an account

Register a new account