# Math Help - Heron's Formula for equilateral triangle

1. ## Heron's Formula for equilateral triangle

Hello everyone and thanks in advance for any help!

So I want to reduce Heron's formula to simplify equilateral triangles. But I'm getting caught up near the end:

Formula:

S=(a+b+c)/2, In this case S=3a/2 since a=b=c

A=(S(S-a)(S-a)(S-a))^1/2

So, ((3a/2)(3a/2-a)^3)^1/2

This is where I seem to lose it.

My calculator and wolfram agree that (3a/2)(3a/2-a)^3=(3a^4)/16
However, I am getting (57a^4)/16. I am cubing everything ((3^3a^3)/2^3-a^3) and multiplying that by 3a/2. I'm positive that my problem lays in cubing.

((3^3a^3)/2^3-a^3)=(27a^3)/8-a^3? wolfram claims a^3/8. I don't understand that at all! Where did that 3^3 go?? It has to be something simple that I'm missing or not doing correctly.

This emote: is definitely the most accurate representation of my current state.

Thanks for the help!

2. Notice that $\displaystyle \frac{3a}{2} - a = \frac{3a}{2} - \frac{2a}{2} = \frac{a}{2}$.

So $\displaystyle \sqrt{\frac{3a}{2}\left(\frac{3a}{2} - a\right)^3} = \sqrt{\frac{3a}{2}\left(\frac{a}{2}\right)^3}$

$\displaystyle = \sqrt{\frac{3a}{2}\left(\frac{a^3}{8}\right)}$

$\displaystyle = \sqrt{\frac{3a^4}{16}}$

$\displaystyle = \frac{\sqrt{3}\,a^2}{4}$.

3. Originally Posted by Bowlbase
So I want to reduce Heron's formula to simplify equilateral triangles.
Why get equilateral triangle area from Heron's? Too much unecessary work...
Divide equilateral triangle in 2 right triangles, each with hypotenuse = a and one leg = a/2 : get it?

4. Thank you! I knew it would be something simple that I wasn't doing correctly.

Wilmer, the point of the exercise wasn't to actually use the formula but to practice such problems. I appreciate the input though!