Our equation was in the form \(\frac{a + b + c}{3}=D\)

By multiplying the fraction by 3, we're able to get rid of our fraction (which makes our calculations easier)

After we multiply both sides by 3, we get: \(a + b + c = 3D\)

Cheers,
Brent

It seems to me, in this case, we could ignore that it's a special triangle and write out that the average of the 3 sides is 12

\(\frac{AB + BC + CA}{3}= 12\)

and then multiply both sides by 3, which gives us:

AB + BC + CA = 36

which is the perimeter, because the perimeter of any triangle is just the sum of it's sides.

I did the same. That seems like a faster approach too and nifty!