2 Answers

Richardtz · Answer 1 · 2019-12-04T00:06:11+0000

Yes. But right here I'm going to give you three formulas for the area of a triangle that are better than Heron's that your teacher doesn't even know.

Heron's Formula says the area

of a triangle with sides
a, b, c

satisfies

$H = √(s(s-a)(s-b)(s-c)) \quad \textrm{ where } \quad s = \frac 1 2(a+b+c)$

Hero of Alexandria was a renaissance man a millennium before the Renaissance. Nonetheless his formula could use some modernization. We start by substituting

and squaring.

$H^2 = \frac 1 2(a+b+c)(\frac 1 2(a+b+c)-a)(\frac 1 2(a+b+c)-b)(\frac 1 2(a+b+c)-c)$

16H^2 =(a+b+c)(-a + b + c)(a - b + c)(a + b - c)

That's our first formula that's better than Heron's. Escape the tyranny of the semiperimeter. This one is about as easy to apply in the case of an even perimeter, and easier in all other cases. But we can go farther.

Very often we don't know the side lengths but we know the square of the side lengths. This happens with coordinates for example. Heron's formula is really awkward when the sides are the square roots of things. But there are a couple of awesome formulas that solve the problem.

We just keeping going, with associativity.

16H^2 =(a+(b+c))(-a + (b + c))(a - (b - c))(a + (b - c))

16H^2 =a^2(b+c)^2 - (b+c)^2(b-c)^2 - a^4 + a^2(b-c)^2

16H^2 =a^2( (b+c)^2 + (b-c)^2) - (b^2-c^2)^2 - a^4

16H^2 =a^2(b^2 + 2 bc +c^2 + b2^2 - 2bc + c^2) - (b^2-c^2)^2 - a^4

16H^2 =a^2( 2b^2+ 2c^2)- b^4 + 2 b^2 c^2 -c^4 - a^4

16H^2 =2(a^2 b^2+ a^2 c^2 + b^2 c^2) - (a^4 + b^4 + c^4)

Now all the variables are in terms of the squared sides like we wanted. Let's simplify by abbreviating
A=a^2, B=b^2, C=c^2

.

We subtract the identity

(A+B+C)^2 = A^2+B^2+C^2+2(AB+AC+BC)

There it is, our second formula that's better than Heron's. We can rewrite it in terms of the sides if we like. But let's find the asymmetric looking form by instead adding the identity

(C-A-B)^2 = A^2+B^2+C^2-2(AC+BC-AB)