Question

How do you use Heron's formula to find the area of a triangle with sides of lengths `6`, `6`, and `7`?

Answer 1

`17.06 \ "units"^2`

Explanation:

Heron's formula states for any triangle, the area is given by:

`A=sqrt(s(s-a)(s-b)(s-c))`

• `s` is the semiperimeter of the triangle, given by `s=(a+b+c)/2`.

• `a,b,c` are the sides of the triangle

Here, `s=(6+6+7)/2=19/2=9.5` units.

And so, the area is:

`A=sqrt(9.5(9.5-6)(9.5-6)(9.5-7))`

`=sqrt(9.5*3.5*3.5*2.5)`

`=sqrt(290.9375)`

`~~17.06`

Answer 2

Archimedes' Theorem is a modern form of Heron's Formula. We set `a=c=6, b=7` in

`16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2 = 4655`

`A = {7 sqrt(95)}/4`

Explanation:

Archimedes' Theorem is usually superior to Heron's Formula.

Given a triangle with sides `a,b,c` and area `A`, we have

`16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2`

`= (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4)`

`= (a+b+c)(-a+b+c)(a-b+c)(a+b-c)`

No semiperimeter, no multiple square roots when given coordinates.

We're free to choose whichever form is convenient and assign `a,b,c` however we see fit. For an isosceles triangle, `a=c` gives nice cancellation in the first form. Let's set `a=c=6, b=7` in

`16 A^2 = 4(6^2)(7^2) - 7^4 = 4655`

`A = \sqrt{4744/16} = {7 sqrt(95)}/4`