Heron's Formula; an Intuitive or Visual Proof

Question

I've found several proofs for Heron's formula for the area of a triangle in term of its sides, but none of them is simple and intuitive enough to show WHY the formula works.

Do you know an intuitive or visual proof for it?

Thanks.

Answer 1

Step 1:

Prove $r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$

Step 2:

Use $A = rs$ and you'll have Heron's formula.

It's helpful to know that tangent lengths from angle A are of length (s-a).

Use the Law of Cosines to determine the length of the third side of the isosceles triangle whose equal sides are of length (s-a) and whose angle is A.

Use the Law of Cosines to determine the length of the third side of the isosceles triangle whose equal sides are of length $r$ and whose angle is $(180^\circ - A)$.

Doing this leads to:

$r^2 = (s - a)^2 \frac{(1 - cos(A))}{(1 + cos(A)}$ which resolves to $r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$

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Intuition:

If one can show that the volume of a triangular prism of height $r$ is the same as the volume of a rectangular solid of sides $(s - a)$, $(s - b)$ and $(s - c)$ then this will lead to Heron's formula in the most efficient manner I've ever seen. Of course, one must first prove the formula for r as detailed above.

Enjoy

Answer 2

I noticed that many proofs for Herons formula use the properties of angles, which is why I wanted to show a proof that doesn't involve any, and seems to be simpler than the usual methods that are applied.

A triangle with side lengths $a, b, c,$ where $c$ is the greatest side length and an altitude($h$) that intercepts $c$ such that $c$ is the sum of two side lengths , $c = m + n$, using Pythagorean theorem and the area formula of a triangle, we can prove Herons formula.

First we find $a$ and $b$ in terms of $m$ and $n$ using Pythagorean theorem:

$a^2 - m^2 = h^2$

$b^2 - n^2 = h^2$

$a^2 - m^2 = b^2 - n^2$

$a^2 - b^2 = m^2 - n^2$

Then using $c = m + n$, eliminate $m$:

$a^2 - b^2 = (c-n)^2 - n^2$

$a^2 - b^2 = c^2 - 2cn$

$n = \frac{1}{2c}(c^2 + b^2 - a^2)$

We can Eliminate $n$ from $b^2 - n^2 = h^2$ to find $h$:

$h = \sqrt{b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2}$

Plugging this into the area formula ($A = \frac{1}{2}ch$) gives:

$A = \frac{1}{2}c\sqrt{ b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2} $

$A = \frac{1}{2}c\sqrt{ \frac{1}{4c^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)} $

$A = \sqrt{\frac{1}{16}(c^2 - (a - b)^2)(( a + b)^2 - c^2)} $

$A = \sqrt{\frac{1}{16}(a + b - c)( a + b + c)( b + c - a)(a + c - b)} $

$A = \sqrt{s(s - a)(s- b)(s- c)}$

Q.E.D.