Ok, you probably know Heron's Formula (if your teacher calls it Hero's formula, it's the same)... Heron of Alexandria, sometimes called Hero, lived around the year 100 AD and is most often remembered for a formula for the area of a triangle. The formula gives a method of computing the area from the lengths of the three sides...no angles required. If we call the sides a, b, and c; then the area is given by \(A= \sqrt {s(s-a)(s-b)(s-c)}\) where the "s" stands for the semi-perimeter, \(s=\frac{a + b + c}{2}\). You can find a nice geometric proof of Heron's formula at this link to the Dr. Math site. The proof was done by Dr. Floor, who credits the method to Paul Yiu of Florida Atlantic University. Documents from the Arabic writers indicate Archimedes may well have known this formula 300 years before Heron. In 1896, a copy of Heron's Metrica was recovered in Constantinople (now Istanbul) that had been copied around 1100 AD. It contains the oldest known demonstration of the formula. Wikipedia also has several nice proofs of the theorem, including one derived from the Pythagorean Thm.

I recently (2024) found a very similar formula for the square of the volume of a Tetrahedron in a paper in Letters to the editor of The Mathematical Intellingencer by Martin Lukarevski. The Formula applies to a tetrahedron with all four faces congruent with edges a, b, c. I did a little simple algebra to enhance the similarity to Heron's formula.

\(V(T)= \sqrt(\frac{( -a^2 + b^2 + c^2)( +a^2 - b^2 + c^2)( +a^2 + b^2 - c^2)}{72})\)

*things to do on a rainy afternoon, list 110 different formulae for the area of a triangle*). One was the well known Heron's formula above and then there was another that looked strikingly similar. If we let M

_{A}be the length of the median to vertex A, and similarly for M

_{B}and M

_{C}. The we can call sigma 1/2 the sum of M

_{A}+ M

_{B }+ M

_{C}. Then we can write.

Here is one more I only learned recently. The triangle at right has the lengths s-a, etc shown, and you realize that they are the radii of Soddy "kissing circles".(below) If we think of the side a, as opposite angle A (as we like to do in geometry) the a= (s-b )+(s-c), and b=(s-a)+(s-c), and you've figured out already that c=(s-a)+(s-b). Now if we use a' for s-a, and b' for s-b... then we have a= b' + c', and b=a'+c' and c=a' + b'.

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