Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

Heron’s Area of a Triangle

Posted by mark schwartz on October 24, 2016


This is a brief story about a fun event that almost always happens when discussing the area of a triangle. The formula for the area is A = 1/2 bh, where ‘b’ is the base and ‘h’ is the height.

The Story

Simple enough if it’s a right triangle and the base and height can readily be seen or calculated from a2 + b2 = c2. But what happens when it’s not a right triangle? Well, one has to wiggle around a bit and do a few more calculations to determine the height, but it can be found.

But being an instructor that likes to stretch students thinking and imaginations, I draw a very scalene triangle with sides of 4, 8, and 10 and present them the task of finding the area. As I roam the room watching them work and listening to their grumblings, I ultimately have them stop and present to them Heron’s formula. I don’t bother with the derivation and for those that are interested in knowing it, I recommend Googling it.

The formula is A = 1/4 the square root of (a + b + c)( ̶ a + b + c)( a ̶ b + c)( a + b ̶ c).

This is a really nifty formula because of the pattern in it – add all the sides together, then in the next three parentheses, just negate each side, one at a time, in order. And, given that there are 4 parentheses, just divide the square root of this product by 4.

In class, I first apply the formula to a 3, 4, 5 right triangle to demonstrate how it works. Then we play with a few other right triangles to get comfortable with the formula. Then we return to that weird scalene triangle. Using Heron’s formula, we get an area of 15.1987 … and now the fun begins!

Someone typically asks “how do you know this is correct?” So, this sets us up to explore how to find the area if we didn’t have Heron’s formula; so we do all the Algebra necessary to demonstrate that indeed 15.1987 is correct. In one class, this whole presentation got applause and I take that as their having had fun with it.


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