Hexagon Diagonals, the Dummy Approach

nine hexagon diagonals

The research team runs across lots of semi-authoritative posts while seeking candidates for the DotD award, many of which involve only small inaccuracies that aren't easily spotted (though we do spot them). We often notice that when writing about mathematics and science, our freelancers fail to get the whole story. They may address a general question with a specific, even unique, answer, for instance. Today's candidate is repeat offender Jess Kroll, who we found attempting to explain "How to Find the Diagonal of a Hexagon"¹ for Sciencing.com. He didn't quite succeed...

Perhaps because he avoided math classes while getting his MFA in creative writing, Kroll committed a logical failure to paper (sort of) in his post. Depending on your opinion of what the diagonal of a hexagon is -- either a line connecting two non-adjacent vertices or a line connecting two vertices that passes through the center of the figure -- there are either nine or three diagonals. Kroll chose the first option:

"The shape has nine diagonals, lines between the interior angles..."

...which is a rather poor rewording of "lines connecting two non-adjacent vertices." But hey: it's "creative," right? The problem with Kroll's solution to this question is that he fails to recognize that six of the nine diagonals connect two vertices separated by a third, while three of them connect vertices separated by two vertices. The latter three pass through the center [note: all discussion is of a regular hexagon]. According to Kroll,

"...the nine diagonals form into six equilateral triangles, making it easy to determine the length of each diagonal line."

Jess then "explains" that the formula for calculating the length of the diagonal is simply,

"d (diagonal) = 2g (given side)."

No kidding: just double the length of the side and you have the diagonal... but wait: that's only true for the diagonals that pass through the center, and those don't form equilateral triangles! At least they don't unless you project the nonadjacent two sides to a point...

In other words, Kroll claimed that his method worked for all nine diagonals, but it doesn't: it only works for three of them (green diagonals shown above). For the other six (blue diagonals shown above), you'll have to either dust off your trigonometry or use the Pythagorean theorem to determine the height of a 30-60-90 right triangle and then double the length of the base. That answer certainly isn't twice the length of the side. For giving less than half an answer, we hereby award Jess Kroll his third Dumbass of the Day citation.

¹ The original was sent to the rewrite team by Leaf Group (we'll get to the "correction" later), but it can still be accessed using the Wayback machine at archive.org. Its URL was   ehow.com/how_8411110_diagonal-hexagon.html

copyright © 2017-2022 scmrak

MM - GEOMETRY