## A Puzzling Heptagon

At somepoint Ed Southall (@solvemymaths) posted this heptagon puzzle:

I saved it in my phone to solve later, and forgot about it until the other day.

At first I looked at it and wasn’t sure where to start. I attacked the problem by sketching what I knew:

The heptagon is split along the base of the red shape into an isosceles trapezium and a pentagon. Because I know the interior angle of a heptagon (900/7) I know enough about this trapezium to work out the other angles (51 3/7).

This means I can deduce the angle BAE (77 1/7):

Because the heptagon is regular I know that AE = BE, and as such ABE = BAE. This means I can use the angle sum of a triangle and the sine rule to calculate each angle and each side of the triangle.

From here I could calculate the area of the triangle ABE either using absin(c)/2 or Heron’s Formula. I chose Heron’s Formula:

I know that triangles ABE and ADE are congruent so the area I’m looking for is double this area subtract the area of the overlap.

I considered the Pentagon above the line AE and how the triangles split it up into 4 triangles and a rhombus. I briefly considered the rhombus:

And quickly realised that using the rhombus properties and opposite angles I had enough information to calculate the area of the overlap:

First calculating the height, then the area:

Then I could find the area I was looking for:

*I thought this was a great puzzle, and it got me thinking about which of my students would be able to attack it. The skills needed are all skills that higher level GCSE students should have, and I think that some of my year ten class would give it a good go, but I also worry that some of my sixth formers may struggle with it. I think by exposing students to these problems early, and by sharing our own thought processes, we can start to build the resilience and mathematical thinking needed to succeed.*

*When I first attempted this problem it was late, and I made some daft errors transfering working from one line to another, and got a very wrong answer, this showed me that it’s easy to do, and we need to make sure we reiterate often the importance of checking our work to our students.*

*There was also the problem of rounding, as we were dealing with angles that didn’t give us nice exact trig ratios I had to round, through my working I rounded where it seemed sensible, but all these errors would have built up so I decided 1dp would be a good limit to round my final answer to, although now I feel nearest whole number would have been better.*

*I’m fairly sure that there is a much better and more concise way of solving this problem, but I currently can’t see it. If you do spot one, or solve it an even more long winded way, please let me know.*