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Area Puzzle 21

I came across the following area puzzle on Ed’s (@solvemymaths) site.

I found it quite an interesting idea and had a little go at solving it.

First I considered an equilateral triangle, all the angles are 60 degrees so we can see the area would be (x^2 sin (60))/2. As sin 60 = 3^(1/2)/2 it was easy enough to solve.

I then realised I could have used Heron’s Formula, so did it that way too. Luckily I got the same answer:

The square was a tough one:

Then I considered a pentagon, I wasn’t sure how to approach it at first,  so I reverted to my favourite shape, the triangle. Split the pentagon into 5 congruent isosceles triangles and solved with trig.

Wolfram alpha gave me the lovely exact answer:

I then considered the hexagonal case, which is really just 6 equilateral triangles.

I approached the decagon in the same manner I had the Pentagon.

And got an equally lovely exact answer:

I enjoyed working through these, and thought it would make a nice lesson to build resilience and cognitive activation. I also thought about what else could be done. What does the sequence of side lengths generated look like? Could an equation be formed to describe it? What if we were looking at the areas of regular polygons with side length six, what would the sequence look like then? All these would make nice investigations.

Categories: GCSE, Maths, Resources, Starters
1. September 23, 2015 at 10:18 pm

(3x/2 – x)^3 = (x/2)^3 = x^3/8

• September 23, 2015 at 10:21 pm

Indeed! Would have saved me a bit of time.