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Quadrilateral Puzzle

Yesterday the fantastic Ed Southall (@solvemymaths) tweeted this brilliant puzzle:


It looked fun, so I thought I’d give it a try. First I sketched it out and gave all the vertices labels. A strategy I advise my students to take.


I then considered triangle wzg, as it was the triangle I knew most about:


My first thought was to find the length of the hypotenuse using Pythagoras’s Theorem. This was something that I didn’t use in the end, but I had yet to really formulate at strategy, and as I tell my students, you can never have too much information. I then thought I’d find the angles, but realised that it is this defaulting to trigonometry that often leads me to overcomplicate matters, so I thought I’d leave that til later, (plus I don’t know tan 2 or tan 0.5 off the top of my head.)

I considered the area of the triangle, then sketched the next triangle I knew stuff about fyz. It was here I saw my strategy.


The right angles meant I could use congruent triangles.


I could work out the area of the triangle yxk, which is congruent to fyz, as half the parallelogram area is 32, which is made up of this triangle and one which has area 8.

Thus the other leg of the right angled triangle must be 6rt2, and so a=12rt2 (as y is it’s midpoint!)

From there it was a question of Pythagoras’s Theorem to find b.


A fantastic little puzzle, one that I enjoyed solving, and one which should be accessible to higher GCSE learners. If you haven’t already do check out Ed’s website.

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