Comments (0)
Trackbacks (0)
Leave a comment
Trackback

### Recent Posts

### Archives

- June 2017
- May 2017
- February 2017
- January 2017
- December 2016
- October 2016
- September 2016
- July 2016
- June 2016
- May 2016
- April 2016
- March 2016
- February 2016
- January 2016
- December 2015
- November 2015
- October 2015
- September 2015
- July 2015
- June 2015
- May 2015
- April 2015
- March 2015
- February 2015
- January 2015
- December 2014
- November 2014
- October 2014
- September 2014
- August 2014
- July 2014
- June 2014
- May 2014
- April 2014
- March 2014
- February 2014
- January 2014
- December 2013
- November 2013
- October 2013
- September 2013
- July 2013
- June 2013
- May 2013
- April 2013
- March 2013
- February 2013
- January 2013
- December 2012
- July 2012

### Categories

### Latest from the twittersphere…

- RT @ColinTheMathmo: "One can measure the importance of a scientific work by the number of earlier publications rendered superfluous by it."… 12 hours ago
- @icecolbeveridge @ProfSmudge It's using the fact that the sine ratio of pi/6 is a half. Definitely trig. 12 hours ago
- @icecolbeveridge @ProfSmudge I would say it definitely is trigonometry. And will let the use if degrees slide.... 12 hours ago
- @ProfSmudge @icecolbeveridge Thinking about it I presume this? Quite nice actually. https://t.co/rVBuhpVAnQ 12 hours ago
- @ProfSmudge @icecolbeveridge I'm not sure why you are hinting at 're the halved equilateral so would love to see what you mean. 12 hours ago

## 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.## Share this via:

## Like this:

Related