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Nice area puzzle

Yesterday evening I came across this lovely area puzzle on twitter:

The puzzle is from Gerry McNally (@mcnally_gerry) he says its his first, and I hope that’s “first of many”.

I reached for the nearest pen and paper and had a quick go:

As you can see, I misread the puzzle originally and thought the lower quadrilateral was a square. The large triangle is isosceles as given in the question. This allowed me to use the properties of similar triangles and the base lengths given to work out the areas of the square, both right angled triangles and the whole triangle. This then allowed me to calculate the area of the shaded quadrilateral and hence that area as a fraction of the whole.

Then I went to tweet my solution to Gerry and realised that nowhere does it say that the bottom quadrilateral is a square. I had added an assumption. This made me ponder the question some more. Instincts told me that it didn’t have to be a square, but that the solution would be the sane whether it was a square or not. But I didn’t want to leave it at that, I wanted to be sure, so I had another go.

I sketched out the triangle again:

This time I called the height of the rectangle x.

This made it trivial to find the area’s of the rectangle and the triangle GCD. Triangle HAB was easy enough to find using similar triangle properties.

and then I found the area of the whole shape again using similarity to discover the height.

This allowed me to find the shaded area:

Then when I put it as a fraction the xs cancelled and it of course reduced to the same answer.

I really like this puzzle, and would be interested to see how you approached it, please let me know in the comments or on social media.

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  1. Mark Bennet
    April 23, 2019 at 6:32 pm

    Use similar triangles. The base of the ‘cap’ isosceles triangle at the top of the quadrilateral is 5 (which you can see by constructing a mirror image of the small triangle of base 2 on the other side). The area of this cap isosceles is therefore 25/81 of the whole.

    Now to deduct the area of the small triangle base 1, which is half an isosceles triangle of base 2 and therefore has area half of 4/81 of the whole ie 2/81.

    The difference is 23/81.

    Of course this depends on understanding dimension – but that is a useful thing to know.

    • April 23, 2019 at 6:34 pm

      Very nice concise solution.

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