Area Mazes - fun puzzles by a master puzzle inventor


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I think you’ve swapped the easy and hard examples…


That was my reaction too. It’s got one more number you need to work out, but all of them are easier. (In a couple more steps you can get the innermost rectangle too.)

Also, until I spotted the little footnote to the instructions, I was thinking the first puzzle must have been in error, since one of the first numbers you need to calculate is a fraction and not a whole number.


I agree. The second one makes it seem like the book is going to just be a lot of adding and multiplying with extremely little thinking. The first one gives me some hope that the book is actually interesting. (In particular, I think first one has a “more elegant solution” as described in the instructions)


The second one, I just literally starting writing in the unknown sides mechanically and the answer popped out less than 10 seconds later. Could you post an actual hard one?


I’m with the “you sure that’s the hard one” crowd?

35 in^2

I didn’t need all the info given and I didn’t double check, but I think the other one is 4"


The website says

Whole numbers are all you need. You can always get the answer without using fractions!

I can’t figure out how to do that with the first.


Ah, I didn’t read that footnote at first. With that, it seems that the rule indicating you should stop if you get something that’s not a whole number means the answer to the puzzle must be a whole number, but the intermediate steps may result in some other length or area that isn’t.


From the book scan:

If your calculation creates a fraction or decimal, STOP and look for another way. Area numbers can be solved using whole numbers only!*
*However do not assume every length or area in the puzzle must be a whole number.

There must be another way to figure out the ‘easy’ 11x13 puzzle? Even knowing the answer, I can’t figure out how else to get it.


Sorry to keep posting, but I finally figured it out!
The 21" and 35" rectangles are both 8" wide. Together, their area is 56". 56/8 = 7. 11-7 = 4…


Nice that you kept searching untill you found it!


Oh, nice! I wasn’t getting that, but you’re right.


The top is 8 long, Go down to the next horizontal line, part of it is 3 so the other part is 5. Follow that line down to the bottom which is 13 long. Since the left portion is 5, the right portion is 8.

The length we are looking for is the total height (11), minus the height of the top and the lower right rectangles. Both of those rectangles are 8 wide. So instead of dividing their areas by 8 to get their heights, we add their areas to get 54, then divide by 8 to get 7, then subtract that from 11 for 4.

ETA: I guess I should read the whole thread before I answer.

Like I said above, this is the one that I thought was interesting and made me think there might be interesting puzzles in the book. Not because solving the initial numbers is hard, but because you look for a simpler way to do it that uses less of the information given.


So, here are my solutions but I’m not convinced that these are minimal, especially the second, in the sense of involving the smallest number of computations.



On the second one, once you have C & F, the unknown area is (14-C)*(11-F).


Yep, that definitely saves a few steps.


I don’t think these are the wrong way round - I think they’re different and both easy.
For the easy one, it’s mostly deduction: looking at the area=27, only possible factors (or sides) are 1 & 27 or 3 & 9, so the long side is either 9 or 27. If 27 (and if the shape drawn misleadingly), next area down must then have a long side of 35 (only possible factor bigger than 27), making left edge of whole shape 9+1+1. area=21 can’t have a short side of 9 if it’s long side is 8 (don’t really need to do the maths…), so area=27 must have long side 9, and short side = 3 (the answer).
For the hard one, as soon as you do 14-6, the rest is simple maths.
(I think anyway! ;))


The correct answer is 4, so this deduction doesn’t work.


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