## An interesting area puzzle

Regular readers will know that I have a tendency to collect puzzles and I like to have a go at them. This evening I had a crack at this while my daughter was playing before bed.

It looked interesting when I saw it on Facebook a while ago and so I thought I’d have a crack. If you haven’t yet, do it now. I want to know if you took the same approach!

I wasn’t too sure where to start, so I drew it out, labelled some stuff and came up with some equations:

I thought if I multiples two of my equations together and rearranged I could get the yellow area as the subject:

Then I needed cx:

I thought “now I need ay”, then realised I had it:

This meant I could.sub back in through the equations:

So the area:

I thought 27 was a nice answer, and I’m fairly sure it’s correct, however I have a feeling that I may have missed something blindingly obvious that would have gotten me there much quicker.

*If you did it, I’d love to hear your approach, especially if you spotted something I missed!*

Express the product of all widths and all height in two ways, by choosing in each case a rectangle from each row and column: 12x20x15x21 = 8x25x14x?

Nice insights. Thanks for the comments. I like your method.

My first thought with these type puzzles is to try a value out and see if it leads to an inconsistency or a valid solution. Either way I’ve found something out about the structure. I also thought of setting up some equations but tabled that for a bit.

So since the 2nd column seemed suggestive I started with setting it to 5 and then followed the chain of needed results and ended up with 45/7 * 21 /5 = 27.

I think my method is mathematically equivalent, but evades the algebra, and can – just about – be done without writing anything down! I used the fact that the ratio of any given pair of rectangle areas would be equal to another pair from the same rows or columns. Not sure that is very clear, but for instance top left is 14/15 of 8 since 14 and 15 share a dimension. From there it just remained to find a path to yellow, looking for three out of four corners known each time. And, yes, I got 27 too, which was reassuring after all those fractions!

Nice! A lovely approach.

The puzzle reminded me of the Japanese “area mazes” that were popular a couple of years ago. So I solved it as if it were one of those, in my head, using logic very similar to @thechalkfaceblog’s:

The area to the immediate left of ‘?’ must be 12/8 of 15. The area immediately to the left again must be 20/14 of that, ie. (20x12x15)/(14×8). And now ‘?’ must be 21/25 of that area, ie. (20x12x15x21)/(14x8x25) = 27.

I do love a good area maze

I like it, thanks, I used a scale factor approach by looking at the ratios between rows and columns, some are simple like top row is 80% of bottom row, others have fractional relationships. I got the same answer as you but it was quite quick as I only needed to work out the two numbers between 20 and 25 and then used the relationship between 2nd and 4th column to multiply up.

Nice approach. Thanks for sharing.

Funnily enough I gave this puzzle to a very Maths-passionate year 10 girl today. I gave her the brilliant.org puzzle about 99^999 and she sorted it in about 5 mins! Don’t know what I’d do with out brilliant actually.

I love brilliant.org I’m unfamiliar with the 99^999 puzzle. What is it?