Home > #MTBoS, Maths, SSM, Starters > An interesting area puzzle

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 dw:

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!

Categories: #MTBoS, Maths, SSM, Starters
1. July 5, 2017 at 6:41 pm

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?

• July 5, 2017 at 8:33 pm

2. July 5, 2017 at 7:47 pm

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.

3. July 5, 2017 at 10:48 pm

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!

• July 5, 2017 at 10:51 pm

Nice! A lovely approach.

4. July 6, 2017 at 8:59 am

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.

• July 6, 2017 at 9:02 am

I do love a good area maze

5. July 6, 2017 at 9:06 pm

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.

• July 6, 2017 at 9:28 pm

Nice approach. Thanks for sharing.

6. July 6, 2017 at 9:28 pm

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.

• July 6, 2017 at 9:31 pm

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

7. March 28, 2018 at 7:42 am

Hi there,

Thanks for your solution. I came across this as I wanted to verify my answer as well.

Here’s my methodology.
First you can’t assume larger rectangles mean larger areas (example: the 14 rectangle is larger than the 15).

Second, since we have 16 smaller rectangles, you can assume that the lines dividing rows and columns are parallel. As such, the ratio of areas of rectangles should be equal.

Third, I used the above for pairs of adjacent rectangles. The first pair I used was the 20 and 14. This means each of the pairs of rectangles in the 2nd and 3rd column should have the same ratio. Thus the “dy” (according to your row and column labels) rectangle has an area of 17.5

Similarly, using the 12 and 8 rectangles yield 22.5 for the “by” rectangle.

Finally, using the same method on the “by”, “bz”, “dy”, and “dz” rectangles gives us 27 for the unknown as required.

Sorry, just noticed someone has already posted a ratio related response and I’m about 7 months too late!