 This is a fun area maze puzzle! I credit Naoki Inaba for inventing these puzzles, and I read about it via The Guardian

Three rectangles are in a row, with areas of 21, ?, and 43 square meters. The second and third rectangles have a total width of 10 meters. A fourth rectangle with area 29 square meters has a height of 4 meters and spans the widths of the top two rectangles. What is the area of the second rectangle in the first row? As usual, watch the video for a solution.

Area maze puzzle

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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

I solved this with algebra. I will also present the creative method from Alex Bellos’ column.

Algebra solution

Let the second rectangle have a width of x and a height of y. We need to solve for its area xy.

Since the second and third rectangles have a width of 10 meters, the third must have a width of 10 – x meters. Since the fourth rectangle’s area is 29 square meters and its height is 4 meters, its width must be 7.25 square meters. Since the top two rectangles have that combined width, the first rectangle must have a width of 7.25 – x meters. Using the known areas of the first and third rectangles’ areas we get the equations:

(10 – x)y = 43
10yxy = 43

(7.25 – x)y = 21
7.25yxy = 21

Subtract the second equation from the first to get:

2.75y = 22
y = 8

Substituting into the first equation gives:

10(8) – xy = 43
80 – xy = 43
xy = 37

But there’s also a fun way to solve this problem.

Creative solution

Shift the first rectangle into the third rectangle, dividing it into two rectangles. Since the third rectangle’s area is 43 square meters, subtracting 21 gives the other rectangle with area 22 square meters. Delete the rectangle in the second row and construct a rectangle with dimensions 4×10 and area 40 square meters. Now re-appear the 29 square meters rectangle and shift it to the 40 square meters rectangle, dividing it into another rectangle of 40 – 29 = 11 square meters. Since the two rectangles of areas 11 and 22 square meters have the same width, the height of the top rectangle must be double. Similarly, the first two rectangles in the top row have the same width as the rectangle with area 29 square meters. Thus their combined areas must be double 29, or equal to 58. We know the first area is 21 square meters, and so the rectangle of unknown area must be 58 – 21 = 37 square meters. Special thanks this month to:

Robert Zarnke
Kyle
Lee Redden
Mike Robertson
Daniel Lewis

Thanks to all supporters on Patreon!

Reference

Alex Bellos article area maze puzzle The Guardian
https://www.theguardian.com/science/2015/aug/03/alex-bellos-monday-puzzle-solution-area-maze-smarter-than-japanese-schoolchild

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