Space Maze Puzzle – Thoughts Your Choices


It is a enjoyable space maze puzzle! I credit score Naoki Inaba for inventing these puzzles, and I examine it through The Guardian

Three rectangles are in a row, with areas of 21, ?, and 43 sq. meters. The second and third rectangles have a complete width of 10 meters. A fourth rectangle with space 29 sq. meters has a peak of 4 meters and spans the widths of the highest two rectangles. What’s the space of the second rectangle within the first row?

As ordinary, watch the video for an answer.

Area maze puzzle

Or preserve studying.

“All might be properly for those who use your thoughts to your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of sport principle and arithmetic. MindYourDecisions now has over 1,000 free articles with no advertisements because of neighborhood assist! Assist out and get early entry to posts with a pledge on Patreon.


Reply To Space Maze Puzzle

(Just about all posts are transcribed rapidly after I make the movies for them–please let me know if there are any typos/errors and I’ll right them, thanks).

I solved this with algebra. I will even current the artistic technique from Alex Bellos’ column.

Algebra answer

Let the second rectangle have a width of x and a peak of y. We have to resolve for its space xy.

For the reason that second and third rectangles have a width of 10 meters, the third will need to have a width of 10 – x meters. For the reason that fourth rectangle’s space is 29 sq. meters and its peak is 4 meters, its width should be 7.25 sq. meters. For the reason that prime two rectangles have that mixed width, the primary rectangle will need to have a width of seven.25 – x meters.

Utilizing the identified areas of the primary 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 primary to get:

2.75y = 22
y = 8

Substituting into the primary equation offers:

10(8) – xy = 43
80 – xy = 43
xy = 37

However there’s additionally a enjoyable strategy to resolve this downside.

Artistic answer

Shift the primary rectangle into the third rectangle, dividing it into two rectangles. For the reason that third rectangle’s space is 43 sq. meters, subtracting 21 offers the opposite rectangle with space 22 sq. meters. Delete the rectangle within the second row and assemble a rectangle with dimensions 4×10 and space 40 sq. meters.

Now re-appear the 29 sq. meters rectangle and shift it to the 40 sq. meters rectangle, dividing it into one other rectangle of 40 – 29 = 11 sq. meters.

For the reason that two rectangles of areas 11 and 22 sq. meters have the identical width, the peak of the highest rectangle should be double. Equally, the primary two rectangles within the prime row have the identical width because the rectangle with space 29 sq. meters. Thus their mixed areas should be double 29, or equal to 58. We all know the primary space is 21 sq. meters, and so the rectangle of unknown space should be 58 – 21 = 37 sq. meters.

Particular thanks this month to:

Robert Zarnke
Lee Redden
Mike Robertson
Daniel Lewis

Due to all supporters on Patreon!


Alex Bellos article space maze puzzle The Guardian


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