Cake
• Here’s my thinking so far.

Hypothesis #1: The area of the square is a multiple of 5

Proof:

Areas of 5 rectangles are equal

Therefore, let area of one rectangle = 🦄

Area of 🔲 = Area of 5 rectangles

Area of 🔲 = 5 🦄

Therefore,

Area of the square is a multiple of 5

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Assume all sides and areas of the rectangles and square are whole numbers, i.e. you don’t have sides of length 3.4

Reason: It’s easier to work with whole numbers and most successful mathematicians are inherently lazy: try the easiest solution or approach first.

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Hypothesis #2: The areas of the rectangles are a multiple of 5

Proof:

Area of 🔲 = 5 🦄 (from above)

If the sides of the 🔲 are whole numbers then

Area of 🔲

= 5 * 5 * Y*Y

= 25 * Y*Y

and each side of the 🔲 = 5Y

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Hypothesis #3: At least one side of each rectangle is even.

Area of top rectangle = 2 * X

X is a whole number (assumption)

Therefore, area of that rectangle is even

Areas of 5 rectangles are equal

Therefore, area of each rectangle is even

Therefore, at least one side of each rectangle is even

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Hypothesis #4: The area of the square is also a multiple of 4.

Area of 🔲 = 25 * Y*Y (from above)

Area of 🔲 = Area of 5 rectangles (from above)

Set equal to each other:

Area of 5 rectangles = 25 * Y*Y

Area of 1 rectangle = 5 * Y*Y

At least one side of each rectangle is even (from above)

Therefore, Y must be even.

So Y*Y is a multiple of 4.

Therefore,

Area of 🔲 = 25 * Y*Y = a multiple of 4

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Hypothesis #5: Area of the square is a multiple of 100.

Proof:

Area of 🔲 = 25 * Y*Y

Y*Y = a multiple of 4 (from above)

25 * a multiple of 4 = a multiple of 100

Area of 🔲 = a multiple of 100

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Hypothesis #6: Area of the rectangle is a multiple of 20.

Area of 🔲 = a multiple of 100 (from above)

Area of 🔲 = Area of 5 rectangles (from above)

Set equal to each other:

Area of 5 rectangles = a multiple of 100

Area of 1 rectangle = a multiple of 20

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Possible solution:

Area of the square is 100.

Problem with this solution:

In the diagram, the sides of the rectangles could not all be whole numbers.

• There can be only one answer. We have a dimension; "2" which allows us to calculate the area of the Square at 64.

I reverse engineered this based on a couple of facts. I also submit to everyone that I mislabeled the X and Y axis on my drawing and so I've probably not only confused myself, but everyone else too.

Fact: Each rectangle is 1/5 the area of the square

Fact: the 'X' side of the square is 2+ Rec2X

Fact: the 'Y' side of the square is the same as 'X' also (Rec1Y + Rec5Y)

The area of each rectangle equals (2+Rec2X * 2+Rec2X)/5

Substituting a number for Rec2X was best done in a spreadsheet, below. I started with 6. Coincidence only, as Rec1 look to be about that dimension.

Now that we have the area of each rectangle, and the 'X' dimension of 3 of them, we can figure the 'Y' dimension of Rec1, Rec2 and Rec5.

Also the dimensions of Rec3&4 as they are easily figured from the dimensions of Rec1&2. (that is Rec3X&4X=Rec1Y - Rec2Y and Rex3Y&4Y=Rec2X/2)

The spreadsheet I built cross references all the numbers in several ways.

There are no other solutions.

• Here's a simple "bottom-up" approach resulting in the same area @skinny_tom calculated:

1. All rectangles have the same area.

2. The area of rectangles 2, 3 and 4 combined has the same width as rectangle 1.

3. Because it has three times the area with the same width, it needs to have three times the height of rectangle 1. This means that the height of this area (and of rectangle 2) is 6.

4. The height of the square is made up by the height of rectangle 1 (2) and 2 (6), so is 8.

5. This means that the area of the square is 8^2=64.

Everything else, especially the fractional widths of some of the rectangles, is irrelevant.

• Here is my solution:
a= side of the square
x= short side of yellow rectangle
Blue rectangle
2(a-x)= 0.2a^2
a-x=0.1a^2
x=a-0.1a^2
Yellow rectangle
ax =0.2a^2
a(a-0.1a^2)=0.2a^2
a^2-0.1a^3=0.2a^2
÷a^2
1-0.1a=0.2
0.1a=1-0.2=0.8
a=8
Area of the square

8^2 = 64 cm2