How can we do that? 5 equal-area rectangles in a square problem. Reddit geometry r/theydidthemath

Here's a fun "find the area" problem from Reddit. We are given a square that has been equally divided into 5 parts. We are only given one side is 2 and asked to find the area of the square. Is this question even possible?
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#reddit #mathbasics #geometry

[Zuftware]

You can also notice that area of top rectangle 3 times less than area of compound rectangle below it. And since both rectangles share same width, bottom rectangle has 3 time height of top rectangle: 2*3=6. So side of big square is 2+6=8.

[incircIe]

Thats too complicated bruh. This is my way.

Let side length of the whole square be *x* and length of orange be *y* . Therefore;

3( area of orange ) = ( total area of yellow green and pink )

3(2y) = (x-2)y
6y = (x-2)y
6 = x-2
x = 8

So, Area of the large square = 8.8 = 64 sq units.

[meurdesoifphilippe]

When algebra complicates everything. Green, pink and yellow have a combined area of 3 times the orange one, and same width, so height of yellow must be 3 times 2 and the square has sides of 2+3*2=8.

[highelectricaltemperature]

I can see why this was so popular on reddit -- there are so many different ways of approaching the same problem.

[DC-FGC]

This was so easy.
You first blocked off 1/5 of the total area, leaving 4/5 blank.
Then next you block off 1/4 of the remaining area(1/4 of 4/5 = 1/5, so the area is still the same as the first piece), and that's as far as we need to go. Yellow, pink, and green are entirely unnecessary to think about.
Since that means you'd be able to lay 4 of those pieces side by side in the remaining area after blocking off the first piece, 2 x 4 = 8. That's one side.
And the whole thing is a square, so 8 x 8 = 64.
ETA: additional context for clarity.

[tontonbeber]

This one is extremely easy ... you only need orange and blue rectangles ...
square side x area x² area of each rect x²/5
Orange : 2a = x²/5 => a = x²/10
Blue : x(x-a) = x²/5 => a = 4x/5
And so x²=8x, x=8
Area 64

[stevenwilson]

when you had 8x = x^2 there was no reason to move x, just cancel the x and note that x = 0 is a solution, and that leaves 8 = x. Much faster that way, and you can still discout x = 0 as a solution.

[d_dstroyer]

After realizing each rectangle is just x²/5 you can solve this with the blue one and the orange one, since the blue has x on one side, the other side will be x/5, while for the same reasoning on the orange one we have x²/10 on one side.
Now we have x = x²/10+x/5, or 5x=x²/2+x, so 4x=x²/2, 8x=x², x=8.
So the area is 64

[沈博智-xy]

I did it such a long way:

I set up some equations.

The left length is partially given as 2. Let the remaining length be z.
So the left side of the square is (2+z)
The area is (2+z)^2

let the top left rectangle have side lengths 2 and x. This rectangle has area 2x.
Let the middle rectangle horizontal length be b.
Notice all the rectangles have same area.
This means the height of the middle rectangle and the height of the middle bottom rectangle are the same. Let this height be h.
Since we already have z and h, we can compare them. that is, z = 2h. (**)


If the whole length (horizontal) of the top left rectangle is x, and the length of the 2 middle rectangles is b, the horizontal width of the bottom left rectangle is x - b.
Area of bottom left rectangle = Area of middle rectangle
z(x-b) = bh
2h(x-b) = bh (from **)
2(x-b) = b (since h can't be 0)
2x - 2b = b
3b = 2x
b = 2x/3

This leaves the width of the bottom left rectangle to be x/3 (since x - b = x - 2x/3 = x/3)

Area of bottom left rectangle = area of top left rectangle
zx/3 = 2x
z/3 = 2 (since x can't be 0)
z = 6
z + 2 = 8
(z + 2)^2 = 64

therefore the area of the square is 64 square units.

[MitchBurns]

That’s one way to do it. The way I did it was that I realized the only way to design this problem would be by making the first rectangle divide the square in fifths like you said. But then take that a step further and see the second rectangle divides the remaining rectangle by 4. So x would have to be 2 times 4, which is 8.

[kajdronm.]

My solution:
The yellow, green and pink rectangle form a greater rectangle with 3 times the area of a single rectangle.
This combined triangle has the same width as the orange rectangle but 3 times the area, so height as to be 3*2 = 6.

The whole square has a side length of 6+2 = 8.
So the size of the rectangle is 8² = 64.

[Khantia]

I took a different approach:
1) I looked at the green and pink rectangles. They had the same width, and since they also had the same area, their height should've also been the same.
2) Then I looked at the yellow rectangle. It had the same height as green + pink, but half of their area. So A (the area of the rectangles) = (x - 2) * y, where y is the width of the yellow shape.
3) Then looking at the orange rectangle, we can also say that A = 2 * 3y, or A = 6y
4) From step (2), we can say that y = A / (x - 2)
5) We can replace y in step (3) with its equivalent from step (4): A = 6A / (x - 2) => 1 = 6 / (x - 2) => x - 2 = 6 => x = 8
6) The area of the square is x^2, so 8^2 = 64

[shadowclone]

You can actually extend the question further by determining the dimension of every colored rectangle using the fact that the side of the big square is 8 and all 5 rectangles' areas are equal and 1/5 of the area of the big square, which is 64 sq. units. So each of the 5 rectangles has an area of 12.8 sq. units.

Solving, you get:

Orange: w = 6.4, h = 2
Yellow: w = 32/15, h = 6
Green and Purple: w = 64/15, h = 3
Blue: w = 1.6, h = 8

[charlescalvin]

The way I did this is as follows:
1: Side length is s
2: The middle two rectangles share their length
3: They must, therefore, share their height as well, as their areas are equal
4: If their heights are equal, and they add up to the left rectangle's height, that must mean they each have half its height
5: To maintain equal areas, their length must be double the left rectangle's length
6: The top rectangle's length is o
7: If the lengths of the bottom rectangles are at a 2 to 1 ratio, and add to create o, the left rectangle's length must be a third of o
8: The left rectangle's height is s-2, as we already know the top rectangle takes 2 away from the side length
9: The left rectangle has the area (s-2)(o/3), and the top rectangle has area 2o
10: The equal areas mean that these quantities wre equal, so (s-2)(o/3) = 2o
11: Simplify, giving you os/3 - 2o/3 = 2o
12: Add the 2o/3 to the other side, giving you os/3 = 8o/3
13: Multiply both sides by 3, giving you os = 8o
14: Divide both sides by o, giving you s = 8. This is a valid step because o is a side length, and if a side length like o is 0, that kinda breaks the whole problem.
15: Square s to give you the total area, which is 64.

[animaltrax]

😮 no teacher could ever explain that in a way that makes any sense to me

[DTN.]

I am wondering your thoughts on collatz conjecture. Recently I have tried to develop a formula for it, but could not go to the depth of it. All I know about it is, it cannot be a power of 2, it cannot make a loop in a way where numbers come odd-even-odd-even... in turn, or it cannot make a loop as this way: odd-even-even-even...
(ax+b)/c is the best I could get, where a, b, c are all positive integers. But we can equal this to x since it is a loop and we get: b/(c-a) = x. (and here b>c>a>1)
We know that "a" must be a power of 3 and it is always odd, "c" must be a power 2 and it is always even. "b" is also always odd but I could not make a formula for finding "b". What I know about "b" is it must not be a prime number.

[CollieDog]

Let x = width of yellow rect.
Thus width of green = 2x
Thus width of orange = 3x.
Thus area of orange = 6x
Thus area of green = 6x
Thus height of green = 6x / 2x = 3
Thus height of whole square = 2 + 3 + 3 = 8
Thus area = 64

[alessiodaini]

solution:
5 rectangols have same area so 5A is the surface of the square, so if square has side equal to l => 5A = l². So it's sufficient that we find l or A to resolve the problem.
we notice that:
- orange rectangle has h = 2 and b = 1/2 A, from that we can recover that
- blue rectangle has b = l - 1/2A and h = l
We can use relation of blue square to find A:
l(l-1/2A) = A
l² - 1/2Al = A => A(2+l) = 2l² => A = 2l²/(2+l)
now, since we know A:
l² = 5A => l² =10l²/(2+l) =>(since l≠0) 1 = 10/(2+l) => l = 8 => 5A = l² = 64

[OptimusPhillip]

My attempt:
Let x equal the side length of the square, α equal the area of each rectangle, and A equal the area of the square.
By definition, we know that A=x^2. And by construction, we can see that A=5α, and that one side of the blue rectangle is x.
Let y equal the other side of the blue rectangle.
α=xy
A/5=xy
(x^2)/5=xy.
Assuming that x, as a measurement, is greater than 0, we can divide both sides by x.
x/5=y
Let o equal the width of the orange rectangle.
By definition, we know that α=2o. By construction, we can see that o=x-y.
o=x-(x/5)
o=(4/5)x.
α/2=(4/5)x
α=(8/5)x
5α=8x
A=8x
(x^2)=8x
Again, if x is greater than 0, we can divide both sides.
x=8
A=(8^2)=64

Wow, that took longer to type than it did for me to do mentally. But I didn't want to miss any rigor, so...

[ryznak]

Here is my reasoning:
The orange rectangle is as wide as the rectangle formed by the green, pink and yellow rectangles. But the 3 color rectangle has trice the area therefore it is trice as long. Since the length of the blue rectangle is 2 the length of the 3 color rectangle is 6. So the length of one side of the big square is 8 and 8^2 = 64.