How smart are you, /b/?

>>716858913
like 10 smart

>>716858913
Too smart for /b/'s autistic questions.

can't be solved, no apex

>>716859108
Wrong.

>>716858913
28 ?

28cm2

>>716858913
is this loss?

>>716859152
right. Unless you just guess and check.

>>716859291
No guessing is required.

>>716858913
28, simultaneous equations and shit

>>716858913
36

32

>>716858913
Dealing with integers.
A = 20 + 32 + 16 + x = length^2

Can't be solved, we have no way of knowing if side length is 9, 10, or any other integer greater than 8.

>>716860117
You don't have to know the side length to solve the puzzle.

I feel like information is missing

but when I think that, I consider that if the point where they all come together was slightly different, then the area of the cells would be different also

so then there would be a definite and predictable change in the area of those three known values, which could be used to determine the unknown value

but I don't know how to math

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20-A+B=X
32-B-C-D=X
16+A+C+E=X
D+F=X
fucked

?=30 IF we assume that the answer MUST be an integer.

23

the answer is 32cm2
/thread

I am creator of The branch.

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>>716858913
20 : 4 = 5
16 : 4 = 4
32 : 4 = 8

8 + 4 + 5 = 16 : 4 = 4

the answer is 26.

I am the stupid one here yet you all amaze me.

>>716858913
Being smart does not mean you have to memorize a hundred useless formulas faggot

32

32 cm is the correct answer, if you assume that it must be an integer.

>>716861292
no. you must be blind.

>>716860980
You're right. "Being smart" means having an intricate understanding of the world around you and know where those formulas come from and why they look the way they do.

You only have to memorize math when you are dumb.

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>>716861292
>>716861244
>>716859874
bullshit must be smaller

>>716861292
no, the answer is between 16 and 26

22.25cm^2

20-30cm^2

>>716860888
>8+4+5 = 16
huehueheuhe

>>716858913
Just give us the damn answer already OP
How do you find it out?

24

so what is the answer?

>>716861779
between 18 and 26

>>716858913

? = 2x - 68

>>716858913
28

>>716860888
>8 + 4 + 5 = 16

>>716858913
So its solvable without knowing any of the angles? (Like not measuring them)

>>716862895
Yes.

>>716862664
the tick marks on the sides indicate all sides are equal. the smallest square that would satisfy that would be 9 x 9

9x9=81
81-20-32-16=13
?=13

>>716862647
Correct post. OP confirm. Or should I confirm?

20

Answer = I Don't Give A Fuck

>>716862974
Ah, and that the ? is larger than 16 by plain eyesight is an optical illusion, of course.

>>716860117
>integers

dude i haven't even heard that word for like 10 years. fucking sophomore year of high school throwback

22,6

>>716858913
28

>>716858913
Is it just 20cm by symmetry, or am I oversimplifying? Dubs if tru

>>716860725
you need two more equations. i would use
20+32+16+A+B+C+D+E+F = 4*X
for one

>>716863545
by plain eyesight, the verticals are shorter than the horizontals...but given the tick marks, you assume they are equal. be less dumb. k. theeeeenks!!!

>>716859178
>>716859186
>>716859521
>>716863707

Correct.

Now, who's got a simple proof?

>>716863707
Because ?+20=16+32

>>716861686
Top kek

>>716858913
32, it's a square therefore the closest squared number that makes sense is 10 for each side. 9^2 gives 81 but that leaves 13 for the remaining area which is by scale impossible.

>>716863791
holy fuck man. Any more?

>>716863981
>>716863991
Since all the quadrilaterals have the same lenghts as base and height, opposite ones must sum half of the square.

>>716858913
26

>>716864100
but by scale 32 is impossible too

>>716864100
this
/thread

>>716864100
which is how I arrived at 22.25, making each side of the main square 9.5cm

>>716863866
The 16cm^2 area is smaller than the ? area. Do you need a C&P pic for that? No way it is 13.
BTW, thanks is written with an a, not e. Be less dump naxt time.

>>716864100
you were close, yet so far away

9^2 is the smallest square that will satisfy this. don't go by visuals...go by what's given :)

>>716864223
Also, the side of the square is sqrt(96)

neverwinter key for some thing im not sure i got it from the humble bundle christmas thing here you guys go

AR2pETya

>>716858913
Assume ratios top same as bottom
Top
2:3 1/5

Bottom
2 * 8 = 16
3.20 * 8 = 25.6

Im guessing about twentyfiddy cm2

>>716864170
nope, sry

the sides cant be integers, 20<X<32

20+16+32+(20)=88 sqrt88=9.3

20+16+32+(32)=100 sqrt100=10,

but X<32

>>716864223
>opposite ones must sum half of the square

This is true

It's not, however, only because the quadilaterals have two sides equal; that's not sufficient.

Answer is 28, I know it was already said in thread but here's my logic, 20 : 32 are related.
Assuming that the relationship is mirrored.
20 : 32
- 4
16 : 28

>>716858913
4

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any other answer is wrong, learn to math retards

>>716864749
bummer

>>716863549
What the fuck, are you a fry cook or something?

Not solvable. Given that it is a square, the side length of the full square is L, and the area in question is x, we have
x = L^2 - 68
There are infinite many solutions, for any L > sqrt(68).

>>716858913
the answer is obviously 10 cm^2

>>716865392
Because your retarded attempt at solving it doesn't work, it's not solvable? kek

>>716858913
?=x²-68 with x>8.24

>>716865392
pretty sure this would be true, if it weren't for the congruency and placement of the lines. I think they give away some information that idfk.

>>716865064
your so stupid that my head actually hurts

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e and f arent the same length, right?

>>716865903
No, they don't need to be.

If you look the Image and Split in half from horizontaly the Image are the mirror of the other so the answer is 32

>>716865754
When L changes, then the position of the center line connection moves, too. It gives a range of possible x's with an upper and a lower bound where the result is similar enough to the given image. Still, infinite many possible L's and x's within this range.

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>>716866314
Utter horseshit.

>>716866308
dont think it mirrors in any way (none of the inner sides are equally long)

>>716866444
Kek has spoken, all math proofs are now utter horseshit.
https://www.youtube.com/watch?v=iKcWu0tsiZM

>>716866904
Only 'proofs' that are clearly invalid and obviously written by autistic retards who know nothing about math.

Ok, I don´t think this equations would help, but it theoretically is corrent, isnt it?
((A + B)^2) - (Area - ((C + X)^2))
Where A, B and C are the given Areas and X is the ?

>>716865764
>Y O U R

>>716858913
It depends, has the joint of the four squares been displaced perfectly diagonally?
If so, 20cm, due to symmetry.
If it has been displaced by an unknown amount, the question is unsolvable, all that can be said for certain is that it's more than 16 and less than 32.

>>716864859
It's 28 and you're a cunt. We're solving this particular problem. In real life we would have more informations. I will not solve unsolvable, because I have more important things to do. Fuck you and your mother.
Loves Boris Babaian

>>716863981
16
20 add 4
28 add (4)*2
32 add (4*2)*2

Not good at explaining things, but just multiply..

>>716858913
could it be 22,6(periodic)?

>>716868264
The joint having been displaced by an unknown amount does not make the question unsolvable.

25.6 faggots

>>716858913
32cm squared

>>716858913
28

>>716864702
That's what I got, I feel validated by the fact, it also works out with ratios left and right

44 cm²

Smart enough to realise that, I CANT BE ARSED TO FIND THE ANGLES OF THESE RETARD SHAPES DURING FUCKING WINTER BREAK

26,3266666666666666666

20:16 = 32:x
X=25.6

>>716864223
I am dumb but willing to learn. Can someone explain this?

16, 20, and 32.

We need a solution that adds to a perfect square.

4,9,16,25,36,49,64,81,100,121, etc. = n^2 for n= 1,2,3,4,...

32 + 20 + 16 = 68

So it must be n >= 9 or greater. Assuming the middle point can be defined arbitrarily, we can have

9 =< n < infinity where n-> infinity is the limit in which the adjoining point goes to the top left corner.

This gives us a solution of

x = n^2 - 68 for all integer n >=9

Retard answer: Z=/={28,32}
Dumb answer: 28cm
Average answer: 28cm^2
Smart as fuck answer: 32cm^2

>>716870200
That's retarded, the puzzle doesn't even say that the side lengths are integers.

>>716870200

Who says the total square surface is an integer???

This is the most correct solution assuming the mid point has no bounds other than outside the square and that the total area must be an integer. However, I realize I made a mistake in that the area need not be an integer inwhich case sqrt(68) < n < infinity where n can be any number from the real group. Let's just assume n is not in the complex plane.

>>716869829
Seems legit!

>>716870447
>>716870461
I accounted for this in my reply, apes.

>>716869034
a+b = 20, b = 20 - a
a+c = 32, c = 32 - a
b+d = 16, d = 16 - b = 16 - (20 - a) = a - 4
c+d = (32 - a) + (a - 4) = 28

>>716870695
Still retarded, though. The puzzle clearly has one and only one solution.

>>716870447
>>716870461

My bad not linking properly

>>716870200
>>716870599

>>716870780
Why would you say that? That's obviously wrong. If you take the limit that the mid point goes to the top left corner, the other area becomes infinitesimal and the bottom right can become infinitely large. I believe you're disillusioning yourself because the physical image is present.

>>716870759

Ding! Ding! Ding! Ding! Ding! Ding! Ding!

>>716869034
> Different surface with the same letters

What?

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hopefully u will understand my way of thinking. many have already failed at it. one orange square must be a number that is G = IN. therefore ''?'' must be an amout that is even.

>>716871144

Triangles with the same base and height have the same area.

>>716871073
Nonsense. I say that the puzzle has only one solution because if you move the joint at which the four lines intersect, the ratio between the three given areas change. Therefore, the joint can only be in one specific place. Then, if you change the side length of the square, the values of the three given areas change. Therefore, there can only be one specific side length. These two facts combined mean that there can be one and only one solution.

>>716870759
In Addition, the question was made on december the 28th! xD.

that was one clue!

>>716858913
It's 20 square cm ok

>>716871404
>Ding! Ding! Ding! Ding! Ding! Ding! Ding!

28 - split each quad into two triangles, with base as edge of square. Each triangle is the same area as the adjacent triangle in the next quad. This gives 4 simple equations: a + b = 20; b + c = 32; c + d = ????; d + a = 16.

Take the first two equations to eliminate b and express in terms of a gives:

a = c - 12

Substitute a in the final equation gives:

d + c - 12 = 16

Rearranging gives c + d = 28

>>716862681
It must be higher than 20

What about this:

20:16 = 32:x
X=25.6

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>>716858913

A=4x^2-68...as per mathbro. Makes sense.

>>716858913
30.4cm2,

>>716860750
>>716872927
This

>>716871322
Fuck, now I get it.

Yeah, in that case 28 seems like the correct answer considering this post:
>>716870759

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>>716872580

>>716873303
> Closing one of the doors
> Thinking we wouldn't notice

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>>716872580
easy

>>716873162
>Fuck, now I get it.

Now you're thinking like a mathematician.

Lel. 21 :)

>>716860888
>>716862888
checkd

I was thinking about..... three fidy

>>716874469
god damned loch ness monster

>>716858913
I'm not sure this can be done. The side segments are all congruent size, but are not a fixed size. Therefore you could drag the centerpoint down into the gray area, effectively shrinking the total area, which would force the sides to become smaller to preserve the individual areas of the other three sides.

Filename is area, but the cm^3 implies volume.

OP is the actual Autist here.

>>716874821
Any change in the centerpoint will result in a change in the ratio between the three given areas.

13
Could be autistic or really smart

>>716858913
>>716871286

is my solution correct?

>>716875425

No. See
>>716869034
>>716870759
>>716871800

>>716873526
the true autist has spoken

>>716873303 also

>>716875425
28 cm^2 is the correct answer, but your way of getting there is wrong. There is no valid reason to assume that one sixteenth of the square's area has to be an integer.

>>716863981
32-20+16=28 aylmao

>>716876564
For that to be a proof, you've got to justify the use of that equation.

the anwser is 36

Everyone in here is autistic, the obvious answer is that the part with a question mark is gray while the rest is white.

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Here is the proof. It's 28

The 2 dark green triangles have same area because of congruent base and height.
So light green area and the sum of the 2 yellow are equal. So 20 + ? = 16+ 32

>>716869034

Here's the final proof!

a+c=32
a+b=20
b+d=16

a=32-c
a=20-b
b+d=16

32-c=20-b
b+d=16

-32+20+c=b
b+d=16

-32+20+c+d=16

then

c+d=28

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>>716874469

>>716878903
>It's a square

It doesn't need to be a square. An arbitrary quadrilateral works as long as all the semichords meet the midpoints of the edges.

>>716878903
Any reason you skipped refuting all the correct answers (28 cm^2)?