How smart is Sup Forums?

How smart is Sup Forums?

24cm^2

24 cm^2

12 cm^2

There's no question...

Problem is indeterminate

21

>How smart is Sup Forums?
There is a question, now answer it.
I say, superst clever on earth.

Average intelligence, i'd wager.

Is there a way of solving it other than

"All sides are equal so it has to be a square. An all the areas are whole numbers so the sides have to be whole numbers. And it's a square so the area has to be 'some number' squared. And 28 + 16 + 32 = 76 so s^2 > 76. So that leaves 9^2, 10^2, 11^2 so on. Well 10^2 is 100 and that looks sort of right so 28+16+32+?=100 so ?=24cm^2!"

I feel like that's not satisfactory
The sides could be 11cm, or 12cm or 13cm or whatever and the drawing is not drawn to scale or something

>all the areas are whole numbers so the sides have to be whole numbers

if you couldnt figure out that op wants you to determine the area inside the shaded polygon labeled "?" then you should just give up on life right now. its gonna get much harder

its 20 cm2 morons

Maybe I'm retarded but is there a case where a non whole number squared gives you a whole number?

root of 2 squared?

Yes, the square root of literally any number that is not a square number.

√3 * √3
i can keep going if you want

and if u fags still don't believe me look at the lines they all come from the middle and come together aswell, so wherever you move the point where they come together the area plus the area diagonal of it together stay the same as the other 2, sorry I'm not native English.

>another "do my homework" thread
I'll help if you explain to me why you're posting on Sup Forums while in summer school

correct

you forgot to times it by root 3 the third time dumbdicks

Do you even know what the difference between square root cube root is?

haha i trolled u

How does that hold correct??

if 1.4 + 1.4 = 2.8
then we round it to 3
does that mean 1+1=3

It's unsolvable, you pathetic Deists.
You just think there is an answer because you can't possibly conceive that there is no answer.
Sound familiar to anything? hehe
Enjoy your thread, simpletons, I have more intelligent matters to attend to.
*teleports to another place with the teleporter I made with my superior intellect*

kek

kek

Fuck off faggot. I'm more mathematic than you asshole

18ish cm^2?

32/28 = x/16
512=28x
x=18.28

(28+16)/2 = 22
(32-28)/2 = 4
4/2 = 2
22+2 = 24cm^2

i had in mind this comment when i prepared the original trolling material

you have been double trolled

trolled in two dimensions

submit, you lost, i fucked your mother and she begged for more

and then i said no

how do you reckon 32/28= x/16?

i submit

retard

I tried to do it like proportions.
The right side is 32 over 28, the left side is x over 16.
Then i cross multiplied, etc etc.

Same thing as like 1/2 -= 2/4 1/2 = 2/x
idk lol

The picture doesn't exactly say the lines come from the middle, but I assume they mean that.
Why does that mean the diagonal areas are equal, though? I guess you're correct but I don't understand the reasoning.

every virgin loser claims he knows math

It doesn't work like that. I think it's unsolveable. If I had two angles in the middle I think I could solve it with some trig. Don't know for sure though, might just be I'm not at that level yet.

It's Sunday you fucking piece of shit nobody cares

28 cm^2

What are you even saying

This is impossible to do, there is no real way of knowing any angles, except for the right angles, and the lengths of the sides.

Thusly it is insolveable, the only fake way is to guess and estimate the area value.

He's telling you that
32+16=28+x
so x=20

I just don't understand why.

How would one go about proving this is unsolveable though? Because that is a big part of math also, being able to definetively prove something is unsolveable or false.

It apparently isn't unsolvable, even though my first guess was it's underdetermined.

Proving it would be easy in this case. Give two different concrete examples that fit the original setup.

Proposal: polygon ACIG is a Rhombus of area 101.33 cm2, and the dark grey section is of area 25.33 cm2

Proof:
Givens: AB = BC, BC = CF, FI = IH, HG = GD, GD = DA
Therefore, AC = CI, CI = IG, IG = GA, by Modus Ponens
Therefore, ACIG is a four-sided symetrically-sided Rhombus.
Given: all four sided symetrically-sided Rhombi can be divided into 4 equally sized parts.
28 + 16 + 32 + x = 76 + x
76 / 3 = 25.33
25.33 x 4 = 101.33
Therefore, Rhomubs ACIG is of area 101.33 cm2
101.33 = 76 +x
101.33 - 76 = x
x = 25.33
Therefore, the dark grey area is of area 25.33 cm2

alternatively, 101.33 / 4 = 25.33 by line 4 of this proof

conclusion: polygon ACIG is a Rhombus of area 101.33 cm2, and the dark grey section is of area 25.33 cm2

user means the line segments meeting in the central area of the square, each one intersects with the midpoint of a side of the square

Because there is not enough data available to formulate an equation to work it out.

The areas shown in the square is not uniform, no-one knows any lengths, only that each side of the areas are equal, because the irregular-apothems are all leaving the centre of each side.

You only know each corner is a right angle which is 90°

Thusly it is in insolveable, the only real thing you could do is measure the equal lengths and formulate the area that way and then perhaps use a ratio method to find the area's size, but again that won't work as the square and area values are not consistent with the size of the square in the picture.

What the fuck am I looking at OP?
Im retarded btw so idk.

Seems like this guy has the right idea even if he has a hard time communicating it. Makes logical sense doesn't it?

No, it's about 23.1 cm2

This guy makes a valid point, we don't know whether the corner angles are right.

nah man that aint right

you arbitrarily decided that the grey area was the average of the 3 others....

>76 / 3 = 25.33
>25.33 x 4 = 101.33

I'm not a math guy but this is wrong. It's making an assumption that the 3 measured areas take up exactly 75% of the total area.

You talk a big game Bill Nye, but I'm not buying it.

4 equally sized parts does not imply 3 equally sized parts. Why'd you divide 76/3 and then multiply by 4

>A = 28 + 16 + 32 + x
>A = 76 + x
>A = (3 * (76/3)) + x
>A = (4 * (76/3))

substandard bait

Yes but he does make the point that we don't know if the corner angles are right.

>You only know each corner is a right angle which is 90°
you dont even know that
it could be a rhombus slightly skewed from 90

Oh hmm
It seems like that is right

Problem is indeterminate autistic morons there are infinite solutions
Use heron's formula and prove my point.

>maybe
Definitely.

Sorry user, pre-uni trig is as far as my geometry knowledge goes I'm afraid.

say the x and y at bottem left are 0, and the distance between corner to line is Z.
then we have the lines straight and come together in (lets call it) point A, so every area is a quarter square.
Then, if you move A 5 pixels down and 4 to the right, you take 5 pixels from 2 squares in the top and give them to the bottem squares, and
take 4 pixels from the left squares and give them to the right. This means that when you add square 1 and 3 as a team and square 2 and 4 as a team their cm2 stays the same everytime you move point a with lines coming from the middle.

Sorry, "all symetrically sided NON-INTERSECTING Rombi..."
Substitute that statement in place of line 4 and we've got ourselves a working proof

good catch

where's you're proof?

see: line 4

>Why'd you divide 76/3 and then multiply by 4
You can set 76/x = 3/4 and see that you're on the right track. I did this whole proof because I couldn't rememer that formula.

>A = (3 * (76/3)) + x
Why are you taking a combinatorial? Where did you pull those 3's from?

It has to be 90° for the simple fact all sides to the point where the irregular-apothems leave are all equal in length.

Thusly for each corner has to be the exact same angle for the fact that each side is the same length exactly, i.e. a perfect square, thusly all corner angles must add to 360° and that means all 4 have to be equal, and it can only be 90°

>tell us to prove his point for him

This is a clever troll...but not clever enough

The problem in OP obviously has a solution
It's just the area of a shape within a bound shape
How could it not?

If all the corners are right angles, it makes sense looking at it and experimenting (say you kept the line on the x axis untouched but dragged the line on the y axis all the way to the right, the same answer would work, as well as at the halfway point, etc.) but I don't know why it works mathematically.

Even if we assumed the corners were all right angles and knew that the bottom right area was 20 cm^2, I have no idea how we'd approach figuring out each separate line's measurements.

Do you even know what a rhombus is, faggot?

oops I meant 1 and 4 and 2 and 3, I thought I did it right in paint. but what I mean is 2 and 3 is 48 so 1 and 4 must be 48 as well, making 4 20.

Nah man that still aint right

You can't assume 76/3 will get you 1/4 of the Rombi

REEEEEEEE problem has infinite solutions

Because you use just a high school formula

In the end you will have 4 equations with 5 uknowns

Second this

Now you listen to me Hannibal Buress, "n/x = (n2 proportional to z as n is to x)/z" is a tried and true formula for figuring out percentages from either just a denominator or numerator. You can use it to find the proportion or percantage of anything.

and now I'm signing out

Move the internal point and change square length and you will always have the same shape

I know what a rhombus is but again in this situation it isn't a rhombus.

Because, since you did not read what I said, the fact an irregular-apothem is leaving each side from the centre and each side of it is of equal length, that means this has to be a square and not a rhombus.

Even if it was a rhombus for some amazing reason, you still wouldn't be able to find an exact value of the missing area, as there is not enough data to formulate a fully working equation to work it out.

The only way as I have said already is by guessing or estimating.

The fact people are trying to see it is a Rhombus when the picture given is a square, is totally pedantic. It is grabbing for any means possible to try and find a solution that will never come.

WHY DOES THAT HOLD TRUE FAGGOT?

Is it true, watches at $ 0?

nah man you didn't use it right

You have no clue what you're talking about, and try to use math terms you don't understand to cover up your ignorance. Firstly, wtf is an "irregular-apothem"? You just made that up. Secondly, finding a solution for this puzzle is fairly trivial, and you don't even need to find out whether the figure is a square or a rhombus, it works either way. To claim that because you're too retarded to find the solution, there is no solution, only highlights the chasm between how smart you think you are and how smart you actually are.

>1+4 = 28 + (4)
>mfw
Why?

Shut up bot, we're trying to figure out a high school math problem.

24

The simplified non full answer to the question would be this..


Assume each side of the square is A,B,C,D.

Area = Length x height

Area = AB * BC or AD * BC (doesn't matter)

So for in this instance will use the unknown area as a variable of y

y = AB*BC > since as side all sides are of equal length, we can simplify to

y = AB2

So lets assume the shaded area to find is variable x and that would equal y

y = 28 + 16 + 32 + x
y - 29 -16 -32 = x
y - 77 = x > substitute AB2 for y

AB2 - 77 = x

Is the best answer you could give.

>the fact an irregular-apothem is leaving each side from the centre and each side of it is of equal length, that means this has to be a square and not a rhombus.
Learn to math, faggot

Basically this, the uknown area changes when you change the length of square so problem is indeterminate.
Thanks user you enlightened me

The fact you have no idea what an Apothem is shows you have no clue about what you are talking about.

In this case they are irregular as a regular Apothem moves from the centre of a side of a shape to the centre, perpindicular to where it left.

In this case that didn't happen but each Apothem left from the centre to another point, thusly they are irregular.

learn some maths kid.

>y = 28 + 16 + 32 + x
>y - 29 -16 -32 = x
>29

fine since I'm autistic and cant handle everybody saying different answers i'll make one final picture

Grabs my planimeter. Problem solved.

Hmm seems the board is changing the Squared ascii to a 2.

It should be AB^2

I genuinely can't tell if you're trolling or simply a retard who thinks he knows geometry.

The polygon marked ? has area A + C.
Each quarter of the big square has area A + B.
B is blatantly larger than C.

Therefore, the area of ? must be smaller than 1/4 of the area of the square itself, not equal to it.

The only reasonable solution to this problem is to flood-fill ? and the rest of the square with different colors, then write a script to count the number of pixels in each area.

Yeah I noticed that typo my bad, but you get the idea.

Would be

AB^2 - 76 = x

Seems the board doesn't allow the Square ascii symbol and changes it to a 2.

You have an entire side there. Squares have equal sides. You can can just work out the area of the entire square and then subtract the known area. Fuck me.

I know very well what an apothem is, what I called bullshit on is "irregular-apothem". An apothem by definition moves from the center of the figure, whereas the lines in OP's figure do not. Hence, they are not apothems at all; there is no such thing as an "irregular-apothem". And even if I accept your homemade bullshit terminology, those lines still do not prove that the figure has to be a rhombus, and the puzzle is still perfectly solvable with very basic math.

Irregular-apothems do indeed exist. Ever studied discreet maths?

Next you will tell me there is no such thing as imaginary numbers.

With A=28+32+16+? you get unfortunately two variables with only one formula.
Estimating with ?=(28+32+16)/3=25,333... does not help without further insturctions.

This seems like a really simple task but the following aproach is advanced mathematics.
Unfortunately i don't see any simple way for sure.

This seems solvable with a series of formulas and equations.
You have 4 Convex Quadrilateral areas with two same sides each and (hopefully) a right angle bedween these sides.

says the guy who is inventing the square is now a rhombus, quid pro quo.

No they don't, and yes I have. Show me any math paper that uses the term "irregular-apothem". Nice going to focus only on the terminological bullshit and ignore the fact that your claims are wrong, though. The figure could still be a rhombus, and the puzzle is still solvable.