How smart is /b/?

24cm^2

>>738337534
24 cm^2

>>738337534
12 cm^2

>>738337534
There's no question...

>>738337534
Problem is indeterminate

>>738337534
21

>>738338660
>How smart is /b/?
There is a question, now answer it.
I say, superst clever on earth.

>>738338776
Average intelligence, i'd wager.

>>738337534
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

File: picard.jpg (90KB, 1440x1080px) Image search: [Google]

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>>738338945
>all the areas are whole numbers so the sides have to be whole numbers

>>738338660
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

>>738337534
its 20 cm2 morons

>>738339025
Maybe I'm retarded but is there a case where a non whole number squared gives you a whole number?

>>738339364
root of 2 squared?

>>738339364
Yes, the square root of literally any number that is not a square number.

>>738339364
√3 * √3
i can keep going if you want

>>738339181
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.

>>738337534
>another "do my homework" thread
I'll help if you explain to me why you're posting on /b/ while in summer school

>>738339701
correct

>>738339675
you forgot to times it by root 3 the third time dumbdicks

>>738339774
Do you even know what the difference between square root cube root is?

>>738339856
haha i trolled u

>>738339701
How does that hold correct??

>>738337534
if 1.4 + 1.4 = 2.8
then we round it to 3
does that mean 1+1=3

>>738337534
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*

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>>738339893
kek

>>738339911
kek

>>738339960
Fuck off faggot. I'm more mathematic than you asshole

18ish cm^2?

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

>>738337534
(28+16)/2 = 22
(32-28)/2 = 4
4/2 = 2
22+2 = 24cm^2

>>738339960
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

>>738340086
how do you reckon 32/28= x/16?

>>738340149
i submit

>>738340134
retard

>>738340163
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

>>738339701
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.

>>738340065
every virgin loser claims he knows math

>>738340278
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.

>>738337534
It's Sunday you fucking piece of shit nobody cares

28 cm^2

>>738339701
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.

>>738340643
He's telling you that
32+16=28+x
so x=20

I just don't understand why.

>>738340683
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.

>>738340852
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.

File: geometry is fun.png (17KB, 750x750px) Image search: [Google]

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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

>>738340426
anon means the line segments meeting in the central area of the square, each one intersects with the midpoint of a side of the square

>>738340852

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.

>>738337534
What the fuck am I looking at OP?
Im retarded btw so idk.

>>738340983

Seems like this guy >>738339701 has the right idea even if he has a hard time communicating it. Makes logical sense doesn't it?

>>738341094
No, it's about 23.1 cm2

>>738341094
This guy makes a valid point, we don't know whether the corner angles are right.

>>738341094
nah man that aint right

>>738341094
you arbitrarily decided that the grey area was the average of the 3 others....

>>738341094
>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.

>>738341094
4 equally sized parts does not imply 3 equally sized parts. Why'd you divide 76/3 and then multiply by 4

>>738341094
>A = 28 + 16 + 32 + x
>A = 76 + x
>A = (3 * (76/3)) + x
>A = (4 * (76/3))

substandard bait

>>738341378
Yes but he does make the point that we don't know if the corner angles are right.

>>738341179
>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

>>738340790
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.

>>738339364
>maybe
Definitely.

>>738341574
Sorry anon, pre-uni trig is as far as my geometry knowledge goes I'm afraid.

File: explanation try.png (13KB, 907x809px) Image search: [Google]

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>>738340643
>>738340790
>>738340426
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.

>>738341335
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

>>738341241
where's you're proof?

>>738341355
>>738341378
>>738341440
see: line 4

>>738341440
>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.

>>738341446
>A = (3 * (76/3)) + x
Why are you taking a combinatorial? Where did you pull those 3's from?

>>738341472

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°

>>738341574
>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?

>>738340790
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.

>>738341795
Do you even know what a rhombus is, faggot?

>>738341741
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.

>>738341744
Nah man that still aint right

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

File: 1490816711150.png (187KB, 758x631px) Image search: [Google]

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>>738341821

REEEEEEEE problem has infinite solutions

Because you use just a high school formula

In the end you will have 4 equations with 5 uknowns

>>738341941
Second this

>>738341941
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.

File: explanation.png (29KB, 1743x809px) Image search: [Google]

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>>738339701
>>738339181
>>738340426
>>738340790
>>738340643
>>738341741
and now I'm signing out

>>738342158
Move the internal point and change square length and you will always have the same shape

>>738341870

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.

>>738342267
WHY DOES THAT HOLD TRUE FAGGOT?

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>>738337534
Is it true, watches at $ 0?

>>738342199
nah man you didn't use it right

>>738342326
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.

File: hold the fuck up.jpg (21KB, 600x576px) Image search: [Google]

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>>738342267
>1+4 = 28 + (4)
>mfw
Why?

>>738342369
Shut up bot, we're trying to figure out a high school math problem.

>>738337534
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.

File: wrong.jpg (51KB, 482x377px) Image search: [Google]

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>>738342326
>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

File: 1491779354045.jpg (54KB, 500x350px) Image search: [Google]

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>>738342286
Basically this, the uknown area changes when you change the length of square so problem is indeterminate.
Thanks anon you enlightened me

>>738342682

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.

>>738342778
>y = 28 + 16 + 32 + x
>y - 29 -16 -32 = x
>29

File: ok.png (34KB, 1743x809px) Image search: [Google]

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fine since I'm autistic and cant handle everybody saying different answers i'll make one final picture
>>738342267
>>738342267
>>738342267
>>738342267

>>738337534
Grabs my planimeter. Problem solved.

>>738342778

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

It should be AB^2

File: 1486704008966.png (27KB, 750x750px) Image search: [Google]

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>>738341744
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.

>>738342937

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.

>>738338945
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.

>>738342931
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.

>>738343275

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

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

>>738337534
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.

>>738343275

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

>>738343347
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.

>>738343225
Doesn't even realize he's retarded

>>738343461

Then solve it. Since you think you know everything.

Go ahead, show us your undeniable proof.

>but but but I can't
>but I know it is solvable

Whatever kid.

>>738343672
Given up your claim that it can't be a rhombus, then? Good, I guess that's progress of a kind.

>>738343540
The lines in the middle are a ruse. Look at the angles.

>>738343154
>flood-fill ? and the rest of the square with different colors, then write a script to count the number of pixels in each area.
Bruh you can't count up the units a shape is made of and call it a day. what if someone handed you this problem on a piece of paper? You gonna count up the atoms? How the fuck you think mathemeticians solved shit before computers? They didn't guess like you just did, they used logic.

>B is blatantly larger than C.
Oh of course, and dark grey area blatantly has an area of x.
I can say that because I eyeballed it and its just blantant.

I can't even begin to figure out your picture, did you try and make it 3D? because that's the only way I can rationalize you trying to add area A to segment C or B. UNITS, FUCKER. Pay attention to them.

Also,
>Therefore, the area of ? must be smaller than 1/4 of the area of the square itself, not equal to it.
is my second given statement. So you should agree with my findings.

>>738343166
Sorry, I was just being a dick. But to be fair you were off by one number. Thats how we lose million-dollar Space Shuttles. You have the second-best proof in the whole thread so far.

File: quadrilaterals-300x230.jpg (18KB, 300x230px) Image search: [Google]

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For all fags that dont know the difference between square and rhombus

>>738343813

It isn't a rhombus that is only assumption on your part, I am going off data present, and again just because you have no clue that non-uniform/irregular apothems exist does not mean you are right.

You claim if it was somehow, by your assumption, a rhombus that it is solvable.

Then explain how it can be solved.

Or we going to get another childish response from you to try and deflect your claim away that nobody will notice?

Little boy, you can do it, show us all.

Also answer this for me

x = SQRT(-100) * SQRT(25)

>>738343886
Still doesn't realize he's retarded

>>738343429
Convex/concave is for curves. But I do see what you're trying to say, and there's only 1 shape with an obtuse angle.

>>738343451
Prove that the angles are all right.

I'll wait.

>>738344211

prove that it is a rhombus

>>738344052
I didn't assume that it's a rhombus, I said that it *could* be a rhombus, in contrast to your claim that it has to be a square. You have yet to prove that claim of yours.

You have yet to link to any math paper that uses the term "irregular-apothem", because that's a bullshit term that you made up.

I also never claimed that the figure has to be assumed to be a rhombus in order to be solvable, on the contrary I said the direct opposite in >>738342682: That you don't have to assume either way, the solution works whether the figure is a square or a rhombus. Clearly you need to learn how to read as well as how to math.

x = -10i * 5 = 50i

>>738344365

Wouldn't waste your breath on this guy, he has no clue, all assumptions, he claims it is solvable if it is a rhombus I have asked him to prove that also and give us the answer, but he is deflecting like all kids do.

Which would mean he may fail my equation also.

>>738343050
What's the area of 2? 0? Neat trick, but now you're working with 2 quarters and a half. This may yet prove useful.

>>738342267
>>738343050
This guy is definitely right, just checked on solidworks and it checks out

>>738344460
I'm not >>738344211, dumbass. And once again, I never said "it is solvable if it is a rhombus". I said it's solvable in any case. Learn to read.

>>738344457
Nigga, stay out of the imaginary plane for this one. You tryna summon /sci/borgs?

File: area3_solution.png (16KB, 750x750px) Image search: [Google]

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>>738344599
Here you go faggot, here is your solution:

a + d = 28, a + b = 16, c + d = 32

b + c = (a + b) + (c + d) - (a + d) = 16 + 32 - 28 = 20

>>738343898
1. The problem is not properly defined. You need more data to solve it. Without that data, the best you can do is an estimate based on counting pixels.
2. Import it into your favourite image editor and measure the triangles' bases and heights.
3. B and C are areas, not segments, you fuckwad.

Basically what I'm saying is congratulations on making me reply, but you can leave us and go sodomize yourself with a baton now.

>>738344602
>The figure could still be a rhombus, and the puzzle is still solvable.

That is your own words.

Then solve it. Hurry up.

You claim it is solvable (even in any case) then go ahead and solve it. Tell everyone how.

Oh wait, you can't.

>but but but I was only trying to troll someone else
>i still don't understand irregular-apothems that only exist when more than 1 apothem sure a single point that is irregular from the centre of a polygon/shape.

File: correct.png (68KB, 931x720px) Image search: [Google]

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>>738344508
>>738342267
>>738342267
>>738342267
This is the answer folks

>>738344794
I already did faggot (>>738344712), once again you need to learn how to read.

>>738344734
This is a sketch, it doesn't need to resemble the reality perfectly.
Besides, we assume the problem is to be solved mathematically, and not by measuring with a ruler or counting pixels.

>>738344794
Also, still waiting for you to post a link to a single math paper that uses your "irregular-apothem" bullshit term.

File: correct 2.png (66KB, 933x700px) Image search: [Google]

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>>738344866

>>738344712
in mathmatics, the image is never 100% accurate unless specified otherwise.
faggot

File: correct 3.png (63KB, 936x688px) Image search: [Google]

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>>738345004

>>738344712

If that area is 20 then all areas add up to 96

SQRT of 96 = 9.80 rounded down to 2 figures.

So you are saying all sides are 9.8 units?

If you applied that to each shape inside the square, then the maths is off for those areas.

So that is wrong.

File: correct 4.png (67KB, 918x706px) Image search: [Google]

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>>738345017

File: 1499608651372.png (45KB, 750x750px) Image search: [Google]

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Bit it for smaller pieces for you guys. This should be much easier from this point. To prove that it really is 20.

>>738344712
Your right a is at least 1.5x longer than your left a my dude.

>>738345018
Assuming that the figure is a square, the sides have lengths of ~9.8, yes. Show how that causes the math to be off for each shape inside the square. Also, show where my solution is wrong.

File: correct 5.png (51KB, 693x653px) Image search: [Google]

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>>738345018
>>738345046
each side of the square is sqrt of 96

>>738345111
The letters don't mark the lines, they mark the areas.

>>738344664
Oh good, add another dimension to the problem, that will make it easier.

>>738344712
>28's a = 16's a
>16's b = ?'s b
>?'s c = 32's c
>32's d = 28's d
ohshitniggerwtfareyoudoing.jpg

>>738344734
What data do you need? I'll find it, and show proof.

>>738345018
I'm confused too, but CAD don't lie.

>>738345156
Then only write one of them in each area, nigger.
I'm also looking forward to an actual proof of how 28 cm^2 + some non-zero area = 28 cm^2, but I guess I shouldn't hold my breath for too long.

>>738344887

And you are wrong, you are using assumptions again.

WRONG WRONG WRONG
ASSUMPTIONS ASSUMPTIONS ASSUMPTIONS

I said it is not solvable due to lack of data available to formulate any type of answer, I gave the best possible simplified answer already (even if it had a type)

Being AB^2 - 76 = x

Without a value of one of the sides or more angles in one of the irregular polygons, it is impossible to solve.

You just threw in assumptions into your little equation.

You are assuming that variables are equal in each smaller irregular polygons, which is wrong, and again assumptive.

Go sit in the corner kid. You solved nothing.

In fact you cannot prove it is a rhombus either, like the other guy asked you.

>>738345320
I *did* write only one of them in each area, are you fucking blind? I guess you're just too retarded to read the figure.

>>738337534
About 22 cm^2

>>738337534
I'm old. I solved it almost immediately. The image looked symmetrical to me so I added 32+16 then subtracted 28. equals 20 square centimeters.
Tits or go home.

>>738345349
Are you blind mate,

look at this
>>738342267
>>738343050

and then this>>738344866
>>738345004
>>738345017
>>738345046

Answer is 20

>>738345349
I *assumed* that medians divide the area of triangle s in half? Yeah, I guess I did assume a fucking obvious thing that everyone learns in fucking elementary school. If you don't even know that, then fuck off. And once again, I never claimed that the figure is a rhombus, merely that it *could* be. The onus is on those who disagree to prove that it must be a square.

>>738337534

Dimensions don't work:
>Height 60
>Width 44
>All equal segments

Troll question is troll

>>738345557

look at this faggot
>>738342267
>>738343050

proof answer is correct
>>738344866

>>738344712
Not OP but I believe this solution is correct

>>738345111
The letters are the areas of each bisected triangle.

>>738345238
It's correct, work it out. The area of a triangle is 1/2 * base * height.

>>738340435
I'm a virgin loser and I don't claim to know math

>>738344712
OP here, this is the correct solution. Props for disproving all the retards

>>738346028
>>738345462
Like I said, symmetrical.

>>738340683
>>738341179
>>738341795
>>738342326
>>738342931
>>738343347
>>738343672
>>738344052
>>738344460
>>738344599
>>738344794
>>738345018
>>738345349
What a shocker, the "irregular-apothem" faggot has disappeared from the thread after being proven to be an absolute retard.

>>738344712
That's it, thanks.
That's the missing rationale for the guy who much earlier claimed (correctly but without proof) that you can move the midpoint around and the sum of opposite areas remains unaffected.

Thanks for being the only non-retard in this thread.

File: MTNmjpe.jpg (2MB, 4505x6000px) Image search: [Google]

2MB, 4505x6000px

>>738344712
ma nigga. tits 4 u.

>>738341741
>>738344712
Math Ph.D here. I'm glad to see that there are people who can catch the essence of problems like these. Two very different approaches, both right.
The first one works amazingly well when you have to explain things like this to kids, the second one is straightforward and only needs people to know what the area of a triangle is.
Math is beautiful when has problems with simple, elegant and creative solutions.
I am very pleased.
Most of /b/ though will refuse to listen, because to them their beliefs are stronger than a proof.

>>738345669
You're not multiplying anything by 1/2.

>>738344712
you have four variables when you should have eight

>>738347625
The triangles with the same letters have the same area because they have equal bases and equal heights

>>738347706
Ah, thank you. A most elegant solution. Kudos to triangle guy.

>>738338694
Your gender is indeterminate

>>738337534
24 cm2

(2x)^2=x^2+16+28+32

3x^2=76

x^2=76/3

>>738348505
wrong, on the right side of the equation the missing part does not have area x^2.
The solution with proof is >>738344712

File: polska.jpg (168KB, 795x1200px) Image search: [Google]

168KB, 795x1200px

>>738348346
nice

high five

File: idiot.jpg (62KB, 600x386px) Image search: [Google]

62KB, 600x386px

>>738345018
>>738345111
>>738345238
>>738345320
>>738345349
kek, dumbasses can't even understand the simple solution

>>738344712
Although I'm not a robot I do have several degrees in math. Correct. Very similar what I did. The heights of the opposite triangles are double the side of the square.