How smart are you, /b/?

24

>>737452427
24cm^2 random guess, didn't do any math

About 4

>>737452427
Twenty-Four

26.489

>>737452427
Total area we know is 76 cm^2. The nearest perfect square that would logically fit would be 100 cm^2.

100 - 76 = 24

>>737452867
Seems totally legit to me.
>It's impossible by the way.

>>737453031
Well, syllogism can work it's way out. Has to be greater than 16 and less than 28. So... 24 seems like it could work.

>>737453135
I really hope you're not serious.

>>737453135
>Has to be greater than 16 and less than 28
Why? It may not be drawn to scale

>>737452427
I'm quite smart, thank you.

Idk, I dropped out of high school

>>737453235
I get that. But then everything else about the problem would be incorrect if you were to not treat it as if it were drawn to scale.

>>737452427
do your own homework, faggot

>>737453563
What?

>>737453563
So you always solve problems by how they look, rather than by using the facts?

>>737453563
You have not enough information to solve this shit. And answering "24" because it could be a valid answer doesn't make it correct.

Not enough data. Next.

>>737453314
High school is overrated

>>737452427
70 nigga

>>737454155
1 times 70 is 70 nigga

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>>737453031
>>737454036
>>737454057
>I'm too retarded to solve it, therefore it's impossible

>>737454346
A math master should be enough to solve that I guess.

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finish it im tired

hangover like fuck

>>737455054
Wait, do you think a=a, b=b, c=c and d=d?

>>737452427
smart enough to not give a fuck.

>>737455301
then a+b=28 etc keep doing till values are found

The answer is 20, because Excel Solver told me the right answer.

>>737455628
Get on my level, fagits

Side size could also be 9.875 for example, and therefore the unkown part area would be 21.515625

9.875^2 - 76 = 21.515625

So 24 might not be the correct answer. Indeed there is no way of knowing what the size of the unkown part is, because you have to compare the size of each known areas real (image) sides, compare it with the area given (28, 16 or 32) and calculate the other area the dame way

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>>737455756

>>737452427
x

>>737455628
Length of each side is 4.90. The center point is at <3.67,2.86>

>>737455756
https://www.youtube.com/watch?v=xnE_sO7PbBs

>>737455902
Faggot

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>>737452427
32 + 16 = 28 + x

x = 20

>>737455779
https://www.duckware.com/tech/worldshardesteasygeometryproblem.html

>>737454036
Except you definitely do.

>>737455891
nevermind. The answer is still 20, but...
side length is 9.80. Center is at <7.35,5.72>
>now get on my level

>>737455054

fuck me. is that legit ?

16/2 = 8
32/2= 16
8+16=24

24 cm^2 confirmed

>>737456062
>winrar

>>737456062
>>737456695
explain this to me please does this assume opposing regions are equal?

>>737456876
Basically, by moving the center point, what you take away from one block you add to the opposite one.

>>737457058
Move the point in one direction at a time to make it easier to visualize.

>>737456876
Opposing regions must combine to half the volume because all lines converge in the middle and bisect the sides.

>>737457107
Lol how do you move it in multiple directions at the same time ? impossible.

>>737457178
but we dont know that this is true. do you have a mathematical proof of this ?

>>737455437
but its not true to start. a does not = a

>>737457058
>>737457107
>>737457178
not disagreeing but like >>737457362
said, is there a proof?

>>737457379
Yes it does.

>>737457178
they do not converge in the middle. If they did, all sections would be equal size.

>>737457514
No it doesnt. The third side is different so the area cannot be the same.

>>737457647
Yes it can (and does), because the angles are different.

>>737457519
wtf is that

>>737457838
what happened why is there a line in my reference link

>>737452427
Room construction fail

>>737452427
21

>>737457922
The guy deleted his post.

>>737457922
guy you responded to deleted his post newfag

>>737458049
>>737458096
so if i delete my own post you cant see it anymore or what ?

but i still see his post.

I'm not going to do it, but assign (x,y) to the center point, write equations for the area of each quadrant knowing that the midpoints are at half the edge length, constrain the sum of all areas to the edge length squared, and solve the system of equations. Pretty trivial, if a bit tedious.

>>737458283
Refresh the page and his post will be gone

>>737457379
a = a because of the formula of the area of a triangle. It's middle school level... The area of a triangle is base*height/2, and both triangles have the same height and their base have the same length.

>>737457527
I meant a single point, so their angles combine to 360.

>>737458315
Unless you can use value mapping, wouldn't that be impossible since you don't know the edge length?

>>737452427
smart enough not to waste time on it

>>737458315
Is this the instruction to the proof we asked for ?

>>737458492
It's just an unknown. You have three unknowns, x and y of the center point and the edge length. Express the area of each quadrant as a rectangle plus/minus two triangles (dimensions the difference of x,y and coordinate midpoints).

Do I really need to do it?

>>737452427
around 25cm, too lazy to plug in all of the decimal numbers

25,72668658067777

>>737458765
That I can understand, but is it solvable? That seems like a largish system of equations that may or may not be solvable.

>>737459292
I just realized I'm a retard at 3 am

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>>737458315
Elaborating a little,

Area of 28 cm^2 quadrant expressed as the rectangle fully containing the polygon less the two right triangles between the nonsquare edges and the rectangle
28 = A1 = x(l/2) - x(y - l/2) - (l - y)(l-x)

Sum of areas must be area of square
A1 + A2 + A3 + A4 = l^2

This is five equations and three unknowns. You have some shitty polynomials in those equations, but its sufficient.

>>737459473
>28 = A1 = x(l/2) - x(y - l/2)/2 - (l - y)(l-x)/2
Forgot the factors of 1/2 on the triangle areas.

>>737459473
The solution in >>737456062 is vastly simpler.

>>737460407
That may be valid but proving it pretty much requires what I outlined.

>>737460583
no

>>737460583
Nah, it's sufficient to use the figure in >>737455054:
a + d = 32, b + c = 16, a + b = 28
c + d = (a + d) + (b + c) - (a + b) = 32 + 16 - 28 = 20

>>737460800
Proving that diagonal quadrants are still half regardless of where the midpoint is? Yes.

>>737455301
Yes, a=a, b=b, and so on. If you rotate both "a" triangles 90 degrees counter-clockwise then they have the same length base and same height. Since the area of a triangle is 1/2 x base x height then they have the same areas. Same goes for the other three sets.

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

Solve for c and d respectively
d=32-a
c=16-b
So 32-a+16-b=?

Rewrite
48-a-b=? to 48-(a+b)=?

Replace a+b
48-(28)=?

And we get 20=?

>>737455054
Even at a glance, they're hardly equal.

>>737462287
Yes they are.

>>737459473
way to overcomplicate shit needlessly, faggot

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>>737452427
I think this is as close as you can get.

>>737463638
Also the only whole number in that range when you graph it would be 5.

So if it has to be a whole number then then its 100cm2 cube so the answer is 24, but there's infinite values if you can use decimal numbers.

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>>737464049
Infinite values between:

sqrt(22) or 4.69041575982343

and

sqrt(30) or 5.477225575051661

that is.

>>737455779
Nice try, this image has been edited. There's no way <EAB is a 70 degree angle.

>>737464863
Read the thread, the correct value (of which there is only one) has already been found.

>>737452427

This equation is simply not solvable due the lack of information given.
It is an equation with two unknown(complete area and the lenght of one side), and that makes it unsolvable.

>>737466009
Imagine if-

Just bear with me for a a second here.

Imagine if...

There were another equation?

>>737466376

Which one?

>>737452427
24

>>737466376
Tell me, you fuckhead!!

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>>737466009
Dumbass

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>>737455054
>>Half of 28 cm^2 = 14 cm^2
>>Half of 16 cm^2 = 8 cm^2
>>b variable used to represent area on both polynomials
>>Mfw you think 14 cm^2 = 8cm^2
>>mfw you think the area is equivalent because they are drawn to an intersect

>>737468202
Not sure if bait or completely retarded. It's not the four given quadrangles that are divided into equal parts, it's the four triangles created by the new red lines.

>>737452427
24cm ^2
both segments of each side of the square are equal

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>>737452427
I'm not going to do the math but find the square area of the whole square, minus the knowns. The remaining will be?

>>737469725
yeah, andthe square area is not given, so how do the math?

>>737467672
so give the explanation, you retard

>>737470319
Read the thread, you lazy fuck

>>737470267
i actually feel like
>>737455054
basically solved it

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>>737466009
this
all these niggers who say 24 are ass pulling or not solving it purely mathematically

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>>737457498
>>737457362
create a new square by connecting the bisections.
we now have 4 new triangles, with a combined area of half the total area of the original square.
consider the two opposing triangles (those that don't share an edge). their combined area remains constant, no matter where we mobe their shared vertex (so long as it stays within the new square):
y + z = x
area = .5 * x * y + .5 * x *z
area = .5 * x * (y + z)
area = .5 * x * x
area = (x^2)/2
independent of y or z, and x won't change no matter where you move the vertex

tl;dr answer is 20

>>737452427
smart enough not to care

>>737471111
seems legit.

>>737471111
nice quads btw

>>737471813
ha, i didn't even notice, thanks. i must be correct, then, right?

>>737471111
The quads confirm he is a smart

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>>737452427
After some here and there i figured out that while pretty much everything is unknown the comparision of pixels per surface gives a good value.
Do i get good boy points for that?

>>737471111
Quads of truth. This solution works even if the central point is outside the tilted square, you just have to put a minus sign in front of the relevant values.

>>737473145
>minus sign
hurray for a more general solution

>>737462287
.5*Base*height

>>737473556
there are no two trianles with matching bash and height

>>737474048
s/bash/base/

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This

>>737475187
that has no meaning, you can't take the square of an area of any rhombus and expect that number to be equal to the length of one of its sides

>>737475187
app's name?

>>737457784
Are you saying a triangle with the disease 3,3,1 has the same area as a triangle with sides 3,3,3? Because that's wrong.

>>737458392
They don't have the same height. They have two sides that ate equal in length. Those are not the same thing.

>>737461117
That's the correct response.

>>737457379
http://math.tutorvista.com/geometry/medians-of-a-triangle.html

>>737476715
>disease
Wut? All I'm saying is that two triangles with the same base and height have the same area, even if one (or two, for that matter) of the sides have different lengths.

>>737452427
answer is 18.2857143

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>>737475187