How smart are you, Sup Forums?

How smart are you, Sup Forums?

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youtube.com/watch?v=xnE_sO7PbBs
duckware.com/tech/worldshardesteasygeometryproblem.html
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24

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

About 4

Twenty-Four

26.489

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

100 - 76 = 24

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

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.

I really hope you're not serious.

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

I'm quite smart, thank you.

Idk, I dropped out of high school

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.

do your own homework, faggot

What?

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

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.

High school is overrated

70 nigga

1 times 70 is 70 nigga

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

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

finish it im tired

hangover like fuck

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

smart enough to not give a fuck.

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

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

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

...

...

x

Length of each side is 4.90. The center point is at

youtube.com/watch?v=xnE_sO7PbBs

Faggot

32 + 16 = 28 + x

x = 20

duckware.com/tech/worldshardesteasygeometryproblem.html

Except you definitely do.

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

fuck me. is that legit ?

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

24 cm^2 confirmed

>winrar

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

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

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

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

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

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

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

not disagreeing but like said, is there a proof?

Yes it does.

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

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

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

wtf is that

what happened why is there a line in my reference link

Room construction fail

21

The guy deleted his post.

guy you responded to deleted his post newfag

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.

Refresh the page and his post will be gone

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.

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

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

smart enough not to waste time on it

Is this the instruction to the proof we asked for ?

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?

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

25,72668658067777

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

I just realized I'm a retard at 3 am

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.

>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.

The solution in is vastly simpler.

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

no

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

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

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=?

Even at a glance, they're hardly equal.

Yes they are.

way to overcomplicate shit needlessly, faggot

I think this is as close as you can get.

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.

Infinite values between:

sqrt(22) or 4.69041575982343

and

sqrt(30) or 5.477225575051661

that is.

Nice try, this image has been edited. There's no way

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

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.

Imagine if-

Just bear with me for a a second here.

Imagine if...

There were another equation?

Which one?

24

Tell me, you fuckhead!!

Dumbass

>>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

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.

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