How smart are you, Sup Forums?

How smart are you, Sup Forums?

Over 9000

I watch Rick and Morty so I must me pretty smart to understand the jokes

Dubs is anwser

smart enough to know this question is meaningless to my existence, if someone knows or needs to know this.... you might likely be poor

36 cm squared.

Neer pedazo de gay.

Assuming whole numbers are what we're looking for they all add to 64 which is 8 squared, it must be a square greater than this. The number needed to get 9 squared is less than 28 but if you use 36 you get 10 squared. Therefore lazy boy no real math answer is 36

proof?

Im not that poster but i did post basically if the overall square has sides of 10cm then the remaining portion is 36cm^2

Cant be solved fag. Not enough information

What if the overall square had size 10.2cm? Why does it have to be 10cm?

Well im asuming we're working with whole numbers this is the smallest possible solution

There's no reason to assume whole numbers, you're basically just guessing here

Theres every reason to assume whole numbers the numbers given are all whole numbers that's standard practice in a maths question. It's also a rule i added to actually get to an answer without it there isnt an answer because there isnt enough information

its definitely more than 1cm3

> without it there isnt an answer because there isnt enough information

Proof?

Its also not a guess, we know it is a regular square. If the gap is 36cm^2 then the sides of the square are 10cm which is a regular shape and so 36cm^2 is fulfills the requirement

literally any number greater than 8cm would give an answer. why choose 10cm?

32 cm^2.

Each of the subsquares has a volume of x^2. Set up a system of three equations with three unknowns and solve for x.

4x^2 is equal to the total area, which comes out to 96cm^2, therefore the unknown area us 32 cm^2

But it may not be the only answer, who's to say it's not a 10.2cm square making the blue area 50.04 cm2

This was my work for it btw.

Finally a legit attempt at the actual question. I'm checking your work now...

Because if the shape was 9cm the gap would be under 28cm when in the diagram the gap is larger than 28cm^2 however 10cm gives an answer of 36cm^2 which is larger than 28cm^2. This means 36cm^2 is the smallest whole number to fulfill the requirement.

You just proved why I said not using the rules 1) whole number and 2) smallest answer gives an infinite amount of answers

Thanks man. My picture is kind of bad, but I used the symmetry of the square to show how all of the triangles were formed.

You still have not proved that there is an infinite number of answers if whole numbers is not a requirement

u dumb fuck, that only aplyes if the subsquares are squares, wich they dont.

looks like its about 32

Trying to understand your proof...
So the three unknowns are:
X: length of the side
Z: offset 1
Y: offset 2?

Am I understanding this correctly?

Let me draw my picture better, my lines are kind of poorly done

100 - 64 = 36cm sq

here's my proof

It's a ratio thing, if you tweak the side length there is always going to be an area that satisfies it above 10cm

why 100, why not 99 or 90 or any other number

Ok but prove this mathematically please

Because 25 is the only squared number between 20 and 28

assume each side is 10 cm because it looks like that so the total area is 100cm squared - 16 - 28 - 20 (assumed five for all the equal sides because that gives you a good bit under 25 for top left a little bit under 25 for top right and more than 25 for bottom left)

There's no reason the length of the side should be a whole number.

Ok now assume each side is 10.2cm.

Your proof looks good to me so far

If we take each side to be x

X^2=28+20+16+y

y>28

X^2=64+y
X^2-64=y
X^2-64>28
X^2-92>0

If x fulfills this requirement the shape is possible therefore 10 is the first whole number that can be used and all whole numbers after can be used, its infinite.

You're not seeing the whole picture.

There are additional constraints that makes the solution unique even if you don't assume whole numbers!

Rather if x^2>92 the shape is possible

Correct

Here's a better picture

according to Blender...

Who uses sq cm?

>autists

My picture :Uses the same methods and variables as this user's much clearer graphic:

???

Actually, it's possible as long as x^2 > 64. If we say x^2 > 92, you're assuming that the unknown area is greater than 28. Without knowing the angles, we don't know that's true. If we're just going by what we see, we can just use to get the answer (which shows that the unknown area is not greater than 28).

But yeah, no possible solution with this many unknowns.

You're assuming the diagram is to scale, which is the wrong way to approach a geometry problem.

I understand! You got it I'm pretty sure

We have 3 unknowns so to get a unique solution we need 3 eqs:

E1) Consider the top half of the square:

The area is 2x * x.
Also the area is 16+20+ the missing triangle
The missing triangle is 2x * Z/2
Thus the area is 16 + 20 + Z*X
Thus

E1: 2x^2 = 16 + 20 + Z*X

Now do the same thing for the left hand half of the diagram to get E2:

E2: 2X^2 = 16 + 28 + Y*X

Now we need one more equation, E3

Pls help whats E3

E3: The entire square's area is (2X)^2 or 4X^2

That is equal to 16 + 20 + 28 + Blue

the Blue area is: ... um help

Blue is the square bit X*X plus two triangles

Tri 1 is 1/2 * X * Z
Tri2 is 1/2 * X * Y

>the Blue area is: ... um help
undefined

Cant believe you are all wasting this much time on an impossible problem

You fail the autism test.

...or pass i guess

So:

E3: 4X^2 = 16 + 20 + 28 + X^2 + ZX/2 + YX/2

You are wrong, I'm defining the blue area in terms of X, Y and Z which is totally possible, see

Apparently I don't have the diagram saved that actually shows the trick to solving this, but I'll give some hints.

You only need one equation and a formula to solve this, but you won't use the formula for any actual calculations.

The area of a triangle is equal to base times height, divided by two (that's the formula). Therefore any two triangles that have the same base length and the same height will have the same area, regardless of if the angles are the same.

The equation you need basically looks like
C + D = (A + D) + (B + C) - (A + B)

Now we have 3 equations and 3 variables so we just substitute and solve

E1) 2x^2 = 16 + 20 + Z*X
E2) 2X^2 = 16 + 28 + Y*X
E3) 4X^2 = 16 + 20 + 28 + X^2 + ZX/2 + YX/2

Im so smart I get other people on Sup Forums to do the work for me.

lol @ all the morons claiming this problem is unsolvable: "I'm too retarded to solve this, that must mean that it cannot be solved!"

I don't have a proof, but it seems to me that each set of diagonally opposite segments will always have the same combined area. 20+28 is 48, and 48-16 is 32, so the answer is 32cm^2

I used equations:

1) The sum of the top two areas:
2x^2 -(1/2)z(2x) = 16 + 20 = 36

2) The sum if the left two areas:

2x^2 -(1/2)y(2x) = 16 + 28 = 44

3) The top left area

(x -z)(x-y) +(1/2)(y)(x-z) +(1/2)z(x-z) = 16

E1: 2X^2 = 36 + ZX
E2: 2X^2 = 44 + YX = 36 + ZX
8 + YX = ZX
E3: 3X^2 = 64 + ZX/2 + YX/2
6X^2 = 128 + ZX + YX
Substitute E2
6X^2 = 128 + 8 + YX + YX
3X^2 = 68 + YX

... tbc

I'm not. I barely passed level 2 math lmao.

/thread

Smart, I'm using a different third area. Yours is easier.

I basically did this too and got same answer. Seems legit.

Seems we're in agreement.

This is basically what you'll eventually find with

We know that 2X^2 = 44 + YX
so YX = 2X^2 - 44

Previously: 3X^2 = 68 + YX
so 3X^2 = 68 + 2X^2 - 44

Solve for X:

3X^2 = 24 + 2X^2
X^2 = 24

Fuck I made a mistake somewhere

Nah thats right.

X^2 is 24, so 4X^2 = 96 = area of the whole square

Z + 16 + 20 + 28 = 96, so z=32, z being the area of the unknown portion

YOU'RE ALL RETARDED

It's 116 because every square has exactly 180cm in it

16+20+28 use PEMDAS to do it in right order, to get 64

180+(-64) = 116

fucking go back to school assholes

36

Ah yes ok cool so the blue area is 32 cm^2

there you go, I am stupid

I like this observation, but how would you prove the assumption that the two opposite corners sum to the same area?

All you guys are making this way harder than it is.
since the sections each are seperated in the middle the opposing sector pair is equal to the other pair.
20 + 28 = 16 + x
x = 32

stuff like this is nice and all, but this took me 10s to figure out.

The two lines mean the sides are of equal length

aparently Im to drunk to upload the actual ohoto

>since the sections each are seperated in the middle the opposing sector pair is equal to the other pair.
Yup you just need to prove this assumption with a bit more detail or a diagram

yup, basic planar geometry probelms

Smart enough to not give a fuck OP.

That's not what I meant. You're assuming you can just measure the areas of the blue region, but you don't know that the diagram was drawn with any accuracy.

Well, we did your homework OP now fuck off

The answer is undefined you fucking autists. Both of these are answers which fit the given numbers.

I couldn't find the diagram I wanted, so I made my own. From here you basically just need addition and subtraction.

Ok, I see what you're saying now

The interior lines and angles

Lol I'm gonna prove you wrong in about 2 mins and then we'll see

its not about measure, you must dig up into geometry to understand some basic concepts about diagrams

So wat is the answer? All i see is walls of equations. Why cant someone just say its 22 sqcm?

Your left hand solution is wrong

Calculate the area of the top-left bit:

(5 - 1.2) * (5 - 2.8) + (5 - 2.8) *(1.2 / 2) + (5 - 1.2) * (2.8 / 2)

Equals.... 15!!! should be 16 brah

it's 32 cm2

lol I'm so used to breaking down shit into right triangles for my job that I forget shit like this.

Oh thank god.

Nice repeating digits btw.

Lol you are wrong buddy whos the autist now?

Because Sup Forums is full of autists. That's what they do.

Also fucking checked

Nice diagram. From here:
16 + 20 + 28 = 2A + 2B + C + D
64 = 2A + 2B + blue
Blue = 64 - 2A - 2B

Ok now what? We're missing something.

Substitute
B=16-A

You're supposed to solve for C+D

You know that:
A+B=16
B+C=20
A+D=28

You can rearrange your last equation to be
Blue = 64 - 2(A+B)

You know that A+B is 16, so you can plug that in
Blue = 64 - 2(16)
Blue = 64 - 32
Blue = 32

Print any number on there and it will be correct