How smart is Sup Forums?

How smart is Sup Forums?

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cam-x.com
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en.wikipedia.org/wiki/File:Rhombus.svg
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Not very.

I dunno, 24cm squared ? Its a random guess

24

18

bumping for interest because I don't know how to solve this and am genuinely curious

you divide it into triangles. since the unknown square ends up sharing sides, you can get a sum of 2 triangles for each of the 8 pieces.

Should be 22cm^2

I figured that much but I still don't know how to solve it. Still interested

very, but lazy too

Fucking everyone in here is a retard.
The answer is obviously
>The Rest

28+32+16+x=4x
28+32+16+x-x=4x-x
28+32+16=3x
66=3x
22=x
22cm^2
That's how you do it

No it's not dumbass

Aren't you assuming the total area is the product of the area in question multiplied by four?

>28+32+16+x=4x
what is the rationale for 4x ?

4 parts of the whole square.
28+32+16+22=22*4 whew

I'm not a math wizard, but here I go

24cm^2

reason: each side is 10cm making the whole area 100cm^2

28+16+32 = 76
100 - 76 = 24

the equal segments cant be 4cm, or else the smallest area 16cm^2 would be smaller because 4x4 = 16

the equal lengths cant be 6cm or else the largest portion, 32cm, would be bigger because 6x6 =36cm^2

so it must be 5cm

Math is hard

28+32+16+22=98
4x22=88

the correct answer is potato

always

that's a really shitty method. There's no reason to suspect that only whole numbers are used and you can't just simply guess that

i see your point but how can u tell that the side is a whole number, i mean using ur reasoning the side may as well be 10.5cm

28+32+16=76
76/4=19
19+4=23cm2

Nice try, satan.

no, I agree. there's absolutely no proofs to back up my guess, but that's the best I've got.

in seriousness, i think the formula to find the missing # is

28^2 + 32^2 + 16^2 + x^2 = y^2

beyond that, i cant be arsed

Why do girls do this with their younger sisters? cam-x.com

I think we're done here.

Bumping for interest

Do you think it's actually solvable?

I don't feel like writing it all out, but you can solve by drawing a series of triangles between the midpoints because it's given that the midpoints are midpoints because they are congruent lengths on either side.

That gives you 8 triangles to solve. They should be solvable with some algebra and maybe some trig ratios and the fact that the cross in the middle must sum to 2pi

>that's a really shitty method
Yeah, in the abstract. In the world of math problems like this they are almost always designed to have an integer answer because the trick is in finding where to start.

Like this basically

It's 24. 5x5x.5x4 is the four triangles from the midpoints and sqrt50 swared is the middle = 100.

The inner square is equal to the sum of the four outer triangles

>the absolute state of (You)

then?

See

I don't see it. Where are we getting that the legs of the four triangles are of length 5?

The intersections on all the sides are bisectors because each side is marked as congruent, so the sum of the triangles and the central square has to be the same and each are therefore 1/2A.

Each of the four triangles are half of the square of the hypoteneuse of the square cut into 4ths.

Put all of this shit into a linear system and the determinant confirms one solution. Try 5. It works.

And we are done.

The proof is trivial and is left to the reader.

Answer: the area cannot be determined from the information given.

What we are trying to accomplish here is determine the area of an irregular polygon.

You cannot determine the area of an irregular polygon without knowing the following:

a. The length of two or more of the sides.
b. The measure of two or more angles within the polygon in question.

Or, most obviously,

c. The total area of the square in which all four polygons reside.

We CAN assume that the larger shape is a square because all sides are equal. However no lengths are given for any of the sides nor any interior angles given. The only angles we can assume to know are the four corners of the square, which equal 90 degrees due to all sides being equal in length.

Additionally, you cannot assume that we are dealing with whole numbers here when it comes to the length of the sides. Just because the areas of the known polygons are whole numbers doesn't mean any other dimensions in this problem are.

Lastly, approaching this figure and literally measuring the sides and angles is not viable since this figure may not be to scale. This figure may have just been draw with arbitrary line segments meeting at a vertex and areas thrown in a random. It's not worthwhile to measure the sides with a ruler or something since we cannot know if the figure is to scale.

>(You)

True, we can break it down as much as we want, but we won't find an answer

This is a paralellogram

24cm. Squares will always have an area that is a perfect square. 90 is the only plausible total. You don't even need math

Write down the fucking linear system you faggot. You don't have enough equations to make it determined.

If you're talking about the host square, true. All squares are parallelograms, but not all parallelograms are squares.

We know it's a square because of the "II" marks indicating all sides are equal.

Yes you do because you have the areas of the parts.

Nah dawg. The problem is one of 3 variables. Length l (which is the half side length) and the position (x and y) of the intersection point. We have 3 unknowns and 3 variables therefore it is solvable.

what's the name of the || symbol? i googled for list of geometry symbols, and the several places i checked didn't list it

26cm^2

Are you saying x equals 28 and 32 and 16 and 22?

So solve it. Show us.

Fuck off avatarfag

Working on it. Give me a moment

The symbol means that the sides are congruent. Technically "equal" is not the mathematically correct term for me to use, but for our purposes of discussion just know what the tick marks indicate all sides are the same length.

100% of the area - 32% - 28% - 16% = 24%

if the units are cm^2, then it's 24 cm^2

24cm^2

Fucking dumbasses

>avatarfag is a brainlet
shocker, I know

I did a reverse image search on it and nothing came up. It's possible OP made this just to troll us

fuck avatarfags

TLDR; answer is 20. Working below.

Alright so attached is how I have my variables set up. l is the length of one equal section and x and y are the horizontal and vertical distances from the centre of the square.

Therefore:
(l-x)(l-y)+(x(l-y))/2+(y(l-x))/2 = 16
-> l(l-x/2-y/2)=16
(l-x)(l+y) - ((l-x)y)/2 +((l+y)x)/2 = 28
-> l(l+y/2-x/2)=28
(l+x)(l+y) - ((l+y)x)/2 - ((l+x)y)/2 = 32
-> l(l+x/2+y/2)=32

Which I then solved using wolfram alpha, which gave l = 2sqrt(6) x = sqrt(2/3) y = sqrt(6)
Then its trivial to find the final area.

Unknowns and variables are the same damn thing you retard. You need 3 unknowns/variables and 3 equations.

Yeah mb typo'd it.

It's sad when the ylyl thread has better puzzles going than a thread like this.

google.ie/url?sa=t&source=web&rct=j&url=#&ved=0ahUKEwiVxq6m7dDVAhUoCMAKHTKmB1EQxa8BCC4wBg&usg=AFQjCNF1sHlk2z9QIx28jtvfnGP_UlkifA

Previous answer was wrong answer is 28cm^2

Thank you based user

Ah, alright, this works. Thanks.

I see it now.

Kinda sucks to have introduced coordinates but it works.

mmm this is just one way to do it. By far an easier way is to just say that the two opposing sections always have half the total area of the shape. Then 32+16 = 28 + ? so ? is 20, but the working is more convincing cause it doesn't assume knowledge of geometry, just of algebra.

seems like the block is divided on % of total
area. even if its not i can visually see that
32>28>?>16
so if my autistic guess is right, ? equals 100 - 28+32+16 (76) = 24
but thats just because i dont wanna go arround using formulas, but it may be near the result

if im correct then ? equals

24 cm^2 It would have to be a number when
added together be equal to n^2. 24 is the most realistic answer in this problem. Mathematically this is impossible to solve

see
or

I GOT IT! Y=X+76 just graph it for all possible answers

I'm invading this thread. If you retards can't solve this then you officially have an iq of a nigger

If you can't solve this then you are retarded.

Jim and Bob wanted to know Fred's birthday.
Fred gives them 10 choices to choose from, which are:

August 13 August 14 August 17 September 15 September 16 October 12 October 14 November 12 November 13 November 15

Fred then tells Jim and Bob the month and day of his birthday separately and respectively

Jim: I don't know Fred's birthday, but I know Bob doesn't know shit either

Bob: I didn't know Fred's birthday at first glance, but I know now

Jim: Now that I know that Bob knows Fred's birthday, I also know.

What is Fred's birthday?

October 14. You're not the only one in the ylyl thread.

Actually, I like the geometric a lot better. I see it now. Join the midpoints as in Now the problem reduces to showing that for any square and any point interior to the square, you have that triangles opposite along the vertex have half the area of the square. Let's consider top and bottom of triangles of height h and h. Their combined area is (sk + sh)/2 = s(h+k)/2 = ss/2.

Alright, crystal clear now. The answer is 20.

Fucking nice problem, OP.

Just split the areas into triangles and do a^2+b^2=c^2 of each triangle connected to the missing area. Then you can figure it out from there.

>being this retarded
Are you saying the 2 smallest areas of this square is equal to the area of the biggest sections?

Fresh off the boat, from reddit, kid? heh I remember when I was just like you. Braindead. Lemme give you a tip so you can make it in this cyber sanctuary: never make jokes like that. You got no reputation here, you got no name, you got jackshit here. It's survival of the fittest and you ain't gonna survive long on Sup Forums by saying stupid jokes that your little hugbox cuntsucking reddit friends would upboat. None of that here. You don't upboat. You don't downboat. This ain't reddit, kid. This is Sup Forums. We have REAL intellectual discussion, something I don't think you're all that familiar with. You don't like it, you can hit the bricks on over to imgur, you daily show watching son of a bitch. I hope you don't tho. I hope you stay here and learn our ways. Things are different here, unlike any other place that the light of internet pop culture reaches. You can be anything here. Me ? heh, I'm a judge.. this place.... this place has a lot to offer... heh you'll see, kid . . . that is if you can handle it...

fuck

No he's not saying that. Your drawing makes his solution even more obvious than ops.

He's saying opposite areas are equal. Like This

oh that makes sense

But there's nothing to indicate the corners are right angles

A simpler solution is to notice that diagonally opposite quadrilaterals always sum to the same area.

Inscribe a square between the midpoints of the outer square and call its edge length L. Then it encloses 4 triangles, one of them orange. Each pair of opposite triangles has area 1/2 L**2, because one's area is 1/2 LH and the other is 1/2 L(L-H). Since the triangles outside the inner square are of equal sizes, the diagonally opposite quadrilaterals are of constant area, i.e. 48 cm2 no matter the location of the central point of intersection. By symmetry, that applies to both pairs.

QED

This assumes that the figure is a square, though. It could be a rhombus.

19

No it can't.

Why not?

The congruency symbols.

How do they show that the four angles are 90 degrees?

The four corners of the square? If all sides are the same length then it's a square which means each corner is 90 degrees. You must be talking about something else because that's 4th grade level stuff.

Those indicate the sides are the same length, there's an entirely different symbol for angles

No, the sides of a rhombus are the same length, just like in a square, but the angles are not all the same, only the opposites.

Hope this is bait. The very definition of a rhombus is a figure where all four sides are the same length, but where the angles aren't 90 degrees.

What are you even talking about?

en.wikipedia.org/wiki/File:Rhombus.svg

The correct answer is pussy. This is Sup Forums and pussy is always the answer. Nuf said?

Answer is 24
- a mathematics major

Show your work

assuming its to scale about 20cm^2