Sup /sci, I saw this a while back on /b and never got the answer. Any theories?

Sup /sci, I saw this a while back on /b and never got the answer. Any theories?

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desmos.com/calculator/oorpenhig1
math.stackexchange.com/questions/2078331/is-this-solvable-geometrical-image-included
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28

But how

32 because the answer has to square

Then couldn't it be 25? That's 5 squared.

68 + 25 isn't a square

It could be more than 32 though

26

And the number doesn't HAVE to be even.

the side length isn't necessarily an integer so the total area may not be a perfect square

How?

Could it be infinite?

Because googling the answer said so.

Drawing in the four red lines divides each quadrilateral in the original shape into isosceles triangles (blue) and a second triangle (white). The four isosceles triangles are identical.

With the inclusion of the red lines you have four triangles with identical base of the red line. The area of a triangle is found by 12×base×height 12×base×height. Using this to calculate the sum of the area of two triangles directly opposite each other you will get 12 times the length of the red lines squared. Then you can add the identical green triangles onto this to get the following: 32+16=20+?32+16=20+? resulting in an answer of 28.

Let P be that interior point of the square where all the segments meet.

Now draw line segments from P to each corner of the square. Say the square has side length 2x.

Now the square is divided into 8 triangles. Each of the 8 triangles has a side of length x.

Now each of the known areas is the sum of two triangles. Use this to set up a system of equations. You should be able to solve for the unknown area.

its 26 brainlet

no it could not.

The top right and bottom left add up to 48. If you move the middle vertex diagonally up-right, into the centre, now you have two identical perfect squares whose combined area is 48 so each is 24cm2.

This also made the 20 area into another identical square. Clearly the whole square is 96 cm2, subtracting 20, 32 and 16 leaves 28

...

There's no reason that this would be true. You seem to be implying that those are isosceles triangles but really they're both just scalene

same base and height...

I should add that I'm making the assumption that each "arm" of the cross meets its "wall" exactly halfway along.

oh i see now

sorry i'm a brainlete

Observation: Top Right + Bottom Left = Top Left + Bottom Right.

Therefore Bottom Right = 28

Do you mind making a visual for this? Just kinda confused

...

corrected sorry

How do you know the purples, oranges, blues, and greens are all equal parts? Is there some geometry rule I'm missing?

same base and height

this is super helpful though thank you

...

No, because the line is slanted one line is going to be longer for the hight.

height is from base to highest point...
slanted doesn't matter

Yeah they are def not equal

The hypotenuse has farther to stretch though

You are misunderstanding what is meant by "height".

In this picture, if all 3 of these triangles had the same "height" and "base" they would have equal volume, because you can always get the volume by doing base*height/2

shape doesn't matter

the height for the two green areas is even coincident

mfw noobs

are you trolling, I learned this in middle school

all triangles can be pieced together with a clone to form a parallelogram

parallelogram = rectangle with a triangle cut off one side (from a corner) and stuck on the other

parallelogram area = base times height

triangle area = the parallelogram formed by adding a clone / 2 = half base times height

so is that wrong then? or are you confirming the drawing is correct

So is this right? Can't tell what you're trying to say is "trolling"

Yeah I don't know either, but they're the same size because they have the same base*height. Anyone saying otherwise is a brainlet or a troll.

desmos.com/calculator/oorpenhig1

slider S
see area not changing

36
because (-2x2+52)/x=0 positive root is length of side, this ~5.09902 times 2 to the second gives a total area of 104. subtract given areas.

We had this thread last week.

damn i went to all that trouble to find the unicode char for superscript two and that sucker don't even work.. should be (-2x^2+52)/2=0

I didn't know Veeky Forums had so many retards, are all of the guys asking stupid questions humanities majors or something?

this is how. enjoy.

Area, not volume.

No, it's 28.

32 + 16 = 20 + 28

math.stackexchange.com/questions/2078331/is-this-solvable-geometrical-image-included

You still don't know how much they are. Nothing guarantees the quadrilaterals are bisected.
That only affects the perimetre. Notice how the triangle becomes thinner as the hypotenuse grows

troll confirmed
see this and move the slider called S
AREA VALUE DOES NOT CHANGE REEEEE

desmos.com/calculator/oorpenhig1

ALSO the quadrilaterals are not bisected, the 4 sides of the square are or no solution would be determinable by any method .......

>If you move the middle vertex diagonally up-right
is there a theorem that moving that point wouldn't change the areas of the quadrants?
cause pic related disproves it

mega brainlet there is no voume in the image
at least he tried to help you

how wouldn't it? each wall chunk is congruent

for the people that do not get triangles still...

you make me sick

how is that the height? brainlet

Well, now I can prove it. But I still don't know how to understand it intuitively...

nigger

It is not solvable I think.
The only information we have is, that the partitioning is bigger than 1/4 of the length of the side, nothing else...

Ive had enough, going to sleep now...

What's with all the pepes? Is it one guy or a new meme to spam them?
That's what I'm saying, idiot. The triangle may become longer but it also becomes thinner. I was offering a more intuitive way to see the constance of the area than "hurr moev de slider"
?
There are 2 solutions itt

Well good, so my technique was valid.

There are no solutions because it's simply not solvable.

Ill just leave this here...

That reminds me, I really must go to the new maritime history exhibition in town...

The solution for non-brainlets.

The height of a triangle is defined as the length of the line segment which is a perpendicular intersector to the base and intersects the vertex opposite the base.

this is why guys like me score 30+ on the putnam while brainlets like you all struggle in linear algebra

The quads don't need to be perfectly halved. It's the triangles from different quads that need to be equal.

32 + 16 = 48
20 + X = 48

X = 28

...

>[math]x=\cos\theta_2[/math]

Best diagram.

>Any theories?

It's unsolvable:

LackOfCriticalDataError # 29

Abort, Retry or Fail?
>>

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

Why wrong?

way overcomplicated

this desu

I never understood why we don't teach kids in elementary geometry that 2 triangles have the same area if 2 of their sides are the same length.

i was making the same mistake as you are, its silly but easy to make

I thought that when we split a section in half, we will get two equal triangles. this isnt true.

what is true however, when we split two sections that share an outside edge, the two adjacent triangles will be equal, as they have the same base, and the same height

so the area of the final section is equal to the lower triangle of the 32 section, plus the right hand triangle of the 16 section. we cant find these areas directly, but we know the other triangles from these two sections combine to make 20.

so we sum 32 and 16 to get 48, then subtract 20

20 + ? = 32 + 16

The areas in opposite corners always add up to 48.

Proof: it holds when the point P (the one on the interior of the square to which the 4 segments are drawn) is in the center of the square.

Now, it remains true if you shift P to the left (this operation reduces the area on the left and adds to the area on the right, but the sums of opposite corner areas remains constant). Similarly it remains true if you shift P to the right, up, or down.

what do you people study btw? these threads are always filledwith brainlet retards.

green equals ere equal...

why would I put the colors if they are random?