Sup /sci, I saw this a while back on /b and never got the

28

>>8677847
But how

>>8677841
32 because the answer has to square

>>8677854
Then couldn't it be 25? That's 5 squared.

>>8677861
68 + 25 isn't a square

>>8677854
>>8677869
It could be more than 32 though

26

>>8677883
And the number doesn't HAVE to be even.

>>8677854
the side length isn't necessarily an integer so the total area may not be a perfect square

>>8677885
How?

>>8677841
Could it be infinite?

>>8677850
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.

>>8677891
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.

>>8677923
its 26 brainlet

>>8677893
no it could not.

>>8677847

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

>>8678031
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

>>8678036
same base and height...

>>8677993
I should add that I'm making the assumption that each "arm" of the cross meets its "wall" exactly halfway along.

>>8678037
oh i see now

sorry i'm a brainlete

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

Therefore Bottom Right = 28

>>8677923
Do you mind making a visual for this? Just kinda confused

File: 1487204222309.png (17KB, 924x375px) Image search: [Google]

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corrected sorry

>>8678070
How do you know the purples, oranges, blues, and greens are all equal parts? Is there some geometry rule I'm missing?

>>8678077
same base and height

>>8678075
this is super helpful though thank you

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>>8678082

>>8678080
No, because the line is slanted one line is going to be longer for the hight.

File: 1487120290107.jpg (15KB, 233x216px) Image search: [Google]

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>>8678086
height is from base to highest point...
slanted doesn't matter

>>8678086
Yeah they are def not equal

>>8678088
The hypotenuse has farther to stretch though

File: image013[1].gif (10KB, 379x256px) Image search: [Google]

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>>8678086
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

File: 1486956141456.jpg (98KB, 655x695px) Image search: [Google]

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>>8678089
the height for the two green areas is even coincident

mfw noobs

>>8678080
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

>>8678120
so is that wrong then? or are you confirming the drawing is correct

>>8678120
>>8678075
So is this right? Can't tell what you're trying to say is "trolling"

>>8678132
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.

File: 1486857345305.png (177KB, 367x321px) Image search: [Google]

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https://www.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.

>>8677841
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 /sci/ had so many retards, are all of the guys asking stupid questions humanities majors or something?

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>>8677850
this is how. enjoy.

>>8678100
Area, not volume.

>>8677932
No, it's 28.

32 + 16 = 20 + 28

>>8678062
http://math.stackexchange.com/questions/2078331/is-this-solvable-geometrical-image-included

>>8678031
You still don't know how much they are. Nothing guarantees the quadrilaterals are bisected.
>>8678092
That only affects the perimetre. Notice how the triangle becomes thinner as the hypotenuse grows

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>>8680767
troll confirmed
see this and move the slider called S
AREA VALUE DOES NOT CHANGE REEEEE

https://www.desmos.com/calculator/oorpenhig1

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>>8680767
ALSO the quadrilaterals are not bisected, the 4 sides of the square are or no solution would be determinable by any method .......

File: Untitled.png (18KB, 800x600px) Image search: [Google]

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>>8677993
>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

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>>8680086
mega brainlet there is no voume in the image
at least he tried to help you

>>8678039
how wouldn't it? each wall chunk is congruent

File: ygge.png (11KB, 508x158px) Image search: [Google]

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

you make me sick

>>8680963
how is that the height? brainlet

>>8680103
Well, now I can prove it. But I still don't know how to understand it intuitively...

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>>8678075
nigger

>>8677841
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...

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What's with all the pepes? Is it one guy or a new meme to spam them?
>>8680932
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"
>>8681070
?
>>8681090
There are 2 solutions itt

>>8680946
Well good, so my technique was valid.

>>8681468
There are no solutions because it's simply not solvable.

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Ill just leave this here...

>>8681736
That reminds me, I really must go to the new maritime history exhibition in town...

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>>8680025
>>8677847
The solution for non-brainlets.

>>8680984
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.

>>8682009
this is why guys like me score 30+ on the putnam while brainlets like you all struggle in linear algebra

>>8680767
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

>>8682087
>[math]x=\cos\theta_2[/math]

>>8678031

Best diagram.

>>8677841
>Any theories?

It's unsolvable:

LackOfCriticalDataError # 29

Abort, Retry or Fail?
>>

>>8678031

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

Why wrong?

>>8682087
way overcomplicated

>>8678100
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.

>>8683205
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

>>8677841
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.

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>>8683205
green equals ere equal...

why would I put the colors if they are random?