/sci/ - Science & Math

[Anonymous No.8498215]

Well?

[Anonymous No.8498222]

bout tree fiddy

[Anonymous No.8498272]

28cm[math]^2[/math]

[Anonymous No.8498302]

Numberd from $A_1$ (top left) to $A_4$ (bottom right), the following proportion always holds, assuming the intersection point remains in the square: $A_1 + A_4 = A_2 + A_3$.

[Anonymous No.8498318]

>>8498302
> assuming the intersection point remains in the square
Unnecessary assumption whenever you allow areas to be signed (which, let's face it, is basically all the time).

[Anonymous No.8498349]

>>8498215
28 cm^2

Take a+d as a pair and b+c as another pair. It's easy to see that any movement of the centrepoint along an axis will cause one quadrant to lose area, but the paired quadrant to gain an equal amount of area, meaning the pair as a whole remains the same size. Any movement of the centrepoint can be represented as a movement on the x-axis plus a movement on the y-axis; therefore the two pairs will always stay the same size; therefore 32+16=20+d; and plugging this into Wolfram Alpha gives d=28.

[Anonymous No.8498356]

>>8498349
>32+16=20+d
>plugging this into Wolfram Alpha

Humanity truly has declined

[Anonymous No.8498409]

>>8498349
This is a poor argument since you can't be sure that the combined effect will still be the same as the sum of the individual effects.

[Anonymous No.8498463]

>>8498409
Yeah, but it's true and I can't be arsed to go into more detail, so checkmate

[Anonymous No.8498528]

>>8498215
Let's look at the change in area of a diagonal pair when moving the center point. The red is just moved from C to B so no change in area. The green is lost to the other pair while the blue is gained.
So we need to show that the green area is as large as the blue area.
G=(x(r-y')+y(r-x'))/2
Now the slope of the border between B and D is y/(r+x) so y'=ry/(r+x) or ry=y'(r+x) and rx=x'(r+y).
Then
G=(xr-xy'+yr-yx')/2
=(x'(r+y)-xy'+y'(r+x)-yx')/2
=(x'r+x'y-xy'+y'r+y'x-yx')/2
=(x'r+y'r)/2=B
It's a bit ugly though.

[Anonymous No.8498542]

>>8498215
I feel as though this is meant to be done via pattern recognition.

32 - 20 = 12
20 - 16 = 4

Therefore, 28cm^2 is the answer.

28 - 16 = 12
32 - 28 = 4

QED

[Anonymous No.8498563]

Simple visual proof of >>8498302:
The triangles with the same color have the same area.

[Anonymous No.8498584]

>>8498463
A=x*y+(1-x)*y/2+x*(1-y)/2
= xy + y/2-xy/2 + x/2-xy/2
= x/2+y/2

B=y+(1-x)*y/2+(2-x)*(1-y)/2
= y + y/2-xy/2 + 1-x/2-y+xy/2
= y/2-x/2+1

C=x+x*(1-y)/2+(1-x)*(2-y)/2
= x + x/2-xy/2 + 1-x-y/2+xy/2
= x/2-y/2+1

D=(2-x)*(2-y)-(2-x)*(1-y)/2-(1-x)*(2-y)/2
= 4-2x-2y+xy - 1+x/2+y-xy/2 - 1+x+y/2-xy/2
= 2-x/2-y/2

=> A+D=2, B+C=2

[Anonymous No.8498610]

>>8498215

just looking at it you can tell all the areas are factors of 4

but each side is greater than 4 because the bottom left one is less than 1/4 of the area

also it is less than sqrt(32) because the top right is bigger than 1/4

so I'm just gonna guess 5

[Anonymous No.8498866]

>>8498215
Jokes on you I'm a fucking freshman.

28cm though because I am a parrot and I know nothing. Thanks for explaining this to me, thread.

[Anonymous No.8498886]

>>8498356
>Can't even do 32 + 16 - 20 in his head.

Top kek. The mental process should go something like:

30 + 10 = 40
40 - 20 = 20
2 + 6 = 8
20 + 8 = 28

We learned this in 4th and 5th grade.

[Anonymous No.8498973]

>>8498886
wat. just do it all in one step, or two, if you want. you're not 6 anymore.

32 + 26 = 48
48 - 20 = 28

[Anonymous No.8499081]

>>8498215
Well if you move the center over the two left will lose as much as the two right gain. This means that if you pair them diagonally the area will stat constant because one will gain as much as the other loses. So 16+32=20+x, x=28

[Anonymous No.8499411]

>>8498973
Didn't know that 32 + 26 is 48.

[Anonymous No.8499588]

>>8498563
Me and Steve like you.

[Anonymous No.8499605]

>>8498886
>>8498973
best way is to do 16-20, get -4. then 32-4 is 28

[Anonymous No.8500055]

>>8498973
>32 + 26 = 48

[None !niqjediPCA No.8501301]

>>8498272
That would give total area of 92cm^2 and make each side a length of 9.591663047....cm

With each unmodified quarter having a area of 23cm^2

odd....

[None !niqjediPCA No.8501311]

>>8498584
That picture appears to be wrong though.

Yours is along the centerline from top right to bottom left...

OP's is not.

[None !niqjediPCA No.8501328]

>>8498563
Yeah, that's not accurate.

[None !niqjediPCA No.8501330]

This is a common core problem, isn't it?

[Anonymous No.8501891]

>>8501311
Nigga, x and y are distinct. Just because they are equal in the pic doesn't mean their values would be equal, it's just the shape to get a rough idea

[Anonymous No.8501896]

Around 81 square cm, I think

Sqrt of 20cm is someshit, then someshit * 2 is almost 9, and then 9 squared is 81 square cm

I didn't complete year 11 maths

[Anonymous No.8501898]

>>8501896
Oh fuck, i just did the area of the whole square. Yeah fuck idk mate

[Anonymous No.8501928]

Does anyone have any more of these anime girl math problem pics?

[Anonymous No.8502420]

i just gambled with my eye, was correct