/sci/ - Science & Math

[Anonymous No.14996414]

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[Anonymous No.14996450]

>>14996414
Let area = x. Therefore, x is the area.

[Anonymous No.14996471]

>>14996414
I’d recreate it out of string and then turn the string into a square

[Anonymous No.14996508]

>>14996471
It’s only made of three pieces of string, you would have to make a triangle, not a square

[Anonymous No.14996522]

>>14996414
How do you expect us to calculate that if we don't know the area of the total square?

[Anonymous No.14996525]

>>14996508
I can bend string however I like thank you

[Anonymous No.14996526]

>>14996522
it's x^2

[Anonymous No.14996530]

>>14996522
Set the side length equal to 1 “mystery square” unit

[Anonymous No.14996556]

>>14996414
just use calculus

[Anonymous No.14996559]

>>14996414
I can’t see a way to do this without grinding through boring integrals but maybe I’m just dumb.

[Anonymous No.14997240]

It's [math]\frac{2-\pi}{4} [/math]
Did it without using calculus

[Anonymous No.14997263]

>>14996414
would describing each area on a system of equations work?

[Anonymous No.14997268]

>>14996414
just count the pixels bro

[Anonymous No.14997291]

All these kinds of problems can be solved by figuring out the corresponding equations on the Cartesian plane, finding intersection points and integrating for the area.

[Anonymous No.14997310]

>>14997240
>Negative answer
Gigachad

[Anonymous No.14997371]

>>14996414
Let L be the length of the sides (I'm assuming they're of similar length based on your picture).

The area of those two circles on the left and top can be calculated. The diameter is our side L, so their radius is r' = 0.5L.

We gather for their areas (note, half circle):
A' = 0.5πr'^2 = 0.5 * π * (0.5L)^2 = 0.125πL^2.

Now for the bigger half circle, the radius is L. It is one fourth of a circle, so its area is:
A'' = 0.25πL^2.

Now to find that faggot area, you take the area of the square L * L = L^2 and subtract the "inverses" of all your faggot circles. So the area not covered by your top and left circle would be L^2 - A'. Of the big one would be L^2 - A''. Etc. not in the mood to do the rest, just notice that the top and left circle have overlapping area, treat it like a venn diagram and you can find the result.

[Anonymous No.14997385]

>>14997240
huh? are you trolling lol? Do you mean pi minus 2?

[Anonymous No.14997620]

Is there actually a way to do it without le calculus?

[Anonymous No.14997638]

>>14997620
Adding and resting circle segments areas.

[Anonymous No.14997644]

>>14997638
you mean this? >>14997371
because that doesn't work, you should try to actually do it...

[Anonymous No.14997645]

>>14996414
Easiest method is here below if you know calculus.

Let 10 be the length of one side of the square, and [math]A[/math] be the area of the shaded region. We choose this length because it simplifies the relevant points in our coordinate system greatly, and we want to find the answer with calculus.

The left side of the square will run along our y axis, and the x axis will biject the left and right sides of the square. Thus the top left corner of the square is at point (5,0) and the center of the square is at the point (0,5). Now we can find the boundary curves of the red shaded region.

The region is bounded on the left by the curve [math]\sqrt{25-{x}^{2}}[/math], from above by [math]\sqrt{100-{x}^{2}}-5[/math], and from below by [math]5-\sqrt{25-{(x-5)}^{2}}[/math].

Set [math]\sqrt{100-{x}^{2}}-5 = 5-\sqrt{25-{(x-5)}^{2}}[/math] to find that [math]x=8[/math], then plug that into [math]\sqrt{100-{x}^{2}}-5[/math] to find [math]\sqrt{100-{8}^{2}}-5=1[/math]. Thus the coordinate of their intersection is [math](8,1)[/math].

Thus [math]A = \int_{0}^8 {\sqrt{100-{x}^{2}}-5}dx - \int_{0}^5 \sqrt{25-{x}^{2}}dx - \int_{5}^8 5-\sqrt{25-{(x-5)}^{2}}dx[/math]

Because we're not doing a homework problem, here's the mathematica code to get the answer.

f[x_] = Sqrt[100 - x^2] - 5
g[x_] = 5 - Sqrt[25 - (x - 5)^2]
h[x_] = Sqrt[25 - x^2]
N[Integrate[f[x], {x, 0, 8}] - Integrate[g[x], {x, 5, 8}] -
Integrate[h[x], {x, 0, 5}]]

This should output is 9.77357, and we are done.

[Anonymous No.14997652]

>>14997645
>this should output is
fuck. whatever.

[Anonymous No.14997657]

>>14997620
Yes there is, but trying to type up the explanation for is as absolute horrible so it's easier to explain with the calculus solution that's easily understood and replicated.

[Anonymous No.14997714]

>>14997645
also if you have mathematica and want to put in a length of your own, here.

length = < type a number in here >
f[x_] = Sqrt[s^2 - x^2] - s/2 /. s -> length
g[x_] = s/2 - Sqrt[{s/2}^2 - (x - s/2)^2] /. s -> length
h[x_] = Sqrt[{s/2}^2 - x^2] /. s -> length
point = Solve[f[x] == g[x], x][[2]]
N[Integrate[f[x], {x, 0, x /. point}] - Integrate[g[x], {x, length/2, x /. point}] - Integrate[h[x], {x, 0, length/2}]]

[Anonymous No.14997728]

>>14997645
lastly if you wanted the exact form for some reason it's
[math]\frac{75}{2}\sin^{-1}(\frac{4}{5})-25\approx9.77357[/math]

[Anonymous No.14997773]

>>14996414
atan(4/3)/2 if you take one side of the square as 2

[Anonymous No.14997792]

>>14997645
I lied, one more thing. The general solution for any given side length [math]s[/math] is
[math](s^2-{(\frac{s}{2})}^2){\tan}^{-1}(\frac{1}{2})-{(\frac{s}{2})}^2[/math].

>>14997773
it's [math]3{\tan}^{-1}(\frac{1}{2})-1[/math]

[Anonymous No.14997861]

>>14996414
If you assume the side of the square is 2x then the shaded area is somewhere between:
((pi - 2) / 4) x^2
and
(((pi - 2) / 4) + ((pi - 3sqrt(3) + 3) / 6)) x^2

[Anonymous No.14997878]

Answer is pi/8

[Anonymous No.14997914]

>>14997645
well, that's one way of doing it, but it strikes me as a problem you should be able to solve by adding and subtracting regions of known area somehow

[Anonymous No.14997926]

>>14997914
You can, that's how I found >>14997792.
The explanation is absolutely horrible without any labeling of points in OP's picture. The main idea is to find the area of the intersection of the top semicircle and the circle that passes through the opposite corners of the square, and then subtract off the area of the intersection of the two semicircles.

[Anonymous No.14997931]

>>14997926
>the circle that passes through the opposite corners of the square
this is unclear
do you mean the bottom left quarter circle?
if so what you're describing is what I concluded with too after thinking about it for a while, but I couldn't figure out how to calculate that area

[Anonymous No.14997938]

Numerical analysis says an area of .095 +/- .008 accounting for nonzero line width

[Anonymous No.14997960]

>>14997926
looking now at how to calculate the intersection area of circles with different radius, I understand the methodology (calculating the area of each sector and subtracting the corresponding triangle for, just as when calculating the simpler case of the vesica piscis), but the resulting formula is more complex and doesn't involve atan (rather it involves acos), so I'm curious to know what allowed you to simplify it so much

[Anonymous No.14997961]

>>14997931
Yeah that's the one. For additional clarity in case other anons are confused, it's the arc in pic related.
For reference, the area of the intersection of two circles like this is called a geometric lens, and you can find a general formula for them here: https://mathworld.wolfram.com/Lens.html

The area of the quarter-circle semi-cirle intersection (aka lens) is
[math](s^2-{\frac{s}{2}}^2)({\tan}^{-1}(\frac{1}{2}))+\frac{\pi s^2}{2} -\frac{s^2}{2}[/math]

[Anonymous No.14997964]

>>14997960
Just saw your reply after I posted mine and the page refreshed lol. I'm writing a response but sending this quickly first in case it seems like the previous one was meant to answer this question as well.

[Anonymous No.14997965]

>>14997961
yeah, like I just mentioned here: >>14997960
I was looking at a similar page with the same formula (the first one, for the asymmetric lens):
https://mathworld.wolfram.com/Circle-CircleIntersection.html
however, I don't see how you've reduced it to just atan, when only the area you're subtracting (the overlap between the semicircles) is symmetric, and not the original area (the overlap between the top semicircle and the larger quarter circle), which is asymmetric

[Anonymous No.14997998]

>>14997792
>it's [math]3{\tan}^{-1}(\frac{1}{2})-1[/math]
dunno
pic related is furthest I could get

[Anonymous No.14998018]

>>14996414
What is the point of solving this? How can we profit from it?

[Anonymous No.14998023]

>>14998018
Nothing. It's maturbatory math. Or, math.

[shmalculus !!9odRrjwHLlO No.14998061]

>>14997965
Apologies for the late reply, I had food delivered and was very hungry.
It may be helpful to first note that [math]\tan^{-1}(\frac{1}{2})[/math] is a simplified form of
[math]\tan^{-1}(\frac{r}{R}))[/math], using variables [math]R[/math] and [math]r[/math] from the mathworld page.
You can either ignore the formula for asymmetric lenses they provide and find the angles of the sectors in terms of tangent, or you can convert it by noting that
[math]\cos(\tan^{-1}(x))=\sqrt{\frac{1}{1+x^2}} \rightarrow \tan^{-1}(x))=\cos^{-1} \left(\sqrt{\frac{1}{1+x^2}}\right)[/math].
Let me know if you'd like the specific relevant case written explicitly.

>>14997998
You're on the right track anon.
We'll start with the larger lens.
Note that the area of the smaller smaller circular sector is [math]\frac{a^2}{2}(2\theta)[/math]
and the area of the larger one is [math]\frac{b^2}{2}(\pi - 2\theta)[/math].
The angle [math]\theta=\tan^{-1}\left(\frac{b}{a}\right)[/math]
Now let [math]a=2[/math] and [math]b=1[/math], and solve for [math]\theta[/math]
Substituting in for [math]\theta[/math], the combined area of the two sectors is thus
[math] A_{sectors} = (2^2-1^2)({\tan}^{-1} \left( \frac{1}{2 }\right))+\frac{\pi}{2} \rightarrow 3{\tan}^{-1}\left(\frac{1}{2}\right)+\frac{\pi}{2}[/math].
We now need to subtract off the triangles outside of the intersection, which each have an area of [math]\frac{ab}{2} = 1[/math]. This gives us
[math] A_{lens1} = 3{\tan}^{-1}\left(\frac{1}{2}\right)+\frac{\pi}{2}-2[/math]

Let's just reuse this formula we found, but setting [math]a=b=1[/math] this time for the smaller lens.
[math]A_{lens2} = (1^2-1^2)({\tan}^{-1} \left( \frac{1}{1 }\right))+\frac{\pi}{2}-1[/math]
[math]\rightarrow A_{lens2} = \frac{\pi}{2}-1[/math]

Proceed as expected.
[math]A_{lens1}-A_{lens2}=3{\tan}^{-1}\left(\frac{1}{2}\right)+\frac{\pi}{2}-2-\left(\frac{\pi}{2}-1\right)[/math]
which simplifies to

[math]A_{final} = 3{\tan}^{-1}\left(\frac{1}{2}\right)-1[/math]

[Anonymous No.14998064]

idk maybe it'll be a question in mr beast's next video or something

[Anonymous No.14998223]

Agent K

[Anonymous No.14998227]

>>14996471
>who was chuck and what did he sell?

[Anonymous No.14998229]

>>14997310
kek

[Anonymous No.14998325]

>>14996530
why don't I just do what: >>14996450 said?

[Anonymous No.14998351]

>>14996414
am I the only one who sees an ass with panties?

[Anonymous No.14998352]

>>14998351
kys coomer

[Anonymous No.14998379]

>>14998061
okay, now I figured it out and arrived at the same result. turns out I was taking 2atan(1/2) for the area of the bigger sector. cool, thanks anon

[Anonymous No.14998398]

>>14998379
No problem friend! nice work, I wouldn't have posted my solution at all if I hadn't seen how close yours was.

[Anonymous No.14998431]

Here's my solution, I made it as clean as possible.

Three variable names:
The intersection of the two semicircles is [math]A[/math].
The area being looking for is called [math]B[/math], so the intersection of the quartercircle and semicircle is [math]A+B[/math].
I called the length of one side of the square [math]2r[/math]

First pic find [math]A[/math], then the second pic isolates [math]B[/math] which is the area being looking for

[Anonymous No.14998623]

>>14997645
This is very peculiar, suddenly I have a strong desire to go running in Asics shoes

[Anonymous No.14999594]

>>14997644
Why not? If you treat it similar to a venn diagram, you can find the overlapping areas independently then subtract those individually to find the area you're looking for.

[Anonymous No.14999934]

>>14999594
you have sentions that cannot be easily calculated to substract them, *yawns* justy try it yourself

[Anonymous No.14999997]

>>14999934
Not him.
From left to right.
1. Area of half circle
2. - 2 simetrical circular segments
3. Area outside both circles by resting 2 circular segments
4. area outside larger circle rested to area of 2
5. (3) + (4)
6. Area red

[Anonymous No.15000102]

>>14996414
Is this possible without calculus? Cause it appears as if it doesnt require it?

[Anonymous No.15000105]

>>15000102
yuppers
>>14998061

[Anonymous No.15000114]

>>14999997
lol you cannot even calculate the second step

[Anonymous No.15000154]

>>15000114
Calculate the area of a circular segment is trivial.

Area of the circle * angle / 360 - area of inner triangle.

[Anonymous No.15001772]

bump

[Anonymous No.15001856]

>>14996414
Nerd sniping is gay and the people that do it should feel bad

[Anonymous No.15001996]

>>14998061
Which triangles

[Anonymous No.15002336]

alright these natty daddies are telling me im too drunk to think about math im going to go watch cops or something

[Anonymous No.15002537]

>>14996414
[math]\frac{3 [\pi - 2]}{16}[/math]