What’s the area?

Рет қаралды 1,255,908

4 ай бұрын

This is a short, animated visual proof finding the area bounded between three mutually tangent unit circles.
Have a different solution? Share it in the comments!
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Check out these related videos:
• What's the area of thi...
• Area of a Regular Hexa...
• Area of an interior he...
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Пікірлер: 1 600

@Zangoose_ 4 ай бұрын

When you colored in the pink, it was over. Good stuff here

@bebektoxic2136 3 ай бұрын

Why?

@davidc756 3 ай бұрын

Gyattt

@eranodelpum9752 3 ай бұрын

@@bebektoxic2136 search the female reproduction sistem

@czar6203 3 ай бұрын

U guys are sick bro

@ranudas8834 3 ай бұрын

Get medication

@dhhfkkjb6891 3 ай бұрын

Now i know how to calculate the area of a g-string thanks

@linuxp00 2 ай бұрын

Only the rear part

@TheMilkyWayGalaxy--he-they-it Ай бұрын

PFF FR

@PR-fk5yb Ай бұрын

😂😂

@nightfall37 17 күн бұрын

@@linuxp00 that's the front part. The rear part is just a line.

@marenrahn774 7 күн бұрын

And thank you!

@user-wp4vs1ug6z 2 ай бұрын

Solving this correctly was the confidence boost I need during my regular 3 am depressive episode.

@ersetzbarescrewmitgliednr.7063 Ай бұрын

I hope you realize how many people feel exactly the same way xd

@LordNaver Ай бұрын

hey, so totally out of context here but.. my usual depressive episodes are during 3 pm 😅 coincidence!!

@ersetzbarescrewmitgliednr.7063 Ай бұрын

@@LordNaver Everyone has their own individual rythm, so beautiful ☺️

@user-wp4vs1ug6z Ай бұрын

@@LordNaver for me it's because I can distract myself with work or something during daytime but I dread the nights because I'm alone with my thoughts

@LordNaver Ай бұрын

@@user-wp4vs1ug6z interesting.. quite the opposite for me.. At night I feel quite calm, and composed, there is a sense of confidence in my thoughts. whereas in the afternoon, it's hot and dry, the most mundane and boring part of the day.. it is somehow low key depressing.. interesting.. to see different perspectives..

@peternicks7049 17 күн бұрын

Straight away. Always good to get a mathematical pick me up when feeling inadequate.

@wermh3719 3 ай бұрын

Integration has never failed me and this time was no exception 🗿

@absence9443 3 ай бұрын

did you actually? You'd have the curve from the upper circle where the integral starts/ends at the two intersections of the upper circle. And the integral interval is split into two since there are two segments with different functions given the two lower circles. However, since they are the same, we only need to compute one half.

@rasuru_dev 3 ай бұрын

based

@rasuru_dev 3 ай бұрын

what did you integrate over

@absence9443 3 ай бұрын

@@rasuru_dev likely x, left to right area determination

@wermh3719 3 ай бұрын

@@rasuru_dev for simplifying I considered the area of the section between the straight line that contains the centre of the triangle and the left tangent point (the point tangent to the upper circle and the left one). Describing those functions with the centre of the left circle un (0,0) would result in the [1/2 , 1] as the set of integration. Then you multiply that area by 3 and you obtain the same result!

@lexacutable 4 ай бұрын

Hrm. I think I do indeed need to investigate this curvy area more closely.

@drenz1523 4 ай бұрын

🤨

@Elfcheg 4 ай бұрын

Yep. M-hm. Definitely. Well. I see it.

@canyoupoop 4 ай бұрын

"Everything reminds me of her"😢

@thesecondderivative8967 4 ай бұрын

No more internet for you.

@estroncio64 4 ай бұрын

Aw man i thought you meant the pizza wedges lol

@GREGGRCO 3 ай бұрын

I've been away 40 years from this ! I was on the right path !! I didn't see the Equal lateral triangle in the middle! Paused the video and finished !! That was great ! Thanks !! I was very interested in the comments ! Gonna look at those methods !

@Ram_Bhakt0777 2 ай бұрын

Not my sister: "Is that my underwear???"

@farhanferdaus8786 Ай бұрын

💀💀

@user-kx5bu4is8f Ай бұрын

Bro 💀 "ram bhakt"

@Sans-the-truely-gamer Ай бұрын

that's what I gonna sa-

@madtscientist8853 4 ай бұрын

That is called a G-STRING

@flmis 3 ай бұрын

It's not a man's g-string, it has no extra circle in the bottom part

@md.sadathassan2407 3 ай бұрын

😂😂😂

@user-yh1nm1vy3i 3 ай бұрын

Not Nicola Tesla 💀

@smikenickleby9839 3 ай бұрын

isn't calculus amazing!? 😍

@alst4817 3 ай бұрын

They use this question to mock mathematics virgins in exams.

@goggypoggy 4 ай бұрын

wow, I actually solved it the EXACT same way! I'm so proud of myself!!!

@savitatawade2403 4 ай бұрын

it's the only way...

@ahmedal-nsour9611 4 ай бұрын

Actually I don't think so, you can divide the shape into two pieces vertically then using integral and the equations of the remaining tangent circle and half one on top, remembering we have the radius of each and the two tangent points. So what about that? Edit: not sure.

@samuelhba8720 4 ай бұрын

Same goggypoggy.

@ShrekPNG 4 ай бұрын

@@savitatawade2403You can also connect the tangent points of the circles to form a triangle and then subtract the area of three segments from its area (as opposed to subtracting the area of three sectors from a triangle formed on center points)

@Danaelivs 3 ай бұрын

SAME BRO

@shanewright4650 3 ай бұрын

Yes, this is exactly how I solved it myself. The general formula would be r^2(sqrt(3) - π/2)

@Edenival.v0x 2 ай бұрын

1. Triangle 2. Area of sector 3. Subtract area of triangle and area of sectors 4. Answer

@farziltheweebo4841 13 күн бұрын

area of sector how?

@sirsamiboi 4 ай бұрын

This solution is so satisfying

@mr.unknown1070 3 ай бұрын

The area will satisfy more💀

@-Cocell 3 ай бұрын

@@mr.unknown1070💀👍

@ShlokParab 19 күн бұрын

I did the same thing, just complicated; taking the bigger triangle.

@srijoyeemisra6175 11 күн бұрын

From commenting on funny and food videos..to commenting on The most educational one..ITS BEEN A LONG JOURNEY 😊❤

@mater5930 4 ай бұрын

The area we want in brackets is exactly the area we want.

@Ruktiet 3 ай бұрын

What?

@munkoman2 3 ай бұрын

😏

@chaprik7480 3 ай бұрын

😅

@SamannayC 3 ай бұрын

Only legends will understand 😏😂

@KojiKazama 3 ай бұрын

Lol, thats what I was thinking. The brackets make up a part of that area too.

@mennolente4807 4 ай бұрын

I drew hexagons around circles, radius equals apothem, area hexagon minus area circles, divide by three to get rid of the unnecessary bits.

@leif1075 3 ай бұрын

Wht made you think of hexagons at all though or remember an apothem even? Thanks for sharing.

@mennolente4807 3 ай бұрын

@@leif1075 that's the only plane filling shape that fits between balls stacked that way. Also, they're the bestagons. And I also know the formula for the area of both a circle and a hexagon, and I have a faint clue of subtraction. About the apothem, I really like Scrabble. And maths. And hexagons.

@hhhpestock951 3 ай бұрын

@@mennolente4807 That's pretty dope!

@Dhruvjindal0426 3 ай бұрын

nice!

@Camolicious 3 ай бұрын

Cool

@alimakhatun5074 3 ай бұрын

People see the area normally. But I was looking in different mind😂😂😂

@XprogaminG.--_--. 2 ай бұрын

😂😏

@physimania 2 ай бұрын

It's not different mind it's dirty mind

@joshsingh1111 2 ай бұрын

Bro reminded of an underwear😂, isn't it?

@nobody_AskedMe 2 ай бұрын

Yes, apparently your mind is not that different

@cze4ever Ай бұрын

@@joshsingh1111 many people did - and some pointed out it gets better when you include the ( )

@PC_Simo 15 күн бұрын

The area is just the area of a unit equilateral triangle ((√3)/4) minus 3 * the area between each circle’s arc and a cord that’s also an edge of an inscribed unit hexagon (A = (6*√3)/4). Thus, the area between the arc and the cord is: (π - (6*√3)/4)/6, π being the area of the unit circle. Then, you just need to subtract 3 times that, from the area of the central unit equilateral triangle (1 for each circle), as mentioned above: (√3)/4) - 3 * ((π - (6*√3)/4)/6) = 0,16125448077398… = √3 - (π/2).

@luciboi8182 4 ай бұрын

This was one of those questions that made me actually pause and solve it in my head because it seems so simple(and it is) and yet quite intriguing

@SWTORLOL87 3 ай бұрын

I was on the math team a lifetime ago in high school. Little things like this let me relive the good ol’ days lol.

@Defeatm1943M 3 ай бұрын

It is because it's class 10 level question Of circles chapter and it is level 5 question😅

@JitendraSingh-lr4nu 3 ай бұрын

Yet you could not solve it. Don't be overconfident.

@luciboi8182 3 ай бұрын

@@JitendraSingh-lr4nu It's okay if you couldn't do it but don't randomly judge people mate. It's a pretty simple question from grade 10

@SWTORLOL87 3 ай бұрын

@@JitendraSingh-lr4nu What makes you think they couldn’t solve it? The answer was provided at the end of the video. Why would they feel the need to include it here?

@deddywijayalim 4 ай бұрын

"can you determine the area of enclosed cheeks" was what i heard 😂

@onradioactivewaves 4 ай бұрын

Area of the thong

@FlowersBeds 2 ай бұрын

Thank you sir, mera 2 hafte me exam hai aur mujhe pura samajh aagya, meine paheli bar aapki video dekhi hai aur ab mujhe pata chala hai aapko log best teacher kyu khete hai

@redtrychguy8676 2 ай бұрын

really nice problem man I loved it. It took me about 10 minutes. I had to draw one more circle to realise something and continue. I found it as a function of the radius: R^2 * (sqrt3 - pi/2)

@lillii9119 4 ай бұрын

Other solution: find equations for the left and top circles, then find the difference of integrals on [0.5;1], double it to find the area

@rubykanima 3 ай бұрын

wtf

@lillii9119 3 ай бұрын

@@rubykanima I have no idea what I was saying, sometimes I get drunk and get especially good at math Glad to know it was apparently relevant though I no longer understand how that works

@waxy1885 2 ай бұрын

@@lillii9119I would definitely buy you a beer if I bump into you in real life

@Maya-ul1rr 2 ай бұрын

Make it polar and integrate from 0 to 2π/3. Triple that for the area.

@urnoob5528 11 күн бұрын

unless u do some quirky math, that s not a simple integral

@benschwartz5854 4 ай бұрын

I solved it in a much more complicated way. Your solution is so elegant!

@MilkGlue-xg5vj 2 ай бұрын

Show us please please

@afrophoenix3111 2 ай бұрын

I had the same methodology (and result), but your explanation and execution were far more elegant than mine. Thanks for the chance to flex the geometry muscles.

@Bubble_Gum7505 2 ай бұрын

At first we need to find area of triangle, then subtract area of circles from it. For example Radius - 5 m 1) Area of triangle: (10×10)/2= 50 m 2) Need to find the area of 3 pieces of circles. Area of one whole circle: 3,14×5²= 78,5 m Angle of piece - 60° because sum of angles in triangle is always equal to 180°. Area of piece of one circle is: 78,5 m×(60°/360°)= 13,083 m. And area of three pieces is equal to 39,249 m. 50-39.249= 10,75; ~11 m

@sperner9069 4 ай бұрын

I solved it in a different, probably less efficient way. I formed a circumscribed hexagon on one of the circles, subtracted the area of the circle from the area of the hexagon, and divided that value by 2

@ThatGuy-yc9yc 4 ай бұрын

Sweet! I had the same idea. Then thought maybe it was a long approach. I'm glad I wasn't mistaken for it to be a solution 😊

@leif1075 3 ай бұрын

Why would younthinknlf hexagons at all just wondering? And not squares or rectangles? I can see maybe forming a hexagon using all three circles..but why only in one..seems kind kf out kf nowhere?

@sperner9069 3 ай бұрын

@@leif1075 I didn’t think hexagons immediately, what I did was take the very center point of the structure, and draw 2 lines to any 2 of the 3 points of tangency between the circles. That forms a shape which is 1/3 of the center area, but also happens to be 1/6 of the area of a hexagon circumscribing the circle which has the 2 chosen points of tangency, minus the area of the circle.

@dyltan 3 ай бұрын

Me when I see the video in my recommendations: *_"Oh math! So cool, OH A BIKINI! SO MORE COOL!"_*

@akultechz2342 3 ай бұрын

Damn u got this

@utkarshjain861 3 ай бұрын

😂😂😂

@udayraj6976 3 ай бұрын

stay strong not HARD

@dyltan 3 ай бұрын

@@udayraj6976 stay hard and strong!

@vishnuchandrabose9875 3 ай бұрын

@@dyltanwhy do we share the same brain cells 😂😂

@ayanacharya9747 3 күн бұрын

The 2 of those equilateral triangles form a square. Two of those semicircles form a circle. So 2 of those G strings can be expressed as Area of square of side 2r minus Area of a circle of radius r. Thus, Area of G string = (4*r*r - pi*r*r) / 2

@tHa1Rune 3 ай бұрын

First time I've been able to solve/figure out the process in one of these videos . Nice!

@nandisaand5287 3 ай бұрын

Havent looked. Here's my solution: Variable: radius, ("R") Constant: Pi, 3.14159 Calculate area of equilateral triangle formed between centers (each side=2R) then subtract 3 60° circle slices(CS). A(T)=1/2×2R▪︎R(sqrt3) A(T)=(R^2)▪︎(sqrt3) A(CS)=60/360•Pi•R^2 =1/6•Pi•R^2 Area (Segment)=A(T)-3▪︎A(CS) =[R^2•(sqrt3)]-3•[1/6•Pi•R^2 =R^2•[(sqrt3)-{Pi/2)]

@bhoomiebhasin 3 ай бұрын

Exactly what I thought!

@tariqwaheed7429 3 ай бұрын

How do you know sectors will form 60°

@nandisaand5287 3 ай бұрын

@@tariqwaheed7429 He literally said "equilateral triangle". Anyway, it can be inferred cuz all 3 circles have same diameter.

@hgb3405 2 ай бұрын

that is exactly what i was thinking in my head

@goodmood2000 Ай бұрын

@@tariqwaheed7429same distance bw center of two each circles

@TransportSupremo 4 ай бұрын

√3 - π/2

@cubicalgamer2402 4 ай бұрын

Funny. I got the same thing about 30 seconds later.

@Xi_Jinping_ 3 ай бұрын

Yes 🎉

@JoJo-jj6id Ай бұрын

You all assumed that r is 1 except me ?😅

@TomSando117 Ай бұрын

@@JoJo-jj6iddw i didnt either r^(2)[3^(1/2)-(π/2)] looks less neat buts much more useful

@Enki._ 2 ай бұрын

The first thing I thought of was an equilateral triangle. Then: "Okay, now what do I do with this?" 😅

@NabiruBogdan 3 ай бұрын

had to watch it twice to understand :)) thanks i enjoyed that !

@mibeutbig8909 4 ай бұрын

Good morning everyone! I hope everyone enjoys their daily dose of mathematics!

@johnathanh2660 4 ай бұрын

I started with a spot in the centre of the light blue area and drew three lines, towards the centre of each circle. I then drew three more lines from the spot to each point where the circles touch. You can then have a circle inside a hexagon. And then you realise that the light blue area is half the area between the circle and hexagon. So you calculate the area of the hexagon, subtract the area of the circle, and then divide by two. Your way is better.

@carsonsheets3480 Ай бұрын

I actually got this one, truly amazing what studying can do

@tecno_andre2752 3 ай бұрын

i actually got this one (in the same way), i'm feeling so good rn

@ethanrenckly787 3 ай бұрын

Even if Project Borealis doesn't get released, at least we got this beautiful soundtrack.

@kalashkiv17 4 ай бұрын

Это решение конкретное решение для кругов радиусом 1. Общая формула этой фигуры будет выглядеть так: √3r² - (πr²)/2

@MattSchoen 2 ай бұрын

I found it easier to think of the three wedges as adding up to a half circle, instead of being 1/6 of a circle * 3

@SAM999XXX 3 ай бұрын

(Area of the triangle) -- (Area of the sectors)

@somniumkr6975 3 ай бұрын

area of triangle - are of unit circle the sectors make a full circle and what you have written will give answer in -ve

@Ruktiet 3 ай бұрын

Wrong, you need to switch those terms.

@AlexTrusk91 3 ай бұрын

@@Ruktiet true, but you can also assume negative areas, like distances, as possitive and kill their minus sign

@maxsuphidden667 3 ай бұрын

The area of bigger triangle connecting centes of circles -- Aera of 3 sectors (how will it give negative answer , you can clearly see the area is small than the triangle)

@afj810 3 ай бұрын

@@Ruktiet it's correct wdym

@ChukwuemekaElijah-vt9sx 3 ай бұрын

I stumbled on this problem sometime ago while playing around with circles. I solved it using the same method

@yettsoman4364 4 ай бұрын

Did it at school MANY years ago. Had completely forgotten how, though! 😂

@moofou1427 3 ай бұрын

At first I thought it would be a video about Dark Samus' field of view, then I saw the Triforce, then I understood nothing and was impressed. Good job!

@MirzaBicer 3 ай бұрын

I have gone through a similar (but overly complicated) method. I chose the corners of the triangle as the touching points of the circles. Then, for each curve within the circle, I have used the y^2 + x^2 = 1 function and manipulated it to get the area under the curve within the triangle. Then I multiplied that by three to account for all the curves and then subtracted that from the overall area of the triangle.

@chicaloca9065 3 ай бұрын

Done exactly as explained 😊

@nilaypathak657 4 ай бұрын

I solved it the EXACT same way, except took radius as R instead of 1.

@rishitsoneja307 3 ай бұрын

He said unit circle so R = 1

@RealJackBolt 3 ай бұрын

Yep, same. The general case would be, Area = R²( √3 - π/2) Where R represents the radius of circle assuming all of them have same radius ofcourse. Edit: if radii of all the circles are different, then doing it by calculus would be better as the method shown in the video would get *very* complex.

@ZeroGDucks 3 ай бұрын

I'm amazed for as long as it's been since studying algebra, I got every step except making a triangle from the center points 😆

@Lolfattenski Ай бұрын

Im out of school now but i remember this was one of the few questions that perplexed me at school before i saw the solution, it used coins instead of the normal circles which didnt affect the calculation

@dannypipewrench533 3 ай бұрын

Initially, I wanted to draw a triangle around all of the circles, but then I realized I only needed to connect the centers. So, I grabbed my TI-84 Plus and did it just like you.

@robertolin4568 4 ай бұрын

You have to prove that the intersections of circles lies on the triangle first, although it’s a rather easy one.

@nandisaand5287 3 ай бұрын

A line from the center is always going to be 90° to a line tangent to a circle at point of tangency. Since 2 circles share point of tangency, a line from one center to another is always a straight line.

@aquss33 4 ай бұрын

I got sqrt(3)-pi/2 in my head before watching, gonna check now lol, this seems very easy

@arnabchatterjee1601 4 ай бұрын

how?

@aquss33 4 ай бұрын

@@arnabchatterjee1601 We did this in middle school so I remembered how to do it (it was for some math competition, pretty low level one, I think almost everyone got this question right, we had like 3h for 5 questions so we had a lot of time to think lol). Of course, the answer was different, but I remembered how I solved it (I actually think we had to find the ratio between the area of that little weird part in comparison with the area of the 3 circles (or just one, no idea really) with respect to r as the radius wasn't given and was just labelled r). As for why I was able to do it in my head: because I know 2 very simple middle school formulas: the formula for an equilateral triangle which is a^2*sqrt(3)/4 and since the side was 2*1=2 I just put that number there instead of a and got 4/4*sqrt(3) which is just the sqrt(3) for the area of the triangle. After that I just simply did the "part of the circle area formula, no idea what the english name is" which is pretty simple: r^2*pi*60/360 (because they were all 60 degrees cuz they're equilateral triangles) since r=1 (cuz unit circle) it's just 6/36*pi which is actually 1/6*pi which is the area of each circle part and after multiplying by 3 you get 3/6*pi which is just pi/2, that's the area of the part of the triangle that you do not need, after that just subtract sqrt(3)-pi/2 and that's the answer I got in about 30 seconds. Again, I did this all in my head, I actually pulled up my scientific calculator app on my phone, but ended up not needing it as it was just so simple. Also, I knew this procedure pretty well so it was nothing special, it was one of my favorite middle school "low level competition" problems, I never really did any high level competitions, I just dabbled in them easy ones, and this thing was damn near the easiest thing they'd ever put on them, though I would have needed a calculator had the numbers not been so nice and clean (which you do not get at our stupid competitions, one of the reasons I didn't like them very much). Sorry for the long reply btw lol.

@bradensorensen966 Ай бұрын

I didn’t do all the work, because KZfaq shorts don’t call for it, but I did figure it would involve “triangle minus (given sector x3).”

@coopernoble6139 2 ай бұрын

I’m so proud of myself for figuring it out before. You can form an equilateral triangle, and then subtract three 60 degree sections of the circles. So 2r^2 - (pi*r^2)/6

@_Loki__Odinson_ 4 ай бұрын

Yeah I did it exactly as you did. Simple enough

@bkmishra95 3 ай бұрын

the right answer is r²/2.(2√3-π), where r is the radius of the circle.

@lazygamerz9145 3 ай бұрын

He already gave radius in the video by a Saying unit circles

@QobelD 3 ай бұрын

I had a hexagon inscribed in the circle. The area we want is an equilateral triangle with sides of length r minus the following value: The circle area minus the hexagon area then divided by 2. I didn't consider that r=1, bc I missed the word "unit" Your way is more concise. I like it.

@therealong 3 ай бұрын

Yes, this explains it quite clearly, while the formula doesn't, for those who aren't used to visualize formulas.

@cammobox 2 ай бұрын

I solved it in my head imagining the 4 equilateral triangles composing the big one you drew at first and the. Subtracted 3x the arc area between one of the three outer triangles and the edge of the circle from the center triangle’s area. After some algebra mine is essentially the same as yours, but a little more convoluted

@bijipeter1471 4 ай бұрын

Thank you, sir

@sagardeswal6465 4 ай бұрын

I solved it just like that in my head while driving

@WarpathMaus 3 ай бұрын

Started to explore new videos, this one is pretty cool

@Mathfan002 2 ай бұрын

Lets freaking goooo i have always made some kind of mistake but i finally got it 😊😊

@Krishna-sn3lj 4 ай бұрын

I aslo grade 10❤😅

@Asa-Ragnarok 4 ай бұрын

I did it by plotting the circles as functions (for example sqrt(1-(x+1)^2)) and then calculated the area between. My answer was 0,1612 which is pretty close but I felt so stupid after seeing the answer 😂😅

@_DD_15 4 ай бұрын

Glad I'm not the only one ha ha

@Grizzly01-vr4pn 4 ай бұрын

Is 1 order of magnitude out considered close these days?

@Asa-Ragnarok 4 ай бұрын

@@Grizzly01-vr4pn oh my mistake I meant 0,1612. I actually did the calculation one more time and got 0,1612544807038 which is only of by 0,0000000000702 👍🏼

@dragonballsupersaiyans4482 3 ай бұрын

Yea my 10th maths book has this ques. Feels nostalgic for me.

@SorLord 2 ай бұрын

I thought of the solution you gave immediately

@kaitlynethylia 4 ай бұрын

i didnt notice you said unit circles so i just did it for the general case and ended up with r²(4√3 - π) / 2.

@GlenMacDonald 4 ай бұрын

If r = 1, your result is 2√3 - π/2.

@CarlosArauz 4 ай бұрын

It's actually: (r²(2√3 - π)) / 2

@brianhull2407 4 ай бұрын

How did you get that 4? The area for an equilateral triangle is √3/4•s² (where _s_ is the length of any of its sides). For the general case for this scenario, s=2r, so we get √3/4•(2r)² = √3/4•4r² = r²√3. Subtract the 3•(π/6)•r² = (π/2)r² for the sectors, and we get r²√3 − (π/2)r² = (√3 − π/2)r² = (2√3 − π)r²/2. Somehow, you got double the actual area for the triangle circumscribed around the space, giving an extra factor of 2 for that particular term in the expression.

@kaitlynethylia 4 ай бұрын

@@brianhull2407 I don't have my notes anymore but as I don't know the formula for the area of an equilateral triangle by heart, I worked it out myself, and either accidentally missed out the /2 in sin(60°) = √3/2, or I forgot to divide by 2 when doing b × h / 2. either way it was some trivial mistake when working out the area of the triangle, my bad

@kaitlynethylia 4 ай бұрын

for the record, I didn't do the same method as in the video, I did a smaller triangle between the contact points of the triangles, and subtracting the areas of the chords

@Azuro123-pj7xj 4 ай бұрын

How do you know the pink area is 1/6 of the circle?

@dwayneorrison1560 4 ай бұрын

Because the angle of the pink wedge is 60° which is 1/6 of 360°

@afrika-karibianaestudionan4050 4 ай бұрын

The triangle is an equilateral triangle. This has equal angles of 60⁰. Thus, the pink areas are equal. This is r²•π•(60⁰/360⁰) = r²•π•(⅙) r = 1 Gives π/6

@neitoxotien2258 4 ай бұрын

Equilateral triangle has Interior angle of 60° each. A circle has 360° , then the sector is 60°/360° = 1/6 of the circle.

@Azuro123-pj7xj 4 ай бұрын

Thanks

@afrika-karibianaestudionan4050 4 ай бұрын

Greetings@@neitoxotien2258 , This part, i.e., "A circle has 360°" I am trying to interpret. How do You measure this 360° angle in this particular case?

@treush8471 3 ай бұрын

I like this elemental idea which is not really difficult to see when you have some expérience but can seeme randomly magic for a beginner

@needamuffin 3 ай бұрын

Divide the region into 3 sections using common tangents. Each of those sections becomes one area defined by a circumscribed hexagon around one circle. 3 of those means the region is equivalent to half the difference between the areas of the circumscribed hexagon and the circle. That makes it a little more complicated to find the hexagon's area considering the unit is altitudes, but a little trig would give the answer.

@mathematician_intermediate 4 ай бұрын

Can Integration be used?

@MathVisualProofs 4 ай бұрын

Probably possible :)

@martinkupke2099 4 ай бұрын

yes for sure.

@davidschneide5422 4 ай бұрын

Unforgettable strategy

@blue-cs3fk 2 ай бұрын

I'm proud to say that I was so focused on the math that I didn't notice the shape until I looked at the comments

@Anastagos 3 ай бұрын

r^2(√3-π/2) area of triangle 1/2 a*h a = 2r h -> sin(60°) = b/c h -> √3/2 = h/2r h = 2r*√3/2 h= r√3 area of triangle = 1/2 *2r * r√3 = r^2*√3 area we have to subtract = 3(πr^2)/6 -> (πr^2)/2 r^2*√3 - r^2*π/2 r^2(√3-π/2) I think it's correct, but I haven't watched the video yet

@Im_Vengeance_11 3 ай бұрын

"The area has a Nostalgic look to it... Oh my..😳!"

@EdKolis 3 ай бұрын

Great, now I know how to determine the number of square inches of fabric used in a swimsuit! 😂

@jefersonfaria7862 3 ай бұрын

Fantastic explanation. Using colours became easier to understand. 🎉

@golovkaanna8757 3 күн бұрын

Nobody told me how to solve this type of riddles. So i learned pattern by watching

@rajkote29 4 ай бұрын

Class 10 NCERT problem Indians like

@anujuniyal4590 3 ай бұрын

Nothing like that even i read external So don't be overconfident on a indian.

@101perspective 2 ай бұрын

I solved it by going to the end of the video. Work smarter, not harder:)

@thediamondleaf5111 2 ай бұрын

Yes, I once heard about this from my Japanese friend, he said people in his country call this area "Pantsu"

@shogrran Ай бұрын

This shape brought up memories in me

@FlawlessAnu 2 ай бұрын

By chemistry I used the void = this triangle called tetrahedral void which also measures as 2×no of atoms(N).

@jamshooter3797 3 ай бұрын

This area can be interpreted as the area of triangle with equal sides minus 3 areas of 1 / 6 circles (as the angle of triangle is 60° is common with angle of circle slice, and 60° / 360° = 1 / 6). Area of triangle: a * h / 2 = 2 * r * √3 * r / 2 = √3 * r^2; Area of 3 "1 / 6" circle slices: 3 * π * r^2 / 6 = π * r^2 / 2; Area of figure: √3 * r^2 - π * r^2 / 2 = r^2 (√3 - π / 2). Unit circles has given r value (r = 1), so answer is √3 - π / 2 ≈ 0,16

@pranavsahu487 2 ай бұрын

Literally i solved it exactly in this way ❤

@lordtomlluckrahthegreat9014 3 ай бұрын

I got to the pie pieces, but I didn't know how to calculate the area of those. It didn't occur to me that they are just portions of a circle; instead, I thought of slicing each piece with a straight line that goes from each point where the circle touches another circle, so I got an isosceles triangle (which I do know how to get the are of) and a little semi-circular shape (which I don't know how to calculate the area of). As always, the best answer is the simplest one.

@mohitkumawat9569 3 ай бұрын

Make equilateral traingle by joining three centre of radius 2R and - area of 3 sectors of radius R

@Convexdhal_23 2 ай бұрын

I solved it a different way: The triangle lengths connecting each center is 2r (2 x Radius) and because it is an equilateral triangle, each angle is 60°. Calculate the area of the triangle (let's call this T) and calculate the area of each sector within the triangle (what was coloured pink (let's call this S). Subtract S from T (T-S) and we get the area of that shape. EDIT: with substitution, the equation is... A=(√3/4 x 2r) - ((60° x r^2)/2 x 3)

@websnarf Ай бұрын

The area is that of the circumscribed hexagon minus the circle divided by two. I was too lazy to work out the actual numbers, since that was just busy work.

@YashPatel-xz5hk 3 ай бұрын

amazing

@SimonsAstronomy 3 ай бұрын

Intresting

@germanmustache 2 ай бұрын

I thought of weird integrals that I dont even remember anymore. Nice shi1t

@chandhrusrinivasan9595 3 ай бұрын

Simply wow😮

@vef444 3 ай бұрын

Draw squares around circles. Get are of squares - area of circles. From there you can get area if the needed part.

@8b25arjunarvind6 3 ай бұрын

Nice

@itsme.01 3 ай бұрын

(2√3 -π)R²/2, where R is the radius of circle.

@samnajian6873 Ай бұрын

Ez, draw a triangle from the middles of the circles. Calculate triangle area. the triangle's inner angles add up to 180 so there is 180 degrees of a circle in there which is half a circle. Subtract half a circle from the triangle area and you have your area