What’s the area?
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- čas přidán 19. 03. 2024
- This is a short, animated visual proof finding the area bounded between three mutually tangent unit circles.
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When you colored in the pink, it was over. Good stuff here
Why?
Gyattt
@@bebektoxic2136 search the female reproduction sistem
U guys are sick bro
Get medication
Integration has never failed me and this time was no exception 🗿
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.
based
what did you integrate over
@@rasuru_dev likely x, left to right area determination
@@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!
Solving this correctly was the confidence boost I need during my regular 3 am depressive episode.
I hope you realize how many people feel exactly the same way xd
hey, so totally out of context here but.. my usual depressive episodes are during 3 pm 😅 coincidence!!
@@LordNaver Everyone has their own individual rythm, so beautiful ☺️
@@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
@@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..
Straight away. Always good to get a mathematical pick me up when feeling inadequate.
Now i know how to calculate the area of a g-string thanks
Only the rear part
PFF FR
😂😂
@@linuxp00 that's the front part. The rear part is just a line.
And thank you!
That is called a G-STRING
It's not a man's g-string, it has no extra circle in the bottom part
😂😂😂
Not Nicola Tesla 💀
isn't calculus amazing!? 😍
They use this question to mock mathematics virgins in exams.
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 !
Yes, this is exactly how I solved it myself. The general formula would be r^2(sqrt(3) - π/2)
Hrm.
I think I do indeed need to investigate this curvy area more closely.
🤨
Yep. M-hm. Definitely. Well. I see it.
"Everything reminds me of her"😢
No more internet for you.
Aw man i thought you meant the pizza wedges lol
1. Triangle
2. Area of sector
3. Subtract area of triangle and area of sectors
4. Answer
area of sector how?
@@farziltheweebo4841 Radius=1 and Angle=60°
Not my sister: "Is that my underwear???"
💀💀
Bro 💀 "ram bhakt"
that's what I gonna sa-
wow, I actually solved it the EXACT same way! I'm so proud of myself!!!
it's the only way...
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.
Same goggypoggy.
@@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)
SAME BRO
The area we want in brackets is exactly the area we want.
What?
😏
😅
Only legends will understand 😏😂
Lol, thats what I was thinking. The brackets make up a part of that area too.
From commenting on funny and food videos..to commenting on The most educational one..ITS BEEN A LONG JOURNEY 😊❤
People see the area normally.
But I was looking in different mind😂😂😂
😂😏
It's not different mind it's dirty mind
Bro reminded of an underwear😂, isn't it?
Yes, apparently your mind is not that different
@@joshsingh1111 many people did - and some pointed out it gets better when you include the ( )
I drew hexagons around circles, radius equals apothem, area hexagon minus area circles, divide by three to get rid of the unnecessary bits.
Wht made you think of hexagons at all though or remember an apothem even? Thanks for sharing.
@@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.
@@mennolente4807 That's pretty dope!
nice!
Cool
This solution is so satisfying
The area will satisfy more💀
@@mr.unknown1070💀👍
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
Feeling like a mathematician, really proud of myself😂💕
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
I was on the math team a lifetime ago in high school. Little things like this let me relive the good ol’ days lol.
It is because it's class 10 level question
Of circles chapter and it is level 5 question😅
Yet you could not solve it. Don't be overconfident.
@@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
@@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?
"can you determine the area of enclosed cheeks" was what i heard 😂
Area of the thong
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
Area of triangle = ½•b•c•sin A = ½•2²sin 60° = √3
Area of sectors = ½•r²•theta = ½•1²•π = π/2
(combining the sectors you got a semicircle with angle 180° or π rad)
Area of concerned region = √3 - π/2
Me when I see the video in my recommendations:
*_"Oh math! So cool, OH A BIKINI! SO MORE COOL!"_*
Damn u got this
😂😂😂
stay strong not HARD
@@udayraj6976 stay hard and strong!
@@dyltanwhy do we share the same brain cells 😂😂
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)]
Exactly what I thought!
How do you know sectors will form 60°
@@tariqwaheed7429 He literally said "equilateral triangle". Anyway, it can be inferred cuz all 3 circles have same diameter.
that is exactly what i was thinking in my head
@@tariqwaheed7429same distance bw center of two each circles
First we will calculate the area of forming triangle (center to center)
That will be 1/2*r√3*2r=. r²√3
(Assuming the radius to be r)
Then we will find the area of those parts in the circle
πr² ----> 360°
πr²/360-----> 1°
πr²(theta)/360-----> (theta)°
Hence the area would be πr²(theta)/360
Now we will add those areas of the circle part and get
πr²(theta)/120
(If you want to find radius just put the value of [(theta)° ] to 180)
Now we will subtract hence
Answer:- r²√3 - πr²(theta)/120
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).
I solved it in a much more complicated way. Your solution is so elegant!
Show us please please
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
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 😊
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?
@@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.
(Area of the triangle) -- (Area of the sectors)
area of triangle - are of unit circle
the sectors make a full circle and what you have written will give answer in -ve
Wrong, you need to switch those terms.
@@Ruktiet true, but you can also assume negative areas, like distances, as possitive and kill their minus sign
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)
@@Ruktiet it's correct wdym
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)
Area of triangle - 3×area of each sector
area of triangle = area of equilateral triangle of side 2×r , where r = radius of each unit circle = 1 unit
i.e. area of (this equilateral) triangle = √3/4×side squared = √3/4 ×2×2 = √3
subscribing angle of each sector to the respective centre of circle is 60°(can be proved by similarity of triangles)
So area of each sector = 60°/360° × area of each circle = 1/6 × π × r^2 = π/6 [ as r = 1 u ]
Thus the area of the highlight region is
√3-3×π/6 = 0.161 sq units (appx)
√3 - π/2
Funny. I got the same thing about 30 seconds later.
Yes 🎉
You all assumed that r is 1 except me ?😅
@@JoJo-jj6iddw i didnt either
r^(2)[3^(1/2)-(π/2)] looks less neat buts much more useful
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
wtf
@@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
@@lillii9119I would definitely buy you a beer if I bump into you in real life
Make it polar and integrate from 0 to 2π/3. Triple that for the area.
unless u do some quirky math, that s not a simple integral
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
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.
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.
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!
It might be a more difficult way, but I’m thinking that you could draw an equilateral triangle using the radius of each circle, and find the area of that. Then, find the angular area of each circle (60 degrees for all 3, again assuming the three circles are the same side) and subtract that. That should be the answer. So, if I’m not high, it should look like this:
The radius of all three circles will be represented with “r”
A triangle ABC where A=B=C and A = 2r.
The area of triangle ABC should be 0.5(base)(height). Since the triangle is equilateral, both the base and height will also be 2r, and the base will be, let’s say, side B. So, in this case, the area of this triangle will be:
0.5(B)(B) = B u^2
Therefore, the area of this shape should be the area equal to B units squared minus the angular area of each circle. 60/360 is about .167, so the area of this portion of one circle will be about .167*[area of the whole circle]. Since A= pi(r^2):
A(partial) = (0.167)(pi)(r^2)
With all that said, the total area will be the area equal to value B minus three times this partial area of one of the circles:
Final solution: B - (0.167)(pi)(r^2)
Edit: Ok, I see now that they’re unit circles, so in that case, side B is 2, so the area B is 2 units squared. The radius is 1, so the final solution is:
2 - (0.167)(pi)(1^2) = 2 - 0.5236 = 1.476
Ok so apparently this is completely wrong…oh well never mind 😂🤦🏻♂️🤦🏻♂️🤡
Yes. Yes, I can determine the area. Triangle minus one half the circle area (3 x 60deg = 180 deg). I leave the details to others. R^2 ( 2 sqrt(3) - pi)
Это решение конкретное решение для кругов радиусом 1. Общая формула этой фигуры будет выглядеть так:
√3r² - (πr²)/2
Even if Project Borealis doesn't get released, at least we got this beautiful soundtrack.
Make equilateral traingle by joining three centre of radius 2R and - area of 3 sectors of radius R
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
Good morning everyone! I hope everyone enjoys their daily dose of mathematics!
Did it at school MANY years ago. Had completely forgotten how, though! 😂
I try to solve it just by mental calculation, and the method I used was the area of the triangle - the area of the 3 sector combined. Since the triangle is equilateral each of the angles are the same amplitude which is 60° so the sum of the three sectors is a sector created by an angle of 180° so his area is half the area of the circumference. For the triangle instead I used the standard formula, paying attention to the fact that in an equilateral triangle the height is the cosine of 30° times the edge which itself is 2 times the radius of one circle. The area of the triangle is therefore 2R*(√3/2)2R all divede by 2 minus the area of the semicircle which is (π\2)R². The answer simplified is (2√3 - π)/2 times R².
Easy. Paused.
Form a triangle with the 3 center points. Since all the circles have the same radius, we know the triangle is an equilateral triangle with sides equal to 2r.
Find the area of this triangle (At). If we make a right triangle to find the height, we get a 30 60 right triangle so the height is the radius (r) times rad 3.
At = r^2*rad(3)
Now find the area of all three sectors
As = 60/360 * r^2 * pi * 3
As = (r^2*pi)/2
The area between the three tangent circles is A = At - As
A = r^2*rad(3) - (r^2 * pi)/2
Edit: he said unit circled not equal circles. So I guess r just = 1
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 😆
The first thing I thought of was an equilateral triangle.
Then: "Okay, now what do I do with this?" 😅
Draw a triangle in the middle. Each triangle side is 2r, each angle is 60deg. Calculate triangle area, and remove the areas of the sectors within it. Done
I actually got this one, truly amazing what studying can do
I solved it the EXACT same way, except took radius as R instead of 1.
He said unit circle so R = 1
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.
I got sqrt(3)-pi/2 in my head before watching, gonna check now lol, this seems very easy
how?
@@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.
I did it mostly the same way except I generalized it to circles with radius r, and used a bit more of a roundabout way to calculate the area:
A(blue) = A(triangle) - A(sectors)
A(triangle) = bh/2 = 2rh/2 = rh
Break the equilateral triangle into two 30-60-90 triangles and use trigonometry to find the height
h = r*tan60° = sqrt3 * r
A(triangle) = sqrt3 * r²
The sectors add up to a semicircle, which has half the area of a circle
.•.
A (blue) = sqrt3 * r² - π/2 * r²
= r²(sqrt3 - π/2) ≈ 0.161r²
I did the same thing, just complicated; taking the bigger triangle.
You have to prove that the intersections of circles lies on the triangle first, although it’s a rather easy one.
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.
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
Lets freaking goooo i have always made some kind of mistake but i finally got it 😊😊
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
Yeah I did it exactly as you did. Simple enough
Thank you, sir
at the pause. It is easy. sqrt(1.5*0.5*0.5*0.5)-pi*r*r/2.
Area of triangle minus area of three sectors po 60°.
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
I solved it just like that in my head while driving
"The area has a Nostalgic look to it... Oh my..😳!"
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
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
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.
If r = 1, your result is 2√3 - π/2.
It's actually: (r²(2√3 - π)) / 2
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.
@@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
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
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 😂😅
Glad I'm not the only one ha ha
Is 1 order of magnitude out considered close these days?
@@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 👍🏼
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)
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
How do you know the pink area is 1/6 of the circle?
Because the angle of the pink wedge is 60° which is 1/6 of 360°
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
Equilateral triangle has Interior angle of 60° each. A circle has 360° , then the sector is 60°/360° = 1/6 of the circle.
Thanks
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?
Can Integration be used?
Probably possible :)
yes for sure.
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.
First time I've been able to solve/figure out the process in one of these videos . Nice!
I aslo grade 10❤😅
Unforgettable strategy
wedge shape area is pi/3 square units... as isosceles triangle arms has 60 degrees of angle among em. so area of sector = (angle in radians) r^2 = pi/3 1*1= pi/3
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.
Yes, this explains it quite clearly, while the formula doesn't, for those who aren't used to visualize formulas.
Great, now I know how to determine the number of square inches of fabric used in a swimsuit! 😂
Class 10 NCERT problem
Indians like
Nothing like that even i read external
So don't be overconfident on a indian.
something like it:
x = middle section
a = diameter of sphere or length of triangle side
x=¼(a²•√3) - ½(π•(½a)²)
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
Yes, I once heard about this from my Japanese friend, he said people in his country call this area "Pantsu"
Use the circles radius to make a square, minus the spaqurws space by the circles space to be left with whats left of the cube. Divide that by four then time the answer by 3 (full disclaimer, i have absolutely no idea if this is correct, just my idea of how it could work)
draw out 3 triangles that meet at the center and each cover one triangle. Remove the area of the circle from the area of the triangle. Divide by 3 for the one corner, multiply by 3 for the 3 circles.
im guessing its gonna be a 60 degree even triangle (whatever its called in english) with a side length of 2 radiuses, minus 3 times the piece of the circle of the triangles angle (60). so thats 180/360.
so in total thats an even triangle a=2r minus half of the area of one circle
I solved it by going to the end of the video. Work smarter, not harder:)
Solving it for any case, A = (sqrt(3) - pi/2) * R^2, being R the radius of each of the three circles
I stumbled on this problem sometime ago while playing around with circles. I solved it using the same method
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.
Draw the three circles as shown on a piece of paper with known thickness. (Can be checked with a square of known area) Cut out the center shape, then weight it. Divide the mass by the density of the paper and its thickness. Will give you an answer that is a good enough estimate.
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.
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.
A question similar to this is in the Class 10 NCERT Mathematics book
Area of triangle (s=d) minus area of 3sectors of 60⁰ of (s=r)
Another way---
Area of triangle= root3/4× (2a)²
Area of 3 sectors= 3×60/360× pi×a²
=> pi/2
Area of shaded region= root3- pi/2
Well if you take the end of the shape and draw straight lines connecting the corners then take the area of the triangle - the area of the arc lenghth ×3
Not sure about area of the shape, but I REALLY loved the SHAPE or the area