Its crazy watching these as someone who so far only knows middle school math and seeing that I *could* actually figure it out with my current information
It's why I like to make the distinction between complex and complicated/hard. Most problems are complex, in that they seem daunting but are trivially decomposable like this one. Some problems however are daunting because they are actually hard and can't be decomposed, at least not without great effort. In that case clever simplifications grounded on good assumptions usually go a long way into turning the problem into a complex one.
what a crazy solution to a problem like this. as someone who doesn't know the formulas to solve these off the top of my head, i see things like this at work all the time (im a contractor and volume and area come up constantly) and i always just end up estimating. but to be able to crank out a real solution would be so satisfying
I am a student in the second year of secondary school, and I solved this problem correctly, but in a different way, and it took me an hour and 45 minutes of my time.🥲
I did it a bit simpler: cut diagonally the shape is made from two white circle segments cut from two red circle segments. The two radius 8 segments are equal, so we can fit them together and the total area is a radius 12 segment, minus a radius 4 segment, or (36pi-72)-(4pi-8)=32pi-64.
You seem pretty good at geometry. I wanted to ask for a few tips. I’m in 7th grade, and I’m taking a geometry class. I’m struggling a lot and so is everyone else. Our last test was somewhat in the middle of difficulty, and it was a derivation of Heron’s formula with no prep, and a few problems from the harder half of AMC10. I just don’t know how I can minimize time wasted. I don’t even erase anymore.
Man, your channel shows exactly what all my math teacher once told me: math is not hard, in fact, it is easy, you just need to decompose the complex processes into simpler ones until you solve everything Really nice content, keep on the great work!
You can also cut the claw along the diagonal of the bottom left square. Then shift the smaller part to the top right corner of the box and turn it by 180 degrees. Then x is just (1-1/3^2)*A, where A is given by the quarter of the area of a disc with radius 12 minus the area of a square with diagonal 24.
If you add and subtract the obvious quarter circles, and use a pen to keep track of double counting for each little region, you find that you simply overcount by exactly 4 of the squares :) Quarter circles: 36pi + 16pi - 16pi - 4pi = 32pi Subtract the squares: 32pi - 64
You can also factor out the 16 and radii, seeing that each quarter has a same-size right triangle subtracted, and the r=2 quarters cancel each other out, thus may be omitted. Then you simply evaluate 16*(3*3-1*1)*(π/4-1/2) = 128*(π/4-1/2) = 32π-64, which is the correct answer I got just by looking at the figure.
This is truly beautiful. Like these kinds of problems, you mostly just need to break it up into nice pieces and then it all comes together (pun intended) in such a beautiful way. Love your explanation too.
I watched the other video in the description and I have another solution as well: 1/ calculate the A of the right claw by substracting the larger 1/4 circle (r=12) from the smaller 1/4 circle (r=8) and 2 squares on the left. 2/ calculate the A of the left claw by substracting the larger 1/4 circle (r=8) from the smaller 1/4 circle (r=4) and 1 square in the bottom mid. 3/ Add the A of the 2 claw and substract 1 extra left corner square. This is fun! Thanks for the vid!
This problem is extremely easy with some basic calculus. Calculating area by integration is one of the most common problems in early calculus courses. For this problem all you need besides integration is the equation for a circle.
I mean, that's essentially what he's doing, he's just using the final value of the integral of the equation for a circle from 0 to r, that being pi*r²/4. And I'm not sure this would actually be easier by explicitly defining a bunch of piecewise functions and doing one big integration; that feels like introducing unnecessary complication when you already know the formula for the area under each individual quarter-circle in the grid.
@@Idran They both accomplish the same thing in nearly the same way. Which you find easier probably depends on what you are more familiar with. Fewer fundamental equations are required with calculus, but as you said, more piecewise functions and you need to know how to do basic integration.
@@DataScienceDIYThis is definitely not as easy as you're describing it. If you'd actually tried to do it with integrals you'd run into trouble. This method is much simpler
tl;dr: You can shift the white area above into four white squares and a white quarter circle, turning this problem into something elementary. It can be done even quicker and in a much simpler way (even simpler than the trick in the other video). Since you have two congruent quarter circles, a lot of symmetry can be used. Start with the area of the biggest quarter circle π*6^2, the white area outside it is not necessary. If you look at the two quarter circles with radius 8 you can actually find two full white squares with them, fit the white area in cell 3 and 6 above the red into the white area above the red in cell 4 and 7. Together with the two cells in the top row you have 4 white squares, so you can subtract 4*4^2 from the total. Now the very middle cell is left. In that cell, if you see that the red part in the bottom right has the same area as the white in the top left (again due to symmetry of the two congruent circles) then it suffices to just subtract the area of the smallest quarter circle for the solution. So π*6^2-8^2-π*2^2= 32π - 64. I think this would be the method with the least amount of calculation.
The formula I used involved cutting x down the two bottom left corners. Then if I match the r=8 circle edges together, I get that the area x is: ( a quarter-circle r=12 minus a triangle b=12 h=12 ) minus ( a quarter-circle r=4 minus a triangle b=4 h=4 ) Since it’s trivial that the two parts are similar, I can just simplify it to (3×3 - 1×1) = 8x the hole. 8 ( a quarter-circle r=4 minus a triangle b=4 h=4 ) = 8 [ (π×4×4/4) - (4×4/2) ] = 8 [ (4π) - (8) ] = 32π - 64 (units²)
Thanks so much for this stepped solution. I struggled with algebra and have not done it for nearly 20 years yet watching this really made my memory trigger with how do all that - i understood it and feel like i could apply those principles in other circumstances.
If you cut the area along the diagonal from the top right to the bottom left, and rotate that piece on the left around the center of the large square by 180 degree, you can simply your calculation. The area would be a large circular segment minus a small circular segment, which is 36pi-72-(4pi-8)
What a great video! I had no clue how the process for those inner sections would be calculated when I started and was shocked at how intuitive it was that making them quarter circles and removing the triangle came right as you started saying the solution! Very well structured and paced!
the curves must be circular if they have constant curvature and intersect the squares at exactly their vertices. Neither of which is specified in the problem. So you're right it's a poorly conditioned problem.
Then you estimate and get as close as is reasonable. Or, you use calculus. You measure the rate of curvature for each section of the curve, represent that with a function, and integrate over the range.
I went by a different route (also I used X for the length instead of 4). I cut the red shape along the diagonal, which meant: Large Shape area = Quarter circle of 3X radius - Quarter circle of 2X radius - 2 squares of length X - 1/2 squares of length X = (Pi*(3X)^2)/4 - (Pi*(2X)^2)/4 - 2(X^2) - (X^2)/2 Small Shape area = Quarter circle of 2X radius - Quarter circle of X radius - square of length X - 1/2 squares of length X = (Pi*(2X)^2)/4 - (Pi*X^2)/4 - (X^2) - (X^2)/2 Add the two together: Total_area = (Pi*(3X)^2)/4 - (Pi*(2X)^2)/4 - 2(X^2) - (X^2)/2 + (Pi*(2X)^2)/4 - (Pi*X^2)/4 - (X^2) - (X^2)/2 Multiply everything by 4 to get rid of the divisors: 4 * Total_area = Pi*(3X)^2 - Pi*(2X)^2 - 8(X^2) - 2(X^2) + Pi*(2X)^2 - Pi(X^2) - 4(X^2) - 2(X^2) Open the squared parentheses 4 * Total_area = 9Pi(X^2) - 4Pi(X^2) - 8(X^2) - 2(X^2) + 4Pi(X^2) - Pi(X^2) - 4(X^2) - 2(X^2) Add up 4 * Total_area = 8Pi(X^2) - 16(X^2) 4 * Total_area = (8Pi-16) X^2 sq. units Divide both sides by 4 Total_area = (2Pi-4) X^2 sq. units Same solution, just plug in whatever value you want for X.
I agree, having to calculate only 4 areas (of the same shape) is much faster (and better) solution. Saving x as symbol to plug it in later is cherry on top
1:33 How do you know the arcs are a quarter circle? Thats an assumption that the drawing doesnt really confirm. It could be a slightly asymptotic line, not a radial.
This is a pretty complex solution. I just thought of moving the smaller offcut by translation and rotation inside the big offcut, making it a segment - a smaller segment. in the end you get smth like (36π - 72) - (4π - 8) getting 32π - 64. How, exciting.
I used to be afraid of these kinds of problems, until I learned double integrals. Now I can probably solve this in somewhere around 15 mins. Your solution, which only includes basic mathematics, and takes no more than 5 mins, is beautiful
Kids, this is exactly what most of the maths is all about. It’s not about remembering the formula for area of cylinder or a scalene triangle, it’s about how do you approach a solution given the formulae. Anyone who doubts their education system should watch this.
This is a crazy problem to solve. I thought by clicking on the video I was going to see this being solved with calculus. However, your problem solving solving method stunned me as you were able to make complete sense out such a odd (and complex) question. This is definitely one of the coolest videos I’ve seen lately.
@@TheSpacePlaceYT That kind of information always has to be given in these kinds of problems. The hardest part of the problem shouldn't be you sitting there wondering if a shape is actually what it looks like. So if you find yourself having to think about that, always go back and check to see if it was already clarified.
The real answer is that insufficient information is provided to answer. One must assume right angles and equidistance of the interior line segments. It may seem overly analytical, but real world applications do not tolerate such assumptions and mathematics should not either. For example, if you cut carpet for a "square" room, you may find one side too long and the other too short.
The assumption that the two areas are segments of a circle could be wrong. He's assuming by inspection that the curves have an eccentricity of zero and are, hence, part of the circumference of circles, but this might not be the case. Other types of curves can also connect the two endpoints. So, interesting solution but based on what could be two faulty assumptions.
@@samsowdenJudging by how many people didn't notice it it probably should have been stated out loud at the beginning of the video. He's simplofying the problems too much for younger viewers and missing the nuance
Idk why my mind went straight to putting this image on a graph, splitting the curves up into different functions, and finding the area under the curve with integrals and adding them together.
@AjayKumar-mg3xc Well, it's a 12 by 12 grid that you can put in the first quadrant. You can separate and label each line as a quadratic. For example, that curve that goes from the bottom left corner to the upper right corner can be labeled y=(x^2)/12.
I tried to to do this but trust me.. it is NOT easy, you will have to find a lot of intersection points and figure what all areas to subtract. Wouldn't recommend.
Recently found your channel and i am absolutely in love with your content. As someone who always used to fear math, i recently began on a journey to like math and i cannot tell you how awesome your content has been in guiding me along that journey. As of this peoblem i came close but i coulsnt find out rhe area pf the segment because i didn't visualise it in that way
Nice! I managed to do it, similar basic idea of subtracting a group of smaller shapes from a square, but didn't choose such clever ones, hence resorted to using an integral.
Yeayyy got it right on my first try! Your questions remind me of the math olympiads I took part in when I was in elementary school anywayyy. More challenging questions please, I'm so curious!
1:11 I think these are called spandrels. At least that's what we called them when dealing with the centroid and the rational inertia about these shapes.
Cool solution, but if you split the shape down the diagonal, you can solve it much easier, because then you can subtract whole quarter circles (plus a small rectangle+triangle shape) from the two larger quarter circles that make up to two arcs we see in the shaded area.
It gets even better when you see that the two intermediate quarter circles have the same area, so in the end you just need to subtract the tiniest segment of a circle from the biggest one
@LukesVGArea it gets even even better if you print out the problem, cut out the area you're trying to find and then weigh the section and compare it to the total weight of the entire square. You have now calculated area as a measure of weight and then take that ratio and match it against the total area of the shape and itl give you your total area
I don’t even watch math videos but for some reason this popped up. I watched the whole thing through, super entertaining which I would have never guessed beforehand.
I just love when you see a strange shape in nature, abstract as it may be. Encase it in a symmetrical construction, and calculate the difference. Symmetry, what a wonderful word!
im a decade past recalling exact formulas for things like area of a quarter circle or circle segment, so the explicit numbers didn't come to me, but i still got some decent problem solving, worked out the clever shortcut you mentioned from the other guys vid on my own (asking myself "why wouldn't you simplify and do it like this" and felt very vindicated when you pointed to that video and the guy presented the same alternate solution) super neat stuff!
@ethanjsegatI watched the other video mentioned and the shape is made using explicitly mentioned quarter circles so there’s no assumption, but if I was just given the shape like that, I wouldn’t be comfortable just assuming they’re quarter circles
If they weren't exactly quarter circles then yeah we'd be effed. Unless they gave enough information that you could work out the function of the curve in which case I think integrating to find the area under the curve would be correct.
Good thing is that in real life, you can estimate and be within a margin of error. We're conditioned so early on that math has to have one singular answer but Calculus teaches you that there are multiple approaches and that you can always be within an error of margin. Dividing a line into infinitely many pieces to guess where it most likely converges is peak guessing game and I love it
A lazy solution: The figure can be cut in two parts with a SW-NE diagonal. SE side: From the disk segment whose chord is the diagonal of the 12 units square, we remove the disk segment whose chord is the diagonal of the 8 units square. NW side: From the disk segment whose chord is the diagonal of the 8 units square, we remove the disk segment whose chord is the diagonal of the 4 units square. The area is: A = [Area of 1/4 of the circle of radius 12 - Area of the half of 12 units square] - [Area of 1/4 of the circle of radius 8 - Area of the half of 8 units square] + [Area of 1/4 of the circle of radius 8 - Area of the half of 8 units square] - [Area of 1/4 of the circle of radius 4 - Area of the half of 4 units square] We simplify (lines 2 and 3 cancel each other): A = [Area of 1/4 of the circle of radius 12 - Area of the half of 12 units square] - [Area of 1/4 of the circle of radius 4 - Area of the half of 4 units square] That's to say: A = (1/4.π.12² - 1/2.12²) - (1/4.π.4² - 1/2.4²) A = 36.π - 72 - 4.π + 8 A = 32.π - 64
I’m mad AF, because there’s no part of the problem telling us that the curve is a quarter circle! Yes, it touches the two corners, but there nothing telling us that each point along its curve is equidistant from its center point. It’s spent a good ten minutes trying to figure out the curvature of the shape.
You can assume that the radius of the circle is 8... and that's how he knows that it's a quarter circle. It's will help you calculate the area within that region
There's a most elegant solution of only sums and subtractions of segments of quarter circle if you look at this problem diagonally (literally) Hence only one area formula is necessary, applied to 3 different radii
I dont know why im addicted to watching these. Like, I know the possible ways of figuring it out, but I dont know any of the formulas. Its like watching a friend play a PlayStation game and you know what to do, but you dont know how to handle the controls hahaha
A polar planimeter! I have a K&E device - of course I bought mine in 1967 as an engineering student. But, today there are some interesting computer programs to integrate the thing! Love the computer!
I got the same result but with a different way of calculating. I calculated the area of the big quarter circle and subtracted the triangle part and the round part of the smaller quarter circle and so on.
Did it in my head, with a simpler decomposition: X is the unknown area, A(r) the area of a quarter disk (=πr²/4) C the area of one cube, You got : X = A(3r) - A(r) - 4C I took r=1 which gives: X = 9π/4 - π/4 - 4 = 2(π - 2) Sawing that the problem gives r=4, multiply the area by the wanted scale (r² = 16): X = 32(π -2) PS : To get to the formula for X I followed the shape from the larger arc then add/remove disks and cubes to fit the shape. Before last simplification, you get : X = A(3) - A(2) - 2C + A(2) - A(1) - C - C
My only issue is that by default you assume the curves are circular if that werent the case itd probably be unsolvable. Either way very impressive its a neat seeing you solve these
if it was unsolvable on purpose, then what would be the point of it? from where I see it, this problem tries to teach you how to approach a problem, and how much easier it is when you consider other alternatives. when the entire point of problems and whatnot is to TEACH, then there is no reason for it to be unsolvable
You’re making a few assumptions here without having any evidence to support your assumptions, such as those circles, being quarter circles, and the value that they take up.
I cut the bottom-left square diagonally, and flipped the left piece 180° so that the two 2-unit quarter circles coincide. Made for a much easier shape to solve!
If you have the eye to identify the "blue" segments I don't know how you could not realize that the red shape is just two segments covered by the two blue segments. In other words the red area is: (big red segment)-(small blue segment) +(small red segment)-(smallest blue segment) The diagonal line helps you see where the segments are and the grid lines help you figure out each radius
1:43 i don't know if i am correct' But according to me if we see precisely those are not quater circles as they are ending before the vertices of the squares
It's funny how much the "made with quarter circles" knocks the difficulty of this problem way down. Funnily enough, I did it a completely different way: I started with the full square, and imagined either cutting the square off or adding to it, keeping track of what was added/subtracted. I got to the correct answer, and I only needed to use the formulas for circles and squares, but I get the feeling my idea was a tad more complex than this. Definently more error-prone.
Based on just that picture, there's no information whatsoever if these are circles and where are their ending points, thats why this method is just based on your interpretation, but has bo approval to this. Tbh it can't be solved due to the lack of information.
I find that for shapes like this, it's usually just easier to write the curves as semicricle equations (y = sqrt(radius^2 - x^2)) and then use integrals to find the volumes of solids. however, the way that you did it is super cool!
Here's my solution: 1. Draw a diagonal from top right to bottom left. Now you are with 3 types of 4 pieces (minor segments of a circle). 1 smaller (white), 2 medium (1 red and 1 white) and 1 bigger (red). 2. Now, by observation, we see that area of red region is: (Bigger - Medium) + (Medium - Smaller) => Area = Bigger - Smaller 3. Area of the shape = Area of quadrant - Area of right triangle = (πr²/4) - (1/2)(base)(height) 4. Area of red region = Big - Small = [π(12)²/4] - (1/2)(12)(12) + [π(4)²] - (1/2)(4)(4) = 32π - 64
It's easy and even possible just mentally if we divide the red figure by the diagonal ／ and rotate one of the parts so that the arcs coincide. And now we find out easy that the area of the red figure equals to ¼πR²-½R²-(¼πr²-½r²) where r equals to 4 and R equals to 3r. So... it equals to ((9/4)π-9/2-¼π+½)r² = (2π-4)r² = (2π-4)*16 = 32π-64 Answer: 32π-64 sq units
TBH this solution is over-complicated the problem by requiring knowledge of the area of a "segment of a circle". I did it, more intuitively IMO, by adding and subtracting quarters of circles. I'll try to explain in a comment: - I split this drawing along the bottom left -> top right diagonal, with "part (a)" of the red shape on the bottom right and "part (b)" on the top left. I'll explain how I calculated "part (a)", with the same methodology being applied for "part (b)" - Note: I'll use "y" to indicate the length of a small square, where y = 4 - "Part (a)" is comprised of four sections: one additive, and three subtractive. -- (1) take a full quarter circle of 3y -- visualize the center of that circle at the top left point of the diagram: coordinate (0, 3y), or (0, 12): + π*(3y)^2 / 4 -- (2) remove a quarter circle of 2y -- visualize the center of that circle at (y, 3y) or (4, 12): - π*(2y)^2 / 4 -- (3) remove two full squares at the top left, between coordinates (0,y) and (y, 3y): - 2y^2 -- (4) remove half of the bottom left square, along the diagonal: - y^2 / 2 This gives you: π*(3y)^2 / 4 - π*(2y)^2 / 4 - 2y^2 - y^2 / 2 If you repeat for the "part (b)", you get: π*(2y)^2 / 4 - π*(y)^2 / 4 - y^2 - y^2 / 2 If you add them all together and simplify, you get: r^2 * (2π - 4) -- which is the same as the solution offered in this video. All with even more basic math
An even faster way to solve this, is to rearrange the picture by nestling the little red horn into the curve of the bigger red horn. that way you notice that the shape is actually one big segment with r =3*4=12 minus one small segment with r=4. this gives us a pretty short equation of x=1/4*π(12)^2-1/2*(12)^2-(1/4*π4^2-1/2*4^2)=144π/4-72-16π/4+8=128π/4-64=32π-64 which is the result you ended up with. When there are multiple ways to cut up a shape like this the first step should be to find the best way to describe your area with the least amount of composited shapes in order to avoid doing as much work as possible.
Make a copy of the shape, and superimpose it on the shape, however, rotated 180°. Then the combined area is a large lense, minus a small lense. The large lense covers 1/2*pi*3^2 - 3^2 squares. The large lense covers 1/2*pi*1^2 - 1^2 squares. Hence the combined shape covers 4*pi - 8 squares. However, since each square has area 16, and we have to halve the solution, we get 32*pi - 64
I love problems like this because it demonstrates how to take what appears to be a complex/difficult problem, and break it down into simple steps.
The hard part is to demonstrate that the simples steps lead to the most complex problem (or the other way around)
Its crazy watching these as someone who so far only knows middle school math and seeing that I *could* actually figure it out with my current information
It's why I like to make the distinction between complex and complicated/hard. Most problems are complex, in that they seem daunting but are trivially decomposable like this one. Some problems however are daunting because they are actually hard and can't be decomposed, at least not without great effort. In that case clever simplifications grounded on good assumptions usually go a long way into turning the problem into a complex one.
if you call this difficult its cos you dont see the smaller steps to begin with
honestly, the initial problem looks horribly hard. But the solution was actually easy haha. Thanks for the solution :)
Yeah, the hardest part was figuring out the method to solve it.
You just figured out *everything in life*, congratulations.
eh, everything seems easy in retrospect
@@dani.munoz.a23 It may be easy, but only when you know how to do it
This video is kinda im sorry to say bullshit because who in their right mind would assume those have to be arcs of circles?
what a crazy solution to a problem like this. as someone who doesn't know the formulas to solve these off the top of my head, i see things like this at work all the time (im a contractor and volume and area come up constantly) and i always just end up estimating. but to be able to crank out a real solution would be so satisfying
Same, except I’m a rocket scientist
@@losthalo428 I find it's always best to just guess when you're making orbital adjustments.
What kind of contractor are you, if you're just estimating?
@@gummel82 kinda a pain to count allat
I don't think it's safe for you to estimate.....@@losthalo428
Usually, students panic on seeing these types of figures and give up.
Thanks for simplifying the seemingly complex problem!
I am a student in the second year of secondary school, and I solved this problem correctly, but in a different way, and it took me an hour and 45 minutes of my time.🥲
And then the teacher give you a zero because "how can you be sure the curves come from a perfect circle?"
and -10 for not writing "cm" even if your country doesn't use cm, and it said nothing about that
It tells you in the question!
@@Sidowse do you really think I'm that dumb ?
Cause I am !
I did it a bit simpler: cut diagonally the shape is made from two white circle segments cut from two red circle segments. The two radius 8 segments are equal, so we can fit them together and the total area is a radius 12 segment, minus a radius 4 segment, or (36pi-72)-(4pi-8)=32pi-64.
Omg i actually understood this. I was sitting reading this shit for like 5 minutes straight and finally understood it
Was looking for some else that did it by arc segments 👍
That's what I did as well.
I simplified it more by drawing the shape in CAD/revit and letting the program do its work. 😭
You seem pretty good at geometry. I wanted to ask for a few tips. I’m in 7th grade, and I’m taking a geometry class. I’m struggling a lot and so is everyone else. Our last test was somewhat in the middle of difficulty, and it was a derivation of Heron’s formula with no prep, and a few problems from the harder half of AMC10. I just don’t know how I can minimize time wasted. I don’t even erase anymore.
Man, your channel shows exactly what all my math teacher once told me: math is not hard, in fact, it is easy, you just need to decompose the complex processes into simpler ones until you solve everything
Really nice content, keep on the great work!
🤓
Blud thinks he's dunny
Funny*
Complexity is nothing but compound simplicity :)
I get what they meant but that's literally how you accomplish any hard thing; by breaking the problem apart into smaller more manageable problems.
Why am I sick and watching math at 9am? I’m almost 30.
it's interesting
I'm 24 and watching it at 9am as well lol
I swear to god I'm also feeling feverish rn, and it's 12 am😂
High five dude
bro i got a cold rn at 12 am aswell
Fever sorta, 4,20 AM , but im 15 lol 😂
You can also cut the claw along the diagonal of the bottom left square. Then shift the smaller part to the top right corner of the box and turn it by 180 degrees. Then x is just (1-1/3^2)*A, where A is given by the quarter of the area of a disc with radius 12 minus the area of a square with diagonal 24.
Silksong when?
fuck sent my sides to orbit with this
I feel like a conspiracy theorist saying this but I think we find out at the game awards
If you add and subtract the obvious quarter circles, and use a pen to keep track of double counting for each little region, you find that you simply overcount by exactly 4 of the squares :)
Quarter circles: 36pi + 16pi - 16pi - 4pi = 32pi
Subtract the squares: 32pi - 64
Life hack 😂
this is the best solution
Easy af
You can also factor out the 16 and radii, seeing that each quarter has a same-size right triangle subtracted, and the r=2 quarters cancel each other out, thus may be omitted.
Then you simply evaluate 16*(3*3-1*1)*(π/4-1/2) = 128*(π/4-1/2) = 32π-64, which is the correct answer I got just by looking at the figure.
How do you decide on what quarter circles to add and subtract?
The shape had me expecting a stealth Silksong announcement
shaw!
🤡
This is truly beautiful. Like these kinds of problems, you mostly just need to break it up into nice pieces and then it all comes together (pun intended) in such a beautiful way. Love your explanation too.
I watched the other video in the description and I have another solution as well: 1/ calculate the A of the right claw by substracting the larger 1/4 circle (r=12) from the smaller 1/4 circle (r=8) and 2 squares on the left. 2/ calculate the A of the left claw by substracting the larger 1/4 circle (r=8) from the smaller 1/4 circle (r=4) and 1 square in the bottom mid. 3/ Add the A of the 2 claw and substract 1 extra left corner square. This is fun! Thanks for the vid!
This problem is extremely easy with some basic calculus. Calculating area by integration is one of the most common problems in early calculus courses. For this problem all you need besides integration is the equation for a circle.
I mean, that's essentially what he's doing, he's just using the final value of the integral of the equation for a circle from 0 to r, that being pi*r²/4. And I'm not sure this would actually be easier by explicitly defining a bunch of piecewise functions and doing one big integration; that feels like introducing unnecessary complication when you already know the formula for the area under each individual quarter-circle in the grid.
I initially thought thats how he was going to solve it
@@Idran They both accomplish the same thing in nearly the same way. Which you find easier probably depends on what you are more familiar with. Fewer fundamental equations are required with calculus, but as you said, more piecewise functions and you need to know how to do basic integration.
@@DataScienceDIYThis is definitely not as easy as you're describing it. If you'd actually tried to do it with integrals you'd run into trouble. This method is much simpler
tl;dr: You can shift the white area above into four white squares and a white quarter circle, turning this problem into something elementary.
It can be done even quicker and in a much simpler way (even simpler than the trick in the other video). Since you have two congruent quarter circles, a lot of symmetry can be used. Start with the area of the biggest quarter circle π*6^2, the white area outside it is not necessary.
If you look at the two quarter circles with radius 8 you can actually find two full white squares with them, fit the white area in cell 3 and 6 above the red into the white area above the red in cell 4 and 7. Together with the two cells in the top row you have 4 white squares, so you can subtract 4*4^2 from the total.
Now the very middle cell is left. In that cell, if you see that the red part in the bottom right has the same area as the white in the top left (again due to symmetry of the two congruent circles) then it suffices to just subtract the area of the smallest quarter circle for the solution.
So π*6^2-8^2-π*2^2= 32π - 64.
I think this would be the method with the least amount of calculation.
The formula I used involved cutting x down the two bottom left corners. Then if I match the r=8 circle edges together, I get that the area x is:
(
a quarter-circle r=12
minus a triangle b=12 h=12
) minus (
a quarter-circle r=4
minus a triangle b=4 h=4
)
Since it’s trivial that the two parts are similar, I can just simplify it to (3×3 - 1×1) = 8x the hole.
8 (
a quarter-circle r=4
minus a triangle b=4 h=4
)
= 8 [ (π×4×4/4) - (4×4/2) ]
= 8 [ (4π) - (8) ]
= 32π - 64 (units²)
Thanks so much for this stepped solution. I struggled with algebra and have not done it for nearly 20 years yet watching this really made my memory trigger with how do all that - i understood it and feel like i could apply those principles in other circumstances.
If you cut the area along the diagonal from the top right to the bottom left, and rotate that piece on the left around the center of the large square by 180 degree, you can simply your calculation. The area would be a large circular segment minus a small circular segment, which is 36pi-72-(4pi-8)
What a great video! I had no clue how the process for those inner sections would be calculated when I started and was shocked at how intuitive it was that making them quarter circles and removing the triangle came right as you started saying the solution! Very well structured and paced!
This assumes all curves are spherical - what if they were aspherical ? ...
That is the right question to ask.
Then it would be impossible to solve thus problem
i think
the curves must be circular if they have constant curvature and intersect the squares at exactly their vertices. Neither of which is specified in the problem. So you're right it's a poorly conditioned problem.
Then you estimate and get as close as is reasonable.
Or, you use calculus. You measure the rate of curvature for each section of the curve, represent that with a function, and integrate over the range.
They do not assume. Read the instructions, it says "made with quarter circles"
Impossible, the Terminids have found their way into our math problems!
I went by a different route (also I used X for the length instead of 4). I cut the red shape along the diagonal, which meant:
Large Shape area
= Quarter circle of 3X radius - Quarter circle of 2X radius - 2 squares of length X - 1/2 squares of length X
= (Pi*(3X)^2)/4 - (Pi*(2X)^2)/4 - 2(X^2) - (X^2)/2
Small Shape area
= Quarter circle of 2X radius - Quarter circle of X radius - square of length X - 1/2 squares of length X
= (Pi*(2X)^2)/4 - (Pi*X^2)/4 - (X^2) - (X^2)/2
Add the two together:
Total_area = (Pi*(3X)^2)/4 - (Pi*(2X)^2)/4 - 2(X^2) - (X^2)/2 + (Pi*(2X)^2)/4 - (Pi*X^2)/4 - (X^2) - (X^2)/2
Multiply everything by 4 to get rid of the divisors:
4 * Total_area = Pi*(3X)^2 - Pi*(2X)^2 - 8(X^2) - 2(X^2) + Pi*(2X)^2 - Pi(X^2) - 4(X^2) - 2(X^2)
Open the squared parentheses
4 * Total_area = 9Pi(X^2) - 4Pi(X^2) - 8(X^2) - 2(X^2) + 4Pi(X^2) - Pi(X^2) - 4(X^2) - 2(X^2)
Add up
4 * Total_area = 8Pi(X^2) - 16(X^2)
4 * Total_area = (8Pi-16) X^2 sq. units
Divide both sides by 4
Total_area = (2Pi-4) X^2 sq. units
Same solution, just plug in whatever value you want for X.
I agree, having to calculate only 4 areas (of the same shape) is much faster (and better) solution. Saving x as symbol to plug it in later is cherry on top
These algebra videos have been great. I learned geometry and such years ago, but forgot the formulas. Nice to have the refresher.
1:33 How do you know the arcs are a quarter circle? Thats an assumption that the drawing doesnt really confirm. It could be a slightly asymptotic line, not a radial.
Written above drawing: "Made with quarter circles" :)
@@dawidouss6333 What if it isnt a quarter circle, which would make this problem 10x harder to solve.
@@johnkuang123 Yes, that would be much harder to solve and I would use integrals to calculate the area.
@@dawidouss6333 What if its a hand drawn shape that's all over the place? random curves like a blob of slime, is it even possible to find the area?
The question looked so hard, but the solution felt like 5th grade. Thanks for this! Subbed.
This is a pretty complex solution. I just thought of moving the smaller offcut by translation and rotation inside the big offcut, making it a segment - a smaller segment. in the end you get smth like (36π - 72) - (4π - 8) getting 32π - 64. How, exciting.
Exactly my way of thinking and the same solution
Or just calculate the big one and multiply by 8/9 since you know that the smaller one is 1/3 in lenght so is 1/9 in area
Can't believe I am actually sitting in front of this video enjoying math...
silksong geometry
I used to be afraid of these kinds of problems, until I learned double integrals. Now I can probably solve this in somewhere around 15 mins. Your solution, which only includes basic mathematics, and takes no more than 5 mins, is beautiful
4:25 game recognise game
Kids, this is exactly what most of the maths is all about. It’s not about remembering the formula for area of cylinder or a scalene triangle, it’s about how do you approach a solution given the formulae. Anyone who doubts their education system should watch this.
This is a crazy problem to solve. I thought by clicking on the video I was going to see this being solved with calculus. However, your problem solving solving method stunned me as you were able to make complete sense out such a odd (and complex) question. This is definitely one of the coolest videos I’ve seen lately.
by assuming that everything is circles or segments of xd
What if they arent?
@@GaSevilhawell it was stated that every arc was that of a quarter circle…
@@Sukkulents_ Bruh I missed that. I would've solved it if I had known.
yeah, but that makes things too easy doesnt it?@@Sukkulents_
@@TheSpacePlaceYT That kind of information always has to be given in these kinds of problems. The hardest part of the problem shouldn't be you sitting there wondering if a shape is actually what it looks like. So if you find yourself having to think about that, always go back and check to see if it was already clarified.
The real answer is that insufficient information is provided to answer. One must assume right angles and equidistance of the interior line segments. It may seem overly analytical, but real world applications do not tolerate such assumptions and mathematics should not either. For example, if you cut carpet for a "square" room, you may find one side too long and the other too short.
"Insuffucient information" bro when he learns what a given is: 🤯🤯🤯🤯🤯
The assumption that the two areas are segments of a circle could be wrong. He's assuming by inspection that the curves have an eccentricity of zero and are, hence, part of the circumference of circles, but this might not be the case. Other types of curves can also connect the two endpoints. So, interesting solution but based on what could be two faulty assumptions.
It clearly isn't the case, at least for me, so his calculations are wrong, the outer parts of the circumferences reach the edges too early.
Who are you man
the problem statement literally says they're quarter circles, so by definition they are.
@@samsowdenJudging by how many people didn't notice it it probably should have been stated out loud at the beginning of the video. He's simplofying the problems too much for younger viewers and missing the nuance
I really love your video, please keep posting video😁
What a cool problem! I just found your channel a couple days ago and it is amazing 😁
Idk why my mind went straight to putting this image on a graph, splitting the curves up into different functions, and finding the area under the curve with integrals and adding them together.
H
How would u write equations of the curves tho?
@@AjayKumar-mg3xcthey are circle parts so I think its possible
@AjayKumar-mg3xc Well, it's a 12 by 12 grid that you can put in the first quadrant. You can separate and label each line as a quadratic. For example, that curve that goes from the bottom left corner to the upper right corner can be labeled y=(x^2)/12.
I tried to to do this but trust me.. it is NOT easy, you will have to find a lot of intersection points and figure what all areas to subtract. Wouldn't recommend.
Recently found your channel and i am absolutely in love with your content. As someone who always used to fear math, i recently began on a journey to like math and i cannot tell you how awesome your content has been in guiding me along that journey. As of this peoblem i came close but i coulsnt find out rhe area pf the segment because i didn't visualise it in that way
Nice! I managed to do it, similar basic idea of subtracting a group of smaller shapes from a square, but didn't choose such clever ones, hence resorted to using an integral.
Yeayyy got it right on my first try! Your questions remind me of the math olympiads I took part in when I was in elementary school anywayyy. More challenging questions please, I'm so curious!
Me seeing a random fugure and trying to create an integral of its function: 🗿
1:11 I think these are called spandrels. At least that's what we called them when dealing with the centroid and the rational inertia about these shapes.
Cool solution, but if you split the shape down the diagonal, you can solve it much easier, because then you can subtract whole quarter circles (plus a small rectangle+triangle shape) from the two larger quarter circles that make up to two arcs we see in the shaded area.
It gets even better when you see that the two intermediate quarter circles have the same area, so in the end you just need to subtract the tiniest segment of a circle from the biggest one
@LukesVGArea it gets even even better if you print out the problem, cut out the area you're trying to find and then weigh the section and compare it to the total weight of the entire square. You have now calculated area as a measure of weight and then take that ratio and match it against the total area of the shape and itl give you your total area
So true! I swear I have seen very similar comments on other videos like this, that you?
Exactly this!
hahaha i envy your minds. how did you even think of splitting it diagonally?
I don’t even watch math videos but for some reason this popped up. I watched the whole thing through, super entertaining which I would have never guessed beforehand.
Was able to do it! How exiting!
I just love when you see a strange shape in nature, abstract as it may be. Encase it in a symmetrical construction, and calculate the difference. Symmetry, what a wonderful word!
im a decade past recalling exact formulas for things like area of a quarter circle or circle segment, so the explicit numbers didn't come to me, but i still got some decent problem solving, worked out the clever shortcut you mentioned from the other guys vid on my own (asking myself "why wouldn't you simplify and do it like this" and felt very vindicated when you pointed to that video and the guy presented the same alternate solution) super neat stuff!
Never thought I'd listen to Anthony jeselnick tutor me in math while I try to sleep
So, if the quarter circles were almost quarter circles but not exactly, we would have been effed? Or maybe with some integrals?
@ethanjsegatI watched the other video mentioned and the shape is made using explicitly mentioned quarter circles so there’s no assumption, but if I was just given the shape like that, I wouldn’t be comfortable just assuming they’re quarter circles
If they weren't exactly quarter circles then yeah we'd be effed. Unless they gave enough information that you could work out the function of the curve in which case I think integrating to find the area under the curve would be correct.
Good thing is that in real life, you can estimate and be within a margin of error. We're conditioned so early on that math has to have one singular answer but Calculus teaches you that there are multiple approaches and that you can always be within an error of margin. Dividing a line into infinitely many pieces to guess where it most likely converges is peak guessing game and I love it
That’s not what calculus teaches. Go review what a limit is.
Props for mentioning the MYD video! How exciting indeed. For me the part I didn't figure out was to cut the top right square in half :)
Silksong….
A lazy solution:
The figure can be cut in two parts with a SW-NE diagonal.
SE side:
From the disk segment whose chord is the diagonal of the 12 units square, we remove the disk segment whose chord is the diagonal of the 8 units square.
NW side:
From the disk segment whose chord is the diagonal of the 8 units square, we remove the disk segment whose chord is the diagonal of the 4 units square.
The area is:
A =
[Area of 1/4 of the circle of radius 12 - Area of the half of 12 units square]
- [Area of 1/4 of the circle of radius 8 - Area of the half of 8 units square]
+ [Area of 1/4 of the circle of radius 8 - Area of the half of 8 units square]
- [Area of 1/4 of the circle of radius 4 - Area of the half of 4 units square]
We simplify (lines 2 and 3 cancel each other):
A =
[Area of 1/4 of the circle of radius 12 - Area of the half of 12 units square]
- [Area of 1/4 of the circle of radius 4 - Area of the half of 4 units square]
That's to say:
A =
(1/4.π.12² - 1/2.12²)
- (1/4.π.4² - 1/2.4²)
A = 36.π - 72 - 4.π + 8
A = 32.π - 64
I’m mad AF, because there’s no part of the problem telling us that the curve is a quarter circle! Yes, it touches the two corners, but there nothing telling us that each point along its curve is equidistant from its center point. It’s spent a good ten minutes trying to figure out the curvature of the shape.
You can assume that the radius of the circle is 8... and that's how he knows that it's a quarter circle. It's will help you calculate the area within that region
"Made with quarter circles" is literally what it says above the square.
Uhh, can u read? Lol
Sometimes, less is more.
How can you be so stupid?
thank you for the solution. i'm sure i'll need this to renovate my house with this shape
I hated problems like this in school 😂 how do we know those are exact quarter circles
There's a most elegant solution of only sums and subtractions of segments of quarter circle if you look at this problem diagonally (literally)
Hence only one area formula is necessary, applied to 3 different radii
What so silksong does to a man
I liked these problems during school because it made me think about something in parts and dissect problems into simpler ones
I am curious about how did you know that each arch is 1/4 circle? Did the question give these premises?
It says “made with quarter circles” above the diagram
I dont know why im addicted to watching these. Like, I know the possible ways of figuring it out, but I dont know any of the formulas. Its like watching a friend play a PlayStation game and you know what to do, but you dont know how to handle the controls hahaha
mmm… how do we know the curves are perfect circle curvature?
It says it at the beginning: "Made with quarter circles"
yea...,yor just dumb didn't literate
The math gods are not Greek gods. They cannot be THAT cruel
Because it is defined in the problem.
IT SAYS IT RIGHT THERE
Unironically reteaching me math, something I never thought I’d had to relearn. Thank you
yo he got a room upgrade
A polar planimeter! I have a K&E device - of course I bought mine in 1967 as an engineering student. But, today there are some interesting computer programs to integrate the thing! Love the computer!
For those wondering, this area is roughly equal to 36.53 sq units
If i didn't check it, i wouldn't know where to search your comment :P Great video tho
I put a box round this channel and put a value to it, and it certainly will be a high value
hey gang, bait used to be believable
🚬🤡
Y-
very undemocratic looking claw
SHAW!
I got the same result but with a different way of calculating. I calculated the area of the big quarter circle and subtracted the triangle part and the round part of the smaller quarter circle and so on.
Shaw?
Skong
Did it in my head, with a simpler decomposition:
X is the unknown area,
A(r) the area of a quarter disk (=πr²/4)
C the area of one cube,
You got : X = A(3r) - A(r) - 4C
I took r=1 which gives:
X = 9π/4 - π/4 - 4 = 2(π - 2)
Sawing that the problem gives r=4, multiply the area by the wanted scale (r² = 16):
X = 32(π -2)
PS : To get to the formula for X I followed the shape from the larger arc then add/remove disks and cubes to fit the shape.
Before last simplification, you get : X = A(3) - A(2) - 2C + A(2) - A(1) - C - C
yoooo that's insane! keep up the good work dude !
bait used to be believable
-I
The only youtube video makes me get up bed, get a pen and take note at 3AM
My only issue is that by default you assume the curves are circular if that werent the case itd probably be unsolvable. Either way very impressive its a neat seeing you solve these
idk man, i personally think "made with quarter circles" is good enough evidence for me
its given in the ques, just read
if it was unsolvable on purpose, then what would be the point of it?
from where I see it, this problem tries to teach you how to approach a problem, and how much easier it is when you consider other alternatives.
when the entire point of problems and whatnot is to TEACH, then there is no reason for it to be unsolvable
It’s literally given in the problem that the area is “made with quarter circles”
What do you think “Made with quarter circles.” means?
Math Teacher: “Why are you watching UA-cam in class?”
Me: *Shows phone*
Math Teacher: “Ah. My bad chief, you good.”
You’re making a few assumptions here without having any evidence to support your assumptions, such as those circles, being quarter circles, and the value that they take up.
it is given that they are quarter circles
Man, I wish these videos existed when I went to school
Why am I watching this during winter break?!
SAME
I cut the bottom-left square diagonally, and flipped the left piece 180° so that the two 2-unit quarter circles coincide. Made for a much easier shape to solve!
If you have the eye to identify the "blue" segments I don't know how you could not realize that the red shape is just two segments covered by the two blue segments. In other words the red area is:
(big red segment)-(small blue segment)
+(small red segment)-(smallest blue segment)
The diagonal line helps you see where the segments are and the grid lines help you figure out each radius
man this is so cool
I never thought of this procedure of extracting area
The only ptoblem I had is I didnt knoe how to get the area of the 2 slices, nice, thank you for the explanation
Нашёл по формуле пика за 2 секунды до большого взрыва
1:43 i don't know if i am correct' But according to me if we see precisely those are not quater circles as they are ending before the vertices of the squares
Except the problem states that they are quarter circles. That's just a representational error in the diagram, the definition takes precedence.
It's funny how much the "made with quarter circles" knocks the difficulty of this problem way down.
Funnily enough, I did it a completely different way: I started with the full square, and imagined either cutting the square off or adding to it, keeping track of what was added/subtracted. I got to the correct answer, and I only needed to use the formulas for circles and squares, but I get the feeling my idea was a tad more complex than this. Definently more error-prone.
Based on just that picture, there's no information whatsoever if these are circles and where are their ending points, thats why this method is just based on your interpretation, but has bo approval to this. Tbh it can't be solved due to the lack of information.
Information is given. "Made with quarter circles" is written above the image.
AMAZIINGG
I did it with my boyfriend and we had a realllyyyy good time, THANKS! My bf's a genius
You are my geniussss! ♡♡♡ thanks for sharing this exercise loveee, lysm ❤
The solution shows up within the first 10 seconds of the video. Thanks man
I find that for shapes like this, it's usually just easier to write the curves as semicricle equations (y = sqrt(radius^2 - x^2)) and then use integrals to find the volumes of solids. however, the way that you did it is super cool!
Here's my solution:
1. Draw a diagonal from top right to bottom left. Now you are with 3 types of 4 pieces (minor segments of a circle). 1 smaller (white), 2 medium (1 red and 1 white) and 1 bigger (red).
2. Now, by observation, we see that area of red region is: (Bigger - Medium) + (Medium - Smaller) => Area = Bigger - Smaller
3. Area of the shape = Area of quadrant - Area of right triangle = (πr²/4) - (1/2)(base)(height)
4. Area of red region = Big - Small = [π(12)²/4] - (1/2)(12)(12) + [π(4)²] - (1/2)(4)(4) = 32π - 64
Can't beleive the Nike's logo threw me off so hard from getting a very easy solution, bravo to whoever made this problem
It's easy and even possible just mentally if we divide the red figure by the diagonal ／ and rotate one of the parts so that the arcs coincide.
And now we find out easy that the area of the red figure equals to ¼πR²-½R²-(¼πr²-½r²) where r equals to 4 and R equals to 3r.
So... it equals to ((9/4)π-9/2-¼π+½)r² = (2π-4)r² = (2π-4)*16 = 32π-64
Answer: 32π-64 sq units
TBH this solution is over-complicated the problem by requiring knowledge of the area of a "segment of a circle". I did it, more intuitively IMO, by adding and subtracting quarters of circles.
I'll try to explain in a comment:
- I split this drawing along the bottom left -> top right diagonal, with "part (a)" of the red shape on the bottom right and "part (b)" on the top left. I'll explain how I calculated "part (a)", with the same methodology being applied for "part (b)"
- Note: I'll use "y" to indicate the length of a small square, where y = 4
- "Part (a)" is comprised of four sections: one additive, and three subtractive.
-- (1) take a full quarter circle of 3y -- visualize the center of that circle at the top left point of the diagram: coordinate (0, 3y), or (0, 12): + π*(3y)^2 / 4
-- (2) remove a quarter circle of 2y -- visualize the center of that circle at (y, 3y) or (4, 12): - π*(2y)^2 / 4
-- (3) remove two full squares at the top left, between coordinates (0,y) and (y, 3y): - 2y^2
-- (4) remove half of the bottom left square, along the diagonal: - y^2 / 2
This gives you: π*(3y)^2 / 4 - π*(2y)^2 / 4 - 2y^2 - y^2 / 2
If you repeat for the "part (b)", you get: π*(2y)^2 / 4 - π*(y)^2 / 4 - y^2 - y^2 / 2
If you add them all together and simplify, you get: r^2 * (2π - 4) -- which is the same as the solution offered in this video. All with even more basic math
This man is a menace, Great video
An even faster way to solve this, is to rearrange the picture by nestling the little red horn into the curve of the bigger red horn. that way you notice that the shape is actually one big segment with r =3*4=12 minus one small segment with r=4. this gives us a pretty short equation of x=1/4*π(12)^2-1/2*(12)^2-(1/4*π4^2-1/2*4^2)=144π/4-72-16π/4+8=128π/4-64=32π-64 which is the result you ended up with. When there are multiple ways to cut up a shape like this the first step should be to find the best way to describe your area with the least amount of composited shapes in order to avoid doing as much work as possible.
Make a copy of the shape, and superimpose it on the shape, however, rotated 180°.
Then the combined area is a large lense, minus a small lense.
The large lense covers 1/2*pi*3^2 - 3^2 squares.
The large lense covers 1/2*pi*1^2 - 1^2 squares.
Hence the combined shape covers 4*pi - 8 squares.
However, since each square has area 16, and we have to halve the solution, we get 32*pi - 64
bro I love this channel