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.🥲
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!
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.
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.
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.
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 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!
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!
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
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.
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)
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
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.
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)
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.
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.
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.
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.
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.
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.
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
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
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.
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!
I was thinking the same thing, you wouldn't be able to use the "area of a quarter circle" formula so it might be impossible if the quarter circles were almost quarter circles
@@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
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
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!
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!
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.
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 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
I did these type of questions as a cakewalk at 12 year age before COVID But now I forget all I start fearing from them Seeing your approach towards ques reminds me of my prime
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!
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.
Finding the quarter circles and unknown shapes are the regions I still have no idea how to get. Because I would never know how to get those from a square, not knowing exactly how much of the square it takes up.
The quarter circles all intersect on the grid giving exact dimensions of the circles :) Just count the grid and get their area. Try it on paper one step at a time :) You'll get it. I just started with the obvious shapes and what you can do with the different sized triangles and squares from the grid intersections. Then link the numbers at the end and simplify the equations if you like. Remember the squares of different sizes can be divided into 2 diagonally to get triangles when you need a segment. Subtract a triangle from a quarter circle to get the remaining part (the segment). Subtract shapes from shapes. Helps to use paper so you can keep the sections separate. Maybe different colour pen for the different parts would help order on your page before simplifying. Practice rearranging equations. Tip: You might find it easiest to remove the white areas from the total area of the square to be left with the red area (x)
I thought it was gonna need some advanced formulas that I don't know, but everything used here was things I learned in 5th or 6th grade. Just used very cleverly. Cool.
This assumes though that said figures are part of a perfect circle, which you can only really guess from my understanding (unless you measure the radius with a ruler). Had the circles been a little imperfect, r^2 * pi wouldn’t apply.
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 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!
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 am obsessed with this sort of stuff. I knew all of those equations but I just did not see the broken down shapes. This is soooo easy once you break it down
It’s funny how different people see solutions to problems like this differently. I would have take the 12x12 quarter circle and subtracted an 8x8 quarter circle, two small squares, and a 4x4 triangle in the bottom left. Then added the 12x12 quarter circle minus the 4x4 quarter circle and subtracted another square and triangle. I’m pretty sure that would have accurately solved this. Please let me know if I missed something
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.
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
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 did this a little differently, took the upside down quarters and subtracted their area which left me with the larger part of the claw and the first column with a partial left bottom corner. Similarly i took the smaller claw from the upright quarter circles and got the smaller claw and part of the first and complete second square from bottom row. Now i subtracted the whole boxes which were not part of the claws and the bottom left corner once to get the desired area
And here i am, randomly getting this video on my feed, looking more dam stupid. Even though i used to go to maths faculty in high school, i feel i know nothing but the basics.. also never have thought maths could be this interesting. great channel fam, best wishes to you!
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
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.
I solved without those semi-circles and triangles somehow - I saw the giant quarter circle and calculated it. Then subtracted 1 white square from it. Then subtracted the area not in the bottom 2x2 quarter circle. Then subtracted the area of the upper 2x2 quarter circle. But since this 1. Doubles white, and 2. Deletes red, you add the area of the space not in the middle 1x1 quarter circle to give back what the white took from the red and what the white doubled up on
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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?
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
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
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 😂
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.🥲
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.
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.
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.
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.
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"
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
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!
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!
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?
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 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)
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?
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.
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)
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.
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.
These algebra videos have been great. I learned geometry and such years ago, but forgot the formulas. Nice to have the refresher.
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.
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.
Was able to do it! How exiting!
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.
I really love your video, please keep posting video😁
Silksong when?
fuck sent my sides to orbit with this
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.
Dear, an ingenius construction. CONGRATULATIONS
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
The shape had me expecting a stealth Silksong announcement
shaw!
🤡
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
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.
yo he got a room upgrade
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!
Why am I watching this during winter break?!
SAME
What a cool problem! I just found your channel a couple days ago and it is amazing 😁
4:25 game recognise game
So, if the quarter circles were almost quarter circles but not exactly, we would have been effed? Or maybe with some integrals?
I was thinking the same thing, you wouldn't be able to use the "area of a quarter circle" formula so it might be impossible if the quarter circles were almost quarter circles
@@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.
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
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!
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
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
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 :)
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!
hey gang, bait used to be believable
SHAW!
The question looked so hard, but the solution felt like 5th grade. Thanks for this! Subbed.
Impossible, the Terminids have found their way into our math problems!
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
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?
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
The only ptoblem I had is I didnt knoe how to get the area of the 2 slices, nice, thank you for the explanation
I liked these problems during school because it made me think about something in parts and dissect problems into simpler ones
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
Can't believe I am actually sitting in front of this video enjoying math...
I did these type of questions as a cakewalk at 12 year age before COVID
But now I forget all I start fearing from them
Seeing your approach towards ques reminds me of my prime
thank you for the solution. i'm sure i'll need this to renovate my house with this shape
I was about to use integrals...
Captain America in another universe where he's a math professor and not an avenger:
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!
Never thought I'd listen to Anthony jeselnick tutor me in math while I try to sleep
Man, I wish these videos existed when I went to school
The 16π-32 is actually the segment that completes a triangle of the 64-16π. Added together, you get 32 sq units
I constructed the red region out of quarter circles minus triangles and got the same answer, how exciting!
I hated problems like this in school 😂 how do we know those are exact quarter circles
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.
man this is so cool
I never thought of this procedure of extracting area
All I wanna say, I clicked on this video mainly because the shape looked like the Nike logo but better. 😅
Finding the quarter circles and unknown shapes are the regions I still have no idea how to get. Because I would never know how to get those from a square, not knowing exactly how much of the square it takes up.
The quarter circles all intersect on the grid giving exact dimensions of the circles :) Just count the grid and get their area. Try it on paper one step at a time :) You'll get it.
I just started with the obvious shapes and what you can do with the different sized triangles and squares from the grid intersections. Then link the numbers at the end and simplify the equations if you like.
Remember the squares of different sizes can be divided into 2 diagonally to get triangles when you need a segment.
Subtract a triangle from a quarter circle to get the remaining part (the segment).
Subtract shapes from shapes.
Helps to use paper so you can keep the sections separate. Maybe different colour pen for the different parts would help order on your page before simplifying. Practice rearranging equations.
Tip: You might find it easiest to remove the white areas from the total area of the square to be left with the red area (x)
I thought it was gonna need some advanced formulas that I don't know, but everything used here was things I learned in 5th or 6th grade. Just used very cleverly. Cool.
This assumes though that said figures are part of a perfect circle, which you can only really guess from my understanding (unless you measure the radius with a ruler). Had the circles been a little imperfect, r^2 * pi wouldn’t apply.
Math Teacher: “Why are you watching RUclips in class?”
Me: *Shows phone*
Math Teacher: “Ah. My bad chief, you good.”
I didn’t even think that I was really a nerd fr but this video’s actually interesting
I put a box round this channel and put a value to it, and it certainly will be a high value
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 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!
This is where calculus starts to be the friendlier option
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.
It's fun to try and estimate it in a few seconds then see how close you were, I guessed 36.
I am obsessed with this sort of stuff. I knew all of those equations but I just did not see the broken down shapes. This is soooo easy once you break it down
I wish youtube would recommend me more content like this!
It’s funny how different people see solutions to problems like this differently. I would have take the 12x12 quarter circle and subtracted an 8x8 quarter circle, two small squares, and a 4x4 triangle in the bottom left. Then added the 12x12 quarter circle minus the 4x4 quarter circle and subtracted another square and triangle. I’m pretty sure that would have accurately solved this. Please let me know if I missed something
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.
Can't beleive the Nike's logo threw me off so hard from getting a very easy solution, bravo to whoever made this problem
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
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 did this a little differently, took the upside down quarters and subtracted their area which left me with the larger part of the claw and the first column with a partial left bottom corner. Similarly i took the smaller claw from the upright quarter circles and got the smaller claw and part of the first and complete second square from bottom row. Now i subtracted the whole boxes which were not part of the claws and the bottom left corner once to get the desired area
very undemocratic looking claw
The only youtube video makes me get up bed, get a pen and take note at 3AM
Pretty elegant solution, I like it. Thanks for sharing
And here i am, randomly getting this video on my feed, looking more dam stupid. Even though i used to go to maths faculty in high school, i feel i know nothing but the basics.. also never have thought maths could be this interesting. great channel fam, best wishes to you!
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
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.
I solved without those semi-circles and triangles somehow -
I saw the giant quarter circle and calculated it. Then subtracted 1 white square from it. Then subtracted the area not in the bottom 2x2 quarter circle. Then subtracted the area of the upper 2x2 quarter circle. But since this 1. Doubles white, and 2. Deletes red, you add the area of the space not in the middle 1x1 quarter circle to give back what the white took from the red and what the white doubled up on
Best challenge I got today
I have now subscribed to you. This is the second video of yours I'm watching.
not the shortest or simplest way to solve this problem but as a teacher myself I know what you are doing here and Ike it. well done.