Puzzling Quarter Circle Area Problem [Geometry]

[@saptaksarkar30468]

But wouldn't you have to prove that the area of half of the symmetric lens is equal to the circle's top area ? Because if I did this in an exam, my teacher would say how the half symmetric lens is equal to the circles partial area

[@dennisvanzandt6579]

the calculation of the areas is all the proof you need to show that the full area of the lens (not half) is equal to the left over area at the top.

[@dougaltolan3017]

Yes, you would, but in this case the proof isn't that complicated. You can show that the curve of the lens is the same as the curve of the bit it fits into (same radius circle) And the length of the lens is the same since the distance from the origin to the other tip of the lens is the same as the distance from that tip of the lens to the intersection of the 1/4 circle on the x axis (provable 45 degree angle from the origin will be one side of an inscribed square in the small circle)

[@thephilosophyofhorror]

Already proven, since if a=b => a/2=b/2.

[@acoupleofschoes]

If you want to do it much more difficultly, try it with square roots and integration by parts. The football area is formed by [ sqrt(4 - (x-2)^2) - (-sqrt(4 - x^2) +2) ]. Just integrate that from 0 to 2.

[@vbaldokaizen8748]

Thanks, this is how I approached at first, but didn't know how to get the function for the quarter of the circle. I didn't even remember the circles function lol

[@st3althyone]

Only a masochist would do that. I like it. 🤓🤓

[@thewierdragonbaby4843]

you've successfully just annoyed the entirety of Europe by calling it a "football shaped region"

[@shehannanayakkara4162]

The white shapes both clearly each have an area of 4 (they are comprised of a quarter circle and the part that completes it into a square, you can arrange them to reveal a square with side length 2). The big quarter circle has area of 4pi so the blue area is just 4pi - white areas = 4pi - 4 - 4 = 4pi - 8.

[@TheDittu]

used this method as well to solve it really easily just from seeing the thumbnail

[@johntovar9325]

Why would you be so sure that the square has side length 2 without prior knowledge?

[@shehannanayakkara4162]

@@johntovar9325 The semicircles have a diameter of 4 (given by the question) meaning that they have a radius of 2. The side length of the square is just the radius of the semicircle so the side length of the square is 2.

[@TimothyTranEnjoysLife]

because I tend to gravitate to analytical solutions, it took me a while to see what you wrote: white shape can be broken down into a quarter circle and a part that completed it into a square. good job for seeing it so easily, though I think it is not obvious

[@zinithin-8208]

Did it a different way. To find the symmetrical lens I just drew a square around it (2x2) and subtracted a quarter circle twice to get a negative value which would be the area of the lens.

[@eduardoledesma-md4xl]

I did this too. It just was the most obvious and easiest way for me. The other ones didn't even crossed my mind.

[@federook78]

Of course. I'm appalled that the guy doing the video allows himself to do it his way

[@methos1999]

It's funny how my ME degree required calculus, differential equations, linear algebra... almost none of which I use in my day to day work. But Geometry? I use that all the time! When doing CAD work a firm understanding of geometry is priceless to make more elegant designs.

[@ggw1776]

quite an unsatisfying solution for me. The first relies on an obscure formula and the second is basically a visual proof.

[@dunda563]

This can be solved by considering small quarter circles. If you subtract a quarter circle from a square you get a "corner". If you subtract two corners from a square you get the football shape. If you combine the corners with the other quarters (the other halves of the semicircles) you get two squares - 2x2 each for an area of 8. Simply find the area of the large quarter circle (pi * 4) and then subtract 8 (pi * 4 - 8).

[@spearmintlatios9047]

I just did this using double integrals. The whole area of the enclosed shape is 4pi, then we are subtracting two equally sized cutouts. The two me closer to the x axis can be defined as the integral: 4sin(theta)

[@acoupleofschoes]

You can do it in a single integral. Not too fun by hand though. The football area is formed by [ sqrt(4 - (x-2)^2) - (-sqrt(4 - x^2) +2) ]. Just integrate that from 0 to 2.

[@vitor900000]

You can also do it using segment of a circle. First find the diagonal length of the 4x4 square. With the diagonal length you can calculate the segment of the r=2 circle. Multiple the segment of the r=2 circle by 2 and you got the area of the symmetric lens. And knowing this 1:22 you can multiply it by 2 again and get the answer. Sounds a lot easier than it actually is because during the the segment circle calculation you will need to calculate angles that aren't whole numbers. I think this is one of the longest ways to do it.

[@emergencyfood3568]

At 2:28, how is it possible for a quarter circle to have a greater area than a semicircle?

[@polymathematic]

the radius of the quarter circle is larger than the radius of the semicircles.

[@batchrocketproject4720]

Triangles and squares are your friend when you spot them. 1:33 is the important starting point - you're after 1/4 large circle - 1 small circle + 2 intersections. The 2 intersections equate to 1 small circle minus 2 squares of side length small circle radius (because half an intersection is a quarter small circle minus a 45 45 90 triangle from its centre). That simplifies to 1/4 large circle minus 2 squares or 4 pi - 8.

[@joachimandersson966]

The area of the lens can be calculated with simple mathematics. If we add 4 - π to the lens we get a quarter of a small circle. 4 - π is just a small square minus a quarter of a small circle. If we then subtract 4 - π from a small quarter circle we get the area of the lens: π - (4 - π) = 2π - 4

[@Ron_DeForest]

splitting the football in half was clever but how did you know those halfs would pop up into the empty are to make the triangle? Is there a property I missed?

[@polymathematic]

just the symmetry of the circle (or, if you prefer, the isosceles right triangles we create when we reflect each half of the football shape across vertical and horizontal lines from the intersection point).

[@cdmcfall]

You can more easily visualize a formal proof of this if you construct a 4x4 square to inscribe the quarter circle, then construct diagonals to that square. From there, we can treat each half of the lens as a segment of congruent semicircles of radius 2, then prove that the height (sagitta) of each circular segment is equal due to the fact that each circular segment, including the concave spaces into which the half footballs can be rotated, has the same radius and chord length. Alternatively, you could construct perpendicular lines at the intersection of the two semicircles and prove that those lines bisect the sides of length 4. Since those would create right angles, the same circular segments mentioned above would be congruent by the central angle theorem.

[@Anonymous-ru1dg]

I came here from your short, You make maths interesting, thanks you for your videos Edit: i have a doubt, if we subtract the two smaller semi circles, aren't we subtracting the football shape twice because they both overlap each other, meaning we would need to add 3 football shape to find the area of the blue region

[@wverms]

To explain, imagine you have the full quarter circle then subtract one semicircle. Now you've subtracted the football once, so you need to add it back. Now subtract the other semi-circle, which means you need to add back the football again. That's how you can tell you only need to add the football back twice, not three times.

[@tomatosauce5775]

I saw this in my recommended and took a crack at it before watching the video. I used neither of the methods you showed, but I still got the right answer!

[@tomatosauce5775]

I used the integral of the left circle relative to the x axis to find the area under the “football” then doubled it and removed that plus two quarter circles of the smaller size from the whole

[@tomatosauce5775]

(y-2)^2 + x^2 = 4 -(y-2) = sqrt(4-x^) (negative for lower portion of circle) y = 2 - sqrt(4-x^2) Integrate from 0-2 relative to x 2 -> 2x and the square root portion is just the area of half of the semicircle since it’s radius is 2. Using the info from the circles we have, you arrive at 4pi -8. This probably wasn’t the easiest way for me to do this, but I was able to solve it with what I already knew which was fun

[@tomatosauce5775]

For clarification, the area under the curve is 4 -pi, so we double that and then remove that plus two quarter circles of r = 2 from the larger quarter circle

[@polymathematic]

very nice!

[@Yesytsucks]

Btw, the way i did the final calculation was: I realize that 2 semicircles combined have the same area as 1 quarter circle, because its half the radius and 4 times more parts and area is calculated with second power. And the part that is weird is equal to the lense because its the only part where they overlap, but also semicircles dont go outside of the quarter circle, which means the only part where they lose their area is overlap. So the qrea we're looking for is just 2 lenses and nothing else

[@mrcampisig]

Great question for my students, thank you. BTW what writing tool do you use? I teach math and notice that your writing with the boxes around parts of the solution makes it very clear for students.

[@polymathematic]

thanks for watching! i use an app on my ipad called goodnotes.

[@rikschaaf]

You don't need to know an obscure symmetric lens formula. You can just calculate the area of the white area on one side of the lens by subtracting the quarter circle from the square (2*2 - pi/4*2^2 = 4 - pi). Then you can find the area of the lens from subtracting the white area twice from the square area (2*2 - 2*(4-pi) = 4 - 8 + 2pi = 2pi - 4). The other area can be calculated by subtracting the square and 2 small quarter circles from the big quarter circle (pi/4*4^2 - 2*2 - 2*(pi/4*2^2) = 4pi - 4 - 2pi = 2pi - 4). So both areas are 2pi - 4, so the sum is 4pi - 8.

[@IoDavide1]

Extremely easy. Fot skeptics: Half lens is a quarter of the small circumference less the right triangle in the quarter with both sides 2. Once you have the lens area, all is a play of subtractions.

[@NavanBethrax]

My solution was for some reason starting with a full quad 4 * 4 with 4 half circles inside and 4 of those lenses, next I subtracted the actual area of the quad from the area of all four half circles (ignoring the overlap) to get the difference, the difference divided by 4 is the size of the lens.

[@boredomgotmehere]

Love this video. Perfect explanation. Thank you.

[@nathanmackay7387]

This could also be solved using integrals. Much harder to get the exact value, but not too bad to solve with technology.

[@balabuyew]

"Lens" area can be found much simpler. Let's focus on a bottom semi-circle: 1) Draw a vertical radius to split the semi-circle into two quarters of circle. 2) Draw a line, which connects two points of the lens. 3) You now have a right-angle triangle which resides inside a quarter of small circle. 4) The triangle area is: S1 = r * r / 2 = 2 * 2 / 2 = 2 5) Quarter of small circle area is: S2 = pi * r^2 / 4 = pi * 2 * 2 / 4 = pi 6) So, the half of the lens area is equal to S2 - S1 = pi - 2 7) Finally, the whole lens area is: S_lens = (pi - 2) * 2

[@MikeTaffet]

That’s how I ended up doing this, and if you double the area of that lens you end up with 4*(pi-2) which is a simplified version of writing out the area if you calculated the same thing but by combining all of the shaded areas into one area between a larger triangle and the quarter circle. In that case you’d end up with (4*pi)-((16*16)/2) If you simplify that to (4*pi)-(8), then you can factor out 4 from both sides and you’re back to 4*(pi-2)

[@balabuyew]

@@MikeTaffet No, you use different, much complicated way.

[@edwarddeng7710]

Thats a very good observations!

[@Yesytsucks]

Thats not really simpler?

[@balabuyew]

@@Yesytsucks Definitely smipler :)

[@ScreamingManiac]

Two main ways of doing it, * (π4²/4)-(4*4/2) = 4π - 8 * (π4²/4)- 2(π2²/2 - 2²(π/2 -1)) = 4π -8

[@thephilosophyofhorror]

Nice video ^^ But you don't need to know the formula for the symmetric lens, so even a young school kid could conclude that since the symmetric lens+the rest of the half of the semicircle= the other half of the semicircle, it follows that all that equals 2pi, and given that the right angled isosceles triangle cutting the lens in half has an area of 2(2)/2=2, you have: area of quartercircle=lens/2+triangle=>area of lens/2=pi-2, and since (as you established already) each of the blue areas is equal to the other, the entire blue area is 4 lens/2= 4p-8. All that said, certainly the graphic you made with the separation of the regions was an excellent touch.

[@golddddus]

Congratulations! Perfect use of geometry and color. Solving a problem is important, but how you come up with an idea for a solution is much more important. You deal with it. Eureka moment or Aha!👍

[@simonkhouryAU]

I went to integration straight away. Then you said 7th grade...

[@HenrikMyrhaug]

Great video, but I kind of feel like it misses one component of the simple answer: a proof that the football/lens actually is exactly the size of the missing region from the larger half- lens shape. It looks like it fits, but just because it looks like it fits, that doesn't necessarily mean it actually does. Simply draw a line connecting the corners/intersections of the curve and radii of the large quarter circle. From this, we see that the blue area is [quarter circle]-[triangle]-2*[white half-lens]+[blue lens], and by drawing a line bisecting the blue lens, as well as a line from each corner of the blue lens to the middle of each large radius of the large quarter- circle, we see that the [blue lens]=2*[blue half- lens]=2*([small quarter circle]-[small triangle]), as well as that the [white half- lens]=[small quarter circle]-[small triangle], hence the two half-lenses of the full lens are equal to the half lenses of the missing region.

[@Drawliphant]

I thought I had a totally different solution, but I really just cut it differently. I proved that both shapes had the same area because the two smaller arcs have the same combined area as the big quarter arc so their intersection must equal the outside. then I just found the size of the almond shape by putting a square around it knowing a squared circle takes up pi/4 the area. doubled the almond area to get the total area.

[@ermilooo906]

1/2r²(@π/180 - sin @), isn't?

[@elderwand8874]

Can we use application of integrals, that would make it much easier right?!

[@ConorChewy]

Without the lens area theorem it's trickier but you can get to the answer by the same logic as the shortcut method once you split the lens in half diagonally.

[@rath60]

I guess 4pi-8. Given the two half circles are congruent they mus intersect at there midpoint. Meaning the the convex segment is composed of two cords of length pi/4. Re orienting them so that instead they fill the part of the half circle in contact with the concave blue area we see that we are left with a triangle of height 4 and base 4 as the white space. subtracting the are of the triangle from the area of the quarter circle we get 4pi-8. Assuming all corves are circle arcs and the angle of the base is right.

[@godowskygodowsky1155]

The symmetric lens is just twice a quarter circle minus a triangle. You don't need the insight you describe in the video. This is perfectly fine for a seventh grader to solve.

[@gingeral253]

How do you know it’ll fill in the gap for sure?

[@polymathematic]

symmetry

[@gingeral253]

@@polymathematic Can clarify what you mean by symmetry? I understand if one fits both will, but how do you know one will fit?

[@polymathematic]

@@gingeral253 So, if you imagine drawing a vertical line down from the intersection point on the semicircles (the point that forms the upper righthand corner of the football shape), that segment is a radius of the lower semicircle. If you do the same, but draw a horizontal line from that point to the left side of the quarter-circle, you've created a 2 by 2 square. Meaning the vertical and horizontal lines we drew also split the bases of the semicircles into two radii. Finally, if you draw in a line from the bottom left corner of the quarter circle to the top right corner of the football shape (which would be a diagonal of that square shape we created) it must be the hypotenuse of two isosceles right triangles. Reflect those to the right and up, and you get the other "half" of the semicircles, with the curved parts matching one half each of the football shaped region. Since circles are the same in every direction, that's enough symmetry to be sure that the curved parts fill in the circular gaps precisely. Does that help?

[@gingeral253]

@@polymathematic Thanks I think I got it. I got it by imagining it as taking one of the semicircles and reflecting it onto the missing part. That was what I was missing.

[@-ZH]

Half leaf =pi -2 Full lead = 2pi - 4 Shades area = 4pi - 2 (2pi) + 2(2pi - 4) = 4pi - 8

[@MikeTaffet]

Alternatively, you can write it as 4*(pi-2)

[@makagame438]

What is curious too is that both shapes (the "football" and the other blue one) have the same exact area.

[@Feeelipeeee]

did it my own way and got 4.56 because I converted every Pi since I like decimal answers.

[@Notsogoodguitarguy]

The problem with trying to use the "lense split fills in the thing to make a triangle" comes when the teacher asks - How do you know that? I dunno if the teacher would allow a 7th grader to just say - I mean, look, it does. But, if they don't, then you basically cannot use this method. Teachers usually don't award points for guessing, especially without proof. I wonder if there's a mathematical way to prove that they complement each other like that.

[@MatteoCampinoti94]

It’s not that complicated for 7th grade. The steps I went through are (with x being 2 and A being the area of one of the smaller circles): * area of the slice that comprises the lens (A/4) * area of the triangle in the slice (x^2/2) From that you substract the triangle from the slice and you get half the lens, multiply by two and you get the full area. At that point it’s only a matter of adding and subtracting.

[@HansLemurson]

Not only does the problem reduce to "Quarter-Circle minus Triangle", but so does the formula for the symmetric lens!

[@rikeshramjuttun4689]

why is the radius of the lens =2 ?

[@Metalman42]

If I tried to do this from scratch, mid calculus knowledge btw. I think that I would have tried to find the area of the lens by defining it as two semicircles with the straight line segment being that of the slope from (0,0). The line segment would be found as being the hypotenuse of the triangle between points (0,0) and (2,2). From here, the area of the two semicircles can be found and that resulting area can be further multiplied by two to get the total area. That's how I think I would have solved it without watching the video.

[@tonelove14]

Find the football shape, just in time for the big game!

[@polymathematic]

ha! you're better at titling the videos than i am. that would've been brilliant!

[@scolee6408]

Or you can simply calculate the area of a circle (a fourth grade level formula) and divide by four. Pi R squared for those without a fourth grade education.

[@vincentsmith5385]

Good information, but you made it way more complicated than it needs to be. The 2 white areas can easily be made in to 2x2 squares. Add them together and subtract that from (π4^2)/4.

[@janverhave]

2*(R/2)^2*(pi/2-1)

[@amgwilly565]

bro imagine being in 7th grade and having to do this

[@puddlejumper3259]

The two white areas are a quarter circle plus a square minus a quarter circle.

[@tommyb6611]

There's just a problem with the visual solution. In most math/geometry problems, these are not to scale, and you are never to use that as a base for your solution. Use proper established methodology. And if this problem in particular was made like this, it's bad. Whomever made it just creates confusion within students, because in the future they will believe this methodology is accurate.

[@niko_walks]

4pi - 2[2pi - 2(pi -2)] = 4pi - 8

[@micksc1]

Gee did i do the long way about this - 4 simultaneous equations. But i did get the right answer.

[@kanlu5199]

4 * Pi -8, got the answer within 30 seconds

[@peterpan408]

There are many ways to get the football region..

[@chessls]

4.56 maybe

[@theAmazingJunkman]

I cheated and used integrals and semicircle functions lol

[@andywhy5879]

Overly complicated way to find the Rugby ball shape. Take the square it is within (2x2) and subtract the quarter circle (pi) to get one of the white pieces of the square (4-pi). Double that to get both of the white pieces of the square (8-2pi). Now subtract this answer from the full square to get the Rugby ball part (4-(8-2pi) = 2pi-4). Done. No rad knowledge required.

[@olegchernysh8391]

I thought 12π-8

[@svenwild813]

Uhmm. ((Pi*4²)/8)-((4*4)/2).

[@99totof99]

No time to pause and you show the solution 😢

[@underfilho]

just integrate it

[@CompuPhysix]

Please warn before spoiling the problem next time!😅

[@RealnoMis]

That is terrible advice. Math is all about precision. If you have to try and feel around to see if any of the pieces match to simplify the problem then you are going to run into situations where you assume they do but they dont, and you will have the wrong answer. Unless there is enough information in the problem to definitely declare that that football shape can fit together with the other shape, dont assume that it does. The first solution is great, but as you point out its not something that a 7th grader should be capable of. And if this solution was given to a 7th grader i would accept the answer "the question is flawed, it assumes that the football shape fits with the other shape but that is not declared".

[@trevorjames4619]

Isn’t geometry wonderful?

[@dp70737]

Spoiler alert man… i paused it a couple seconds too late & know idk if i would have solved it😅

[@user-qb8fp8oj1p]

🤩

[@SekadarPendapat]

too easy