The first comment is from 1 month ago because I posted it to Math StackExchange where I received great help: math.stackexchange.com/questions/3016718/a-way-to-find-this-area-without-calculus Early comments also come from Patreon, as I do my best to give early access to supporters on Patreon: www.patreon.com/mindyourdecisions And finally, thanks for 1 million subscribers! We are building one of the best mathematical communities in the world. Thank you.

Just check the resolution of picture and count the red pixels. It will be surprisingly accurate.

Another solution: print it on a paper, cut out the red part, weight it, divide by the total weight.

-"the first method is simple" me: "alright, this might be the first time where i can solve a hard Math problem on my ow..." -"shows space rocket schematic" me: "okay, next video"

I'm proud of myself for this one. It is the first time I solved one of your math puzzles. It took me like an hour but I got the right result of 9.77 I'm only 14 years old and I've never had anything like this in school before.

you know i've been watching a couple of your vids and there's usually a common theme in finding the correct answer. Either you use coordinates or geometry. When using geometry there's always one rule... find triangle and you'll find the answer. Just throwing that out. It's amazing to how much extent you can use this simple concepts to solve something like this

Next video: Put many circles inside a square; make most of them pass from intersections of other circles. Find the most absurd looking piece of area. And then solve for it.

For those who are wondering where π came for θ, it is the radian system or circular system for angles. 180 degree = π radian. And 1 radian = 57 degree (approx). *And 1 radian can also be represented by the angle formed at the centre of a circle by 2 radii separated by an arc of 1 unit length (on the circumference of the circle)*

Any mathematics enthusiast will be blessed by finding this channel...Hats off to the creator❤️🎉🙏👍

I’m majoring in Mechanical Engineering and the calculus method seemed to be the easiest method. 😂

I paused the video and solved the problem on my own before watching the full video. I approached it geometrically. I did it in my head without even using paper and pencil. I thought it was quite simple. I continued to watched the full video just to make sure I got the same answer. I was surprised that all the 3 methods in the videos are so long. Below is the outline of my method. Using the same labelling as 00:34 of the video, Area of shape DPQ = Area of shape DPQC - Area of shape DQC. It is easy to see that the shape DPQC repeated 4 time will fill the square. So the area of DPQC is just 25. Next step is to solve for area of shape DQC. Let midpoint of DC be E. Area of DQC = Area of sector QEC + Area of quadrilateral DEQA - Area of sector DAQ. It can be easily checked that Angle QEC = 2 * Angle QDC = Angle QAD = 2 * tan^-1(0.5). Therefore, Area of shape DQC = pi * 5^2 * (2 * tan^-1(0.5)/2pi) + 2 * 1/2 * 5 * 10 - pi * 10^2 * (2 * tan^-1(0.5)/2pi) =50 - 75*tan^-1(0.5) Finally, Area of DPQ = Area of shape DPQC - Area of shape DQC = 25 - (50 - 75*tan^-1(0.5)) = 75*tan^-1(0.5) -25 This is the same answer. I didn't finish reading all the 1.5k comments. But so far among the comments I've read, I haven't seen a similar method. This is the first time I see this problem. I'm not sure whether or not this method above has been posted before elsewhere on the internet.

Me: I know calculus well Me after watching this: I know nothing about calculus

i have enough problems in my life

If I only had 5min to solve, I would answer "definitely less than 100" & walk away

As an alternative, one might use symmetry and scale to simplify the solution. Steps: 1) 'Symmetrize' the picture by drawing another quarter circle radius 10, centered at the TOP, RIGHT vertex of the square. This divides the square into 8 regions, of which 3 are pairwise identical, thereby resulting in 5 distinct shapes ( with areas x1 thru x5). Our target region (shown in red in video) is one of the paired regions (say x2). 2) Express the areas of the small semicircle, the large quarter circle, and the square as linear equations in the variables x1 thru x5. 3) Thus far we have 3 linear equations, (and 5 unknowns) so we need 2 more. These are a) The area of the small 'lens' region formed by the intersection of the two semicircles, which can be easily computed as described in the video, or otherwise. and b) The area of the larger 'lens' formed by the intersection of the two large quarter circle. Since the larger lens has the same shape and twice the linear dimensions as the smaller lens, its area is 4 times as large. 4) Now we have a complete set of equations, from which x2 can be easily obtained.

9:00 instead of using trig functions, we can also apply the similar triangles rule. Find the ratio of the kite triangle and smaller triangle hypotenuse, then their areas have ratio to the square

Option 4 - Depending on how many sig figs you need. Step 1: Find a sheet of graphing paper. Step 2: Sse a compass and rulers to draw out the figure. Step 3: Spend the last 3 hours of the exam to count the number of little squares.

I used the 4 option.... I watched the video

I Just loooooove the feeling when you can use so many different solutions to find the same answers. Keep up the good work!

The amount of work that goes into your videos is truly phenomenal 🙌

It's funny how calculus is 10 time easier than geometriy

I literally looked at the thumbnail and tried to solve it for 10 mins. My friend said “are you looking for something on the screen?”

Area of intersection of two semi circles can be calculated by drawing all four semi circles inside a square. Suppose area of semi circle is A, area of intersection of two semi circles is B and area of the square is C than 4*A - 4*B = C. Because when you add areas of four semi circles together, each intersection will be counted twice so you have to subtract four times of the area of intersection in order to make it equal to the area of squire. For a square of side 10, the value of B is comes out to be 14.26. It is first half of the solution.

I’m just gonna pretend I understand what he said

Your first "little lens" can be calculated much simpler, namely as the 5x5 square without the two other shapes, so that's 25 - (25 - quarter circle) - (25 - quarter circle), so that is a half circle of radius 5, minus 25 all in all, making it 12.5pi - 25. That's a whole lot faster than your "lens" calculation.

I put this in AutoCAD 😎😎😎 Ans - 9.7736

I literally just finished calc 3 this morning so it was still very fresh in my head. I was thinking "oh, you could do polar coordinates and use single or double integrals (whatever you fancy), broken up by the intersection point, with the origin being in the top-right corner" I actually hate that these solutions were significantly easier and used a much more basic version of math

I'm proud to report I sat through the entire video without skipping. That alone is an accomplishment given how little I understand anything that was said.

I just recently discovered your channel. I did the problem with both methods, the area subtraction using geometry (since that stood out to me) and the calculus way (because that just made sense integrals for the win). I set up the problems in my head and didn't particularly solve for an exact solution. Once that was done I finished watching the video. I m glad my physics/math brain is always on the ball great problem. Math is just awesome!

Thanks, dude you made me realized I need to retake Calculus and Integration.

Something tells me we may be able to solve this with a system of equations. I named each section with constants a --> f (with b being the objective), and got to b + e = pi*(10)^2/8. Only worked on it for a couple of minutes but I'm sure I can represent e (which would the the lower right section of the quarter circle) in terms of the total area of the square then I'm good! Thanks for sharing this Presh!

I ended up a lot closer than I expected. I've been challenging myself recently to not use a calculator while solving problems, so when I tried using the geometric solution mentioned here, I ran into a major snag when having to calculate the inverse tangent. I ended up taking a best guess on it based on what looked right for the problem and ended up with a final answer of "9.9 or a little less". I'm surprised that I ended up using the correct method though. Usually geometry is my Achilles Heel for mathematics. I didn't even consider using calculus to solve for the answer, which surprises me given how much I tend to default to it.

One of the best math channels of KZfaq, greetings from Brazil!

I am still struggling with this problem so I haven‘t watched the solution- it is driving me nuts! I am not yet ready to give up. Wish me luck. 🤔🙄😳😠😩

LETS GOOOO I ACTUALLY GOT ONE OF YOUR VIDEOS RIIGHTT ✊🏼😤 Damn I've never been so proud of myself lmao

I used the 2nd method with some modifications. When calculating the blue shape I first calculated both of the sectors, and then I took the total "kite" area (= 50) and substracted one of the sectors from it to get the part of the other sector that is outside the blue lens. And then I simply subtracted this area from that other sector to calculate the blue shape. Seems to be easier.

And i was there thinking i can do it in my head with my calc 🤣

I solved this problem using calculus, it was the fastest and easiest way, in my opinion. I really liked the other two geometric forms of solving it!

Hi Presh, many thanks for this gem. It's a lot more simple though. Focus on area PDA (via inner arc). I call it the battle axe. It's area is obviously 25. You instantly also know the area of PD, the double-sided symmetrical konvex lens which I call the fish. (done via computing semi-circle minus 25). Now work out area BCD (via point Q), I call it the "kite", like a hangglider (done by the "square" minus "quartercircle"). Thereafter you work out area QBC using year 10 high school maths (drawing a line from Q orthogonal to line BC) followed by integrating separately on both sides of this line "under" (here to the right of) the two arcs of the respective circles. Take QBC away from the "kite" and you get QCD, known as the cleaver. Take away cleaver and fish from already known semi-circle and yield PQD being 9.7735... Woody

1. 3D print the shape but give every edge the same height wall 2. Measure how much water it hold and how much red area holds 3. Red water divide by all water X100 4. Area of Square X Step 3 = Area of red

Wow! You really had a lot of work with this one! Great job! I used a mix of methods. For the 5x5 lens, I used geometry b/c its very easy to calculate: Area of 1/2 circle with radius 5 - 1/4 Square area. For the 5x10 lens I used calculus, but I turned the original pic upside down. So then it became integral of smaller circle from x=0 to 8 minus the integral of the larger circle from x=0 to 8. To solve the "sqrt's" integrals I had to look it up in a table of common indefinite integrals. PS: I also did it in the geometrical way, but my final solution uses arcCos function b/c I used the cosine law to determine the angle of the circumference arcs.

I did not calculate it, I guesstimated 10%, being a building contractor has it's perks ! ;)

I solved this when I was a fetus in India.

Thanks for making me feel like I learned both something and nothing in this video

I liked it at the end when he said "did you figured out this problem ?" XD Please have mercy

As soon as I saw this picture, the word "calculus " came to my mind, even though I had no idea how exactly...

I used method 1, except: first, I was too lazy to do the math for the larger lens, so used an online calculator for that (area=24.043), and second, I found the area of the smaller lens by a simpler method. Not having the graphics available, I describe it thus: the smaller lens is found by calculating 1/2(area of left-side semicircle) - 1/2( 1/2 the box area - area of upper semicircle), or 1/2(39.270) - 1/2(50-39.270) = 14.270. So large lens less small lens is 24.043 - 14.270 = 9.773.

I've graduated high school 2 years ago and I almost forgot those math elements.. Now I remember that I was genius in math when I was a student and now I realized that my brain is rotten since graduation... Better study again. I've been wasting my time since graduation playing games and internet surfing... Thank you so much.

I literally wrote a python code to do this computationally: start with an nxn grid. Imagine a circle at point (0,0) of radius n. (circle 1) Imagine a circle at point (0,n/2) of radius n/2. (circle 2) Imagine a circle at point (n/2,0) of radius n/2. (circle 3) Loop through literally every point in the nxn grid. Add to variable A if the point is: inside circle 1 AND outside circle 2 AND inside circle 3. The ratio A/n^2 is the percentage of the square containing the area we want. Multiply by 100 square units and you have the answer to the question. I blew n up to 5000 and waited about 45 seconds for my code to run and I got 9.77, so there is a "Red Green" solution lol.

This is solvable using geometry and algebra only. Algebra can find the coordinates of the right-most circular intersection using the general equation for circles and the known centers. Knowing that, a sufficient system of triangles can be constructed such that the desired area is obtained purely by addition / subtraction of triangular areas and arc segment areas. I will confess it took 2 hrs to come to this conclusion.

Same with me. As soon as I saw the picture I thought this is to complicated to think about. Let's just integrate and get it over with. But I knew for sure there must be a geometric solution.

I got the area of the yellow area a differrent way: Area of the semicircle = 12.5π (Solve with πr^2 × 1/2) Multiple it with four 12.5π × 4 = 50π It is like you draw four semicircles inside the square. Or imagine as mirroring the two given semicircles. Now you can see it covers the whole square and also the yellow area 4 times overlaying. Which means 50π = area of the circle + 4 × yellow areas 50π = 100 + 4y (y=yellow area) y = (50π - 100)/4 y = 25/2(π-2) y = 12.5π - 25

2:02 Solution using geometry and trigonometry 6:49 Longer solution using geometry and trigonometry 12:06 Solution using calculus (integrals)

I did integrals but I’m finding a common theme. Whenever looking for an area of a circular shape, find a right angle triangle that splits the needed area into a right triangle and a proportion of the circle

1. Suppose it was given the rectangle has length 10. First we have to prove the rectangle is a square. 2. Also the yellow lens area is not necessarily symmetric is it if it is a rectangle? 3. This applies only for the square case.

You can divide the square in to 4 smaller squares each with a area of 25 and assume the circles radius to be 5 and 10 by completing them, the result is similar and can be calculated by normal math.

As an observation, the formula found for the area of a lens in the video is for the special case of a lens where the two circles meet at right angles - in general, you won't have right-angled triangles, and end up with a longer general formula (the standard version is given in terms of r and R, the two radii, and d, the distance between the two centers)

When you showed that the length of the side of square is 10, my first guess was "the area is somewhere around 10". And then you showed that it is around 9.77. :D

My hearty thanks:from India

i just drew this up in Fusion360 and had it measure the area for me.

"So let's simplify" Me: Yes, please *complex equation turns into another complex equation* Me: ok

My ignorant younger self took AP Stats instead of AP Cal..so I did this the long geometric way, and essentially just nodded my head as you narrated the on-screen calc operations, like, "Yes, the audio appears to be in sync with the video here," so, long story short... 🙏 I appreciate your videos, which make me wish I'd taken more maths courses at university, because I always enjoyed the subject and I ought to have continued to be good at it (Well, that wasn't very short, was it?) ✌️

I used hyperoperated calculus with stable equality with a trig failsafe, also including operation successorship. This method is unbeatable.

Ngl its an emotion After 2.5 hrs of constant doing math i did itttt Yess i did it

Wait a second, this video was a month ago. Was it only for patreons or something?

“Uploaded January 10, 2019” 1 month comments W A I T

I remember a few years ago I wasn't able to solve this. I was able to solve it today having learned calculus. The geometric way requires less calculations, but the critical thinking involved in that method would have taken me longer to be honest. So the calculus way was the fastest way to solve it for me.

I've never felt more smooth brained in my life.

*3 Ways to Get Depressions (Insanely Work)*

After watching the solutions, I realize that I'm simply mathematically undereducated, I didn't have the tools to solve it with what little trigonometry I remember from school. I guess, I'll have to start from scratch with basic maths if I want to be able to tackle problems like these... Oh well, it was fun trying to derive what I could with what little I could remember.

🤯 I get the second solution best out of all of them but I wouldn't have got there in a million years and I wouldn't be able to do it even now without watching this again!

The way I did it was drawing the diagram and putting A, B, C, D, E and F into each of six parts of the jigsaw puzzle, with B being the shaded bit. If we take half of one side of the square as r, the area of the square is 4r2 then express B = 4r2 - A - C - D - E - F and you can find other relations where two of the letters add together to make half an area of a circle etc etc and eventually all other variables cancel out to give the answer.

The second method seemed to me to just be the first method worked out twice with specific values for a and b. In both cases you find all the triangles etc.

Could you work out the area of the semi circles and the quarter circle, then add them up, then take away the area of the square, then since you have two layers of the required area then dive it by 2?

0:53 yeah

Hello! I am from Brazil and I love your channel. Thanks for everything

MCAT skill: “idk looks like around 10% or so” Mathematician / physics skills * 25 minutes of work later = 9.77 *

I looked at that and thought: It is pretty straightforward with calculus, though it would take a while with multiple integrals. With geometry, that's tough. EDIT: I wrote that before watching. That makes me feel better.

For the calculus solution, I used polar coordinates. Rotated the figure and shifted the origin to the point where the left, top point (0,5) is now located at. Doing so, all the curves pass through the origin and r-theta functions are easy to set up. Used A= 0.5 x integral of r^2 d theta. All the cartesian equations of circles can be converted easily to polar x = r* cos(theta) etc. This is a shorter and less painful method than using cartesian system shown here. All the angular limits are easy to find using regular trig/geometric formulae.

The method by the integrales is great !!! And it's a good mean to understand how powerful are the integral concept and tool !!! Very nice video !!!

I used an calculator. Definitely the best method.

Here is another way of showing the geometrical method: For the semi-circles, we have the circle area formula of πr², so: half circles area = 25π/2 = 12.5π quarter circle area = 100π/4 = 25π Considering that the small lens goes between two identical circles, we can claulcat it as: smalllens = semi_circle1 + semi_circle2 - kite smalllens = 12.5π + 12.5π - 25 smalllens = 12.5 x (π-2) or smalllens = (π-2) x side² / 8 And for the biglens we need the lens formula: biglens = side² x (6 x arctan(0,5) - 4 + pi) / 8 f = side² - 12.5π - 25π + biglens Then we can solve: Answer = biglens - smalllens

okay, at the end of the video, you asked if i got this question and what method I used, but I did Not in fact get an answer, but I had the thought process of what methods I would have used which were all correct. I haven't been in math for two going onto three years, but I knew and followed everything you were explaining. Made me smile

For solving for that yellow portion you can take the area of a square 10x10 then subtract the area of a circle r = 5 divide the num by 2 it will equal 10.75 then take the half-circle givin and subtract the 10.75 this will give the area of 2 of those yellow shapes so divide by 2. I think this may be easier to follow example that more people would be able to understand. I realize that what I said above may be hard to follow but with pictures, everyone would be able to follow. However, I have been unable to solve for the area of the blue shape using only basic shape areas and/or without advanced geometry knowledge. Originally, when I tried solving the problem I was trying to take the half circle and subtract the yellow shape and DQC. Would be super interested if anyone were to solve this using only the basics of geometry.

I approached it with calculus. BUT the geometric solutions were elegant.

Love your vids! Please don't stop!

I think there is a simpler method that uses no trigonometry, pls. correct me if I miss anything: 1. Compute the yellow area as in method 2 2. We can compute DPA area as the semicircle minus the lens we already calculated 3. We can compute DPC area in the same way (also they are equal - symmetry) 4. We can compute APCB area as the square minus lens minus DPA minus DPC 5. We can compute DBC as square minus DAB 6. We can compute DPABC as square minus left semicircle 7. Now we can form 4 equations with 4 unknowns: - DPQ + DQC = DPC (known from point 3) - DQC + CQB = DBC (known from point 5) - APQB + CQB = APCB (known from point 4) - DPQ + DQC + APCB + CQB = DPABC (known from point 6)

- Draw it on CAD software - select the boundary - Find the area

Took me a full hour, now i finally can go to bed

It’s been 20 years and I still remember soh cah toa.

Thank you for your helpful lecture. The method of subtracting area to calculate the area of ​​the figure to find.

I did't solve it but just watched the solutions. Very interesting, and of the three I think the second solution is simpler and better. The Calculus solution is very drawn out and would need a person who is good at it - we all can be good at Calculus, just keep practicing and memorize first pricipals. Thanks for posting and really enjoy your maths videos.

*10* *9* *8* *7* *6* *5* *4* *3* *2* *1* *0* *and that ends the official countdown of my brain cells*

He said "simplify" Me : "How?"

I know I am late to the party, but I have one small suggestion. You could make your life a little bit easier during the trigonometric solutions when you assume the square to have a length of 1 and at the end multiply the solution for B by 100. This way you get rid of a few r^2 since r=1.

I gave my math teacher this problem at the end of last lesson and she, her sixth form class and the head of math spent their whole lunch and the entire sixth form period attempting it, she had to watch this video to figure it out

My head hurts now.....

I feel like if I just studied, revisited and rediscovered the maths I learned in polytechnic I'd be able to solve using both trigo & calculus. I really need a jumpstart course on that.

The geometric stuff is so confusing, calculus is so much easier.

Intersection between two equal circles in that position will always be (pi/2 - 1)*rˆ2. You can easily deduce it drawing the combination of the two circles into a square of its diameter.