Well this kind of calculation is pretty easy using integration (Calculus), but it would be pretty satisfying to find an elegant geometric solution. I wonder if there is more than one elegant geometric solution.
He makes it look much worse than it actually is. Simply draw a few straight lines, and you see the blue area is equal to a much simpler shape. The lines needed are: 1. A diagonal line down the middle of the lens shape 2. A vertical line down from the top right corner of the lens. 3. a diagonal line between the top left and bottom right corners of the big quarter- circle. With these three lines drawn on top of the figure, you can see the small lens-shaped part of the blue area is equal to two small half lenses. You can also see the long, thin part of the blue area is equal to a big half lens minus two small half lenses. This means the sum of the blue area is a big half lens minus two small half lenses plus two small half lenses, which is simply equal to a big half lens. You should also be able to see a big half lens is equal to the big quarter circle minus a big triangle. Since the quarter circle is πr^2/4 with r=4, we get a big quarter circle is 4π. Since the big triangle has a base and height both being the radius, we get the area of a big triangle is 4*4/2=8. The area of a the big half lens, which is equal to the area of the blue shape, is therefore equal to 4π-8.
It's a bit confusing given that you have to make a bunch of assumptions, eg. the circles being perfect. I think that it makes the problem look much more difficult than it actually is. the picture should have listed the dimensions of the smaller shapes that are not meant to be derived through calculations but are essential to actually solving the problem.
its pretty simple. i tried to figure it out for myself before watching, but i got stuck on the two x’s up until he used a right triangle, then it all came together for me
@@HellecticMojo Coping? Dawg hop off the internet lmao. I literally said all it does it make it look more difficult, not that I couldn't solve it lol. You probably never studied past pre algebra so I totally understand your inability to understand my point.
@@FirstPersonLife before watching I thought the 2 Xs, added to the Zs, would have completed the "long side" of a square triangle with two sides of 4. So the Blue area would be: A quarter area of a circle with radius 4 - the area of a square triangle with sides of 4 (which is 4*4/2). I think it works
It his weird way of doing maths. You move the triangle to the ither side of the equal sign you dont cancel both sides by the variable, triangle in this instance. Thats retardes.
nothing has ever blown my mind harder than the sentence "we can subtract triangle from both sides"! I absolutely love the way you think about these problems and these videos always make me smile.
Another method could be to ”move” the x-areas into the ”citcular hole” in the z-area. Doing this on both x-areas creates a non-shaded triangle of b=h=4 which means an area of 4x4/2=8. The area of the quarter circle is 4pi meaning the shaded area is 4pi - 8
both white areas can be cut and moved around to create simple 2x2 squares.. so the white area is 8 sq units, and the whole quarter circle is 4pi. There you have it, the blue area is 4pi-8.
exactly - I spotted that in about 20 seconds and thought "gee I hope doing it with the algebra is just a precursor to showing how much easier it is if you do a simple bit of geometry first" ... but I checked the whole video and no he never mentioned the much easier method - disappointing as I don't mind most of this guy's work.
Since both 2x + y and x + y + z were equal to 2pi it was probably easier to just determine that x and z were equal to begin with, and that 2x + 2z can be renamed to 4x equals the area. Then once getting that pi - 2 is x, you can just multiply out to get 4pi - 8 is the area.
Exactly, i watched it twice cuz i thought i somehow missed the reason this solution wasn't viable. But it is. So i suppose the guy just did an extra 3 min of impressive graphics and basic geometry moves for no reason. Still very interesting choice of solution by him...
How exciting! I did it slightly differently, I took the bottom left part to be a square with sides 2. With that area being 4 square units, if I took away 1/4 of the circle with radius 2 from the 4 I would have a variable I called x. I saw that if you had a square of area 4 and you took away 2 of those x’s you’d be left with the blue value between the smaller circles. If you then took the whole quarter circle, took away 2 quarter circles of radius 2, the square of 4 and added back on the value found earlier you’d arrive at the same conclusion.
I did it as the following: label the top blue part “A”, the bottom blue part “B”; take the 2x2 square around B. Area of B is the overlap between the two white parts in that square, since there is no empty parts, its just the combined area of the white parts in the square, mod the area of the square. A_whiteareas is just pi x d^2/8 = 2pi. Now we have area of B as 2pi mod 4 or 2pi-4. Now for the area of A. This is the area of the whole quarter circle, - the areas of the white + back the overlap. A_qcircle is pi x r^2 /4 = 4pi. Substituting with all the values we have gives 4pi - 4pi + 2pi - 4, canceling out the 4pi’s, we have area A as 2pi-4. Adding this with the other one gives us the final area of the blue as 4pi-8 u^2. Quite lengthy but thats just what came to mind when i saw it.
Andy your the greatest teacher at math in youtube, please teach us how to make it simpler because alot of people are having a hard time, only few can do this like you.
It's faster to recognize that the two x regions can be moved to mate with the z regions, creating a single circular segment of radius 4 and angle 90. Then it's just the area of the full quarter circle minus the area of the right equilateral triangle.
I did this one a bit differently. Rather than focusing on the blue area, I found it by working out the white area and subtracting it from the area of the total shape. If we break it down, the white area is made up of two quarter circles plus an extra bit that we will get to later. To find the area of the quarter circles, each one has a radius of 2. Since the area of a quarter circle is πr²/4, we substitute in 2 to get: 2²π/4 = 4π/4 = π Due to there being two of these quarter circles their total area would be 2π. If we drew a square of radius 2 around the pointed oval type shape in the bottom left, it becomes apparent that the white area within that square is equal to two of the remainding area when you draw a quarter circle of radius 2 within the square. Using the same formula, πr²/4, we can work out: 2 * (4 - 2²π/4) = 2 * (4 - 4π/4) = 2 * (4 - π) = 8 - 2π Thus, the total white area is equal to: 2π + 8 - 2π = 8 Using πr²/4 once again, we can work out that the area of the total shape, as it has a radius of 4, is equal to: 4²π/4 = 16π/4 = 4π Therefore, as it is equal to the total area minus the white area, the blue area is equal to 4π - 8 square units.
If you rotate the two blue circular segments until they are adjacent to the two uppermost blue pieces, the white and blue areas retain the values of their areas but with this maneuver the white area becomes an isosceles right triangle with legs equal to 4 (therefore area 8) and the blue area is a circular segment that has an area equal to the difference between the area of the quarter circle you calculated at the beginning (4pi) and the area of the white triangle (8). So the final area is 4pi-8.
Hardly, when compared to the geometric solution (which someone else has already pointed out.) A simple rearrangement turns it into a really easy puzzle.
quick question, how do u prove that intersection between x areas and z areas is located exactly half of the two semi sphere, making 90 degree angle, which ultimately leads to a conclusion that height of the triangle is 2?
Assuming the two half circles are proper half circles, their "centers" if they were whole circles would be exactly in the middle of the lines they're on, labeled 4 units long. Radius from that point going directly up and from the other half circle going directly to the right would get you the coordinate point (2,2), if you imagine the bottom right being the origin. So the triangles would be 2 units high. 🤙
Draw a line between the center of both semi circles. Because the two semi circles have the same radius, and the bottom left corner is a right angle, it has to be a 45-45-90 triangle. Both semi circles intersect st the corner of that triangle, and by symmetry, both semi circle has to intersect at the same point again opposite to the line. Combining these two triangles create a square with side length 2 with corners at the intersections and the centers of both circles.
I solved this by finding the areas of x and the semicircles so I could use subtraction to find the area of y and then z, but I like the methodology you used better because it is generalizable. 1. Label the areas with N unknowns 2. Find N linearly independent equations 3. Row reduce the system of equations Thanks for sharing!
The semicircles are the same because their diameters are the same. When you divide the large quadrant in half, the right and left are divided equally in half. Imagine bringing the two x's formed to the point where y and z intersect with a little imagination. The question is now like this: (quarter circle with radius 4) - (isosceles triangle with side 4) = (4π) - (8)
I also like the geometry approach to this one. The quarter of the larger circle - both semi circles = 2z-2x. However the quarter of the larger circle (x2 radius = x4 area) is the same area as the smaller circle. So smaller circle - smaller circle = 0 = 2z-2x. Then you calculate 4x as the smaller circle - the square inscribed in the smaller circle. All together circle - circle +(circle - inscribed square) = circle - inscribed square = 2z-2x + 4x = 2x+2z. Beyond the joy of adding and subtracting the same circle, I like how it reveals 2x=2z.
A little geometry makes this so much easier. Bisecting the quarter circle leaves you with the 2 x sections, which fit exactly into the scalloped section from the 2 z areas, which leaves the quarter circle with a redistributed shaded area in the curve, and an unshaded half square internally. This can be done because the two semicircles that make up the scalloped portions are exactly proportional to the larger circle. Bisecting the larger circle through the two smaller semi-circles leaves us with a tangent that is exactly half the length of the secant connecting the two points forming a hypotenuse with the radii making the right triangle. The area of the entire quarter circle is (pi*r^2)/4 with a radius of 4. This becomes (pi*4*4)/4, with a four cancelling above and below, leaving us with 4pi. The half square has an area of lw/2, which equals 4*4/2=8. The shaded area is therefore 4pi-8 sq unites.
Great way to do for the sake of completeness and understanding. You can find out that z and x are equal from the start if you just consider the sum of the areas of the two semicircles being equal to the sector itself. From that you can just call both shaded parts X. From there I just drew 2 lines as you did creating a box and solved for X.
The area of the big quarter-circle is the same as the area of two small half-circles, subtract and get 0, carry the blue bean because it's subtracted twice. The area of the blue bean is 2x area of small quarter-circle minus half area of square (cut diagonally), so 1/4π2² - 2²/2 = π - 2, times two. Carry the blue bean, the answer is 4π - 8.
If you draw a radius from each of the 2 semi-circles (one horizontal, one verticle), the lower left blue area can be seen as the overlap area of 2 small quarter circles contained within a square. The lower left blue area can be calculated by summing the areas of the 2 overlapping quarter circles (𝝅2²/4 + 𝝅2²/4) and subtracting the area of the shape formed by the perimeter of their outer (non-overlapping) edges which is the square that they are contained within (2²). The 2 radii drawn to create the square also form small quarter circles at the top left and bottom right. The upper right blue area can be calculated from the difference between the area of the large quarter circle (𝝅4²/4) and the sum of 3 areas: the 2 small quarter circles (top left and bottom right) and the area of the square (𝝅2²/4 + 𝝅2²/4 + 2²).
I saw 2x + y = 2π and x + y + z = 2π, and from that got x = z. Wouldn’t save you all that much work, but it would mean you could work in two variables from then on.
I used the square corner and its two quarter circles to get the shape area of 4+2pi. Then the area of the two semicircles is 4pi, and the area of the large quarter circle is also 4pi. The blue eye is obtained by subtracting the main shape from the semicircles, and the blue spectre is obtained by subtracting the main shape from the large quarter. They both result in 2pi-4. Added together its 4pi-8.
I like this solution. I did mine a slightly different way: Got the full area as 4π and the smaller semi-circles as 2π. I then split the semi-circles in half and subtracted that from the area (4π - 2π = 2π) I then took the total area of the overlap of the semi circles (2²=4) and subtracted half of a semi circle (4-π) giving me the remaining non-blue area in the second half of one of the semi-circles Finally, I subtracted twice the second calculation from the first one and got my answer (2π - 2(4-π)=2π - 8 + 2π=4π - 8) Hope that makes sense,
As a Calc student who just learned how to find area so I would find x using the top and bottom bounds. Add the semi circles together and subtract x and then subtract what's left from the quarter circle to get z
If you draw in two more semi circles to turn the image into a square it’s easy to see visually that if you split the lower left blue area in two, and “slide” these mentally clockwise and anti clockwise, you’ll create a white right-angle triangle of area 2*2*0.5= 8, and a blue semi circle of unknown size, but whose size equals the initial two blue areas summed. So just subtract 8 from the area of the quarter circle (4pi). Also the two blue shapes have the same area; you can prove that by subtracting the area of the semi circles from the larger quarter circle. This gives you the area of the upper right blue shape minus the area of the lower left blue shape, which turns out to be 0 ie they have the same area.
Another fun solution is drawing a rectangle around the white area, size 2x4. Cut it in half, size 2x2. The “positive space” in one is equal to the negative space in the other, which means you can put the two half’s together and get one whole square of area 4
*More simply* just draw a diagonal line from top-left to bottom-right. By symmetry the 2 half white lens are equal to the 1 full blue lens. So, the blue area answer = Area of the large quarter circle - Area of the large isosceles right triangle = pi / 4 (4 x 4) - (1/2 x 4 x 4) = 4 (pi - 2) More generally, the blue area = (pi/4 - 1/2) R^2 = *Area of the large quarter circle - Area of the large isosceles right triangle* .
Very interesting, the comments have all found in many different ways. Mine was: If you make your small 2 x 2 square as you did surround the x sections. You can solve for the white of a single side and add it to half semi circle to get one of your Y area's in the video. Then can multiply be 2 to get all area of white and minus from blue quarter circle to get answer. Area of small white section in small square =( 2*2- ((π(2^2))/4) = 4-π Area of white quarter circle =(π(2)^2)/4 = π Total area white sections ( your Y multiply by 2) =2*(4-π+π) = 8 Blue quarter circle minus white = (π(4)^2)/4 - 8 = 4π -8
I used a much simpler solution. First I found the area of the large quarter circle 4pi. I drew two lines going from the center point of the intersection to create 2 quarter circles which have a radius of 2 and an area of pi. Instead of finding x, you can easily find the white area inbetween the midpoint and the weird curve that bounds x. Just use the square which has an area of 4 then subtract the area of the semi circle now its an area of 4 - pi. Subtract 2pi, for each quarter circle, and 4 - pi, for each weird curve region, from 4pi. You get 4pi - 8.
How do you know for sure that when you split the semi circle in half that the line you made connects with the edges of the blue section you marked as x?
If you’ve already accepted that it IS a semicircle, the intersection being the midpoint is a given. But this particular image is missing measurement marks that indicate it’s a semicircle, that the larger arc is a quarter circle, and even that the corner is a right angle. On the other hand, the point of the video isn’t proper labeling technique, but how to approach complex shapes analysis.
Umm... The diagonal helped, but after that, I did it this way. 1) Draw the diagonal (like he did). 2) Due to all the shapes being circles (and the way the lenses connect 90 degree radii), when the biconvex (the thinner blue section) is cut in half, you can "shift" those pieces to fit into the larger blue area's concave sections 3) You now have one white right triangle and one blue plano-convex lens area. Simply subtract the area of the triangle from the area of the area of the quarter section of the larger circle. Alternatively, if you think of the smaller circles as rotate-able, you will see how I got the blue sections together more clearly. Also, his cutting of the sections at 1:36 will help see the triangles (and a leftover square) that you will have to subtract from the larger circle to get the remaining blue area.
I did it a bit differently but got the same answer. I drew a 2x2 square and subtracted a quarter-circle with radius 2 to get the section of the white part of y that would be inside of said square and outside said arc (a). Then I doubled it noting that the same deduction is made on the other side of the arc (b). I subtracted that number from the 2x2 square which returned 2Pi-4. I then noted that everything outside of z within the 4 radius quarter-circle can be described by a 2x2 square plus 2 quarter-circles. In other words, Z could be described by subtracting the r=4 quarter-circle with a 2x2 square and 2 r=2 quarter-circles. So, z could therefore be described as Z= ((4x4xPi)/4) - 2(2x2xPi)/4 - 4. This came out to 2Pi-4. Add Z and X together and you get 4Pi-8.
My approach was this: To calculate the area of the eye shape (your 2 x'es) simply do 2 quarter circles with radius 2 minus 1 square with an edge length of 2, so: 2 * (π * 2² /4) - 2² = 2π - 4 => the "eye" shape Then calculate the area of the large quarter circle like you did and subtract 2 half circles with radius 2, then add 2 "eyes", so: π * 4² / 4 - 2 * (π * 2² / 2) + 2 * (2π - 4) = 4π - 4π + 4π - 8 = 4π - 8 This also means, that the white area is exactly 8.
I believe the way I calculated, is a little easier if you like thinking in shapes. Seeing that 1/4 * Pi4^2 = Pi2^2 ( a quarter big circle = 2 half small circle = small circle) you can see the both blue areas are the same size ( Since the overlapping 2 small circles form a blue "leaf", but that leaf should've filled the rest of the big quarter circle ). Now you just try to find out what that size of the leaf is. If you imagine 4 half circles in a 4x4 box you can see it forms 4 identical leafs. Meaning 4 half circles minus 4^2 (the box the half circles are in) equals 4 leafs. ( 2Pi2^2 -4^2) . Now just divide that by 4 and you get the area of a leaf. Double that and you get the whole blue area. It sounds complicated written out like this but once you imagine it, it's so much easier. Hf
2(x+z) makes a segment of the whole quarter circle, so you can take the area of the quarter circle 4π and substract the area of the right triangle formed by both radii, which is 4x4/2, obtaining the same solution in like 5 seconds
I have no idea why do how this clip ended up in my feed but all I know is I’m genuinely saddened I gave up mathematics in school. When you showed that x=z I paused and looked hard at the shapes and thought ‘yes that looks right’. I don’t know where to go from here. Happy Easter I suppose.
Simple visual solution: Draw the line from (0,4) to (4,0), the triangle formed has an area of (BH/2) = 4x4/2 = 8. Note that the area "x" can be rotated outside the triangle to fill the space between the diagonal and the area "z". This means that the area of the quarter circle minus the area of the triangle is the area in blue. The area of the quarter circle is easily calculated as 4pi so the difference in areas (blue area) is 4pi-8.
I'm 26 and you're getting me excited about math again for the first time since I graduated highschool years ago. It's kinda crazy to think stuff like that used to be my bread and butter when I was 15 but now I'd have to google the formula for the area of a triangle. You're helping me get back in touch with who I was before the 9 to 5 grind took over :) it makes me feel smarter and flexible in my way of thinking. Like hitting a mental gym. Turns out I can still run! Yay!
If you cut the white shapes along what would be their radius perpendicular to their base (e.g. into 2x2 squares), you'll get two pieces. If you put the two curved parts of the pieces together, you get a 2x2 square. Thus, [A] = 4^2*pi/4-2*(2*2)=4pi-8.
Split the quadrant into an octant along the symmetey axis and note that the blue area (x) matches rhe convext component of y. Swap those two areas to show that the octant contains a white triangle and a blue convex element. Reverse the symmetry of the octant to reform the quandrant as a white triangle with a blue convext element. Total area = 4pi. Triangle = 0.5 x 4 x4 = 8 Blue area = 4pi-8
I used a different approach, where i calculated the area of the overlapping small half circles and of the big quarter circle separately. I first treated the half circle as a full circle and constructed a square inside it. Then took half of (the volume of the circle minus the volume of the square) to get the overlapping area. Then, I took the volume of the big quarter circle minus the volume of the 2 small half circles {which is just the area of a small full circle) plus the already calculated overlapping area. I got the same result.
Cool, I did something similar and got the same answer! I found the 4π of the quarter circle with a radius of four, and then the area of the two semi-circles with radius of 2, which also comes out to 4π, so I knew the area where the two semi-circles overlap is equal to the other blue region. I then did the same thing you did to find x, then multiplied by four to get the answer
The two semi circles are identical, and the quadrant has a 90° angle. I don't know how to explain this, but i feel like these two facts can prove that x=2
You could also just subtract the semi-circle from the entire shape twice (PI*4^2/4-PI*2), then since we subtracted the 2*x blue zone twice we calculate x same as in the video, add it 4 times to the shape and boom we got the result
Since we quite early on established that the big quarter circle's area is 4π, and later established that y = 4, we could have stopped there, after a single substraction. The blue area is 4π - 2y; or 4π - 8.
If you see that the area of the large quarter circle with radius 4 is equal to the area of the two half circles (with radius 2) combined, you know that the the area where the two half circles overlap is (the eye shaped blue area) equal to the area where neither of them overlap with the large quarter circle (the "weird" shaped blue area). Another way of thinking about that is that, since you need both half circles to cover the entirety of the bigger quarter circle, every piece that is covered twice by those half circles will by definition remain uncovered someplace else. When I saw that, I calculated x from Andy's explanation and multiplied it by 4 to get the answer.
If we take one x and rotate it by 90 degree taking center of semicircle to put it close to near z. Similarly do it for second x for second semicircle we will get triangle of 4×4 because lines of will be connected to become straight line. Then blue area will be= area of quadrant with radius 4 minus area of triangle of both base 4. 1/4 circle of 4 - square triangle of 4 = 1/4 π 4×4 - 1/2 ×4×4 = 4(π-2)
That really threw me off as well - I'd assumed going in that would be one of the main challenges of the problem. One can easily imagine a shape that to the naked eye looks nearly identical to this one, but that has a slightly more concave/convex arc and therefore a different area. I wonder if there is some reason this assumption could safely be made? I don't know enough about geometry to say, though that doesn't sound right to me. Even if that is the case, I wish he'd mentioned how he knew that everything was perfectly circlular.
The way I did it started with finding the area of the quarter circle (4 pi) and trying to break that down until we isolated the blue area. The two semicircles have areas of 2 pi, meaning that they can be divided into two quarter circles of area pi each. Each semicircle has a half that is wholly in the white area, meaning we know that there is 2pi not in the blue area, meaning that the remaining sections, the blue area that is outside of the semicircles and the area where the semicircles overlap, is equal to 2pi as well. You can then cut out a section where the other half of the semicircles are, making a square equal to area 4, meaning that the upper blue area is equal to 2pi - 4. If we call the lower blue section X, we also know that each of those white sections have an area of Pi (the area of the small quarter circle) minus X (the area of overlap), meaning that that total square is 2(pi - x) + x = 4. When you solve for X, it simplifies to x = 2pi - 4. So, the two areas of blue both equal to 2pi - 4, meaning combined they are 4pi - 8 :) So rare for me to be able to find the solution for one of these!
Something really interesting about this example: the whole quarter circle's area is equal to 4pi and the two semi-circles are equal to 2pi. Since the two semi-circles add up to 4pi, that means that whatever area they are overlapping must be the same as the excess are of the quarter circle. So x+x=z+z, and therefore x=z
There is an easier way if you look at it you can easily prove that x fits into y creating a segment and then solve from there by getting area of sector (quarter circle) minus area of the triangle
Call the radius of the big circle as R (4 in the example) The radius of the small circle is then R/2 The blue area consists of two parts: the overlap between the small circles, X, and the outer part Z. First we show that X=Z Area big circle = pi R² => area of the quarter shown = pi R²/4 Area small circle = pi (R/2)² => sum of areas of the two half circles shown = pi R²/4 Compared to each other, the big quarter has Z in excess, the sum of the two half circles has X in excess. Their area being equal, X=Z Next we concentrate on the overlap X and look at the square with side R/2 which has total area R²/4 It consists of the X and two equal leftover pieces Y. So X+2Y = (R/2)² In the quarter of the small circle: X + Y = pi (R/2)²/4 or 2X + 2Y = pi (R/2)²/2 subtracting one of the other gives X = (pi/2 - 1) (R/2)² The blue part being X+Z it is 2X = (pi - 2) (R/2)². Fill in R=4 to get the result 4pi-8
or....you could just find x first....and then removing the area of two semicircles from the bigger quarter circle excluding the 2x area as it would be a repeating area for both semicircles.....much simpler.....but i found this one interesting too.....tysm for yr content❤
I thought that too. Someone else made the same comment that he should have elaborated on that crucial point. If it was assumed knowledge he should have said so
I saw your thumbnail image and got this equation (π*r^2 / 4) - (π*(r/2)^2) + ( (π*(r/2)^2)/2 ) - ( (1/2) * (r * (r/2)) ) (1/4 of area of big circle) - (area of smaller circle) + (half of area of smaller circle) - (area of base triangle of width=r and height=r/2)
How do you know that when you divide the semi circle that where the x and z pieces meet is exactly the centre of the semi circle? Visually it looks avout right, but how is it proved?
How exciting! I calculated y (by calculating x and subtracting 2x from a semicircle of radius 2), then subtracted 2y from the total area of the shape, because the two white areas are equal and any area in the shape that isn't white must be blue
I wouldve added 2 more sides to the shape to make it a 4x4 square, added 2 more semi circles inside the square, this will make it so there is 4 intersections of 2x (2x from video’s method), i will add all semicircles and substract by the square to get 8x, calculate the large quarter circle, then the answer is quarter- 2 semicircles + 4x
I got to the answer with a slightly different method, I first determined that x and z were equal by dividing the quarter circle in half (kinda like you did) and then dividing 4pi in half to find that the area of the new 8th circle was 2pi, which is also the area of the half circles, so if y + x = 2pi, and y+ z = 2pi, then z = x, then I found the value of x the same way you did. And bc x = z, I did 4(x-2) to find the answer. It’s interesting how there are many different ways of solving the same problem.
If the inner half circles are added together to make 1 circle, it's 4Pi, same area as the bigger circle. Because of that, you know x and z are equal without doing anything else, because z literally is the space that x doesn't take up, because of the overlap
In fact, if you only want to know the blue area, once you know the value of y you just need to do area of the quarter circle - 2y. You don't even have to bother calculating z since x and y are the only relevant values (you can't know y without using x)
I was thinking of eliminating the areas of each semicircle And hold onto whats common in them. Then Calculate the area of the Big Quarter and then subtract the area of both semi circles from it. Then add the value we held onto. nd that's our answer to the area of the blue circle. Thought I like your way as well. Appreciated
how exciting
tbh that's the first thing that came to my mind after looking at the thumbnail
Yeah, I think Andy may have conditioned us, like Pavlov's puppies
it's really disturbing
😂😂😂
Well this kind of calculation is pretty easy using integration (Calculus), but it would be pretty satisfying to find an elegant geometric solution.
I wonder if there is more than one elegant geometric solution.
Using the geometric shapes as placeholders for the numbers in the equation is a really great way to help people visualize what the numbers mean.
He makes it look much worse than it actually is. Simply draw a few straight lines, and you see the blue area is equal to a much simpler shape. The lines needed are:
1. A diagonal line down the middle of the lens shape
2. A vertical line down from the top right corner of the lens.
3. a diagonal line between the top left and bottom right corners of the big quarter- circle.
With these three lines drawn on top of the figure, you can see the small lens-shaped part of the blue area is equal to two small half lenses. You can also see the long, thin part of the blue area is equal to a big half lens minus two small half lenses.
This means the sum of the blue area is a big half lens minus two small half lenses plus two small half lenses, which is simply equal to a big half lens.
You should also be able to see a big half lens is equal to the big quarter circle minus a big triangle.
Since the quarter circle is πr^2/4 with r=4, we get a big quarter circle is 4π. Since the big triangle has a base and height both being the radius, we get the area of a big triangle is 4*4/2=8.
The area of a the big half lens, which is equal to the area of the blue shape, is therefore equal to 4π-8.
I'm 35 years old and mad that math wasn't taught this way when I was in school
@@HenrikMyrhaug womp womp
Is this possible using integration
Not as straight forward as I thought it will be but still exciting
It's a bit confusing given that you have to make a bunch of assumptions, eg. the circles being perfect.
I think that it makes the problem look much more difficult than it actually is. the picture should have listed the dimensions of the smaller shapes that are not meant to be derived through calculations but are essential to actually solving the problem.
its pretty simple. i tried to figure it out for myself before watching, but i got stuck on the two x’s up until he used a right triangle, then it all came together for me
@@kuil123bruh, that's just coping.
"What if the circle wasn't a circle" my ass
@@HellecticMojo Coping? Dawg hop off the internet lmao.
I literally said all it does it make it look more difficult, not that I couldn't solve it lol. You probably never studied past pre algebra so I totally understand your inability to understand my point.
@@FirstPersonLife before watching I thought the 2 Xs, added to the Zs, would have completed the "long side" of a square triangle with two sides of 4.
So the Blue area would be:
A quarter area of a circle with radius 4 - the area of a square triangle with sides of 4 (which is 4*4/2).
I think it works
First time I ever canceled a triangle
It's the next phase of woke culture.
Lol @@drmodestoesq
It his weird way of doing maths. You move the triangle to the ither side of the equal sign you dont cancel both sides by the variable, triangle in this instance. Thats retardes.
Triangle’s remarks must be understood within the social context of his time
You beat me to it, @@drmodestoesq
nothing has ever blown my mind harder than the sentence "we can subtract triangle from both sides"! I absolutely love the way you think about these problems and these videos always make me smile.
Another method could be to ”move” the x-areas into the ”citcular hole” in the z-area. Doing this on both x-areas creates a non-shaded triangle of b=h=4 which means an area of 4x4/2=8. The area of the quarter circle is 4pi meaning the shaded area is 4pi - 8
How exciting
Very clever
This is really elegant.
I came here to comment the same thing but saw that you beat me to it hahaha
This is what I expected from the video, I wonder why he did not show this much simpler solution
both white areas can be cut and moved around to create simple 2x2 squares.. so the white area is 8 sq units, and the whole quarter circle is 4pi.
There you have it, the blue area is 4pi-8.
wait thats so smart
Damn that's smart
exactly - I spotted that in about 20 seconds and thought "gee I hope doing it with the algebra is just a precursor to showing how much easier it is if you do a simple bit of geometry first" ... but I checked the whole video and no he never mentioned the much easier method - disappointing as I don't mind most of this guy's work.
he literally says in the title "an algebraic approach"
how do you cut it?
Since both 2x + y and x + y + z were equal to 2pi it was probably easier to just determine that x and z were equal to begin with, and that 2x + 2z can be renamed to 4x equals the area.
Then once getting that pi - 2 is x, you can just multiply out to get 4pi - 8 is the area.
He's stupid
Exactly, i watched it twice cuz i thought i somehow missed the reason this solution wasn't viable. But it is. So i suppose the guy just did an extra 3 min of impressive graphics and basic geometry moves for no reason. Still very interesting choice of solution by him...
Womp womp
I never visualized the equation like this ever in my life. This way of teaching makes it so simple and interesting.
I think the most important think to solve this kind of question is knowing how to breakdown the shape into more workable form
how exciting
These problems are indeed fun and your enthusiasm is infectious. ❤
What’s most fascinating about this is that z and x are the same area, despite not looking anywhere near the same area visually!
I love how you don't waste anytime over explaining things. Subscribed
How exciting! I did it slightly differently, I took the bottom left part to be a square with sides 2. With that area being 4 square units, if I took away 1/4 of the circle with radius 2 from the 4 I would have a variable I called x. I saw that if you had a square of area 4 and you took away 2 of those x’s you’d be left with the blue value between the smaller circles. If you then took the whole quarter circle, took away 2 quarter circles of radius 2, the square of 4 and added back on the value found earlier you’d arrive at the same conclusion.
Creative method 👍
Even I did it the same way.
I did it as the following: label the top blue part “A”, the bottom blue part “B”; take the 2x2 square around B. Area of B is the overlap between the two white parts in that square, since there is no empty parts, its just the combined area of the white parts in the square, mod the area of the square. A_whiteareas is just pi x d^2/8 = 2pi. Now we have area of B as 2pi mod 4 or 2pi-4. Now for the area of A. This is the area of the whole quarter circle, - the areas of the white + back the overlap. A_qcircle is pi x r^2 /4 = 4pi. Substituting with all the values we have gives 4pi - 4pi + 2pi - 4, canceling out the 4pi’s, we have area A as 2pi-4. Adding this with the other one gives us the final area of the blue as 4pi-8 u^2. Quite lengthy but thats just what came to mind when i saw it.
Andy your the greatest teacher at math in youtube, please teach us how to make it simpler because alot of people are having a hard time, only few can do this like you.
It's faster to recognize that the two x regions can be moved to mate with the z regions, creating a single circular segment of radius 4 and angle 90. Then it's just the area of the full quarter circle minus the area of the right equilateral triangle.
I was so excited around 1:40 when I saw Z and X were the same. I was like "I KNEW IT!"
I did this one a bit differently. Rather than focusing on the blue area, I found it by working out the white area and subtracting it from the area of the total shape.
If we break it down, the white area is made up of two quarter circles plus an extra bit that we will get to later. To find the area of the quarter circles, each one has a radius of 2. Since the area of a quarter circle is πr²/4, we substitute in 2 to get:
2²π/4 = 4π/4 = π
Due to there being two of these quarter circles their total area would be 2π.
If we drew a square of radius 2 around the pointed oval type shape in the bottom left, it becomes apparent that the white area within that square is equal to two of the remainding area when you draw a quarter circle of radius 2 within the square. Using the same formula, πr²/4, we can work out:
2 * (4 - 2²π/4) = 2 * (4 - 4π/4) = 2 * (4 - π) = 8 - 2π
Thus, the total white area is equal to:
2π + 8 - 2π = 8
Using πr²/4 once again, we can work out that the area of the total shape, as it has a radius of 4, is equal to:
4²π/4 = 16π/4 = 4π
Therefore, as it is equal to the total area minus the white area, the blue area is equal to 4π - 8 square units.
I also did it like this and got the same result as you 👍
If you rotate the two blue circular segments until they are adjacent to the two uppermost blue pieces, the white and blue areas retain the values of their areas but with this maneuver the white area becomes an isosceles right triangle with legs equal to 4 (therefore area 8) and the blue area is a circular segment that has an area equal to the difference between the area of the quarter circle you calculated at the beginning (4pi) and the area of the white triangle (8). So the final area is 4pi-8.
Quite a nice solution
beautiful solution!
Hardly, when compared to the geometric solution (which someone else has already pointed out.) A simple rearrangement turns it into a really easy puzzle.
quick question, how do u prove that intersection between x areas and z areas is located exactly half of the two semi sphere, making 90 degree angle, which ultimately leads to a conclusion that height of the triangle is 2?
Assuming the two half circles are proper half circles, their "centers" if they were whole circles would be exactly in the middle of the lines they're on, labeled 4 units long.
Radius from that point going directly up and from the other half circle going directly to the right would get you the coordinate point (2,2), if you imagine the bottom right being the origin. So the triangles would be 2 units high. 🤙
You need to prove that before going on to use it. You can't just use intuition. Not how math works
Draw a line between the center of both semi circles. Because the two semi circles have the same radius, and the bottom left corner is a right angle, it has to be a 45-45-90 triangle. Both semi circles intersect st the corner of that triangle, and by symmetry, both semi circle has to intersect at the same point again opposite to the line. Combining these two triangles create a square with side length 2 with corners at the intersections and the centers of both circles.
Amazing, great question and an elegant solution!
This was on my math homework(7th grade advanced) and it was a fun problem. It was satisfying once everything lined up.
Just wanted to say I really like your channel! Keeping math fun!
This channel is great. Love it all
I have no idea why my brain loves these videos so much. I loved geometry in HS. Everyone else hated proofs
I solved this by finding the areas of x and the semicircles so I could use subtraction to find the area of y and then z, but I like the methodology you used better because it is generalizable.
1. Label the areas with N unknowns
2. Find N linearly independent equations
3. Row reduce the system of equations
Thanks for sharing!
The semicircles are the same because their diameters are the same. When you divide the large quadrant in half, the right and left are divided equally in half. Imagine bringing the two x's formed to the point where y and z intersect with a little imagination. The question is now like this: (quarter circle with radius 4) - (isosceles triangle with side 4) = (4π) - (8)
I also like the geometry approach to this one. The quarter of the larger circle - both semi circles = 2z-2x. However the quarter of the larger circle (x2 radius = x4 area) is the same area as the smaller circle. So smaller circle - smaller circle = 0 = 2z-2x. Then you calculate 4x as the smaller circle - the square inscribed in the smaller circle. All together circle - circle +(circle - inscribed square) = circle - inscribed square = 2z-2x + 4x = 2x+2z.
Beyond the joy of adding and subtracting the same circle, I like how it reveals 2x=2z.
I like how you smile while teaching, keeps the problem upbeat and exciting! :D
A little geometry makes this so much easier.
Bisecting the quarter circle leaves you with the 2 x sections, which fit exactly into the scalloped section from the 2 z areas, which leaves the quarter circle with a redistributed shaded area in the curve, and an unshaded half square internally. This can be done because the two semicircles that make up the scalloped portions are exactly proportional to the larger circle. Bisecting the larger circle through the two smaller semi-circles leaves us with a tangent that is exactly half the length of the secant connecting the two points forming a hypotenuse with the radii making the right triangle.
The area of the entire quarter circle is (pi*r^2)/4 with a radius of 4. This becomes (pi*4*4)/4, with a four cancelling above and below, leaving us with 4pi.
The half square has an area of lw/2, which equals 4*4/2=8.
The shaded area is therefore 4pi-8 sq unites.
Great way to do for the sake of completeness and understanding. You can find out that z and x are equal from the start if you just consider the sum of the areas of the two semicircles being equal to the sector itself. From that you can just call both shaded parts X. From there I just drew 2 lines as you did creating a box and solved for X.
The area of the big quarter-circle is the same as the area of two small half-circles, subtract and get 0, carry the blue bean because it's subtracted twice. The area of the blue bean is 2x area of small quarter-circle minus half area of square (cut diagonally), so 1/4π2² - 2²/2 = π - 2, times two. Carry the blue bean, the answer is 4π - 8.
If you draw a radius from each of the 2 semi-circles (one horizontal, one verticle), the lower left blue area can be seen as the overlap area of 2 small quarter circles contained within a square.
The lower left blue area can be calculated by summing the areas of the 2 overlapping quarter circles (𝝅2²/4 + 𝝅2²/4) and subtracting the area of the shape formed by the perimeter of their outer (non-overlapping) edges which is the square that they are contained within (2²).
The 2 radii drawn to create the square also form small quarter circles at the top left and bottom right.
The upper right blue area can be calculated from the difference between the area of the large quarter circle (𝝅4²/4) and the sum of 3 areas: the 2 small quarter circles (top left and bottom right) and the area of the square (𝝅2²/4 + 𝝅2²/4 + 2²).
I love your “how exciting” in the end
I saw 2x + y = 2π and x + y + z = 2π, and from that got x = z. Wouldn’t save you all that much work, but it would mean you could work in two variables from then on.
This channel is so wholesome
I used the square corner and its two quarter circles to get the shape area of 4+2pi.
Then the area of the two semicircles is 4pi, and the area of the large quarter circle is also 4pi.
The blue eye is obtained by subtracting the main shape from the semicircles, and the blue spectre is obtained by subtracting the main shape from the large quarter.
They both result in 2pi-4. Added together its 4pi-8.
I like this solution. I did mine a slightly different way:
Got the full area as 4π and the smaller semi-circles as 2π.
I then split the semi-circles in half and subtracted that from the area (4π - 2π = 2π)
I then took the total area of the overlap of the semi circles (2²=4) and subtracted half of a semi circle (4-π) giving me the remaining non-blue area in the second half of one of the semi-circles
Finally, I subtracted twice the second calculation from the first one and got my answer (2π - 2(4-π)=2π - 8 + 2π=4π - 8)
Hope that makes sense,
As a Calc student who just learned how to find area so I would find x using the top and bottom bounds. Add the semi circles together and subtract x and then subtract what's left from the quarter circle to get z
If you draw in two more semi circles to turn the image into a square it’s easy to see visually that if you split the lower left blue area in two, and “slide” these mentally clockwise and anti clockwise, you’ll create a white right-angle triangle of area 2*2*0.5= 8, and a blue semi circle of unknown size, but whose size equals the initial two blue areas summed. So just subtract 8 from the area of the quarter circle (4pi).
Also the two blue shapes have the same area; you can prove that by subtracting the area of the semi circles from the larger quarter circle. This gives you the area of the upper right blue shape minus the area of the lower left blue shape, which turns out to be 0 ie they have the same area.
This one was weirdly satisfying. How exciting.
Another fun solution is drawing a rectangle around the white area, size 2x4. Cut it in half, size 2x2. The “positive space” in one is equal to the negative space in the other, which means you can put the two half’s together and get one whole square of area 4
*More simply* just draw a diagonal line from top-left to bottom-right. By symmetry the 2 half white lens are equal to the 1 full blue lens.
So, the blue area answer = Area of the large quarter circle - Area of the large isosceles right triangle = pi / 4 (4 x 4) - (1/2 x 4 x 4) = 4 (pi - 2)
More generally, the blue area = (pi/4 - 1/2) R^2 = *Area of the large quarter circle - Area of the large isosceles right triangle* .
I had this very question today on my math test and with ur video was able to solve it!
Very interesting, the comments have all found in many different ways.
Mine was:
If you make your small 2 x 2 square as you did surround the x sections. You can solve for the white of a single side and add it to half semi circle to get one of your Y area's in the video. Then can multiply be 2 to get all area of white and minus from blue quarter circle to get answer.
Area of small white section in small square
=( 2*2- ((π(2^2))/4) = 4-π
Area of white quarter circle
=(π(2)^2)/4 = π
Total area white sections ( your Y multiply by 2)
=2*(4-π+π) = 8
Blue quarter circle minus white
= (π(4)^2)/4 - 8 = 4π -8
I used a much simpler solution. First I found the area of the large quarter circle 4pi. I drew two lines going from the center point of the intersection to create 2 quarter circles which have a radius of 2 and an area of pi. Instead of finding x, you can easily find the white area inbetween the midpoint and the weird curve that bounds x. Just use the square which has an area of 4 then subtract the area of the semi circle now its an area of 4 - pi. Subtract 2pi, for each quarter circle, and 4 - pi, for each weird curve region, from 4pi. You get 4pi - 8.
How do you know for sure that when you split the semi circle in half that the line you made connects with the edges of the blue section you marked as x?
If you’ve already accepted that it IS a semicircle, the intersection being the midpoint is a given. But this particular image is missing measurement marks that indicate it’s a semicircle, that the larger arc is a quarter circle, and even that the corner is a right angle.
On the other hand, the point of the video isn’t proper labeling technique, but how to approach complex shapes analysis.
Umm... The diagonal helped, but after that, I did it this way.
1) Draw the diagonal (like he did).
2) Due to all the shapes being circles (and the way the lenses connect 90 degree radii), when the biconvex (the thinner blue section) is cut in half, you can "shift" those pieces to fit into the larger blue area's concave sections
3) You now have one white right triangle and one blue plano-convex lens area. Simply subtract the area of the triangle from the area of the area of the quarter section of the larger circle.
Alternatively, if you think of the smaller circles as rotate-able, you will see how I got the blue sections together more clearly. Also, his cutting of the sections at 1:36 will help see the triangles (and a leftover square) that you will have to subtract from the larger circle to get the remaining blue area.
I did it a bit differently but got the same answer.
I drew a 2x2 square and subtracted a quarter-circle with radius 2 to get the section of the white part of y that would be inside of said square and outside said arc (a). Then I doubled it noting that the same deduction is made on the other side of the arc (b). I subtracted that number from the 2x2 square which returned 2Pi-4.
I then noted that everything outside of z within the 4 radius quarter-circle can be described by a 2x2 square plus 2 quarter-circles. In other words, Z could be described by subtracting the r=4 quarter-circle with a 2x2 square and 2 r=2 quarter-circles. So, z could therefore be described as Z= ((4x4xPi)/4) - 2(2x2xPi)/4 - 4. This came out to 2Pi-4. Add Z and X together and you get 4Pi-8.
God it felt so good that I could still do this in my head after being out of school of 5 years now
My approach was this:
To calculate the area of the eye shape (your 2 x'es) simply do 2 quarter circles with radius 2 minus 1 square with an edge length of 2, so:
2 * (π * 2² /4) - 2² = 2π - 4 => the "eye" shape
Then calculate the area of the large quarter circle like you did and subtract 2 half circles with radius 2, then add 2 "eyes", so:
π * 4² / 4 - 2 * (π * 2² / 2) + 2 * (2π - 4) = 4π - 4π + 4π - 8 = 4π - 8
This also means, that the white area is exactly 8.
How exciting! Makes total sense after all.
I believe the way I calculated, is a little easier if you like thinking in shapes.
Seeing that 1/4 * Pi4^2 = Pi2^2 ( a quarter big circle = 2 half small circle = small circle) you can see the both blue areas are the same size ( Since the overlapping 2 small circles form a blue "leaf", but that leaf should've filled the rest of the big quarter circle ).
Now you just try to find out what that size of the leaf is. If you imagine 4 half circles in a 4x4 box you can see it forms 4 identical leafs. Meaning 4 half circles minus 4^2 (the box the half circles are in) equals 4 leafs. ( 2Pi2^2 -4^2) .
Now just divide that by 4 and you get the area of a leaf. Double that and you get the whole blue area.
It sounds complicated written out like this but once you imagine it, it's so much easier.
Hf
Did it the same way, searched '4x4' in the comments and found your response xD
2(x+z) makes a segment of the whole quarter circle, so you can take the area of the quarter circle 4π and substract the area of the right triangle formed by both radii, which is 4x4/2, obtaining the same solution in like 5 seconds
I have no idea why do how this clip ended up in my feed but all I know is I’m genuinely saddened I gave up mathematics in school. When you showed that x=z I paused and looked hard at the shapes and thought ‘yes that looks right’.
I don’t know where to go from here.
Happy Easter I suppose.
🗣️🗣️🗣️We are getting excited to this one🗣️🗣️🗣️🔥🔥🔥
Simple visual solution: Draw the line from (0,4) to (4,0), the triangle formed has an area of (BH/2) = 4x4/2 = 8. Note that the area "x" can be rotated outside the triangle to fill the space between the diagonal and the area "z". This means that the area of the quarter circle minus the area of the triangle is the area in blue. The area of the quarter circle is easily calculated as 4pi so the difference in areas (blue area) is 4pi-8.
Lunes are so cool. And how they equate to basic right triangles
Maravilha. Obrigado pela partilha. Wonder. Thanks for sharing.
I'm 26 and you're getting me excited about math again for the first time since I graduated highschool years ago. It's kinda crazy to think stuff like that used to be my bread and butter when I was 15 but now I'd have to google the formula for the area of a triangle. You're helping me get back in touch with who I was before the 9 to 5 grind took over :) it makes me feel smarter and flexible in my way of thinking. Like hitting a mental gym. Turns out I can still run! Yay!
If you cut the white shapes along what would be their radius perpendicular to their base (e.g. into 2x2 squares), you'll get two pieces. If you put the two curved parts of the pieces together, you get a 2x2 square. Thus, [A] = 4^2*pi/4-2*(2*2)=4pi-8.
Split the quadrant into an octant along the symmetey axis and note that the blue area (x) matches rhe convext component of y.
Swap those two areas to show that the octant contains a white triangle and a blue convex element.
Reverse the symmetry of the octant to reform the quandrant as a white triangle with a blue convext element.
Total area = 4pi.
Triangle = 0.5 x 4 x4 = 8
Blue area = 4pi-8
Brilliant, simple to follow.
I used a different approach, where i calculated the area of the overlapping small half circles and of the big quarter circle separately.
I first treated the half circle as a full circle and constructed a square inside it. Then took half of (the volume of the circle minus the volume of the square) to get the overlapping area.
Then, I took the volume of the big quarter circle minus the volume of the 2 small half circles {which is just the area of a small full circle) plus the already calculated overlapping area.
I got the same result.
This is just beautiful. But how did you get the idea of that first line that cuts everything in two? I think I would've never thought of that.
because of symmetry
It’s because the two semicircles have the same diameter so it’s symmetrical.
Cool, I did something similar and got the same answer! I found the 4π of the quarter circle with a radius of four, and then the area of the two semi-circles with radius of 2, which also comes out to 4π, so I knew the area where the two semi-circles overlap is equal to the other blue region. I then did the same thing you did to find x, then multiplied by four to get the answer
Lame Question: At 1:31, how do we know that the blue areas, x and z exactly meet at the centre? Isn't that an assumption critical to make y equals 2?
The two semi circles are identical, and the quadrant has a 90° angle. I don't know how to explain this, but i feel like these two facts can prove that x=2
You could also just subtract the semi-circle from the entire shape twice (PI*4^2/4-PI*2), then since we subtracted the 2*x blue zone twice we calculate x same as in the video, add it 4 times to the shape and boom we got the result
Since we quite early on established that the big quarter circle's area is 4π, and later established that y = 4, we could have stopped there, after a single substraction. The blue area is 4π - 2y; or 4π - 8.
If you see that the area of the large quarter circle with radius 4 is equal to the area of the two half circles (with radius 2) combined, you know that the the area where the two half circles overlap is (the eye shaped blue area) equal to the area where neither of them overlap with the large quarter circle (the "weird" shaped blue area). Another way of thinking about that is that, since you need both half circles to cover the entirety of the bigger quarter circle, every piece that is covered twice by those half circles will by definition remain uncovered someplace else.
When I saw that, I calculated x from Andy's explanation and multiplied it by 4 to get the answer.
If we take one x and rotate it by 90 degree taking center of semicircle to put it close to near z. Similarly do it for second x for second semicircle we will get triangle of 4×4 because lines of will be connected to become straight line.
Then blue area will be= area of quadrant with radius 4 minus area of triangle of both base 4.
1/4 circle of 4 - square triangle of 4
= 1/4 π 4×4 - 1/2 ×4×4
= 4(π-2)
This makes me thankful for calculus
I'm in love with math after watching you❤❤❤🇧🇩
This problem assumes the shapes to be a quarter circle and two semi circles. Either state those assumptions or properly label the diagram.
That really threw me off as well - I'd assumed going in that would be one of the main challenges of the problem. One can easily imagine a shape that to the naked eye looks nearly identical to this one, but that has a slightly more concave/convex arc and therefore a different area.
I wonder if there is some reason this assumption could safely be made? I don't know enough about geometry to say, though that doesn't sound right to me. Even if that is the case, I wish he'd mentioned how he knew that everything was perfectly circlular.
Wow, kids have no idea how lucky they are.
The way I did it started with finding the area of the quarter circle (4 pi) and trying to break that down until we isolated the blue area. The two semicircles have areas of 2 pi, meaning that they can be divided into two quarter circles of area pi each. Each semicircle has a half that is wholly in the white area, meaning we know that there is 2pi not in the blue area, meaning that the remaining sections, the blue area that is outside of the semicircles and the area where the semicircles overlap, is equal to 2pi as well.
You can then cut out a section where the other half of the semicircles are, making a square equal to area 4, meaning that the upper blue area is equal to 2pi - 4. If we call the lower blue section X, we also know that each of those white sections have an area of Pi (the area of the small quarter circle) minus X (the area of overlap), meaning that that total square is 2(pi - x) + x = 4. When you solve for X, it simplifies to x = 2pi - 4.
So, the two areas of blue both equal to 2pi - 4, meaning combined they are 4pi - 8 :)
So rare for me to be able to find the solution for one of these!
Marvellous and many thanks.
That was fun!! I really enjoyed that!
Something really interesting about this example: the whole quarter circle's area is equal to 4pi and the two semi-circles are equal to 2pi. Since the two semi-circles add up to 4pi, that means that whatever area they are overlapping must be the same as the excess are of the quarter circle. So x+x=z+z, and therefore x=z
I dont even like math, but still, subscribed
How exciting
There is an easier way if you look at it you can easily prove that x fits into y creating a segment and then solve from there by getting area of sector (quarter circle) minus area of the triangle
Call the radius of the big circle as R (4 in the example)
The radius of the small circle is then R/2
The blue area consists of two parts: the overlap between the small circles, X, and the outer part Z.
First we show that X=Z
Area big circle = pi R² => area of the quarter shown = pi R²/4
Area small circle = pi (R/2)² => sum of areas of the two half circles shown = pi R²/4
Compared to each other, the big quarter has Z in excess, the sum of the two half circles has X in excess. Their area being equal, X=Z
Next we concentrate on the overlap X and look at the square with side R/2 which has total area R²/4
It consists of the X and two equal leftover pieces Y. So X+2Y = (R/2)²
In the quarter of the small circle: X + Y = pi (R/2)²/4 or 2X + 2Y = pi (R/2)²/2
subtracting one of the other gives X = (pi/2 - 1) (R/2)²
The blue part being X+Z it is 2X = (pi - 2) (R/2)².
Fill in R=4 to get the result 4pi-8
or....you could just find x first....and then removing the area of two semicircles from the bigger quarter circle excluding the 2x area as it would be a repeating area for both semicircles.....much simpler.....but i found this one interesting too.....tysm for yr content❤
How did you know the interesection between to the two semi circles is exactly half way??
I thought that too. Someone else made the same comment that he should have elaborated on that crucial point. If it was assumed knowledge he should have said so
I saw your thumbnail image and got this equation
(π*r^2 / 4) - (π*(r/2)^2) + ( (π*(r/2)^2)/2 ) - ( (1/2) * (r * (r/2)) )
(1/4 of area of big circle) - (area of smaller circle) + (half of area of smaller circle) - (area of base triangle of width=r and height=r/2)
Area of quater circle- (π4²/4)
Area of trangle - (½ * 4*4)
Now substract the trangle area from the quater circle area.
(π4²/4)-(½ * 4*4)= 4(π-2)
How do you know that when you divide the semi circle that where the x and z pieces meet is exactly the centre of the semi circle? Visually it looks avout right, but how is it proved?
Moving the two inside x. It’s a quarter circle minus a triangle base 4 height 4.
it will take me 10 years to understand what just happened
How exciting! I calculated y (by calculating x and subtracting 2x from a semicircle of radius 2), then subtracted 2y from the total area of the shape, because the two white areas are equal and any area in the shape that isn't white must be blue
I wouldve added 2 more sides to the shape to make it a 4x4 square, added 2 more semi circles inside the square, this will make it so there is 4 intersections of 2x (2x from video’s method), i will add all semicircles and substract by the square to get 8x, calculate the large quarter circle, then the answer is quarter- 2 semicircles + 4x
Loving your videos! Could you share which tool you use for these explainers. Thanks!
I got to the answer with a slightly different method, I first determined that x and z were equal by dividing the quarter circle in half (kinda like you did) and then dividing 4pi in half to find that the area of the new 8th circle was 2pi, which is also the area of the half circles, so if y + x = 2pi, and y+ z = 2pi, then z = x, then I found the value of x the same way you did. And bc x = z, I did 4(x-2) to find the answer. It’s interesting how there are many different ways of solving the same problem.
If the inner half circles are added together to make 1 circle, it's 4Pi, same area as the bigger circle.
Because of that, you know x and z are equal without doing anything else, because z literally is the space that x doesn't take up, because of the overlap
idk how i got here but this is lit
In fact, if you only want to know the blue area, once you know the value of y you just need to do area of the quarter circle - 2y. You don't even have to bother calculating z since x and y are the only relevant values (you can't know y without using x)
I fear the day andy says "This doesn't look like a fun one"
I was thinking of eliminating the areas of each semicircle And hold onto whats common in them. Then Calculate the area of the Big Quarter and then subtract the area of both semi circles from it. Then add the value we held onto. nd that's our answer to the area of the blue circle. Thought I like your way as well. Appreciated
blue area*