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 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.
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.
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...
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.
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 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.
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.
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 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.
Hardly, when compared to the geometric solution (which someone else has already pointed out.) A simple rearrangement turns it into a really easy puzzle.
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!
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.
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
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 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 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.
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.
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
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.
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'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 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.
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.
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.
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 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 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.
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
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
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
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.
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 basically divided the two overlapping semicircles in half, calculated the white-only half of both of them (pi sqU) and made the sum of them (2pi sqU); then I calculated the square formed by the semicircles' radii (4 sqU). The area of the big quarter circle (4pi sqU) minus the sum of the three aforementioned areas (2pi + 4) equals one blue area (4pi - (2pi + 4) = 2pi - 4). By other calculations, you proove that both blue areas have the same value. So the total blue area is 2(2pi - 4) = 4pi - 8. I took more than I would've liked to to find the answer. My math is quite rusty...
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,
There's a little trick in there as well if no one has already mentioned it, being the the two smaller half circles both equal the area of the larger quarter circle. This means that the lens, is equal in area to the other blue area, and need only solve for one.
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 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.
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.
*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 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.
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❤
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.
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 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.
