the line divide the rectangle by equal half actually means to be use like this?! And, people keep removing more red parts to 'make' the question more difficult seems defeating the purpose...
Hi Presh, For the original question, just draw a straight line cutting the rectangle into 2 squares. Then it is easy to substitute pieces from one square to the other to end up with 100 - 25Pi.
That is absolutely brilliant! At first I smirked and kept going, then I slowly put it together in my head, 'hey, that actually works! So cool😎 tx Maki... heh heh heh i'm going to go build a person, now🤪
It's not about weight, it's about area. Though you can arrive at result by finding ratio of total cardboard to cutout weight and using this ratio to find out area of cutout from area of rectangle. With small assumption of equal mass distribution in cardboard.
Makioka, I thought of something. Math is perfect. The cutting could not be so this only works in theory and we couldn't use it for something that required a given degree of accuracy.... such as a question on an exam, I guess. hmm life sure is beautiful. peace
Should it not be the sum of the area of the two circles subtracted from the area of the rectangle and then halved. ((10 * 20) - 2 * (3.14 * 25)) / 2 = 21.5
There is another way to solve for the smaller area under the circle (SA) on the left. Since you know the equation of the line, you can fine the intersection of the line to the circle, which is (2, 1). From there, you can find various areas of the arcs and angles. This is actually a bit simpler, IMO.
It can be solved that way, but it wouldn't really be strict mathematics as You would have to use some approximations (at some point you run into sin(alpha)=0,8)
+Aaron Earl in most countries, even sin cos tan is taught at middle school, and it's only the basic of it. like, what is actually a sin in a 90° triangle. Not all that special formula. And that's even in many Asian country. I mean, let the 11 yo plays~
Many people suspected the second problem (with the missing spot) was too hard for 6th graders. I have received updated information about this. It appears the first problem was the one assigned to 6th graders, and the second problem (with the missing spot) was a misprint that went viral on social media. Jin Yang provided me with a link to the original problem assigned (in Chinese): wenku.baidu.com/view/de4510b0770bf78a652954a9.html Here is the Twitter thread about it: twitter.com/jzy/status/763406201468047360
That makes so much sense. I'm going onto grade 7, with straight As, and a 100% total average, but I couldn't understand 2nd problem when it came to finding the area of the small part of the circle, aroun 5:48
I would do differently, at 6:28, I would calculate the top angle of the big triangle 5.10, and I know that the angle "theta" is 90 minus the angle of this triangle using this, I would calculate all the angles of the orange triangle and split the orange triangle in the middle, from top to bottom(a little to the right), until the line, to form two right triangles then just use sine and cosine to know the size of the sides of this half orange triangle and calculate its area, and multiply by 2.
Well you got to figure that when you had to use an inverse tangent that your now in trigonometry. I know the US gets hammered on it's math skills, but the soonest that gets taught is 9th or 10th grade.
do you guys understand what i said in my previous comment ? i know my english sucks, but i think the math is right, and is much easier do the way i told than do the way the video says
That was what i could'nt understand about most comments. Like yeah first problem is easy and you Can use BASIC geometric formulas. But the other problem is impossible to do without advenced knowlade in geometry. In my European school we learned most of the formulas used for 2nd problem in 9th grade and up
Edgars Stalts in Singapore I learnt the basics of trigo (sin is opp/hyp, cos= etc), trig circle, half ab c = area of triangle etc in grade8. I’m sure advance students will learn this in grade 6. The question just said grade 6, they didn’t mention whether those students were gifted, in higher level programs, in Olympiad training etc. So this is entirely plausible but definitely too hard for the average grade 6 student in China...maybe
@@SahilGhosh use the initial area and subtract the small portion. the small area can be found by looking at the two lines as two functions, y = 0.5x & y = -sqrt(25-(x-5)^2)+5. then you can solve for x where they both intersect by equating them. After that, break it into to parts; a small right angle triangle, and a circle function with the integral of y = integral of -sqrt(25-(x-5)^2)+5 dx from x=2 to x=5.
That Chinese comment in English ""I saw these similar questions in elementary school. Later I learned that only the knowledge of elementary school has certain limitations. The teacher did not come up with the questions.
@Jia.X - I was considering circle math as necessary for trig functions, therefore the beginning of trig, but that's probably not true. I did learn them in the same class though, in grade 8, 30 yrs ago. Nowadays, grade 7 is first yr of middle school and I know it is mandatory to have a scientific calculator, so they're doing trig in 7 for sure. Idk about 6
I found a much simpler and straightforward way to calculate the first problem. You can take the red colored sliver at the bottom left and put it at the upper right side, inverted. Likewise you can transfer the left half of the central area, also red, turn it and fit it in the right half of the rectangle. So we end up with a square minus the inscribed circle, or (2r)2 - pi.r2 which equals r2(4-pi). I enjoy your videos a lot! Many thanks
I think they gave the wrong answer. I have 21.460. The red portions add up to all of the pieces around 1 circle 10 10 unit diameter in a square 10 units to a side.
Actually i found a very nice solution to the harder version using geometry Geometry is extremely intuitive and interconnected and can provide really elegant answers
I used calculus and the answer from the first problem to solve the second problem. I saw the removed area as an area under a curve. There were two functions here creating this shape. f(x)=0.5x g(x,y)=(x-5)^2 + (y-5)^2 = 25 Solve for y in the g function and you get y = 5+-sqrt(25-(x-5)^2) For the square root, choose the negative sign since it’s the left side of the circle. Set the two functions equal to each other and solve for x. This should be 2 or 10. Use 2 since it’s within the base length of the shape. Now take the integral of f(x) from 0 to 2 and take the integral of g(x) from 2 to 5. Add the two integrals together to get about 1.956, which is the area of the shape. Now subtract the answer from problem 1 by 1.956 and you will get approximately 19.504.
Actually the Part 2 (which did go viral in Chinese social networks) is misprinted, and the correct problem for 6th graders is Part 1. Even Chinese students are absolutely impossible to solve this at 6th grade. Most students (those who follows the progress of common math textbooks) will not be able to solve it until grade 10. Disclaimer: I'm Chinese
I thought there was something fishy when he started mentioning advanced trigonometry and decomposing the shapes into smaller and smaller pieces with more and more complicated solutions. Children at that age (10-11) typically have trouble with prolonged concentration, and decomposition of problems into more manageable portions, so the more complicated it gets, it exponentially gets less likely to be solved. Of course this ignores the most likely scenario (and what most of these "a child can solve this" stories completely forget about) is that teachers will, more often than not, show a similar problem beforehand, give the solution, then give a separate problem to be solved with the techniques used in the previous one. This makes the children in question more aptly equipped to tackle the problem, since the solution is still fresh in their mind.
That makes more sense. I didn't learn trig till I was a sophomore in high school. The first problem seems to be quite appropriate for 6th grade or even 5th grade problem.
@@TheEddagosp The main thing that bugs me about those 'are you smarter than a 12 year old' format TV shows. They're actively _learning_ that stuff prior to being asked about it. If we could check wikipedia first I'd imagine we'd do just fine answering the question too.
Trivial if you consider the symmetry: (Area of rectangle - area of 2 circles)/2 = (2r*4r - 2*pi*r^2 ) / 2 = (4 - pi)r^2 and r = 5, so = (4-pi)*25 = 21.46018...
its basically all about that small piece in the corner, which leads to this geometrical nightmare hes trying to sell. With a simplified triangle its a 3 line equation. As you equate the area, equate the area of the cut piece and subtract the radial area of the circle. But that way that wouldnt have been such an entertaining/shocking video
This is wrong....did you watch passed the first few seconds of the video? That solution doesn't account for the continued problem that was actually presented. But yes, that would be the solution if it were the entirety of the triangle-the area of the circle.
I was 0.13 off, not quite sure how I'm so close yet so far. My method was to use the aspect ratios. The rectangle is 2:1, so the bottom left corner should, I figure, be split into areas with an aspect ratio also of 2:1. So we just calculate the area of the entire corner (easy), then divide by three. I can't figure out why I'm wrong.
It's wrong because the shape of the corner isn't linear. You can't just take 1/3 of it to calculate the area. Obviously, the middle part of the corner has more area than the outer parts, so you can't use ratios here.
@@ongbonga9025 Even if the shape of the corner was linear this would still be wrong. You're making the assumption that because the sides of this right triangle have a ratio of 2:1, their corresponding angles must also have a ratio of 2:1. This is actually slightly off; the ratio of the angles' _sines_ is 2:1.
Answer they figured is: drink several times a day (5- 6 ) boiled water but to cooled tp suitable bearable temperature. Inhale the steam as necessary too.
(The area of the rectangle - the area of both circles) / 2 Much easyer, as well as here you dont need to prove the two pices fit to gether as a whole circle (a step you skipped)
that's what I thought too but if you look closely you can see that in the bottom left there is a part which is white not red which complicates things and makes this invalid.
So I am particularly proud! I got the same answer a different way. I took the lower odd shaped section and divided it into 2 portions separated by a line passing though the origin of the left square at (5,5) and the left intersection of the diagonal and the circle at (2,1). [(2,1) was found by simultaneous equations of the circle and the diagonal.] I then got the area of the triangle formed by the origin, the intersection and the point this new boundary line crossed the x-axis. I then got the area of the region bound by the x-axis, the curve and my boundary line. This was the area of the triangle between the axis, the boundary and the vertical radius less the area of the sector. Then as per the video, large triangle less circle, less little triangle less triangloid. Far less trigonometry - I only used it to calculate the portion of the circle in the sector - but I did use co-ordinate geometry to keep track of everything.
Area of rectangle: 20x10 = 200m Area of circle : 5^2 x Pi = 78,54m There are 2 circles so, circles = 157,08m 200m-157,08m=42,92m 42,92m/2= 21.46m Close?
You can also doing it by considering the diagram on a graph, with the lower left corner of the rectangle being the origin. From there, you know that the diagonal is y=x/2 , and that the first circle is given by (x-5)^2 + (y-5)^2 = 25. You can solve these equations to give the points of intersection between the circle and line at (2,1) and (10,5). You can then use a similar method to the one you used to find the segment and subtract the isosceles triangle.
The harder problem is NOT a geometry problem, it is a trigonometry problem. I spent a bunch of time trying to figure it out, while purposefully not using trig, as that was presumably not allowed.
Hi Presh, 21,4602 and 19,5039 are the respective solutions. First problem has a total area of 21,4602 and simply resembles: 10^2 - 5^2*Pi by symmetry, that's all. Second problem: I need to work out the area of the small red piece in the top right hand corner, looking like a skateboard ramp standing on its point. Focusing on the top right hand corner, draw a horizontal line from the intersecting point of the diagonal and the circle extending to the right hand side of the rectangle, thus bisecting the "skateboard ramp", left to right. The "ramp" will be divided into an upper small triangle and the area located below, i.e. between circle and right hand side of rectangle. Continue as follows: i) Work out the coordinate of the intersection of diagonal and circle (located in the top right corner), using some basic algebra [see Footnote]. It conveniently results in (18|9). The area of the small triangle (red area) is therefore exactly 1. ii) Work out area below the small triangle, between circle and right hand side of the rectangle, by using CALCULUS. TIP: Compute INTEGRAL from 0 to 4 using 5 - Sqrt(5^2 - x^2). Upper boundary "4" is given via coordinate (18|9). This will results in 2.4088. [Note: For clarity I turned the coordinate system anti-clockwise and integrated left to right] iii) Add the two areas above, yielding 3.4088 iv) Add up 3 times "Hang glider" (corner pieces outside both circles), and also add the little piece we just stitched together. Done! Final answer: 19,5039. No need for any angles, trigonometry nor working out the "lens". This method allows you to solve both problems lightning fast. 10 minutes or less. [Footnote:] Getting the coordinate can be done in different ways. I simply equated two functions, circle and diagonal. That eliminates "y" and I solve for "x". Careful here, both functions need to use the same point of origin, hence: +/- Sqrt(5^2 - x^2) = x/2 + 2.5 . I solve for x and done. I eventually found point (3|4) with respect to the first quadrant.
easy. the bottom half of the rectangle is shaded red, except where the circles are. since the circles are the same size, you can take the smaller part of one with the bigger part of another to get a whole cirlce. that is exactly what you see in both halves, so finding the area of one circle and subtracting it from the area of half the rectangle give the area of the red. .5(20*10)-pi*5^2 .5*200-pi*5^2 100-pi*25 100-78.53 21.46 is the final answer
You haven't proved to me that the smaller part of the circle and bigger part of the circle both add upto make a whole circle. You must be able to prove that without the use of intuition, or apparent symmetry.
Nice soluton to Part II! I ended up using an integral to solve the problem, but as the integral for the area below the circle was a bit hard, I had to use WolframAlpha. Here's how I did it anyway: The first tiny triangle bit where the triangle line intersect the first circle has an area of 1 (2*1/2 = 1). The equation for a circle with radius 5 with the center in (5, 5) is (x-5)^2 + (y-5)^2 = 5^2, or just y = 5 - sqrt((10-x)x) for the part that we are interested in. The intersection point is when the triangle line and the circle line intersect; x/2 = 5 - sqrt((10-x)x), which can be written as 1.25x^2 -15x + 25 = 0. This yields x1 = 2 and x2 = 10. x = 2 is the first intersection point, and is what we need to use as a lower limit when calculating the circle integral. I then used WolframAlpha to integrate this equation from 2 to 5, added it to the area of the tiny triangle (1), and got ~1.96. Subtract this from the answer in part I, and you have the answer. :-)
Nice one. I was thinking of doing something like that at first, but i couldn't remember how to write a circle as a function. It's been too long since i've been in school. Eventually i solved it like the video, but i had to look up how to calculate a circle segment.
I intended to do that too... But I failed too hard at figuring out the integral from 2 to 5 of "x/2 - 5 + sqrt(25 - (x - 5)^2)" so I took a little over 1h... I can even explain what I did wrong, for your information I tried to solve everything with a pen and paper :^ ) When I had to integrate the part "sqrt(25 - (x - 5)^2)", I ignored the silent voice in my head telling me I had to use the Integral table ;-; Secondly I forgot that the formula "integral of u(x)-v(x)" for the area between two functions doesn't work if they intersect in the given domain, so I used [0;5] at first... Yeah and then things slowly went uphill again; at some point I realized I had to use [2;5], then I realized that the integral of sqrt(a^2 - x^2) was some crazy arcsin stuff so I had to start from scratch again... And lastly, after I still had false results I just decided to use my calculator and I finally solved it :) Please end my life
You are making it way, way too hard for yourself ;-) You know the areas of the rectangle and the two circles - then find the area outside the circles, and finally use symmmetry to find the requested area. Very simple ... once you know it ;-)
I found a very simple way to do the second problem. To find the area of that small bit on the bottom left, you can do this: get the right angled triangle at the bottom left that has base half of the length of the rectangle and height half of the height of the rectangle aswell. Find the area using bh/2. Then the small bit at the bottom left (we’ll call x) is equal to that big triangle -that sagement and that curved thing. To get the curved thing, just divide the answer to part 1 by 4. To get the segment (this is where my strategy is different) you do the following: take the circle on the left, say it has centre 0,0, it’s equation is x^2 + y^2 = 25. Since you know one of the points that the diagonal intersects this circle and another pint which would be (-5,-5) because it touched the corner of the rectangle, you can find a formula for the equation of this line and then solve it simultaneously with the circle to get the other point of intersection which is (-3,-4) using this, you can find the chord length on the circle, then use cosine rule to find the angle of the sector, then find sector, then minus the triangle area, then u have the segment, and then u can just do the big right angled triangle that we talked about earlier - curvy bit that we found earlier - segment and that gives x, then just minus x from answer to part 1
I rounded to get 19.672. Two circles each with a radius of five make an combined area of 25π. Subtracted from half of the rectangle area is (100-25π). Each "diagonal triangle" has the same area (the center one can be two triangles), but the corners are split by the diagonal. Looking at the bottom right triangle, the diagonal line with a slope of 1/2 means that the shaded half is half the size of the other part, so x + 2x = area of the other triangles (x being the area of the shaded region, the 2x being the non shaded part). That means that the shaded area in the bottom right is 1/3 the area of the other diagonal triangles, and the shaded one in the top right is 2/3 of the area. Since there are 3 triangles (the center counts as two) and two incomplete triangles, you can make an equation 1/3 + 2/3 + 3. Multiply by three and you get 1 + 2 + 9, or 12. Removing the shaded region in the bottom corner would remove the one from the twelve, making 11. The numbers mean that 11/12 of the red is still there after the bottom corner is removed. (11/12) • (100-π) = 19.672 (rounded).
There's a 3,4,5 triangle hiding in the right circle that greatly simplifies solving this problem and doesn't require dragging in so many equations. If you place the origin (0,0) at the lower left, the equation for the right circle is (x-15)^2 + (y-5)^2 = 25, and the diagonal line is y = x/2. The intersection of this line with the right circle occurs at (10,5) and (18,9). Draw a second diagonal from the top left to the lower right, and connect the two right hand intersections with the right circle with a vertical line. You can solve for the areas of the three segments outside this circle-filing triangle and the rest is easy.
Jeff Bowermaster Are you sure that the three segments outside the triangle but inside the right circle are identical? Because I do not arrive at the right solution, and line piece (18,9) (18,1) is shorter than both (10,5) (18,9) and (10,5) (18,1). Or is there another way to calculate the surface of those different segments?
You're correct they're different sizes. I wish I could post pictures this would simplify everything. The three points {15,5), (18,9) and (18,5) form a 345 triangle. That sweeps out 53.13º which is the only place angles are needed. Double that to 106.26º, divide by 360º and the fraction of the circle swept out is 0.2951666. Now you have the area of the sector, it's easy to find the area of the segment to the right of the line (18,1)-(8,9). Add this to the triangle area with vertices (10,5), (18,9) and (18,1), subtract that from the entire circle and you have 2 times the area of the two left hand segments. One of those plus (area of square - area of circle)/4 gives the area of everything except that annoying little missing piece. (10*5)/2 is the area of the triangle containing all three. So the little piece is just the difference.
Just saw a typo in my explanation - It should read "right of the line (18,1)-(18,9). This is a brute force problem, you just have to go in and solve for whatever sectors and segments you can, then combine those together to come up with the answer.
Jeff Bowermaster Your solution still is relying on an inverse trig ratio -- to know the angle 53°+ and its double, you need to use invtan 4/3. So you are not really escaping using trigonometry for Part II, but you do need less math -- just area of triangle formula, area of rectangle, and area of circle sector.
I'm not reading the previous 1,948 comments to see if someone already said this or not. But your solution for the harder version is overly complex. Using basic rules of geometry you can find that the obtuse angle of that isosceles triangle is 120 degrees, and the sector is exactly one third of the circle. If you split the isosceles triangle into two right triangles you'll see that the height of the isosceles is 2.5, and its base is 5 * the sqrt of 3. So the area of the circle cut off by that cord is Pi * 25 / 3 - 6.25 * sqrt of 3. If the hypotenuse line of the triangle is mirrored above, and then the points where those two lines cross the circle are connected, you'll see an equilateral triangle whose corners intersect the circle. That is how you know the sector is 1/3 of the circle, it is also how you know that the isosceles triangles obtuse angle is 120 degrees.
However, it isn't /quite/ an equilateral triangle; if it was, then the grey region would be exactly 1/3 of the area of the full circle, but it's actually .352 of the area. That's not a rounding error, I calculated from scratch.
Hi George, eye balling it will only get you so far! The isosceles triangle is more than one third of a circle (120degrees). theta = tan-1(5/10) = 0.4636rad The angle in isosceles triangle = pi-2*theta = 2.6779rad = 126.87degrees
Yeah, Nat pointed out my mistake. I was confusing 1 2 3 right triangles with 1 sqrt3 2 right triangles. The first not being a 30 60 90, while the second is. A triangle with a ratio of 1 2 3... I don't really know what I was thinking there.
I was thinking this, too. But you get a 30, 60, 90 triangle with sides 2, 1, and sqrt(3), from bisecting an equilateral triangle with sides 2. Or just using the Pythagorean theorem. Anyway, here you have sides 1, and 2, and hypotenuse sqrt(5). In the former case, you have sides 1, sqrt(3) and hypotenuse 2. Clearly different but apparently the same until you look at them.
I am a Chinese and this should be something like an advanced problem for some brilliant students in 4 or 5th grade primary school and of course the expected solution should have nothing to do with trigonometry, calculus or anything too academic. The idea of such problems is to test whether a child is clever enough to figure out the tricky solution to solve the problem in a much easier way.
I doubt 4th or 5th even if brilliant. I would say more like brilliant 9th or 10th. This required a well developed frontal lobe and as such there is not short-cut to human development physically.
1.calculate the area of rectangle 2.calculate the area of circle (radius=5) and double the answer(as there are two circles) 3.subract the area of circles by area of rectangle 4.divide the difference with 2
Could we not for the more difficult part of the equation: 1) 5/10=rise/run=1/2 2) Line 1 eq: y=.5x 3) Line 2 eq: (x-5)^2 + (y-5)^2 = 25 4) Use typical system of equations strategy to reveal two coordinates, with the lesser being the intersection of interest: (2,1) 5) With the intention of doing an integral and to simplify, we set the circle equation to x^2 + (y-r)^2 = r^2 and the new x value of interest to 5-2=3 and integrate from 0-3 from this new equation. 6) Solving for y yields two functions with the one of interest being y = r - sqrt(r^2 - x^2) 7) Integrating from 0 - 3 for 5 - sqrt(25 - x^2) or taking antiderivative yields: 0.956 8) 0.956 + ((1 x 2)/2) = 1.956 for the total area of that shape! 9) Then putting it all together: ((10*20)/2) - pi5^2 - 1.956 = 19.504 It ends up being similar in complexity, but I never showed up to Trig and got your "lazy student" C in the course.
That method should work, but I would disagree that it is of similar difficulty, not even counting that people usually learn calculus after geometry and trig. I think most people who still remember learning geometry could have given you all of the basic area formulas used in the video's solution. I don't think most people who learned calculus ever knew the antiderivative of sqrt(r^2-x^2) off the top of their head. Sure, you can plug the formula into a numerical integrator and get the right number, but you could also just cut the pieces out and weigh them to also get the right answer. Additionally, neither of those answers is exact like the answer in the video is. Additionally, my experience is that you would not have a calculator for a question like this. Personally, I don't think this is a question that most 11-12 year old kids, or most educated people in general, would be able to answer reasonably quickly in an exam-like situation.
Gordon Walsh Why are you comparing Geometry to Calculus? The basic area formulas and thus easier steps are doable by any reasonably intelligent young child. As for difficulty, the number of steps is similar and require a similar spacial level of understanding. I'm with you on people learning trig before calc though. Personally, I slacked through trig and just got my BS in Biochemistry and so know calculus like the back of my hand. For >90% people, the trig method is the way to go. The calc way though worked for those who don't know trig anymore/well, and wasn't rendered arbitrary by a significant difference in number of steps or complexity.
That's funny cuz I started to do the same thing then I was like "ain't no 6th graders know calc and trig!" I spent like 20mins trying trying to find a solution without using them lol
You should realise that the it has a perpendicular line of y=x/2 pass through center so you find the eq of normal and find the distance(l) between centre and the intersection of normal and y=x/2,then let angle t between the radus and this found length of line,cos t =l/r,the following step should be easy
Fun fact - you don't need all the tan/cos/angle stuff. I did this without a calculator. Three steps: (1) subtract one circle from half the rectangle. (2) find the area of one corner ((half rectangle - circle)/4) (3) use ratios to find the portion of the corner that's left out and subtract it from the total - in this case the slope is "2 horizontal for every 1 vertical" so the ratio will be thirds and it's the smaller portion so it's 1/3 of the corner area. Subtract this from the area found in step 1 and wham bam thank you ma'am :)) It's all about looking for the relationships!!! And trust that there IS a solution, and you're meant to find it. Like a little archaeologist with a dusting brush. Good luck friends
The first half puzzle I found a lot easier by realizing that the 'red' areas are congruent to a single square (10x10 - half the rectangle) minus one of the circles.
I also done the second part by a small method without this video help. Solution to second one 👇👇👇 First calculate the area of the edges by taking a square of 10 with a circle in it now subtract the circle area from square which gives area of remaining 4 parts as they are identical divide that by 4 so area of one edge is 5.365.now the actual problem we have to find out the area of left bottom edge below the diagonal for that I consider that the diagonal has bisects the edge in to 1:2 ratio we also know the edge area divide that with 3 so we can get the required area substrate that with first solution we can get around 19.5 as simple as that
This was basically what I thought but the problem i ran into was that 5.365 / 3 is approximately 1.788, and then I made the mistake of adding (3 x 5.365) to 1.788, but eventually realized I wanted the 2/3rds of the corner in and the 1/3rd out, which meant (5.365 x 4) - 1.788 or (2 x 1.788) + (3 x 5.365). Ultimately this comes out to 19.672 instead of Presh’s 19.504.
The area of the rectangle is 10*20=200. The area of each circle is 25pi, so the remaining area in the rectangle is 200 - 50pi. The red area is half of that, so 100 - 25pi, which is approximately 21.46 units squared.
No, he doesn't use it as the base of the isosceles triangle, he uses the bigger triangle (10, 5, sqrt(125) ) to calculate the tan, sin and cos of theta, because their value is the same for both triangles (its the relation between the sides, which is the same because the angle is the same, theta).
another approach would be, total volume of the rectangle minus total volume of circles divided by 2. (20*10) - 2*pi*(r squared), divided dy 2-> 200 - 157 = 43 / 2 = 21.5 Go back to school guys! No hard mathematics included, just a bit of logical thinking!
Watch the damn video, you idiot. It contains TWO SEPARATE questions. ~21.46 is the answer to the first question not the same as the second question which this video explains.
yeah that's what i was thinking too, but h e's explaining an other problem, one were you are missing the area of the bottom left corner. (as in that corner is not red), otherwise, yes you are correct,
czierwo Next time think before you go jumping down people's throats. My first comment, on another comment thread, I just pointed out the second question. Then, you had to (try to) act smart and say "Pi is not exact number." and that I was wrong.
First of all, I did not jump at anyone. Second, The video title does not indicate that it contains 2 questions. Third, PI is NOT exact number unless you consider 13.3 trillion digits an exact number... Have a nice day!
homi do azulejo You mean in china where children don't graduate 6th grade until they can calculate the appropriate course and trajectory to send a rocket to the moon and back? Some people will believe anything if it supports there preconceived biases.
@@sevsesi7866 how hard am I supposed to think it is? This is an iq type of problem. Maybe they have the gifted 6 graders but normal human 6 graders' brains don't work like that no matter which country they're from. It's no coincidence that the conceptual difficulty is roughly the same per age group everywhere in the world.
I did it! 1). Find area of 2 circles and Rectangle. Area of 2 circles : 2.phi.r^2 = 2.phi.25 = 50phi Area of rectangle : 20.10 = 200 2). Substract area of circles and rectangle 200-50phi 3). Devide subtract result by 2 (diagonal line made recangle slice into 2 triangle similar) (200-50phi)/2 = 100-25phi
An overly complex explanation. The diagonal line clearly cuts the rectangle in half because it goes from corner to corner. Therefore the area above and below the line is the same. Thus the area under the line is the same area you get if you divided the rectangle vertically down the middle into two halves, each with a circle inside a square. You then have a whole circle (d=10) inside a whole square (10x10). The red areas simply is the area of the square (10 squared) minus the area of the circle (5 squared by pi.)
I thought exactly the same. You just need to prove the areas correspond (by demonstration and laws) and then it's really easy. But the result isn't the same as in the video, gives about 21 but in the video it's around 19... Edit : Ok I watched the useful parts in the video, and in fact 19.5 is the answer to the smaller red region x). The 6th grade exercice might be the first variation that gives 21. But we never know, after all they "are asian".
Hi Ian McFadyen, it looks like there are two questions in this video. Your solution solves only for the first and simpler problem. Just thought you'd like to know.
Ian McFadyen, Mustang Rider: This solution doesn't work for the real problem. It is obvious that it works for the easy introductory problem, but if you didn't notice: The video is mainly about the problem where the lower left part of the red area is missing. 0:30
I solved the first problem by finding the area of the circles and subtracting this by the area of the rectangle then finally dividing this by 2. Area of circles: 5^2 x pi=78.539x 2 =157.079 Space out of circle = 200-157.079= 42.290 I then took half of this number since we are only measuring the red to get 21.460 (These are not the full numbers but rough estimates)
Sorry fellows but you missed the point... there is an area in the left bottom corner that is not red, in fact I did the same but yeah... there is this little part that do not count so you have to go through many calculs
Area of the triangle is 20(10)/2 = 100. Area of each circle is π5² = 25π. The red area is equal to the area of the triangle minus one of the circles. Ignoring the color, the figure is symmetrically mirrored about the diagonal. The area of the circle on the right that is _not_ overlapping with the red triangle is exactly the same as the area that _is_ overlapping on the left. A = 100 - 25π sq units As for the second question, oof... The equation for the left circle is (x-5)²+(y-5)² = 25. The equation for the doagonal is y = x/2. Equating them we get: (x-5)² + ((x/2)-5)² = 25 x² - 10x + 25 + x²/4 - 5x + 25 = 25 5x²/4 - 15x + 25 = 0 5x² - 60x + 100 = 0 x² - 12x + 20 = 0 (x-10)(x-2) = 0 x = 10 ❌ | x = 2 (2-5)² + (y-5)² = 25 9 + y² - 10y + 25 = 25 y² - 10y + 9 = 0 (y-9)(y-1) = 0 y = 9 ❌ | y = 1 So the circle and diagonal intersect at [2,1]. If we draw a line from the bottom point of tangency to that point of intersection, we get a triangle of base 5 and height 1, so area 5/2. If we subtract that from the previous total we get 97.5 - 25π. Now we just need to add back the area of the circular segment created by the chord we just drew. The length of the chord is √(1³+3²) or √10. cos(θ) = (5²+5²-(√10)²)/2(5)5 cos(θ) = (50-10)/50 = 40/50 = 4/5 If cos(θ) is 4/5, we've got a 3-4-5 triangle, so sin(θ) will be 3/5. A = 5(5)sin(θ)/2 A = 25(3/5)/2 = 15/2 And the sector area is: A = (cos⁻¹(4/5)/360)π5² A = (36.870/360)25π ≈ 8.044 Circular segment: A = 8.044 - 15/2 = 0.544 So total area is ≈ 98.044 - 25π ≈ 19.504 sq units. Hey! I got it right!
You know you are in too deep when you use calculus to solve the second one xD. Calculate the intersection point, and then integrate the area for that section outside of the circle
yeah... I tried solving with just geometry but my brain kept saying, " WTF ARE YOU DOING? You know the easy way to solve this!" Calculus is a cruel mistress
72golfcrazy dunno, but if 6th grade China students can figure this out with just geometry then they'll do spectacularly in Calculus. That's a lot of problem solving power there to do it the basic way.
Finding the exact intersection points to establish limits of integration is not a trivial exercise either; thus, even if one were to use calculus, the amount of work involved is approximately the same.
Much, much easier way to solve the 2nd part, I think. Starting from the solution of the 1st problem we know that the area of the first red sections is ~21.460. Now looking at the diagram we can see that this is a rectangle with side lengths 10 and 20. If we draw a line straight down the center of the rectangle (making two identical 10 by 10 boxes) this allows us to see that if we flip the left side of the triangle (the smaller red area side) ontop of the other part of the red area triangle. This means that in one 10 by 10 square the red area takes up ~21.460 excluding the circle. Now to find the smaller section we can divide by 4 which gives us the area of just one of those sections. Looking back at the overall triangle we can see that it is a 2 to 1 ratio in the side lengths. Meaning that the line going down the center has one angle of 30* while the other is 60*. That means that while looking at any square or symetrical meaurment the line divides the area up inbetween 1/3 and 2/3. With this we know the area of one of the small red sections and we know what fraction of it the area we need to find is. so we take 5.365 (~21.460 / 4) and divide it by three. This gives us the area of ~1.78 which is the missing small section. Then we just subtract this from the original red area, giving us ~19.6716666 (video answer = ~19.504). There is a slight variation of the answer due to my use of the estimated values displayed in the video.
This is super easy you take (20*10=200) witch is the volume of the rectangle. Then to find the volume of the circle is ((Pi*r2)*2 for the number of circles) so (3.14*(5)2*2)=157 so then you 200-157=43 then you divide by 2 to get rid of the remaining white area which is 21.5. The answer is 21.5. Sorry about the squared part stupid iPad don't have square root or even Pi.
I haven't done highschool maths problems in 5 years so I'm very happy I was able to answer that. I didn't need to use the sector area formula or find the area of the little piece in the bottom left.
there is much easier reasoning, by observation, one can see that the area is equal to a square with a side length of 10 withdrawn by the circle contained in this one, the area is then very simple to calculate : x = side of the square area of the square - area of the circle = x^2 - pi.(x/2)^2 = (1-pi/4).x^2 problem solved ;-)
+Don the first the other corner on the left side fills it by using the fact that thre diagonal is not curved (problem would be solvable if it was but you see that it's very simplified with this assumption)
+koolkitty8989 oh yeah that's true ... I didn't see that one ^^ ok ... so I don't have any paper here but I'd say that it's equal to 3/4 of previous answer + value of the area of the remaining area at the top right this area can be computed by dividing it in two parts (you must turn the surface by -pi/2 radians (or 90 deg horlogically) to visualize easier) : - left part, where upper part is curved, which is calculated by integrating the function of this curve until the intersection with the diagonal (function and point of intersection must be figured but I don't have tools to do it now) - right part : upper part is linear, which makes the surface extremely easy to compute (triangle area with height of the image of the finction mentioned above) Thus, that's the idea, hope it's not too complicately explained xD
Even more simply by saying that the area is the total area of the big rectangle subtract the area of the two circles and then divide that in half. (1/2)[B*H - (2pi*r^2)] where b is 20, h is 10, and r is 5 However it does not help (nor does your comment) with the 2nd one (at least not with the new challenge of it) ... I probably would of ended up integrating it myself to be honest. This video is pretty good as far as the 2nd is concerned.
Lots of them do competition math to get a chance for a better middle school(you have to take a test to see what middle school you go to, kinda like college)
Robert Zhang It should not be a surprise American students can't answer questions like this one. With our "No Child Left Behind Act" mean no child get's ahead either. Our schools are being run by MBA's who are treating education like a production line. Treat students like ram material, educate all of them exactly the same and you will have an educated student. The thing is educating doesn't work that way. The raw material, (the students) are all different. To expect the same outcome when the raw material are people and different is just silly. If one looks at the vast majority of students in our universities in the sciences, engineering, and computer science advanced degree programs it's hard to find an American student. Close to 97% of the students in those programs are from foreign countries. Our country and politicians do not value educating our youth. One of the by-product of this is that Russian, Indian, and Chinese computer hackers are stealing billions of dollars from American citizens and America companies. We just don't have people educated enough to protect our computer systems and the Internet, (something America created). Until we fix our education system Americans will continue to get ripped off financially and technologically by people in countries where education is valued.
Subtract the area of one circle from half the area of the rectangle and divide the result by 2. Half area of rectangle = 10*20/2 = 100. Area of circle = Pi*5^2 = 78.54. 100-78.54 = 21.46
It's true, this specific problem I've seen in my friends 6'th homework assignment. China is militant about their maths education department. With a solid grasp of intermediate calculus and she is only a freshman in college. You also have to factor in that they attend 12 hours of school a day.
This could easily be for sixth graders. This guy made it way harder than it needs to be. If you can use percentages, and calculate the are of each of the shapes, and have like, a few days of trig experience, you can figure this one out. He makes it seem wayyyy harder than it has to be. Maybe a few weeks to teach sixth graders basic trig, but it's still fairly simple
My method was using basic trig to find the angle of the diagonal. Then I found how much of the corner the line intersected (the percentage of the angle from 90 degrees). Using that percentage you can do the same for area. If you need more info for how I did it, I'll gladly try to explain it better, just let me know. By the way, my answer came out to be about 19.914. So it was a bit off, but was really close considering I used a much, much simpler method. Here is a copy and paste of another comment I made with a more in depth explanation: Dude! You made the harder problem WAYYYYYY harder than it needs to be. I just calculated the degree of the diagonal using some basic trig (only had to use inverse tangent once). I then calculated the percentage of the degree compared to the corner (i.e. the corner is 90 degrees, and the diagonal has an angle of 26.6 degrees from the bottom of the rectangle, meaning that the diagonal divides the corner piece into roughly 29.5% and 70.5%). SO, if we look at one half of the rectangle (divided of course by the diagonal), we can see that it is made up of 3 whole "corner" pieces, plus 70.5% of a corner piece. (TL;DR) I used inverse tangent only once, and nothing else, and then found some degrees and percentages, and that was it. I can completely see how sixth graders could do this, so long as they knew some very basic trig. Honestly, your super complicated method is like using calculus to find the area of a hexagon. Like, why make it so hard dude? Good vid though
I think I know:(its easier than you think!) For first problem: first we substract area of circle from the rectangle: l x b(10x20) - πr^2(r=5) we get the remaining area without circles. Then we know a diagonal divides the rectangle in two equal parts so... area left unoccupied by circles divided by 2 gives the answer. Basic geometry. so final answer here is->60.75 For the second question: we can go by substracting the areas one by one. The area of the edge will be the same on all the corners so find the area of the lower right red part and then you know area in middle is 2 times the same if seen as two congruent squares however its challenging to go for right top but again these are congruent triangles(SAS) so we can count that part as one whole edge and substract the area of one edge from the remaining so we get the un red chip in lower left and substract the value from one of the four equal edge area and there you are with the answer. EASY! please do not hesitate to ask doubts, Thank you.
" its challenging to go for right top but again these are congruent triangles(SAS) so we can count that part as one whole edge" Which triangles are you referring to? I got lost here.
@@chickencyanide9964 I was considering the triangles cut by diagonals of the rectangle so I suggested that the area in the upper half is equal to the other half of the rectangle so that the unequal pieces at top right red and unshaded down left fill as one side then we might solve this same as 1st
Slade951 The area of the little difficult piece is the double integral of dydx from y=0 to y=x/2 and from x=0 to x=2 plus the double integral of dydx from y=0 to y=5 -sqrt(10x-x*x) and from x=2 to x=5
Before watching this is how I would go: You see diagonal line is half of rectangle so ab÷2. When you have that you focus circle the upper one is missing small part. Small part is on the bottom one so small part + big part= whole circle we can see. So to solve this you have to substrat the are of the diagonal minus one whole circle I think
The area of the rectangle excluding the circles: 0a) W * H so 10 * 20 = 200 0b) circle is diameter/2 to get radius, pi * Radius squared for area so 10/2 = 5. pi * 25 ~= 78.539816339744830961566084581988. 1a) two circles is about 157.07963267948966192313216916398., so 200 - 157.07963267948966192313216916398 is 42.920367320510338076867830836025. 1b) half of 42.9 is about 21.460183660255169038433915418012 Easy part answered. For the hard version: 2a) the rectangle is 2 X 1, so the angle of the line is 26.6 degrees. 2b) therefore, the missing slice is 26.6 / 90 * 100 = 29.6 percent of the area of the corner. 2c) the rectangle has 8 such "corners" 2d) so the missing slice is ~0.296 of 0.125 of the rectangle outside the circles, 42.9, so 42.9 * 0.296 * 0.125 = 1.5873 23) 21.45 - 1.5873 = 19.8625 EDITED: corrected rise/run angle and recalculated. There must be something I'm doing slightly wrong, but I think the correct answer could be gotten this way, which seems more like what a middle school child would be doing, not your way.
***** if the rectangle was a 10 x 10 square, the line going through from bottom left corner to top right would be at a 45 degree angle, and the missing red piece in the hard version would be half of the corner. So we divide the angle by 90 degrees to find what percent of the corner is the formerly red slice. Multiplying by 100 is just to make the percent more readable.
Dan Kelly I was trying to see if rounding or what was causing my answer to be slightly different from the video, maybe someone can see what is wrong here.
I was thinking about it and his solution for the harder version is more complicated than it needs to be. All you need is basic rules of geometry, no tangents, sine, or cosine are required.
This is the reason why so many eastern school kids are so bad at maths. This is Asia buddy all western countries are 4 years behind in everything. I live in Australia and am in grade 10. I solved the 1st problem very easily, but the 2nd was much harder but it was manageable. I bet you 100% that half my maths class couldn't solve the first problem very easily. In Asia everything is hard. They start doing calculus at grade 10. So plz if u think they don't teach trig at grade 6 then maybe you should travel to ANY Asian country and ask there teachers what they are learning!
I wouldn't be surprised that other countries are ahead of the US, or other western countries in what they teach in school. That said, even the hard problem here doesn't require trigonometry, or any fancy math beyond Pi. When the diagonal line crosses the circles it is marking one leg of an equilateral triangle in each circle. Once you realize that you don't need to calculate any of the angles with fancy math. The sector defined by the points where that line crosses the circle is exactly 1/3 of the circle. To get the area of the slice defined by that same cord you'll have to figure the area of that isosceles triangle like the video says. But since you know the arc is 1/3 of the circle you know that the obtuse angle of that triangle is 120 degrees. If you then split that isosceles triangle into two right triangles that are 30 60 90's. From there you can just rely on the 1:2:3 ratio of 30 60 90's to determine the dimensions of the isosceles triangle.
If it's from United States, it will definitely break because yo mama with her cowboy jeans is sitting on it with her 300lb fat ass filled with Cheeseburger meat. I am just going to sit here an watch.
First try before watching the rest of the video: The area of the rectangle is 200 the area of each circle is piR ^2 = pi(5)^2 ~ 78.54 The diagonal perfectly bisects the rectangle, and the red spots are all beneath it, so you can reduce the potential area by 100. The diagonal also perfectly cuts the circles so that there is in total the equivalent of 1 circle of white space under the line. 100 - 78.54 = 21.46 So, the area that is red is ~ 21.46 You can also treat the diagonal as if it were vertically down the middle and solve it that way. This works because the red area in the bottom left is equal to the white space in the top right, and the red area in the bottom middle is equal to the white area in the top middle. Now it's just a circle in a box, and that is simply the area of the square minus the area of the circle. 100 - 78.54 = 21.46 The variation however, is more complicated.
+Rafael Cruz even if it is 6th grade, it isnt a challange at all, it has nothing to do with knowledge just sense. The 2nd is a little tricky but alright. Dont feel offended please.
+emfederin Well I agree I would much rather choose this Chinese curriculum over what is happening back in the States with them Creationist and Flat Earthers.
Dark Phoenix how does this comment make any reference as a reply to the original comment? You are doing replies to comment wrong. There are rules my friend.
As others have said, the first part is relatively easy. In fact I didn’t even bother with the triangle area thing - because quite clearly the line diagonally across the rectangle will describe the same area if it was straight across the middle of the rectangle, and it’s easy to then see that it will also bisect the two circles. Once I discovered there was a second part to the problem, I simply gave thanks that I don’t have to do mathematical solutions and bailed out at that point.
I think there's an easier way for second part: Having the area of the first part, we know that it's composed by 4 times the shape on the right bottom corner, so we can calculate it by dividing 21.46 by 4 . Then we do a construction: we draw the other diagonal so that it forms on both sides an isosceles triangle. We can easily find the area of one of it and then we subtract it from the initial right triangle. What we get is the sum of 2 times the little shape on the corner left bottom that we are searching and 2 times the shape that we calculated before. Thus the little shape is this last result minus 2 times the shape of before all divided by 2. Then just subtract it from 21.46.
What we get is the sum of 2 times the corner part part, 2 times the small corner thingy and 2 times the unknown circular segment* Yes, I know. Answering to 5 month old comments is my life
@@Revo5660 area of the space between the large triangle and the two circles = 100-25pi area of square for one circle = half rectangle = 10^2 =100 area of each isosceles triangle = 100/4 = 25 area of quarter circle = (5^2pi)/4 = 25pi/4 everything between the triangle and circle = isosceles triangle - quarter circle = 25 -25pi/4 one piece = (isosceles triangle - quarter circle)/2 = (25 -25pi/4)/2 area of the red part besides the little corner piece = (100 - 25pi) - (25 -25pi/4)/2 I am getting 18.78 whereas the video shows 19.504
@@asiyaehsan1682 1. Isosceles triangles that the original comment meant are twice bigger than yours, for they are (20×10)/4 = 50 each. I'll assume you meant dividing half this rectangle into four right triangles, for this is the closest thing I can imagine from your enigmatic machinations. 2. Even if that's the case, we can't just subtract a quarter of that triangle. Because that's clearly not a quarter there - it's still an unknown segment that we haven't found. Second: because that way we ignore area of "that concave corner thingy".
you wouldn't need trigonometry to figure out the orange triangle's area, though. Apply similarity triangle, you'd get: height = √5 base = 2 × 2√5 = 4√5 What is way out of your math's level is when it comes to figuring out the segment's area, or just simply when the "inverse tan" comes in.
Doing so is a good habit since you can more easily check that you have obtained a result that makes sense. Say for example that the expression evaluates to a negative number, you know immediately that you have made a mistake in your calculations; this can be hard to spot if you're just staring at the expression.
It can be solved it in two steps 1)red spot area is 100-25π-x Where x is that small curved triangle area For finding that we should know intersection point of diagonal line and first circle. Here take left vertical as y -axis and down horizontal as x-axis then U will get equation of diagonal line by passing through (0,0) and (20,10) as x-2y=0 First Circle centre will be (5,5) and radius as 5, u will get circle equation as (x-5)^2+(y-5)^2=5^2. For getting coordinates substitute x=2y (from diagonal line equation) in circle equation then u will get quadratic equation in y solving that We will get y=1,5.(consider small) Now y=x/2,and Y=5+√(5^2-(x-5)^2) here these equations are from diagonal line and first circle equations by bringing y in left hand side Keeping limits of y from 0 to 1 and for Y from 1 to 5. Small curved triangle area =integration (0,1 as limits ) of x/2 +integration (1,5 as limits) of 5+√(5^2-(x-5)^2) on integrating and calculating we will get x=1.9561 And substituting in 100-25π-1.9561=19.504
Let's see if I can work this out before watching. Looking at this, I would say it's (area of the rectangle minus area of the 2 circles)/2. So, (10*20 - 2*25*pi)/2= 10(20 - 5*pi)/2= about 21.5 For the HARD version, you need to subtract that bottom corner somehow or work your way up without it. We know that one circle has radius 5, and that the corner ends where the circle touches the bottom boundary of the rectangle. We can't just halve the area difference like we did in the easy version. Instead, split the rectangle down the middle so you get two identical squares with holes. With the split, you now see that there are 4 separate red areas, 3 are indentical. We have 3*(area square - area circle)/4 for the total area of the 3 identical red slices. The last one, from the top right corner, will be harder. Unfortunately, this is as far as I got. If I knew where the lines intersected the circles, I could do an integral. I'm sure there's a geometry trick to this I've forgotten.
Yes, they do teach trigonometric functions and inverse trigonometric functions in sixth grade. In many more countries than you think. I remember learning trigonometric functions and inverse trigonometric functions in sixth grade myself. And that certainly wasn't in China. It was in Europe in fact.
7:10 (10/cos(theta))/5=5cos(theta)/10 10/(5cos(theta))=5cos(theta)/10 25cos^2(theta)=100 cos^2(theta)=1/4 cos(theta)=1/2 theta=pi/3 Actual height of the original triangle which you labeled as 5 is: 10tan(pi/3)=10sqrt(3) Area of orange triangle=(5×5×sin(pi-2pi/3))/2=(25sin(pi/3))/2=25sqrt(3)/4 Green shaded area=25pi/6 Blue shaded area=25(pi/6-sqrt(3)/4)
Sides of rectangle are 2 r and 4 r. Red area has 4 identical shapes Each of this shape equals to (Square of side r) - (quarter circle of side r) Hereby RED AREA = 4 (r^2) ( 1 - π/4) = (2r)^2 - π r^2
I thought same thing, was gonna comment on it, but look at bottom left corner of the triangle, that parts not red, gotta account for that, im in fucking 10th grade honors mathematics, finished geometry last year, and this is news to me.
I got the last question right! You know, the one where he asked, "Did you figure this out?" and I answered, "No!"
At least you are honest Sam.
I got it too!!
Lmfao
For the first part, I just subtracted the two circles from the area of the rectangles, and just divided by 2
Was the obvious thing to do
You could actually take the area of a 10x10 square and subtract that by the area of one circle
Yep, same.
That's what I did...... easy!
the line divide the rectangle by equal half actually means to be use like this?! And, people keep removing more red parts to 'make' the question more difficult seems defeating the purpose...
Holographic arithmetic -
1 980 000 pixels
424 908 appear red (I counted)
red area = 21.46%
tyvm
My friend talked about the same thing. 😂
Not bad
>1 980 000 pixels
>424 908 appear red (I counted)
>red area = 21.46%
You forgot step 0. Use the correct figure. :)
When the quarantine hits too hard
@@ahmedabdulraheem7103
Z
Ayush
Hi Presh,
For the original question, just draw a straight line cutting the rectangle into 2 squares. Then it is easy to substitute pieces from one square to the other to end up with 100 - 25Pi.
i did the same
Me too!
1. draw the figure on a board.
2. cut the board into pieces.
3. weigh them.
That is absolutely brilliant! At first I smirked and kept going, then I slowly put it together in my head, 'hey, that actually works! So cool😎 tx Maki... heh heh heh i'm going to go build a person, now🤪
It's not about weight, it's about area. Though you can arrive at result by finding ratio of total cardboard to cutout weight and using this ratio to find out area of cutout from area of rectangle. With small assumption of equal mass distribution in cardboard.
Makioka, I thought of something. Math is perfect. The cutting could not be so this only works in theory and we couldn't use it for something that required a given degree of accuracy.... such as a question on an exam, I guess. hmm life sure is beautiful. peace
@@kalpeshpatel03 r/woooosh
makioka is a scientist at maths- thats called practical math respect
2:21 it was all going well till there was a part 2
I agree
Just using symmetry
Up to 2:05 and finish
i mean, if you can derivate, then it's much, much more easier.
Yeah. Calculus parts are bit boring
Yeah, seeing others to trigonometry is boring, rather solve it in my own to find it than watching.
@@4zdr456 I still don't get integration
Should it not be the sum of the area of the two circles subtracted from the area of the rectangle and then halved.
((10 * 20) - 2 * (3.14 * 25)) / 2 = 21.5
I did the same!
This is exactly I was about to comment...thanks for saving my time ...
two ways of reaching the exact same answer. both are identical except you change when you halve it
That was my first reaction - started watching the video and thought wow this is a really round about way to solve a simple problem smh
wrong, left conner is not red
There is another way to solve for the smaller area under the circle (SA) on the left. Since you know the equation of the line, you can fine the intersection of the line to the circle, which is (2, 1). From there, you can find various areas of the arcs and angles. This is actually a bit simpler, IMO.
Wow, this is the answer which should be in the video 🤩
Coordinate geometry is so much nicer I agree
This is what I did.... its always cool to see your answer pop out correct through a completely different method....
It can be solved that way, but it wouldn't really be strict mathematics as You would have to use some approximations (at some point you run into sin(alpha)=0,8)
"supposedly assigned to 6th graders in China" sounds 5000% LEGIT
Thats because US education is so behind the rest of the world.
I agree, the first part should have been fairly easy for a sixth grader here in the US. Simple formulas coupled with deductive reasoning.
Trigonometry is hardly "simple triangle math"--there's no way the unit circle and trig formulas are a part of 6th grade math in ANY nation. Nice try.
They teach basic geometry in China by the 6th grade. In fact, they teach it in the US here too. This is actually a very easy problem.
+Aaron Earl in most countries, even sin cos tan is taught at middle school, and it's only the basic of it. like, what is actually a sin in a 90° triangle. Not all that special formula. And that's even in many Asian country. I mean, let the 11 yo plays~
Many people suspected the second problem (with the missing spot) was too hard for 6th graders. I have received updated information about this. It appears the first problem was the one assigned to 6th graders, and the second problem (with the missing spot) was a misprint that went viral on social media. Jin Yang provided me with a link to the original problem assigned (in Chinese): wenku.baidu.com/view/de4510b0770bf78a652954a9.html
Here is the Twitter thread about it: twitter.com/jzy/status/763406201468047360
That makes so much sense. I'm going onto grade 7, with straight As, and a 100% total average, but I couldn't understand 2nd problem when it came to finding the area of the small part of the circle, aroun 5:48
Congratulations.
Social media is definitely the best source from which one can quote.
I would do differently, at 6:28, I would calculate the top angle of the big triangle 5.10, and I know that the angle "theta" is 90 minus the angle of this triangle
using this, I would calculate all the angles of the orange triangle and split the orange triangle in the middle, from top to bottom(a little to the right), until the line, to form two right triangles
then just use sine and cosine to know the size of the sides of this half orange triangle and calculate its area, and multiply by 2.
Well you got to figure that when you had to use an inverse tangent that your now in trigonometry. I know the US gets hammered on it's math skills, but the soonest that gets taught is 9th or 10th grade.
do you guys understand what i said in my previous comment ? i know my english sucks, but i think the math is right, and is much easier do the way i told than do the way the video says
this comment section makes me sick. So many people trying to be intelligent, but they can't even realize there're two different problems here
King 445 Basically people that click the video, not to watch it, but to seem ridiculous by trying to look clever.
King 445 And people who try to be smart by solving the second problem with geometry techniques But in reality it is impossible to do that.
That was what i could'nt understand about most comments. Like yeah first problem is easy and you Can use BASIC geometric formulas. But the other problem is impossible to do without advenced knowlade in geometry. In my European school we learned most of the formulas used for 2nd problem in 9th grade and up
Edgars Stalts in Singapore I learnt the basics of trigo (sin is opp/hyp, cos= etc), trig circle, half ab c = area of triangle etc in grade8. I’m sure advance students will learn this in grade 6. The question just said grade 6, they didn’t mention whether those students were gifted, in higher level programs, in Olympiad training etc. So this is entirely plausible but definitely too hard for the average grade 6 student in China...maybe
@@jowsonjgong8303 not really. in my counntry, we are taught about trigs basics in 9th grade
1st part: easy, 2nd part: oh 💩
Same
Hedef 2520 sen devamet belki gelir he bide burda sedef demi vardi ortasini bulmadik ? D hdar dumen
Oh3dozami
No it's easy thé 2nd part
Facts, I could solve the first one but not the harder version
The second half of the problem felt like watching a really scary horror movie
😅😅 the best comment
….stomp……stomp…….stomp
oaoaowoaowoawoaoaowoowAAOAOWOAOWOAOWO
is everything alright?
*I N V E R S E T A N G E N T OF N E G A T I VE O N E*
scarier than any horror movie tbh
IKR it was like I've seen and learned all of this but HOWWW would I ever think of this all by myself jfjsjsnsbksk
Me at the first 30 seconds : This will be 21.46. That was easy.
Me after 30 seconds : *Wait, wat?*
Yea forgot the little area on the lower left
Yep
As a 13 year old I can confirm part one is extremely easy for us as well.
Omg didn’t notice
Yeah it was kinda easy
Me as an intellectual: using the formula of the circumference to calculate the area of the circle
Sounds about right ✅🤔
Lol. I have done similar things too bro
yeah i thnki about you take over the area of 2 circles from the rectangle and divide by 2.
sry for bad english
sounds right. .. wait whattt
That sounds correct
This became extemely easy when i used integration. I hope 6th graders in china know calculus
我小学时候见过类似的这些题目，后来知道只是小学的知识有一定的局限性，老师出题目时候没有考虑。
Fried Liver I know calculus but in this case we just need logical thinking and knowledge of basic geometry. Integrating just makes things harder
@@SahilGhosh use the initial area and subtract the small portion. the small area can be found by looking at the two lines as two functions, y = 0.5x & y = -sqrt(25-(x-5)^2)+5. then you can solve for x where they both intersect by equating them. After that, break it into to parts; a small right angle triangle, and a circle function with the integral of y = integral of
-sqrt(25-(x-5)^2)+5 dx from x=2 to x=5.
@Kevin John Duerme you should translate it don't scold him
That Chinese comment in English ""I saw these similar questions in elementary school. Later I learned that only the knowledge of elementary school has certain limitations. The teacher did not come up with the questions.
2nd part is not even solved by many 12th passed students in china.
Or
Trigonometry is not taught in class 6th
Trigonometry is taught in grade 9 in China.
@@ilovenature-r8f - Canada also, grade 8 at least. Probably 7 or 6 actually, because I learned circle area, and pi in Grade 8 that was in the 80's!
@@ilovenature-r8f bt isn't that only the introduction of trigonometry only... In which inverse functions are not included!!!
@A k lol I'm also from India 😂
@Jia.X - I was considering circle math as necessary for trig functions, therefore the beginning of trig, but that's probably not true. I did learn them in the same class though, in grade 8, 30 yrs ago. Nowadays, grade 7 is first yr of middle school and I know it is mandatory to have a scientific calculator, so they're doing trig in 7 for sure. Idk about 6
I found a much simpler and straightforward way to calculate the first problem. You can take the red colored sliver at the bottom left and put it at the upper right side, inverted. Likewise you can transfer the left half of the central area, also red, turn it and fit it in the right half of the rectangle. So we end up with a square minus the inscribed circle, or (2r)2 - pi.r2 which equals r2(4-pi). I enjoy your videos a lot! Many thanks
Yeah, there were a number of different methods to solving the first one.
I think they gave the wrong answer. I have 21.460. The red portions add up to all of the pieces around 1 circle 10 10 unit diameter in a square 10 units to a side.
This is why Calculus is invented. Integrals are much better here than Geometry.
Actually i found a very nice solution to the harder version using geometry
Geometry is extremely intuitive and interconnected and can provide really elegant answers
@@giuseppedeluca4465 can you tell the easy answer please?
I'm not very good at geometry so I just used calculus lol, 19.5038 good enough for me.
I also want to know how can I resolve it easier using integrals
@@pikachuelpokemonsalvaje9653 plz tell me to how using integral??
I used calculus and the answer from the first problem to solve the second problem. I saw the removed area as an area under a curve.
There were two functions here creating this shape.
f(x)=0.5x
g(x,y)=(x-5)^2 + (y-5)^2 = 25
Solve for y in the g function and you get y = 5+-sqrt(25-(x-5)^2)
For the square root, choose the negative sign since it’s the left side of the circle.
Set the two functions equal to each other and solve for x. This should be 2 or 10. Use 2 since it’s within the base length of the shape.
Now take the integral of f(x) from 0 to 2 and take the integral of g(x) from 2 to 5. Add the two integrals together to get about 1.956, which is the area of the shape.
Now subtract the answer from problem 1 by 1.956 and you will get approximately 19.504.
exactly how i did it, tho i feel you worded it far better
I don't think calculus is part of the tools they have in 6th grade
Where did the second function come from can you show how you got it
Actually the Part 2 (which did go viral in Chinese social networks) is misprinted, and the correct problem for 6th graders is Part 1.
Even Chinese students are absolutely impossible to solve this at 6th grade. Most students (those who follows the progress of common math textbooks) will not be able to solve it until grade 10.
Disclaimer: I'm Chinese
That's a relief. Thought I was dumber than a 12 year old Asian there.
I thought there was something fishy when he started mentioning advanced trigonometry and decomposing the shapes into smaller and smaller pieces with more and more complicated solutions.
Children at that age (10-11) typically have trouble with prolonged concentration, and decomposition of problems into more manageable portions, so the more complicated it gets, it exponentially gets less likely to be solved.
Of course this ignores the most likely scenario (and what most of these "a child can solve this" stories completely forget about) is that teachers will, more often than not, show a similar problem beforehand, give the solution, then give a separate problem to be solved with the techniques used in the previous one. This makes the children in question more aptly equipped to tackle the problem, since the solution is still fresh in their mind.
That makes more sense. I didn't learn trig till I was a sophomore in high school. The first problem seems to be quite appropriate for 6th grade or even 5th grade problem.
Not a surprise. I've done maths at uni and struggled with that. Admittedly I didn't do geometry, but still.
@@TheEddagosp The main thing that bugs me about those 'are you smarter than a 12 year old' format TV shows. They're actively _learning_ that stuff prior to being asked about it. If we could check wikipedia first I'd imagine we'd do just fine answering the question too.
Trivial if you consider the symmetry: (Area of rectangle - area of 2 circles)/2
= (2r*4r - 2*pi*r^2 ) / 2
= (4 - pi)r^2
and r = 5, so
= (4-pi)*25 = 21.46018...
yes, now continue watching when he takes that bottom left corner away ^^
its basically all about that small piece in the corner, which leads to this geometrical nightmare hes trying to sell.
With a simplified triangle its a 3 line equation. As you equate the area, equate the area of the cut piece and subtract the radial area of the circle.
But that way that wouldnt have been such an entertaining/shocking video
Outstanding solution. You can learn this tricks if you practises.
v
This is wrong....did you watch passed the first few seconds of the video? That solution doesn't account for the continued problem that was actually presented. But yes, that would be the solution if it were the entirety of the triangle-the area of the circle.
Almost everyone missed the second part of the problem which was much harder than the first.
I was 0.13 off, not quite sure how I'm so close yet so far.
My method was to use the aspect ratios. The rectangle is 2:1, so the bottom left corner should, I figure, be split into areas with an aspect ratio also of 2:1. So we just calculate the area of the entire corner (easy), then divide by three.
I can't figure out why I'm wrong.
It's wrong because the shape of the corner isn't linear. You can't just take 1/3 of it to calculate the area. Obviously, the middle part of the corner has more area than the outer parts, so you can't use ratios here.
Fear of the dark
@@ongbonga9025 Even if the shape of the corner was linear this would still be wrong. You're making the assumption that because the sides of this right triangle have a ratio of 2:1, their corresponding angles must also have a ratio of 2:1. This is actually slightly off; the ratio of the angles' _sines_ is 2:1.
@@MrHatoi Thanks, a good explanation. Also thanks wictor.
4 years after this video: "How to solve for the pandemic? A viral homework from China."
Answer they figured is: drink several times a day (5- 6 ) boiled water but to cooled tp suitable bearable temperature. Inhale the steam as necessary too.
XD
it's the area of the rectangle, minus two times the area of a cirle, and that divided by two. It was blatantly obvious.
Ared = 0.5*(b*h - 2*pi*r^2)
*****
yeah the second one wasn't so easy though...
This is what I thought of as well.
That's how I did it too but I got 21.5 and the answer in the video is 19.5
.5*(200-2*(3.14*25)) 200-157 = 43 43*.5 = 21.5 How'd I mess up?
Blatantly obvious you imbeciles !
Tubewatcher, I understand the need to put others down to feel better about yourself, but over a freaking math problem? Really?
(The area of the rectangle - the area of both circles) / 2
Much easyer, as well as here you dont need to prove the two pices fit to gether as a whole circle (a step you skipped)
please look at the second problem
that's what I thought too but if you look closely you can see that in the bottom left there is a part which is white not red which complicates things and makes this invalid.
Albert Kong It was for the first problem... That is quite obvious...
that's how I did it
...didn't watch more than 5 secs did ya?
Part 1: Um~ Piece of cake!
Part 2: 2 times of theta and inverse of tangent... What?!?!
FishingDoing haha good one
Diffon Finsson deduce the fingertips of waltz
WAT?!
ruclips.net/video/hMbSL92jeTE/видео.html
Don't worry, the second problem took me 32 minutes to solve, and I finished trig 2 years ago.
So I am particularly proud! I got the same answer a different way. I took the lower odd shaped section and divided it into 2 portions separated by a line passing though the origin of the left square at (5,5) and the left intersection of the diagonal and the circle at (2,1). [(2,1) was found by simultaneous equations of the circle and the diagonal.]
I then got the area of the triangle formed by the origin, the intersection and the point this new boundary line crossed the x-axis.
I then got the area of the region bound by the x-axis, the curve and my boundary line. This was the area of the triangle between the axis, the boundary and the vertical radius less the area of the sector.
Then as per the video, large triangle less circle, less little triangle less triangloid.
Far less trigonometry - I only used it to calculate the portion of the circle in the sector - but I did use co-ordinate geometry to keep track of everything.
Area of rectangle: 20x10 = 200m
Area of circle : 5^2 x Pi = 78,54m
There are 2 circles so, circles = 157,08m
200m-157,08m=42,92m
42,92m/2= 21.46m
Close?
nvm I saw the whole video
miguelxlx :3 dude, that's exactly what I got too. I think this is the correct answer.
fmachado509 , you didn't watch the whole video.
Miguel M , the units of meters aren't actually assigned in this video. You do the math without them.
I don’t know who you learned multiplication from but 5^2X Pi isn’t 7,800 and you don’t know where commas belong in big numbers either.
You can also doing it by considering the diagram on a graph, with the lower left corner of the rectangle being the origin. From there, you know that the diagonal is y=x/2 , and that the first circle is given by (x-5)^2 + (y-5)^2 = 25. You can solve these equations to give the points of intersection between the circle and line at (2,1) and (10,5). You can then use a similar method to the one you used to find the segment and subtract the isosceles triangle.
The harder problem is NOT a geometry problem, it is a trigonometry problem. I spent a bunch of time trying to figure it out, while purposefully not using trig, as that was presumably not allowed.
I was thinking the same thing. In the US, Geometry classes come before Trig.
I was in 9th grade Honors Geometry which my be the reason why.. but I learned basic trig in Geometry
Geometry also has a tradition of restricting the means that can be used. "Using only a straight edge and compass..."
At that point, you may as well just use an integral.
+CorwynGC No, that's constructions. You are not limited to a straight edge and compass in geometry.
Hi Presh,
21,4602 and 19,5039 are the respective solutions. First problem has a total area of 21,4602 and simply resembles:
10^2 - 5^2*Pi by symmetry, that's all.
Second problem: I need to work out the area of the small red piece in the top right hand corner, looking like a skateboard ramp standing on its point.
Focusing on the top right hand corner, draw a horizontal line from the intersecting point of the diagonal and the circle extending to the right hand side of the rectangle, thus bisecting the "skateboard ramp", left to right. The "ramp" will be divided into an upper small triangle and the area located below, i.e. between circle and right hand side of rectangle. Continue as follows:
i) Work out the coordinate of the intersection of diagonal and circle (located in the top right corner), using some basic algebra [see Footnote]. It conveniently results in (18|9). The area of the small triangle (red area) is therefore exactly 1.
ii) Work out area below the small triangle, between circle and right hand side of the rectangle, by using CALCULUS.
TIP: Compute INTEGRAL from 0 to 4 using 5 - Sqrt(5^2 - x^2). Upper boundary "4" is given via coordinate (18|9). This will results in 2.4088. [Note: For clarity I turned the coordinate system anti-clockwise and integrated left to right]
iii) Add the two areas above, yielding 3.4088
iv) Add up 3 times "Hang glider" (corner pieces outside both circles), and also add the little piece we just stitched together. Done!
Final answer: 19,5039. No need for any angles, trigonometry nor working out the "lens". This method allows you to solve both problems lightning fast. 10 minutes or less.
[Footnote:] Getting the coordinate can be done in different ways. I simply equated two functions, circle and diagonal. That eliminates "y" and I solve for "x". Careful here, both functions need to use the same point of origin, hence:
+/- Sqrt(5^2 - x^2) = x/2 + 2.5 . I solve for x and done. I eventually found point (3|4) with respect to the first quadrant.
easy. the bottom half of the rectangle is shaded red, except where the circles are. since the circles are the same size, you can take the smaller part of one with the bigger part of another to get a whole cirlce. that is exactly what you see in both halves, so finding the area of one circle and subtracting it from the area of half the rectangle give the area of the red.
.5(20*10)-pi*5^2
.5*200-pi*5^2
100-pi*25
100-78.53
21.46 is the final answer
dang it theres a second part. im not entirely sure how to deal with it.
Nathan Lang i got the at too
Manny after watching it, that math was far beyond my education :/
Tup, that´s it. Simple.
You haven't proved to me that the smaller part of the circle and bigger part of the circle both add upto make a whole circle. You must be able to prove that without the use of intuition, or apparent symmetry.
Nice soluton to Part II! I ended up using an integral to solve the problem, but as the integral for the area below the circle was a bit hard, I had to use WolframAlpha. Here's how I did it anyway:
The first tiny triangle bit where the triangle line intersect the first circle has an area of 1 (2*1/2 = 1).
The equation for a circle with radius 5 with the center in (5, 5) is (x-5)^2 + (y-5)^2 = 5^2, or just y = 5 - sqrt((10-x)x) for the part that we are interested in. The intersection point is when the triangle line and the circle line intersect; x/2 = 5 - sqrt((10-x)x), which can be written as 1.25x^2 -15x + 25 = 0. This yields x1 = 2 and x2 = 10. x = 2 is the first intersection point, and is what we need to use as a lower limit when calculating the circle integral.
I then used WolframAlpha to integrate this equation from 2 to 5, added it to the area of the tiny triangle (1), and got ~1.96. Subtract this from the answer in part I, and you have the answer. :-)
Nice one. I was thinking of doing something like that at first, but i couldn't remember how to write a circle as a function. It's been too long since i've been in school.
Eventually i solved it like the video, but i had to look up how to calculate a circle segment.
Me too
I intended to do that too... But I failed too hard at figuring out the integral from 2 to 5 of "x/2 - 5 + sqrt(25 - (x - 5)^2)" so I took a little over 1h...
I can even explain what I did wrong, for your information I tried to solve everything with a pen and paper :^ )
When I had to integrate the part "sqrt(25 - (x - 5)^2)", I ignored the silent voice in my head telling me I had to use the Integral table ;-;
Secondly I forgot that the formula "integral of u(x)-v(x)" for the area between two functions doesn't work if they intersect in the given domain, so I used [0;5] at first...
Yeah and then things slowly went uphill again; at some point I realized I had to use [2;5], then I realized that the integral of sqrt(a^2 - x^2) was some crazy arcsin stuff so I had to start from scratch again... And lastly, after I still had false results I just decided to use my calculator and I finally solved it :)
Please end my life
You are making it way, way too hard for yourself ;-)
You know the areas of the rectangle and the two circles - then find the area outside the circles, and finally use symmmetry to find the requested area. Very simple ... once you know it ;-)
Bjowolf2 Erm, you can't do that in part II of the problem (which is what we're talking about).
Before watching:
10 x 20 = 200
2 x Pi x 5^2 = 50pi
(200 - 50pi)/2 = roughly 21.46
Easy
After watching 40 seconds:
Oh.
lol
lol likewise.
That easy for me,I just using 20 seconds.
I automatically did this lol
I found a very simple way to do the second problem. To find the area of that small bit on the bottom left, you can do this: get the right angled triangle at the bottom left that has base half of the length of the rectangle and height half of the height of the rectangle aswell. Find the area using bh/2. Then the small bit at the bottom left (we’ll call x) is equal to that big triangle -that sagement and that curved thing. To get the curved thing, just divide the answer to part 1 by 4. To get the segment (this is where my strategy is different) you do the following: take the circle on the left, say it has centre 0,0, it’s equation is x^2 + y^2 = 25. Since you know one of the points that the diagonal intersects this circle and another pint which would be (-5,-5) because it touched the corner of the rectangle, you can find a formula for the equation of this line and then solve it simultaneously with the circle to get the other point of intersection which is (-3,-4) using this, you can find the chord length on the circle, then use cosine rule to find the angle of the sector, then find sector, then minus the triangle area, then u have the segment, and then u can just do the big right angled triangle that we talked about earlier - curvy bit that we found earlier - segment and that gives x, then just minus x from answer to part 1
Well, I am in that part of RUclips again.
Pretty tough question for grade six.
Tough? It is easy. Logic, plus trigonometry.
It could be easy if you know trigonometry, I'm an 8th grader (graduating in a few days) and was never taught trig. (America *wink*)
Really? Check out my answer and how I got to it
Absolutely no trigonometry needed. Area of square, area of circle, subtraction -- wham, bam, thank you ma'am, done.
WARNING⚠️ don't try this at home school or any other place.
ههههه
@@open-minded2242 اهلا بك يا أعرابي
@@abdulrahmanmohamed4131 بالمهلي
@@open-minded2242 انا عربي
@@abdulrahmanmohamed4131 انا أيضا
I have discovered a truly remarkable solution to this problem which this comment section is too small to contain.
Fermat's Last Theorem?
yes.
You didnt get the joke
neks you're an idiot. He is referencing Fermat's last theorem.
Sam Wilt nice try
I rounded to get 19.672. Two circles each with a radius of five make an combined area of 25π. Subtracted from half of the rectangle area is (100-25π). Each "diagonal triangle" has the same area (the center one can be two triangles), but the corners are split by the diagonal. Looking at the bottom right triangle, the diagonal line with a slope of 1/2 means that the shaded half is half the size of the other part, so x + 2x = area of the other triangles (x being the area of the shaded region, the 2x being the non shaded part). That means that the shaded area in the bottom right is 1/3 the area of the other diagonal triangles, and the shaded one in the top right is 2/3 of the area. Since there are 3 triangles (the center counts as two) and two incomplete triangles, you can make an equation 1/3 + 2/3 + 3. Multiply by three and you get 1 + 2 + 9, or 12. Removing the shaded region in the bottom corner would remove the one from the twelve, making 11. The numbers mean that 11/12 of the red is still there after the bottom corner is removed. (11/12) • (100-π) = 19.672 (rounded).
There's a 3,4,5 triangle hiding in the right circle that greatly simplifies solving this problem and doesn't require dragging in so many equations. If you place the origin (0,0) at the lower left, the equation for the right circle is (x-15)^2 + (y-5)^2 = 25, and the diagonal line is y = x/2. The intersection of this line with the right circle occurs at (10,5) and (18,9). Draw a second diagonal from the top left to the lower right, and connect the two right hand intersections with the right circle with a vertical line. You can solve for the areas of the three segments outside this circle-filing triangle and the rest is easy.
Jeff Bowermaster Are you sure that the three segments outside the triangle but inside the right circle are identical? Because I do not arrive at the right solution, and line piece (18,9) (18,1) is shorter than both (10,5) (18,9) and (10,5) (18,1). Or is there another way to calculate the surface of those different segments?
You're correct they're different sizes. I wish I could post pictures this would simplify everything. The three points {15,5), (18,9) and (18,5) form a 345 triangle. That sweeps out 53.13º which is the only place angles are needed. Double that to 106.26º, divide by 360º and the fraction of the circle swept out is 0.2951666. Now you have the area of the sector, it's easy to find the area of the segment to the right of the line (18,1)-(8,9). Add this to the triangle area with vertices (10,5), (18,9) and (18,1), subtract that from the entire circle and you have 2 times the area of the two left hand segments. One of those plus (area of square - area of circle)/4 gives the area of everything except that annoying little missing piece. (10*5)/2 is the area of the triangle containing all three. So the little piece is just the difference.
Don't know your explanation ...but anyways interesting
I love solving mathematics problems
Just saw a typo in my explanation - It should read "right of the line (18,1)-(18,9). This is a brute force problem, you just have to go in and solve for whatever sectors and segments you can, then combine those together to come up with the answer.
Jeff Bowermaster Your solution still is relying on an inverse trig ratio -- to know the angle 53°+ and its double, you need to use invtan 4/3. So you are not really escaping using trigonometry for Part II, but you do need less math -- just area of triangle formula, area of rectangle, and area of circle sector.
I solved it and there's no one here to congratulate me and no way to prove I did it :(
I did it totally different
I didn't use any trig
+Dappernaut Maybe the first one, but the second one you need to.
then pls show us the the way u did, then we ll believe sth..
He weighed the amount missing from each of the crayons used to draw the figure.
I'm not reading the previous 1,948 comments to see if someone already said this or not. But your solution for the harder version is overly complex. Using basic rules of geometry you can find that the obtuse angle of that isosceles triangle is 120 degrees, and the sector is exactly one third of the circle. If you split the isosceles triangle into two right triangles you'll see that the height of the isosceles is 2.5, and its base is 5 * the sqrt of 3. So the area of the circle cut off by that cord is Pi * 25 / 3 - 6.25 * sqrt of 3.
If the hypotenuse line of the triangle is mirrored above, and then the points where those two lines cross the circle are connected, you'll see an equilateral triangle whose corners intersect the circle. That is how you know the sector is 1/3 of the circle, it is also how you know that the isosceles triangles obtuse angle is 120 degrees.
However, it isn't /quite/ an equilateral triangle; if it was, then the grey region would be exactly 1/3 of the area of the full circle, but it's actually .352 of the area. That's not a rounding error, I calculated from scratch.
Hi George, eye balling it will only get you so far!
The isosceles triangle is more than one third of a circle (120degrees).
theta = tan-1(5/10) = 0.4636rad
The angle in isosceles triangle = pi-2*theta = 2.6779rad = 126.87degrees
Yeah, Nat pointed out my mistake. I was confusing 1 2 3 right triangles with 1 sqrt3 2 right triangles. The first not being a 30 60 90, while the second is.
A triangle with a ratio of 1 2 3... I don't really know what I was thinking there.
I was thinking this, too. But you get a 30, 60, 90 triangle with sides 2, 1, and sqrt(3), from bisecting an equilateral triangle with sides 2. Or just using the Pythagorean theorem. Anyway, here you have sides 1, and 2, and hypotenuse sqrt(5). In the former case, you have sides 1, sqrt(3) and hypotenuse 2. Clearly different but apparently the same until you look at them.
For the easy version: 21.46 square units (rounded)
For the hard version: I have this clump of my own freshly torn hair
I am a Chinese and this should be something like an advanced problem for some brilliant students in 4 or 5th grade primary school and of course the expected solution should have nothing to do with trigonometry, calculus or anything too academic. The idea of such problems is to test whether a child is clever enough to figure out the tricky solution to solve the problem in a much easier way.
Yeah. This feels like one of those math competition trick questions.
Ravie Yeah, now try solving for the area of one individual piece without using trigonometry or calculus
Jason Yang 深圳学生表示毫无压力😂
No, only the first question is easy. The majority of the video is about the harder question, which he uses trig to solve.
I doubt 4th or 5th even if brilliant. I would say more like brilliant 9th or 10th. This required a well developed frontal lobe and as such there is not short-cut to human development physically.
1.calculate the area of rectangle
2.calculate the area of circle (radius=5) and double the answer(as there are two circles)
3.subract the area of circles by area of rectangle
4.divide the difference with 2
The challenging part was actually calculating the small area in the bottom left part of the figure.
Could we not for the more difficult part of the equation:
1) 5/10=rise/run=1/2
2) Line 1 eq: y=.5x
3) Line 2 eq: (x-5)^2 + (y-5)^2 = 25
4) Use typical system of equations strategy to reveal two coordinates, with the lesser being the intersection of interest: (2,1)
5) With the intention of doing an integral and to simplify, we set the circle equation to x^2 + (y-r)^2 = r^2 and the new x value of interest to 5-2=3 and integrate from 0-3 from this new equation.
6) Solving for y yields two functions with the one of interest being y = r - sqrt(r^2 - x^2)
7) Integrating from 0 - 3 for 5 - sqrt(25 - x^2) or taking antiderivative yields: 0.956
8) 0.956 + ((1 x 2)/2) = 1.956 for the total area of that shape!
9) Then putting it all together: ((10*20)/2) - pi5^2 - 1.956 = 19.504
It ends up being similar in complexity, but I never showed up to Trig and got your "lazy student" C in the course.
That method should work, but I would disagree that it is of similar difficulty, not even counting that people usually learn calculus after geometry and trig. I think most people who still remember learning geometry could have given you all of the basic area formulas used in the video's solution. I don't think most people who learned calculus ever knew the antiderivative of sqrt(r^2-x^2) off the top of their head. Sure, you can plug the formula into a numerical integrator and get the right number, but you could also just cut the pieces out and weigh them to also get the right answer. Additionally, neither of those answers is exact like the answer in the video is. Additionally, my experience is that you would not have a calculator for a question like this. Personally, I don't think this is a question that most 11-12 year old kids, or most educated people in general, would be able to answer reasonably quickly in an exam-like situation.
Gordon Walsh Why are you comparing Geometry to Calculus? The basic area formulas and thus easier steps are doable by any reasonably intelligent young child. As for difficulty, the number of steps is similar and require a similar spacial level of understanding.
I'm with you on people learning trig before calc though. Personally, I slacked through trig and just got my BS in Biochemistry and so know calculus like the back of my hand. For >90% people, the trig method is the way to go. The calc way though worked for those who don't know trig anymore/well, and wasn't rendered arbitrary by a significant difference in number of steps or complexity.
Patrick Connolly oh, I hate calculus
That's funny cuz I started to do the same thing then I was like "ain't no 6th graders know calc and trig!" I spent like 20mins trying trying to find a solution without using them lol
You should realise that the it has a perpendicular line of y=x/2 pass through center so you find the eq of normal and find the distance(l) between centre and the intersection of normal and y=x/2,then let angle t between the radus and this found length of line,cos t =l/r,the following step should be easy
Fun fact - you don't need all the tan/cos/angle stuff. I did this without a calculator. Three steps:
(1) subtract one circle from half the rectangle.
(2) find the area of one corner ((half rectangle - circle)/4)
(3) use ratios to find the portion of the corner that's left out and subtract it from the total - in this case the slope is "2 horizontal for every 1 vertical" so the ratio will be thirds and it's the smaller portion so it's 1/3 of the corner area. Subtract this from the area found in step 1 and wham bam thank you ma'am :))
It's all about looking for the relationships!!! And trust that there IS a solution, and you're meant to find it. Like a little archaeologist with a dusting brush. Good luck friends
The first half puzzle I found a lot easier by realizing that the 'red' areas are congruent to a single square (10x10 - half the rectangle) minus one of the circles.
Everytime he says "approximately", engineers become happiest
I also done the second part by a small method without this video help.
Solution to second one 👇👇👇
First calculate the area of the edges by taking a square of 10 with a circle in it now subtract the circle area from square which gives area of remaining 4 parts as they are identical divide that by 4 so area of one edge is 5.365.now the actual problem we have to find out the area of left bottom edge below the diagonal for that I consider that the diagonal has bisects the edge in to 1:2 ratio we also know the edge area divide that with 3 so we can get the required area substrate that with first solution we can get around 19.5 as simple as that
Thank you so much for this solution. You made it so simple.
This was basically what I thought but the problem i ran into was that 5.365 / 3 is approximately 1.788, and then I made the mistake of adding (3 x 5.365) to 1.788, but eventually realized I wanted the 2/3rds of the corner in and the 1/3rd out, which meant (5.365 x 4) - 1.788 or (2 x 1.788) + (3 x 5.365).
Ultimately this comes out to 19.672 instead of Presh’s 19.504.
The area of the rectangle is 10*20=200. The area of each circle is 25pi, so the remaining area in the rectangle is 200 - 50pi. The red area is half of that, so 100 - 25pi, which is approximately 21.46 units squared.
8:21 you use sqrt(125) as the base of the isoceles. You can't though, since the diagonal extends past the limit of the circle.
Wow good point. I'm surprised he didn't notice that.
Noo, he used the triangle below to calculate sin and cos of the angle.
No, he doesn't use it as the base of the isosceles triangle, he uses the bigger triangle (10, 5, sqrt(125) ) to calculate the tan, sin and cos of theta, because their value is the same for both triangles (its the relation between the sides, which is the same because the angle is the same, theta).
That's what I said +Melvin No, or at least what I meant :P
Can't find your answer on the hole page... Where exactly?
another approach would be, total volume of the rectangle minus total volume of circles divided by 2. (20*10) - 2*pi*(r squared), divided dy 2-> 200 - 157 = 43 / 2 = 21.5
Go back to school guys! No hard mathematics included, just a bit of logical thinking!
Watch the damn video, you idiot. It contains TWO SEPARATE questions. ~21.46 is the answer to the first question not the same as the second question which this video explains.
yeah that's what i was thinking too, but h e's explaining an other problem, one were you are missing the area of the bottom left corner. (as in that corner is not red), otherwise, yes you are correct,
Ahh, I did not watch the video at all... just gave it a second of thinking, checked the answers then an posted results...
czierwo
Next time think before you go jumping down people's throats. My first comment, on another comment thread, I just pointed out the second question. Then, you had to (try to) act smart and say "Pi is not exact number." and that I was wrong.
First of all, I did not jump at anyone. Second, The video title does not indicate that it contains 2 questions. Third, PI is NOT exact number unless you consider 13.3 trillion digits an exact number... Have a nice day!
The second part is NOT 6th grade
Bamberghh 6th grade in China, not in the rest of the world
Agree. The double angle formula is something in senior high school
homi do azulejo You mean in china where children don't graduate 6th grade until they can calculate the appropriate course and trajectory to send a rocket to the moon and back? Some people will believe anything if it supports there preconceived biases.
First part is simple, the second part isn't
@@sevsesi7866 how hard am I supposed to think it is?
This is an iq type of problem. Maybe they have the gifted 6 graders but normal human 6 graders' brains don't work like that no matter which country they're from. It's no coincidence that the conceptual difficulty is roughly the same per age group everywhere in the world.
I did it!
1). Find area of 2 circles and Rectangle.
Area of 2 circles : 2.phi.r^2 = 2.phi.25 = 50phi
Area of rectangle : 20.10 = 200
2). Substract area of circles and rectangle
200-50phi
3). Devide subtract result by 2 (diagonal line made recangle slice into 2 triangle similar)
(200-50phi)/2 = 100-25phi
🙃🙃🥳🥳😎😎👽👽👽👽👽👽
An overly complex explanation. The diagonal line clearly cuts the rectangle in half because it goes from corner to corner. Therefore the area above and below the line is the same. Thus the area under the line is the same area you get if you divided the rectangle vertically down the middle into two halves, each with a circle inside a square. You then have a whole circle (d=10) inside a whole square (10x10). The red areas simply is the area of the square (10 squared) minus the area of the circle (5 squared by pi.)
Ian McFadyen answer doesn't come by that donno why
I thought exactly the same. You just need to prove the areas correspond (by demonstration and laws) and then it's really easy. But the result isn't the same as in the video, gives about 21 but in the video it's around 19...
Edit : Ok I watched the useful parts in the video, and in fact 19.5 is the answer to the smaller red region x). The 6th grade exercice might be the first variation that gives 21. But we never know, after all they "are asian".
I
Hi Ian McFadyen, it looks like there are two questions in this video. Your solution solves only for the first and simpler problem. Just thought you'd like to know.
Ian McFadyen, Mustang Rider:
This solution doesn't work for the real problem. It is obvious that it works for the easy introductory problem, but if you didn't notice: The video is mainly about the problem where the lower left part of the red area is missing. 0:30
I solved the first problem by finding the area of the circles and subtracting this by the area of the rectangle then finally dividing this by 2.
Area of circles: 5^2 x pi=78.539x 2 =157.079
Space out of circle = 200-157.079= 42.290
I then took half of this number since we are only measuring the red to get 21.460
(These are not the full numbers but rough estimates)
thats what i did too, but apparently its not right which is what im confused about
Exectly same. You did correct
Sorry fellows but you missed the point... there is an area in the left bottom corner that is not red, in fact I did the same but yeah... there is this little part that do not count so you have to go through many calculs
Look at 0:43
you dont see their are two different tasks? solution 1 is that what you mean... at 2:10 ... second task begins at 2:27
I'm not Chinese, but this problem should be a 10th grade problem in Vietnamese high schools.
I'm a vietnamese and i approve.
Area of the triangle is 20(10)/2 = 100. Area of each circle is π5² = 25π.
The red area is equal to the area of the triangle minus one of the circles. Ignoring the color, the figure is symmetrically mirrored about the diagonal. The area of the circle on the right that is _not_ overlapping with the red triangle is exactly the same as the area that _is_ overlapping on the left.
A = 100 - 25π sq units
As for the second question, oof...
The equation for the left circle is (x-5)²+(y-5)² = 25. The equation for the doagonal is y = x/2. Equating them we get:
(x-5)² + ((x/2)-5)² = 25
x² - 10x + 25 + x²/4 - 5x + 25 = 25
5x²/4 - 15x + 25 = 0
5x² - 60x + 100 = 0
x² - 12x + 20 = 0
(x-10)(x-2) = 0
x = 10 ❌ | x = 2
(2-5)² + (y-5)² = 25
9 + y² - 10y + 25 = 25
y² - 10y + 9 = 0
(y-9)(y-1) = 0
y = 9 ❌ | y = 1
So the circle and diagonal intersect at [2,1]. If we draw a line from the bottom point of tangency to that point of intersection, we get a triangle of base 5 and height 1, so area 5/2. If we subtract that from the previous total we get 97.5 - 25π. Now we just need to add back the area of the circular segment created by the chord we just drew. The length of the chord is √(1³+3²) or √10.
cos(θ) = (5²+5²-(√10)²)/2(5)5
cos(θ) = (50-10)/50 = 40/50 = 4/5
If cos(θ) is 4/5, we've got a 3-4-5 triangle, so sin(θ) will be 3/5.
A = 5(5)sin(θ)/2
A = 25(3/5)/2 = 15/2
And the sector area is:
A = (cos⁻¹(4/5)/360)π5²
A = (36.870/360)25π ≈ 8.044
Circular segment:
A = 8.044 - 15/2 = 0.544
So total area is ≈ 98.044 - 25π ≈ 19.504 sq units.
Hey! I got it right!
You know you are in too deep when you use calculus to solve the second one xD. Calculate the intersection point, and then integrate the area for that section outside of the circle
yeah... I tried solving with just geometry but my brain kept saying, " WTF ARE YOU DOING? You know the easy way to solve this!"
Calculus is a cruel mistress
Yeah, because who doesn't know integral calculus by sixth grade?
72golfcrazy dunno, but if 6th grade China students can figure this out with just geometry then they'll do spectacularly in Calculus. That's a lot of problem solving power there to do it the basic way.
That's what I had to do.
Finding the exact intersection points to establish limits of integration is not a trivial exercise either; thus, even if one were to use calculus, the amount of work involved is approximately the same.
Much, much easier way to solve the 2nd part, I think. Starting from the solution of the 1st problem we know that the area of the first red sections is ~21.460. Now looking at the diagram we can see that this is a rectangle with side lengths 10 and 20. If we draw a line straight down the center of the rectangle (making two identical 10 by 10 boxes) this allows us to see that if we flip the left side of the triangle (the smaller red area side) ontop of the other part of the red area triangle. This means that in one 10 by 10 square the red area takes up ~21.460 excluding the circle. Now to find the smaller section we can divide by 4 which gives us the area of just one of those sections. Looking back at the overall triangle we can see that it is a 2 to 1 ratio in the side lengths. Meaning that the line going down the center has one angle of 30* while the other is 60*. That means that while looking at any square or symetrical meaurment the line divides the area up inbetween 1/3 and 2/3. With this we know the area of one of the small red sections and we know what fraction of it the area we need to find is. so we take 5.365 (~21.460 / 4) and divide it by three. This gives us the area of ~1.78 which is the missing small section. Then we just subtract this from the original red area, giving us ~19.6716666 (video answer = ~19.504). There is a slight variation of the answer due to my use of the estimated values displayed in the video.
This is super easy you take (20*10=200) witch is the volume of the rectangle. Then to find the volume of the circle is ((Pi*r2)*2 for the number of circles) so (3.14*(5)2*2)=157 so then you 200-157=43 then you divide by 2 to get rid of the remaining white area which is 21.5. The answer is 21.5. Sorry about the squared part stupid iPad don't have square root or even Pi.
Did you watch the whole video?
I got the same method and answer with you, it is really a super simple way!
this is a solution to the first part, isnt it -_-
You only solved the first, easier problem. Go to 0:30 to see the more difficult problem.
+David Corbo no stopped it at 5 seconds in, got bored after my answer to it
I haven't done highschool maths problems in 5 years so I'm very happy I was able to answer that. I didn't need to use the sector area formula or find the area of the little piece in the bottom left.
there is much easier reasoning, by observation, one can see that the area is equal to a square with a side length of 10 withdrawn by the circle contained in this one, the area is then very simple to calculate :
x = side of the square
area of the square - area of the circle = x^2 - pi.(x/2)^2
= (1-pi/4).x^2
problem solved ;-)
Yeah he way over complicated it.
Oooh yes indeed x)
+Don the first the other corner on the left side fills it by using the fact that thre diagonal is not curved (problem would be solvable if it was but you see that it's very simplified with this assumption)
+koolkitty8989 oh yeah that's true ... I didn't see that one ^^
ok ... so I don't have any paper here but I'd say that it's equal to 3/4 of previous answer + value of the area of the remaining area at the top right
this area can be computed by dividing it in two parts (you must turn the surface by -pi/2 radians (or 90 deg horlogically) to visualize easier) :
- left part, where upper part is curved, which is calculated by integrating the function of this curve until the intersection with the diagonal (function and point of intersection must be figured but I don't have tools to do it now)
- right part : upper part is linear, which makes the surface extremely easy to compute (triangle area with height of the image of the finction mentioned above)
Thus, that's the idea, hope it's not too complicately explained xD
Even more simply by saying that the area is the total area of the big rectangle subtract the area of the two circles and then divide that in half.
(1/2)[B*H - (2pi*r^2)] where b is 20, h is 10, and r is 5
However it does not help (nor does your comment) with the 2nd one (at least not with the new challenge of it) ... I probably would of ended up integrating it myself to be honest. This video is pretty good as far as the 2nd is concerned.
Sixth graders in China know algebra, geometry and trig? Hard to believe. Maybe a few,, but that would be the same in most countries.
Lots of them do competition math to get a chance for a better middle school(you have to take a test to see what middle school you go to, kinda like college)
Robert Zhang So this isn't really isn't something most Chinese students would know. Kind of like spelling bee.
You should come to Singapore (not in China) and see how hard algebra is.
Passed High School Physics but on the tests, they usually give hard questions like these for the talented math students
Robert Zhang It should not be a surprise American students can't answer questions like this one. With our "No Child Left Behind Act" mean no child get's ahead either.
Our schools are being run by MBA's who are treating education like a production line. Treat students like ram material, educate all of them exactly the same and you will have an educated student.
The thing is educating doesn't work that way. The raw material, (the students) are all different. To expect the same outcome when the raw material are people and different is just silly.
If one looks at the vast majority of students in our universities in the sciences, engineering, and computer science advanced degree programs it's hard to find an American student. Close to 97% of the students in those programs are from foreign countries.
Our country and politicians do not value educating our youth. One of the by-product of this is that Russian, Indian, and Chinese computer hackers are stealing billions of dollars from American citizens and America companies.
We just don't have people educated enough to protect our computer systems and the Internet, (something America created).
Until we fix our education system Americans will continue to get ripped off financially and technologically by people in countries where education is valued.
I'm going to guess approximately 21.46 for the first problem.
Watching the second problem, I was not prepared!
It’s 121,46 sir
BayAlexander nope, you forgot to divide the square by two for the triangle
It looks like you need calculus for the second part
Total area 20 x 10 = 200
Area of circles pi x 5 x 5 x 2 = 157.08
Area of red spots = (200 - 157.08)/2 = 21.46
Subtract the area of one circle from half the area of the rectangle and divide the result by 2. Half area of rectangle = 10*20/2 = 100. Area of circle = Pi*5^2 = 78.54. 100-78.54 = 21.46
It's these kind of problems that made me fall in love with geometry
- Hey, you forgot to paint that piece right there
- Oh, that's not a prob... Holy, turns out it is!
oxford maths grad here. a lot of the people on my course were Chinese and there's no way this was set for 6th graders.
Too easy or too difficult? Too difficult I imagine...right?
Calculus at 11th, Trig at 10th, and Geom at 9th. You are right.Caltech at 12th, BTW.
It's true, this specific problem I've seen in my friends 6'th homework assignment. China is militant about their maths education department. With a solid grasp of intermediate calculus and she is only a freshman in college. You also have to factor in that they attend 12 hours of school a day.
This could easily be for sixth graders. This guy made it way harder than it needs to be. If you can use percentages, and calculate the are of each of the shapes, and have like, a few days of trig experience, you can figure this one out. He makes it seem wayyyy harder than it has to be.
Maybe a few weeks to teach sixth graders basic trig, but it's still fairly simple
My method was using basic trig to find the angle of the diagonal. Then I found how much of the corner the line intersected (the percentage of the angle from 90 degrees). Using that percentage you can do the same for area. If you need more info for how I did it, I'll gladly try to explain it better, just let me know. By the way, my answer came out to be about 19.914. So it was a bit off, but was really close considering I used a much, much simpler method.
Here is a copy and paste of another comment I made with a more in depth explanation:
Dude! You made the harder problem WAYYYYYY harder than it needs to be. I just calculated the degree of the diagonal using some basic trig (only had to use inverse tangent once). I then calculated the percentage of the degree compared to the corner (i.e. the corner is 90 degrees, and the diagonal has an angle of 26.6 degrees from the bottom of the rectangle, meaning that the diagonal divides the corner piece into roughly 29.5% and 70.5%). SO, if we look at one half of the rectangle (divided of course by the diagonal), we can see that it is made up of 3 whole "corner" pieces, plus 70.5% of a corner piece.
(TL;DR) I used inverse tangent only once, and nothing else, and then found some degrees and percentages, and that was it.
I can completely see how sixth graders could do this, so long as they knew some very basic trig. Honestly, your super complicated method is like using calculus to find the area of a hexagon. Like, why make it so hard dude?
Good vid though
I think I know:(its easier than you think!)
For first problem:
first we substract area of circle from the rectangle:
l x b(10x20) - πr^2(r=5)
we get the remaining area without circles. Then we know a diagonal divides the rectangle in two equal parts so...
area left unoccupied by circles divided by 2 gives the answer. Basic geometry.
so final answer here is->60.75
For the second question:
we can go by substracting the areas one by one. The area of the edge will be the same on all the corners so find the area of the lower right red part and then you know area in middle is 2 times the same if seen as two congruent squares however its challenging to go for right top but again these are congruent triangles(SAS) so we can count that part as one whole edge and substract the area of one edge from the remaining so we get the un red chip in lower left and substract the value from one of the four equal edge area and there you are with the answer. EASY!
please do not hesitate to ask doubts, Thank you.
" its challenging to go for right top but again these are congruent triangles(SAS) so we can count that part as one whole edge"
Which triangles are you referring to? I got lost here.
@@chickencyanide9964 I was considering the triangles cut by diagonals of the rectangle so I suggested that the area in the upper half is equal to the other half of the rectangle so that the unequal pieces at top right red and unshaded down left fill as one side then we might solve this same as 1st
Can it be solved without using trigonometry.
yes, by using calculus XD and its way harder
Slade951 for the trig one I instead imagined a square on the lower left corner and subtracted the fourth of the circle
Slade951 The area of the little difficult piece is the double integral of dydx from y=0 to y=x/2 and from x=0 to x=2 plus the double integral of dydx from y=0 to y=5 -sqrt(10x-x*x) and from x=2 to x=5
According to WolframAlpha that's 1.956236
Sebastián López integral (2 to 5) - sqrt(25-(x-5)^2) + 5 dx = 25 - (16 + 25/2 arcsin (3/5)) = 0.95624, soo 1+ 0.95624 = 1.95624
第一问所求面积可以看作是四块一样的红色区域，可以求出这四块每一块的面积。第二步求出被切割剩下的弓形的面积。求法其实不用那么复杂。设左下角一点为坐标原点，向右，向上为正方向建立直角坐标系，得割线方程为y=1/2x，可以求的左边圆的圆心为(5,5)，圆的方程为(x-5)^2+(y-5)^2=25,求的割点坐标，搞定。
方法二：如法一建立坐标系后，求出圆的方程，在[5,5]内对该函数积分，求出面积。
wu peter you must be a 6th grader of china
wu peter 可以順便寫出答案嗎？
I wasn't thought how to do is in 6th grade.
You obviously weren't thought how to use grammar either...
+Brendan Bush I hope your use of thought was sarcasm. 😉
My grammar was intentional.
Yeh intenshena;l
relish?
Before watching this is how I would go:
You see diagonal line is half of rectangle so ab÷2. When you have that you focus circle the upper one is missing small part. Small part is on the bottom one so small part + big part= whole circle we can see. So to solve this you have to substrat the are of the diagonal minus one whole circle I think
The area of the rectangle excluding the circles:
0a) W * H so 10 * 20 = 200
0b) circle is diameter/2 to get radius, pi * Radius squared for area so 10/2 = 5. pi * 25 ~= 78.539816339744830961566084581988.
1a) two circles is about 157.07963267948966192313216916398., so 200 - 157.07963267948966192313216916398 is 42.920367320510338076867830836025.
1b) half of 42.9 is about 21.460183660255169038433915418012 Easy part answered.
For the hard version:
2a) the rectangle is 2 X 1, so the angle of the line is 26.6 degrees.
2b) therefore, the missing slice is 26.6 / 90 * 100 = 29.6 percent of the area of the corner.
2c) the rectangle has 8 such "corners"
2d) so the missing slice is ~0.296 of 0.125 of the rectangle outside the circles, 42.9, so 42.9 * 0.296 * 0.125 = 1.5873
23) 21.45 - 1.5873 = 19.8625
EDITED: corrected rise/run angle and recalculated. There must be something I'm doing slightly wrong, but I think the correct answer could be gotten this way, which seems more like what a middle school child would be doing, not your way.
Please explain step 2b
*****
if the rectangle was a 10 x 10 square, the line going through from bottom left corner to top right would be at a 45 degree angle, and the missing red piece in the hard version would be half of the corner.
So we divide the angle by 90 degrees to find what percent of the corner is the formerly red slice. Multiplying by 100 is just to make the percent more readable.
Awesome thanks man, I knew it would be the case with half of the angle but I didn't know the same would count for other values.
Unnecessary to put actual numbers in your solution. Just use w,l, r, etc.
Dan Kelly
I was trying to see if rounding or what was causing my answer to be slightly different from the video, maybe someone can see what is wrong here.
Are you saying the chinese are teaching trig. to all their 6th graders? Yeah, I'm calling bullshit on whomever is spreading that rumour.
I was thinking about it and his solution for the harder version is more complicated than it needs to be. All you need is basic rules of geometry, no tangents, sine, or cosine are required.
This is the reason why so many eastern school kids are so bad at maths. This is Asia buddy all western countries are 4 years behind in everything. I live in Australia and am in grade 10. I solved the 1st problem very easily, but the 2nd was much harder but it was manageable. I bet you 100% that half my maths class couldn't solve the first problem very easily. In Asia everything is hard. They start doing calculus at grade 10. So plz if u think they don't teach trig at grade 6 then maybe you should travel to ANY Asian country and ask there teachers what they are learning!
Ah, well no, I have not. But, with folk like you around, that thought is all too close to truth.
Jim Parsons Ok.
I wouldn't be surprised that other countries are ahead of the US, or other western countries in what they teach in school. That said, even the hard problem here doesn't require trigonometry, or any fancy math beyond Pi.
When the diagonal line crosses the circles it is marking one leg of an equilateral triangle in each circle. Once you realize that you don't need to calculate any of the angles with fancy math. The sector defined by the points where that line crosses the circle is exactly 1/3 of the circle. To get the area of the slice defined by that same cord you'll have to figure the area of that isosceles triangle like the video says. But since you know the arc is 1/3 of the circle you know that the obtuse angle of that triangle is 120 degrees. If you then split that isosceles triangle into two right triangles that are 30 60 90's. From there you can just rely on the 1:2:3 ratio of 30 60 90's to determine the dimensions of the isosceles triangle.
If it's from China it will most likely break in about 3 days time. I'm going to wait it out.
If it's from United States, it will definitely break because yo mama with her cowboy jeans is sitting on it with her 300lb fat ass filled with Cheeseburger meat. I am just going to sit here an watch.
Where did that come from?! ^
Does not matter where it was from. The fact that yo mama was sitting on it, anything will break.
internet is just too salty
woah man thats pretty *salty*
First try before watching the rest of the video:
The area of the rectangle is 200
the area of each circle is piR ^2 = pi(5)^2 ~ 78.54
The diagonal perfectly bisects the rectangle, and the red spots are all beneath it, so you can reduce the potential area by 100.
The diagonal also perfectly cuts the circles so that there is in total the equivalent of 1 circle of white space under the line.
100 - 78.54 = 21.46
So, the area that is red is ~ 21.46
You can also treat the diagonal as if it were vertically down the middle and solve it that way.
This works because the red area in the bottom left is equal to the white space in the top right, and the red area in the bottom middle is equal to the white area in the top middle.
Now it's just a circle in a box, and that is simply the area of the square minus the area of the circle.
100 - 78.54 = 21.46
The variation however, is more complicated.
Next thing you know the grade 6 schoolchildren in China will be proving Euler's Identity next.
Smart asses in the comments section forgot this is a Chinese 6th grade problem.
+Rafael Cruz even if it is 6th grade, it isnt a challange at all, it has nothing to do with knowledge just sense. The 2nd is a little tricky but alright. Dont feel offended please.
+Phoenix ok
+emfederin Well I agree I would much rather choose this Chinese curriculum over what is happening back in the States with them Creationist and Flat Earthers.
Actually, in China, it is known as the "dead European stole great ideas from Chinese mathematician" identity.
20(10)=200
pi(5)^2=78.5
200-78.5=21.5 roughly
You did not watch the full video did you?
can you calculate ?200-78.5＝21.5？
+Nicholas R.M. No my phone died while answering the second part
+Kerry Brennan LOL. ok
+Ming Feng Zheng forgot to note I divided 200 in half, and rounded pi hence roughly
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Dark Phoenix how does this comment make any reference as a reply to the original comment? You are doing replies to comment wrong. There are rules my friend.
will u like to report this to Microsoft?
As others have said, the first part is relatively easy. In fact I didn’t even bother with the triangle area thing - because quite clearly the line diagonally across the rectangle will describe the same area if it was straight across the middle of the rectangle, and it’s easy to then see that it will also bisect the two circles.
Once I discovered there was a second part to the problem, I simply gave thanks that I don’t have to do mathematical solutions and bailed out at that point.
I think there's an easier way for second part:
Having the area of the first part, we know that it's composed by 4 times the shape on the right bottom corner, so we can calculate it by dividing 21.46 by 4 . Then we do a construction: we draw the other diagonal so that it forms on both sides an isosceles triangle. We can easily find the area of one of it and then we subtract it from the initial right triangle. What we get is the sum of 2 times the little shape on the corner left bottom that we are searching and 2 times the shape that we calculated before. Thus the little shape is this last result minus 2 times the shape of before all divided by 2. Then just subtract it from 21.46.
What we get is the sum of 2 times the corner part part, 2 times the small corner thingy and 2 times the unknown circular segment*
Yes, I know. Answering to 5 month old comments is my life
@@Revo5660
area of the space between the large triangle and the two circles = 100-25pi
area of square for one circle = half rectangle = 10^2 =100
area of each isosceles triangle = 100/4 = 25
area of quarter circle = (5^2pi)/4 = 25pi/4
everything between the triangle and circle = isosceles triangle - quarter circle = 25 -25pi/4
one piece = (isosceles triangle - quarter circle)/2 = (25 -25pi/4)/2
area of the red part besides the little corner piece = (100 - 25pi) - (25 -25pi/4)/2
I am getting 18.78 whereas the video shows 19.504
@@asiyaehsan1682
1. Isosceles triangles that the original comment meant are twice bigger than yours, for they are (20×10)/4 = 50 each. I'll assume you meant dividing half this rectangle into four right triangles, for this is the closest thing I can imagine from your enigmatic machinations.
2. Even if that's the case, we can't just subtract a quarter of that triangle. Because that's clearly not a quarter there - it's still an unknown segment that we haven't found. Second: because that way we ignore area of "that concave corner thingy".
Once you said sin I realized that this is way out of my math level
Why
you wouldn't need trigonometry to figure out the orange triangle's area, though.
Apply similarity triangle,
you'd get: height = √5
base = 2 × 2√5 = 4√5
What is way out of your math's level is when it comes to figuring out the segment's area, or just simply when the "inverse tan" comes in.
Can we have more geometry shaded area puzzles like this plssssssssssssss!!!! 😆😆I love those ! Those are fun !
Were you able to solve this one though? Be honest!
Nasir Akhter yea. Light work
Try to get (success SPM )by Wang Wei and Dr. Wong Sin Mong
Me: thats easy, get the area of the rectangle then minus the circles, then divide by 2
30 seconds later: I can approximate it but screw solving it
China kids are being tortured. This havent been taught to me until grade 10
in china , sin,cos,tan ,is in grade 9!!!! not in grade 6!!!!
but some teachers may carry some test from grade 9 to grade 6, which is sick!!!!
@@中国青年 so basically presh is exaggerating. Got it
You are insulting India
Even I am from India but I knew basic trigonometry in grade 9 icse.
@@funganker1079 No country is teaching trigonometry in 6th, especially inverse formulas. 6th graders could only solve the first problem
If you read the updated question, the first part was actually asked and the second was a misprint that people thought was the actual question.
Every time you express an irrational number in decimals I cringe a little
Some people like doing that. Cause, I dunno it feels nice I guess.
But where does it end, like, he could stop at the 6th place, or 20th place, or even the 10000th place!
How else are you getting an idea how big a complex expression approximately is?
+Teach'EM wherever you feel like stopping really.
Doing so is a good habit since you can more easily check that you have obtained a result that makes sense. Say for example that the expression evaluates to a negative number, you know immediately that you have made a mistake in your calculations; this can be hard to spot if you're just staring at the expression.
It can be solved it in two steps
1)red spot area is 100-25π-x
Where x is that small curved triangle area
For finding that we should know intersection point of diagonal line and first circle.
Here take left vertical as y -axis and down horizontal as x-axis then
U will get equation of diagonal line by passing through (0,0) and (20,10) as x-2y=0
First Circle centre will be (5,5) and radius as 5, u will get circle equation as
(x-5)^2+(y-5)^2=5^2.
For getting coordinates substitute x=2y (from diagonal line equation) in circle equation then u will get quadratic equation in y solving that
We will get y=1,5.(consider small)
Now y=x/2,and
Y=5+√(5^2-(x-5)^2) here these equations are from diagonal line and first circle equations by bringing y in left hand side
Keeping limits of y from 0 to 1 and for Y from 1 to 5.
Small curved triangle area =integration (0,1 as limits ) of x/2 +integration (1,5 as limits) of 5+√(5^2-(x-5)^2) on integrating and calculating we will get x=1.9561
And substituting in 100-25π-1.9561=19.504
Let's see if I can work this out before watching.
Looking at this, I would say it's (area of the rectangle minus area of the 2 circles)/2. So, (10*20 - 2*25*pi)/2= 10(20 - 5*pi)/2= about 21.5
For the HARD version, you need to subtract that bottom corner somehow or work your way up without it. We know that one circle has radius 5, and that the corner ends where the circle touches the bottom boundary of the rectangle. We can't just halve the area difference like we did in the easy version. Instead, split the rectangle down the middle so you get two identical squares with holes.
With the split, you now see that there are 4 separate red areas, 3 are indentical. We have 3*(area square - area circle)/4 for the total area of the 3 identical red slices. The last one, from the top right corner, will be harder.
Unfortunately, this is as far as I got. If I knew where the lines intersected the circles, I could do an integral. I'm sure there's a geometry trick to this I've forgotten.
Do they really teach trigonometric functions and inverse trigonometric functions in sixth grade?
I don't think so.
10/11th grade
I learned them in 8th grade (US public school btw), but I imagine China is 3 or even 4 years ahead of us in math.
Yes, they do teach trigonometric functions and inverse trigonometric functions in sixth grade. In many more countries than you think. I remember learning trigonometric functions and inverse trigonometric functions in sixth grade myself. And that certainly wasn't in China. It was in Europe in fact.
+noex100 so many countries are ahead of you (example Germany where I am from)
Hold up, didn't the thumbnail show the left bottom corner to be missing red.
Also, 21.46.
Oops. Hold on.
7:10 (10/cos(theta))/5=5cos(theta)/10
10/(5cos(theta))=5cos(theta)/10
25cos^2(theta)=100
cos^2(theta)=1/4
cos(theta)=1/2
theta=pi/3
Actual height of the original triangle which you labeled as 5 is: 10tan(pi/3)=10sqrt(3)
Area of orange triangle=(5×5×sin(pi-2pi/3))/2=(25sin(pi/3))/2=25sqrt(3)/4
Green shaded area=25pi/6
Blue shaded area=25(pi/6-sqrt(3)/4)
9:29 "Did you figure this out?"
Yeah....................., I figured it out.
9:30
Easy, if done by integration. Considering the shapes in Cartesian plane.
true
5th graders wouldn't know integration. That's the point of this mess lol
@@kaos7012 5th graders know all the other stuff in this video??
Before watching: (20*10-pi*5^2*2)/2=100-25pi
After watching: dammit, I didn't notice that lower part wasn't red
You are right. Original problem had that lower red piece.
I think it was 100-50pi
Of course it was red,it was a Chinese problem!
Sides of rectangle are 2 r and 4 r.
Red area has 4 identical shapes
Each of this shape equals to
(Square of side r) - (quarter circle of side r)
Hereby RED AREA
= 4 (r^2) ( 1 - π/4)
= (2r)^2 - π r^2
100 minus the area of a circle with a radius of 5
Yes, or 200 minus the area of two circles with a diameter of 10, that was my immediate idea. Did we think too simple?
Occam's razor, my friend.
No, you just didn't watch the whole video.
I thought same thing, was gonna comment on it, but look at bottom left corner of the triangle, that parts not red, gotta account for that, im in fucking 10th grade honors mathematics, finished geometry last year, and this is news to me.
right