This can be simplified by drawing a line between the 2 endpoints of the quarter circle. The nonblue area above the hypotenuse happens to be same area as x. Now you only need to find the area of the quarter circle minus the area of the nonblue triangle. Once again we end up with the blue area = 4π-8
That's great! Maybe your approach is the one that is expected to be considered by the primary school students if they see the figure as in Tangram... Kind regards.
This is one of those situations where you can "train" primary school students into solving seemingly high-level problems. The issue is that they don't actually understand the concepts and can only solve specific subsets of problems in the specifically trained way. This allows the schools, principals, and teachers to look good because "look how advanced our students and teaching methods are!" but in reality, the students didn't actually learn anything.
no one's gonna remember specific concepts after not practicing them for some years, but the problem-solving skills remain with us for the rest of our lives. perhaps you're reading too much into it while chasing conspiracies, which causes you to miss the obvious benefits.
Due to symmetry, you can split the lens-shaped part in half and rotate it to fill in the gaps in the circular edge. That just leaves you with finding the area of a circular segment: Area of quarter circle minus area of right triangle = 0.25*pi*4^2 - 0.5*4*4 = 4*pi - 8
That's easy to prove though. The coordinates can be calculated. +/- right angle rotation, of (0,0) about (2,2) gives (4,0) and (0, 4). The three points, (4,0), (2,2), (0,4) are colinear. By inspection. @mvkishoresr
Area of quater circle- (π4²/4) Area of trangle - (½ * 4*4) Now substract the trangle area from the quater circle area. (π4²/4)-(½ * 4*4)= 4(π-2) Don't make it complicated..
Hey man, your method is unknown to me, I didn't understand it at all. I am not familiar to trick you used here. Can you explain me how did you know subtracting the triangle area from the quarter circle area will give you the exact answer (area of the blue region)?
@@SourasisDebnath-yk6fe Actually it's very difficult to explain without draw the diagram, but I'll try... First forget the sade , now it's just a quater of a circle of radius 4 , now if we connect the ARC start and end point means CHORD , then it will become the hypotenuse of Right Isosceles Trangle. Now draw a median over the hypotenuse to get equal part of the given CONVEX shaded area, now if you impose all the given shade and the Right Isosceles Trangle and the Quater circle on each other then you will notice the Convex shaded area is equal , to that two segments area which are comes over the Hypotenuse of the Right Isosceles Trangle. So, if we deduct the area of Right Isosceles Trangle from the Quater of Circle , get the answer.
@SourasisDebnath-yk6fe it is simple rotate two x's and fix it to two z's so you get a triangle inside a quadrant and hence find the difference of areas!!!
I solved it the following way: Imagine square of length 2 at corner Calculate half of a semicircle Subtract it from square Subtract difference from square twice Thats the first blue area Calculate total area Subtract both semicircles Add first blue area Thats your second blue area Add them. Done.
Same. These kinds of problems always involve making use of flat-sided shapes, so I reckon if you just start by drawing square or triangular frames that box in or intersect the curves at key points, you are likely to find more than one way to solve it.
As an indian I may assure I literally solved this problem in my mind withing seconds of looking at it easy like I got a similar question in class 6 ICSE school question
@@V.444AzlI think he is saying that sarcasticly😅. I am from India passed ICSE primary in 2008. I don’t remember this and I am sure wasn’t there. This is JEE level stuff for sure.
How I did : Took a square of 4x4 in my head. If I draw semi circle of radius 2 on every side of the square I can write: Surface of square - semi-circle - semi-circle + area I look for ( 4 times ) =0. I got the rugby ball = 2PI-4. I took a paper and a pen at that moment.... Then, I took the quarter circle of radius 4 which surface is 4PI, remove semi-circle on a side ( minus 2PI ) , remove semi-circle on the other side ( minus 2PI ) and put back the rugby ball that I remove twice ( + 2(2PI-4 ) ) : gives : 4PI-2PI-2PI + 4PI - 8 = 4PI-8
So much simpler: Draw line between endpoints of quadrant, making a prime right triangle with measurements 4,4, 4sqrt 2. Bisect quadrant, creating 2 similar triangles and unshaded lens halves which are congruent to the shaded lens halves labeled x. Transposing those lens halves results in the shaded area being the area of the quadrant outside the prime triangle. Quadrant measure 4pi and prime triangle=8. 4pi-8.
0:35 at this point you can rotate the x surfaces so they fit the z surfaces, and then the y surfaces become just triangles. Then you see the blue area is just a quarter circle minus a quarter square
Solution: You can divide the blue "rugby ball" at the bottom left into 2 circular segments of a quarter circle, then place them at the top and bottom in the white bulge of the respective semicircles, then you get the blue area of a circular segment of a quarter circle with a radius of 4. Then the blue area is = the area of a quarter circle - the area of a right-angled, isosceles triangle with the two legs of length 4 = π*4²/4-4²/2 = 4π-8 ≈ 4.5664
Can't use that unless you also prove that placing the rugby ball segments results in a straight line and it's a white triangle (and not a quadrilateral)..
@@mvkishoresr The rugby ball segments are segments of a quarter circle with the radius 2 and fit exactly in the white bulge of the semicircle with the radius 2.
This is a primary school question in Singapore, actually not very tough. This can be solved easily by moving tje smaller blue area adjucent to bigger blue area. Afyer moving u have 1/4 th of the circleinus aclean white traingle... i. E, 4pi - 1/2*4*4 which is 4pi- 8 = 4*(pi-2) = 4*1.14 = 4.56
@@Tedde-bi9yp equation of the top left circle is x^2 + (y-2)^2 = 4 which I shall call A. Bottom circle is (x-2)^2 + y^2 = 4 wheich I shall call B and the large circle is x^2 + y^2 = 16 Which I shall call C. When you solve all equations for y, you get functiona that cannot exist in the negative y domain, so the top left curve must be negated to accommodate the lower blue shaded region. You then integrate the bottom circle - (x-2)^2 + y^2 = 4 from 0-2 and then subtract the same integration of the top curve, after solving both for y in terms of dx and of course remembering the top left integration value must be negated. That gives you the value for the bottom left shaded region. For the second blue section on the top right you must reconsider the positive value for the function of the A. You shall integrate C from 0-2 and subtract A from 0-2. Then integrate C from 2-4 and subtract the integral of B from 2-4. Add the values for both shaded regions and you have your result. I imagine there's a better way using polar coordinates, but I'm too lazy for that.
Note that the two half circles equal the area of the Quarter circle. Note that If we subtract the areas of the hald circles in a geometric sense we remove the inner blue area twice, while the outer blue area is untouched. Therefore the inner and outer blue areas are equal. Then it becomes easy because the outer area is very simple to compute.
The normal solution for primary school student should use rotation to make blue part as the difference of quarter circle and a right triangle. This is normal traditional segmenting and patching method in grade 5 to 6. In grade 5, circle concept is not fully introduced; but the students has 3-round learning for shape movement; have one round of segmenting and patching method.
When I was in primary school, I can answer it with linear equations. But now I'm college student and I only think eating two meals or three meals today.
They would totally use this to teach area, negation, etc to the children. Because otherwise they’ll have to teach circle and angle theorem. But seeing the children try to solve this with their own ways would be rewarding, because in mathematics there are a lot of ways to solve questions. I encountered this type of problem in middle school, maybe the primary school ones are for test or olympiad preparation? My way is weird as I imagined them as overlapping half circles. But that’s the charm of mathematics! 1/4 π*4^2-π*2^2+2*2*(1/4*π*2^2-1/2*4)= 4π-4π +4(π-2)= 4π-4π + 4π-8= 4π-8
Solving this with calculus is actually quite intuitive equation of the top left circle is x^2 + (y-2)^2 = 4 which I shall call A. Bottom circle is (x-2)^2 + y^2 = 4 wheich I shall call B and the large circle is x^2 + y^2 = 16 Which I shall call C. When you solve all equations for y, you get functions that cannot exist in the negative y domain, so the top left curve must be negated to accommodate the lower blue shaded region. You then integrate the bottom circle - (x-2)^2 + y^2 = 4 from 0-2 and then subtract the same integration of the top curve, after solving both for y in terms of dx and of course remembering the top left integration value must be negated. That gives you the value for the bottom left shaded region. For the second blue section on the top right you must reconsider the positive value for the function of the A. You shall integrate C from 0-2 and subtract A from 0-2. Then integrate C from 2-4 and subtract the integral of B from 2-4. Add the values for both shaded regions and you have your result. I imagine there's a better way using polar coordinates, but I'm too lazy for that.
U can equate both the areas of the blue region with being equal to each other and then alternatively draw perpendicular bisectors from both the lines of the quarter circle forming a square that encompasses the lower blue part. Now take the sides of the square as the radius of quarter circles (with the line of the blue part as part of its circumference) multiply that by two and subtract the area of the square from that. You got the area of one of the blue parts and double it! (2*2*1/4*2pi) - 2*2 = 2pi - 4 = area of one blue part. Double it u get 4pi - 8
Another method I came up with: Find area of quarter circle Find area of smaller circle Subtract smaller circle from quarter circle = the weird top shape, PLUS a debt of the leaf shape Find area of leaf shape = smaller square - 2(smaller square - smaller quarter circle) Multiply by two to cover previous debt Answer = 4.566... = 4pi - 8
I "drew" a line between the endpoints of the circle, put the 2x therein and solved the remaining triangle ( 4mx4m/2 = 8m² ) to substract from the quarter circle. Always interesting to see, how the real solution looks like.
We can divide the figure y (a half of the white area) into a quarter circle of radius 2 and the remainder. From these two parts we can make a square with side 2. Then the blue area is equal to area of a big quarter circle minus two small squares: (π4²/4) - 2*(2²) = 4π - 8
How I solved it: - Each of the white areas consists of a quarter-circle of radius 2 and a 2x2 square with a quarter-circle of radius 2 taken out of it. Thus, the white areas must each be equivalent in size to a 2x2 square. - The area of the large quarter circle is easy to calculate, it's pi * 16 / 4, or 4pi. - Now it's just a subtraction problem, and the answer is 4pi-4-4, or 4pi-8.
It found it pretty easy to figure that areas of x and z are equal. So I calculated the area of x and multiplied it by 4. The other method I found below (which involves flipping the x portion until its convex bowed side mates with the concave side of z, making it obvious that the answer is a quarter of the circle of radius 4 minus a right angle triangle that has is legs' lengths equal to 4) is even nicer.
My solution:. first establish that areas x and area z are equal. To do that, build a figure with four times the original figure by making mirror images of the original figure - one by vertical mirror reflection, one by horizontal mirror reflection, and one by applying both mirrors (same as rotation by 180˚). Piece it all together to make a figure with one big circle of radius 4, with four inner circles with radius 2. The summed areas of the four individual circles (which double-counts blue overlapping areas, x) is the same as the area of the big outer circle. The summed overlapping areas in the big circle, 8x, must be equal to the summed areas 8z for this to be true (think about it for a minute ...). All we need is the area x then. in one of the small circles, draw a square (diamond) that connects the center of the big circle to the points where areas x and z touch and the point on the big circle where the tip of two areas z meet. It's easy to see that the diamond can be made entirely white by translating and rotating two of the areas x inside of it to the white bulges outside of the diamond. Therefore, the white area of a small circle is (4/√2)² = 8. The area of an entire small circle is π(4/2)² = 4π. So the blue area of a small circle is 4π - 8. EDIT: Even easier: Once you build the big circle I described above, you can move ALL the blue areas, 8x, outside of a square diamond connecting z area touchpoints in the big circle ... and it doesn't change the total blue area in the big cicle at all. Now the big circle has a totally white diamond within it, with all the remaining are 8x + 8z outside of it.
I noticed that when arranging 4 semicircles overlapping each other within a square 4x4, The desired area can be calculated by subtracting the area of the square from the total area of 4 semicircles, then dividing the result by 4.(Sr this is so hard to explain in english since i'm not good at it)
My Goodness, such a difficult explanation. the two ’x, areas added to the 'z’ areas makes a triangle. substract the triangle from the quarter circel and you have the blue area surface ! That easy, you can do it without paper or calculator, by the way the answer is 4.56m2 (if it's in meters)
This is too algebraic, geo problem shall be solved in geom way ____ Note the blue "tri intersection" pt, use it as rotation ctr, rotate the two blue "chord areas" 90° cw & ccw respectively, to fill the symm white area of small semi circles, then you get a larger chord area of quter circle, so blue=4pi -8
if i consider the symmetry i will have to solve 2 equations with 2 unknown numbers, and if i don't, i will have to swap x and y and numbers, see line 100 and line 150: 10 vdu 5:gcol8:print "brainstation-find blue area" 20 for a=0 to 15:gcola:print a;:next a:vdu0:gcol 8:r1=4:r2=r1/2:xml=0:yml=r2 30 xmr=r2:ymr=0:zoom%=1.4*zoom%:nu1=155:nu2=nu1:masy=1200/r1:masx=850/r1 40 if masx run in bbc basic sdl and hit ctrl tab to copy from the results window
Using integration for this problem would only make it at least ten times harder 😂 I wouldn't use integration in an integration exam either and just tell the instructor to deal with it.
It seems to me that there is an easier and more straightforward way to solve this problem. To begin, you need to draw a chord. Along with the radii, this creates a right-angled triangle whose area is 8. Clearly, the desired area is equal to the area of the quarter circle minus the area of this triangle, which equals 4π - 8 = 4(π - 2).
The result is correct, but if you don't explain your formula clearly, you may misunderstand it. As a result, you won't receive any marks from the examiner
This problem is actually very easy! My sister is 10 years old and she could solve it in a few minutes! You just need to observe the diagram and by "observe" I mean to ACTUALLY OBSERVE it. Then you can solve it in a jiffy. (We aren't even Chinese. But yeah, we're Asian)
If you look closely it's 2 semi circles and one quarter circle, findjng the area of the weird blue shape above is easy, but the oval like shape is harder
Imagine cutting the bottom white shape with a vertical line starting at the top corner, the two pieces recombine themseves in a. Square of side 2, there are two of such white shapes so the area of the blue shape is just the area of the big quarter circle minus the area of two squares:( pi×4^2)/4-2×2^2=4pi-8
Or just move the x areas to fit z. Then the figure will become as a triangle and a segment of 90° together making a quarter circle. Then area of blue area= area of segment = area of quarter circle - area of triangle. As it is a quarter circle, the angle of sector is 90° and hence you can simply use the formula 1//2×b×h for calculating the area of the triangle. Solving: πr²/4 - 1/2 × b × h 3.14 × 4 × 4/4 - 1/2 × 4 × 4 12.56-8 4.56 #MATHS IS FUN
Split the lower part diagonaly and you'll see that they each flip perfectly into the top and bottom parts, you're left with a 1/4 circle minus the right angle triangle so 4pi-8
IMHO would be much easier if you first substract the A of the two half circles from the A of the big quater circle to get (2z) and then add deal with (the two) x, no?
Task: Find blue area. China primary school: Does highly complex math procedure to calculate surface of the blue area. US primary school: There it is! 👉
What do I think is highly problematic in this task is that it isn't worded correctly. We can only conclude that two "x" areas are equal if the angle between that line and bottom side (where 4 is written) is 45°. Since this isn't defined, one can only assume two "x" areas are equal. In math, tasks need to be properly worded and leave nothing to assumption.
I did not use any triangles instead I drew perpendicular radius from semi circles to get the region in semi circle not in x region that is separated by the the y region with the perpendicular radius, can found out by visualizing as the semicircle trapped in a rectangle leaving that region between the perimeter of rectangle and semicircle which can be solved just by subtracting the area of rectangle by the area of semicircle
Having to learn the formulas by heart will not foster any conceptual thinking nor creativity. You encounter that pattern later on (when they’re much older) where students are asked to solve complex problems that are outside of any teaching books that require systems thinking. I fear most educational systems still rely on old teaching methods
whats the proof, that the triangle has right angle? Also, why can we be sure that the line splits the quarter curcle in half? (I know these things are obvious on first sight, but I cant find a proof why it MUST be like that)
Ok, it's easy but looks difficult. Your method is unnecessarily complicated. You can flip the parts around to make the area a quadrant less a triangle which is much simpler. The quadrant is Pi x R x R/4 which is 4Pi. The triangle is 4 x 4/2=8. So the blue area is 4Pi-8. Done in two lines.
I know just by rotating x we can get the answer by ar(quadrant) -ar(triangle) but i have done a different method. I divided the semicircle into squares such that there would be 4 squares in the quadrant then i took a big square of side 4 which is bigger than quadrant the only problem i had was thr square which has both x inside so i made a circle with it and a square outside a circle and subtracting the area and divide into two parts and hence adding all these and subtracting from the quadrant area i got the answer as 4π-8sq unit!
It's much simpler than that. Each white part surface is 4, you don't need to calculate anything to see that. Just take one white part, cut it in two pieces (that fit into a 2x2 square), one concave, the other convex. Stick the two part on their round section, you get a 2x2 square. Then the final result 4pi-8 is straightforward. No equation. Still the title was clickbait. There is no way it can be asked to primary school kid, unless this is a kind of contest for child prodigy.
принципиально не вижу проблем. по сути, знание формул по кругу и проблема решена. сплошные половинки и четвертинки. Tсть даже вариант "решения" через координаты x^2+y^2=r^2, через вычисление площади треугольника через шаг dX. - используем компьютер, считать очень много. это как если бы была теория о соотношении площадей по приведённому примеру, и её надо было проверить таким извращённым способом. :)
This can be simplified by drawing a line between the 2 endpoints of the quarter circle. The nonblue area above the hypotenuse happens to be same area as x. Now you only need to find the area of the quarter circle minus the area of the nonblue triangle. Once again we end up with the blue area = 4π-8
Nice approach
Very nice approach
That's great! Maybe your approach is the one that is expected to be considered by the primary school students if they see the figure as in Tangram...
Kind regards.
Holy shit nice
Can't use that unless you also prove that the non-blue area above the hypotenuse is equal to x..
You'd be surprised how much kids can learn when you're not passing into them your own fears.
Or deliberately made making them dumber...
You think Chinese kids don't have their parents' fears rammed into them?
Great concept
This is one of those situations where you can "train" primary school students into solving seemingly high-level problems. The issue is that they don't actually understand the concepts and can only solve specific subsets of problems in the specifically trained way. This allows the schools, principals, and teachers to look good because "look how advanced our students and teaching methods are!" but in reality, the students didn't actually learn anything.
You say this based on what?
@@caixo5604 me for example.... they teach us linear equations without teaching algebra lmao
no one's gonna remember specific concepts after not practicing them for some years, but the problem-solving skills remain with us for the rest of our lives. perhaps you're reading too much into it while chasing conspiracies, which causes you to miss the obvious benefits.
@@_JoeVeryou don't get problems solving skills if you don't understand the problem
@@_JoeVer that’s not true. The fundamental understanding of the laws makes them a part of you. It’s impossible to forget those if you get an idea
Due to symmetry, you can split the lens-shaped part in half and rotate it to fill in the gaps in the circular edge. That just leaves you with finding the area of a circular segment: Area of quarter circle minus area of right triangle = 0.25*pi*4^2 - 0.5*4*4 = 4*pi - 8
That approach could be the expected for the primary school students? Is this approach similar to a chinese game called Tangram?
Can't use that unless you also prove that rotating the half lenses results in a straight line and it's a triangle (and not a quadrilateral)..
That's easy to prove though. The coordinates can be calculated. +/- right angle rotation, of (0,0) about (2,2) gives (4,0) and (0, 4). The three points, (4,0), (2,2), (0,4) are colinear. By inspection. @mvkishoresr
He did the same thing.😂
@@mvkishoresr I think that's exactly the point of this problem.
Area of quater circle- (π4²/4)
Area of trangle - (½ * 4*4)
Now substract the trangle area from the quater circle area.
(π4²/4)-(½ * 4*4)= 4(π-2)
Don't make it complicated..
Underrated man!
I do the same way as you, it is much more simple and easy to understand.
Hey man, your method is unknown to me, I didn't understand it at all. I am not familiar to trick you used here. Can you explain me how did you know subtracting the triangle area from the quarter circle area will give you the exact answer (area of the blue region)?
@@SourasisDebnath-yk6fe Actually it's very difficult to explain without draw the diagram, but I'll try...
First forget the sade , now it's just a quater of a circle of radius 4 , now if we connect the ARC start and end point means CHORD , then it will become the hypotenuse of Right Isosceles Trangle.
Now draw a median over the hypotenuse to get equal part of the given CONVEX shaded area, now if you impose all the given shade and the Right Isosceles Trangle and the Quater circle on each other then you will notice the Convex shaded area is equal , to that two segments area which are comes over the Hypotenuse of the Right Isosceles Trangle.
So, if we deduct the area of Right Isosceles Trangle from the Quater of Circle , get the answer.
@SourasisDebnath-yk6fe it is simple rotate two x's and fix it to two z's so you get a triangle inside a quadrant and hence find the difference of areas!!!
There it is. I point at the blue area. Found it!
Yea, seems like proper answer.
In the US. But it's okay, you at least have freedom.
@@kpic9054do they have sense of humor where you're at?
@@airtioteclint We do. That's why we can differentiate proper joke from failed attempt. U?
@@kpic9054 i would say he did a better job at being funny than you did trying to sound smart. Lmao
@@airtioteclint Remind me, who asked?
I solved it the following way:
Imagine square of length 2 at corner
Calculate half of a semicircle
Subtract it from square
Subtract difference from square twice
Thats the first blue area
Calculate total area
Subtract both semicircles
Add first blue area
Thats your second blue area
Add them. Done.
No integration? Damn! I don't even know how to solve it with integration. Now do it with no algebra. Cuz that's what schools in Singapore are asking.😊
Same. These kinds of problems always involve making use of flat-sided shapes, so I reckon if you just start by drawing square or triangular frames that box in or intersect the curves at key points, you are likely to find more than one way to solve it.
They do these in Chinese primary school? That's nothing. In Peru, they do these problems in kindergarten.
fr?
thats nothing, in north korea, they do these problems in the womb.
Hah we do these in ball
As someone who lives in Perú I agree
that's nothing, in india we do these before reincarnating
I would have failed Chinese primary school because I don't have any logic
As an indian I may assure I literally solved this problem in my mind withing seconds of looking at it easy like I got a similar question in class 6 ICSE school question
Can you give any book reference u used?
You really need to mention that you're Indian to look more intelligent?
@@V.444AzlI think he is saying that sarcasticly😅. I am from India passed ICSE primary in 2008. I don’t remember this and I am sure wasn’t there. This is JEE level stuff for sure.
@abhimanyusingh4281 not really I got the same answer with some simple maths
Congrats lil bro, want a medal?
How I did : Took a square of 4x4 in my head. If I draw semi circle of radius 2 on every side of the square I can write: Surface of square - semi-circle - semi-circle + area I look for ( 4 times ) =0. I got the rugby ball = 2PI-4. I took a paper and a pen at that moment....
Then, I took the quarter circle of radius 4 which surface is 4PI, remove semi-circle on a side ( minus 2PI ) , remove semi-circle on the other side ( minus 2PI ) and put back the rugby ball that I remove twice ( + 2(2PI-4 ) ) : gives : 4PI-2PI-2PI + 4PI - 8 = 4PI-8
So much simpler:
Draw line between endpoints of quadrant, making a prime right triangle with measurements 4,4, 4sqrt 2. Bisect quadrant, creating 2 similar triangles and unshaded lens halves which are congruent to the shaded lens halves labeled x. Transposing those lens halves results in the shaded area being the area of the quadrant outside the prime triangle.
Quadrant measure 4pi and prime triangle=8. 4pi-8.
4sqrt2*
@@지오5 Thanks typo. Yes, 4 times the square root of 2; whereas a 45 triangle has sides with proportions of 1,1, sqrt2
0:35 at this point you can rotate the x surfaces so they fit the z surfaces, and then the y surfaces become just triangles. Then you see the blue area is just a quarter circle minus a quarter square
American kids: struggling to figure out whether they're boys or girls
Chinese kids:
💀
Do you watch UA-cam Shorts sigma edits by any chance
@@HermanDinkins No haha why
London kids : trying to figure out who they murdered
The only people that are confused about gender are transphobes. Everybody else is perfectly comfortable and understands it.
"Teacher, my answer is X, you can find what the true value is by calculating your own exercise"
Love the confidence 😅
Well, I did this once on an exam 😂
@@LabArlyn Lol what was their reaction
Solution:
You can divide the blue "rugby ball" at the bottom left into 2 circular segments of a quarter circle, then place them at the top and bottom in the white bulge of the respective semicircles, then you get the blue area of a circular segment of a quarter circle with a radius of 4.
Then the blue area is
= the area of a quarter circle - the area of a right-angled, isosceles triangle with the two legs of length 4
= π*4²/4-4²/2 = 4π-8 ≈ 4.5664
This could be the primary school approach? Is it related to the Tangram game?
Can't use that unless you also prove that placing the rugby ball segments results in a straight line and it's a white triangle (and not a quadrilateral)..
@@mvkishoresr The rugby ball segments are segments of a quarter circle with the radius 2 and fit exactly in the white bulge of the semicircle with the radius 2.
This is a primary school question in Singapore, actually not very tough. This can be solved easily by moving tje smaller blue area adjucent to bigger blue area. Afyer moving u have 1/4 th of the circleinus aclean white traingle... i. E, 4pi - 1/2*4*4 which is 4pi- 8 = 4*(pi-2) = 4*1.14 = 4.56
When I saw this figure, I immediately thought of integration. But you know your viewers quite well!
How would this be done by integration
@@Tedde-bi9yp equation of the top left circle is x^2 + (y-2)^2 = 4 which I shall call A. Bottom circle is (x-2)^2 + y^2 = 4 wheich I shall call B and the large circle is x^2 + y^2 = 16 Which I shall call C. When you solve all equations for y, you get functiona that cannot exist in the negative y domain, so the top left curve must be negated to accommodate the lower blue shaded region. You then integrate the bottom circle - (x-2)^2 + y^2 = 4 from 0-2 and then subtract the same integration of the top curve, after solving both for y in terms of dx and of course remembering the top left integration value must be negated. That gives you the value for the bottom left shaded region.
For the second blue section on the top right you must reconsider the positive value for the function of the A. You shall integrate C from 0-2 and subtract A from 0-2. Then integrate C from 2-4 and subtract the integral of B from 2-4.
Add the values for both shaded regions and you have your result. I imagine there's a better way using polar coordinates, but I'm too lazy for that.
Note that the two half circles equal the area of the Quarter circle. Note that If we subtract the areas of the hald circles in a geometric sense we remove the inner blue area twice, while the outer blue area is untouched.
Therefore the inner and outer blue areas are equal. Then it becomes easy because the outer area is very simple to compute.
Mathematics creates moulds in our brains into which to fill other intellectual matter like arts, quantum theory etc
Underrated comment
moulds...
:D
The normal solution for primary school student should use rotation to make blue part as the difference of quarter circle and a right triangle. This is normal traditional segmenting and patching method in grade 5 to 6. In grade 5, circle concept is not fully introduced; but the students has 3-round learning for shape movement; have one round of segmenting and patching method.
When I was in primary school, I can answer it with linear equations. But now I'm college student and I only think eating two meals or three meals today.
They would totally use this to teach area, negation, etc to the children. Because otherwise they’ll have to teach circle and angle theorem. But seeing the children try to solve this with their own ways would be rewarding, because in mathematics there are a lot of ways to solve questions.
I encountered this type of problem in middle school, maybe the primary school ones are for test or olympiad preparation?
My way is weird as I imagined them as overlapping half circles. But that’s the charm of mathematics!
1/4 π*4^2-π*2^2+2*2*(1/4*π*2^2-1/2*4)=
4π-4π +4(π-2)=
4π-4π + 4π-8=
4π-8
Solving this with calculus is actually quite intuitive
equation of the top left circle is x^2 + (y-2)^2 = 4 which I shall call A. Bottom circle is (x-2)^2 + y^2 = 4 wheich I shall call B and the large circle is x^2 + y^2 = 16 Which I shall call C. When you solve all equations for y, you get functions that cannot exist in the negative y domain, so the top left curve must be negated to accommodate the lower blue shaded region. You then integrate the bottom circle - (x-2)^2 + y^2 = 4 from 0-2 and then subtract the same integration of the top curve, after solving both for y in terms of dx and of course remembering the top left integration value must be negated. That gives you the value for the bottom left shaded region.
For the second blue section on the top right you must reconsider the positive value for the function of the A. You shall integrate C from 0-2 and subtract A from 0-2. Then integrate C from 2-4 and subtract the integral of B from 2-4.
Add the values for both shaded regions and you have your result. I imagine there's a better way using polar coordinates, but I'm too lazy for that.
I was asked exactly the same question on my first-day of kindergarten schooling.
I’m guessing you flunked kindergarten
we deal with heuristics since 3rd grade
i had a different but long solution
Got it right, but it did take me a good 5 minutes which would not be enough time on a test
U can equate both the areas of the blue region with being equal to each other and then alternatively draw perpendicular bisectors from both the lines of the quarter circle forming a square that encompasses the lower blue part. Now take the sides of the square as the radius of quarter circles (with the line of the blue part as part of its circumference) multiply that by two and subtract the area of the square from that. You got the area of one of the blue parts and double it! (2*2*1/4*2pi) - 2*2 = 2pi - 4 = area of one blue part. Double it u get 4pi - 8
Another method I came up with:
Find area of quarter circle
Find area of smaller circle
Subtract smaller circle from quarter circle = the weird top shape, PLUS a debt of the leaf shape
Find area of leaf shape = smaller square - 2(smaller square - smaller quarter circle)
Multiply by two to cover previous debt
Answer = 4.566... = 4pi - 8
Nice brain teaser. Thx for the upload
I remember this exact problem from my middle school a couple years ago
I "drew" a line between the endpoints of the circle, put the 2x therein and solved the remaining triangle ( 4mx4m/2 = 8m² ) to substract from the quarter circle. Always interesting to see, how the real solution looks like.
We can divide the figure y (a half of the white area) into a quarter circle of radius 2 and the remainder. From these two parts we can make a square with side 2.
Then the blue area is equal to area of a big quarter circle minus two small squares:
(π4²/4) - 2*(2²) = 4π - 8
How I solved it:
- Each of the white areas consists of a quarter-circle of radius 2 and a 2x2 square with a quarter-circle of radius 2 taken out of it. Thus, the white areas must each be equivalent in size to a 2x2 square.
- The area of the large quarter circle is easy to calculate, it's pi * 16 / 4, or 4pi.
- Now it's just a subtraction problem, and the answer is 4pi-4-4, or 4pi-8.
We do this in College
It found it pretty easy to figure that areas of x and z are equal. So I calculated the area of x and multiplied it by 4.
The other method I found below (which involves flipping the x portion until its convex bowed side mates with the concave side of z, making it obvious that the answer is a quarter of the circle of radius 4 minus a right angle triangle that has is legs' lengths equal to 4) is even nicer.
I envy anyone who can solve math problems like this. I can't even divide without the help of a calculator. Math is like sorcery to me.
literally i've got the right answer and i finished primary school half a year ago, and i'm not inteligent, i'm usually getting c's on tests
You are too modest
My solution:. first establish that areas x and area z are equal. To do that, build a figure with four times the original figure by making mirror images of the original figure - one by vertical mirror reflection, one by horizontal mirror reflection, and one by applying both mirrors (same as rotation by 180˚). Piece it all together to make a figure with one big circle of radius 4, with four inner circles with radius 2. The summed areas of the four individual circles (which double-counts blue overlapping areas, x) is the same as the area of the big outer circle. The summed overlapping areas in the big circle, 8x, must be equal to the summed areas 8z for this to be true (think about it for a minute ...). All we need is the area x then. in one of the small circles, draw a square (diamond) that connects the center of the big circle to the points where areas x and z touch and the point on the big circle where the tip of two areas z meet. It's easy to see that the diamond can be made entirely white by translating and rotating two of the areas x inside of it to the white bulges outside of the diamond. Therefore, the white area of a small circle is (4/√2)² = 8. The area of an entire small circle is π(4/2)² = 4π. So the blue area of a small circle is 4π - 8.
EDIT: Even easier: Once you build the big circle I described above, you can move ALL the blue areas, 8x, outside of a square diamond connecting z area touchpoints in the big circle ... and it doesn't change the total blue area in the big cicle at all. Now the big circle has a totally white diamond within it, with all the remaining are 8x + 8z outside of it.
Wow! That's a really detailed and creative approach
I solved it both ways. I think it is actually easier without using calculus. The take away is that the two blue areas are equal.
In the UK we don’t have to do this type of equation… we just outsource it to the lowest bidder…
I noticed that when arranging 4 semicircles overlapping each other within a square 4x4, The desired area can be calculated by subtracting the area of the square from the total area of 4 semicircles, then dividing the result by 4.(Sr this is so hard to explain in english since i'm not good at it)
My Goodness, such a difficult explanation. the two ’x, areas added to the 'z’ areas makes a triangle. substract the triangle from the quarter circel and you have the blue area surface ! That easy, you can do it without paper or calculator, by the way the answer is 4.56m2 (if it's in meters)
This is too algebraic, geo problem shall be solved in geom way ____
Note the blue "tri intersection" pt, use it as rotation ctr, rotate the two blue "chord areas" 90° cw & ccw respectively, to fill the symm white area of small semi circles, then you get a larger chord area of quter circle, so blue=4pi -8
if i consider the symmetry i will have to solve 2 equations with 2 unknown numbers, and if i don't, i will have to swap x and y and numbers, see line 100 and line 150:
10 vdu 5:gcol8:print "brainstation-find blue area"
20 for a=0 to 15:gcola:print a;:next a:vdu0:gcol 8:r1=4:r2=r1/2:xml=0:yml=r2
30 xmr=r2:ymr=0:zoom%=1.4*zoom%:nu1=155:nu2=nu1:masy=1200/r1:masx=850/r1
40 if masx
run in bbc basic sdl and hit ctrl tab to copy from the results window
U hv written a code for it! 🔥👏
Using integration for this problem would only make it at least ten times harder 😂 I wouldn't use integration in an integration exam either and just tell the instructor to deal with it.
😂
I actually learned this in Junior High School in China.
It seems to me that there is an easier and more straightforward way to solve this problem. To begin, you need to draw a chord. Along with the radii, this creates a right-angled triangle whose area is 8. Clearly, the desired area is equal to the area of the quarter circle minus the area of this triangle, which equals 4π - 8 = 4(π - 2).
The result is correct, but if you don't explain your formula clearly, you may misunderstand it. As a result, you won't receive any marks from the examiner
This is why I love Algebra
Singaporean Primary students are also learning this skills. I'm a Math teacher btw
Thanks for letting me know!
Outstanding
This problem is actually very easy! My sister is 10 years old and she could solve it in a few minutes! You just need to observe the diagram and by "observe" I mean to ACTUALLY OBSERVE it. Then you can solve it in a jiffy.
(We aren't even Chinese. But yeah, we're Asian)
Blue axe shape = A
Blue leave shape = L
1/4 Big circle = Q
1/2 Small circle = H
Total blue area = A + L = Q - H + L - H + L = Q - 2H + 2L
How to make this animation,
Please help
It's common in East and SEA not just in China . They gave primary school kids these questions in Top school and private tutoring.
x+y+z=2x+y=2pi
x+z=2x, and…2x+2z=4x
So the question becomes simple, we need to solve the x, and then we’ll get the answer
Solved in mind... though I simplified the answer as 32/7..(4 pie - 8)..
Yes, solved it orally. Also note: the area x and z will come out to be equal.
i would never be able to do this without integration
If you look closely it's 2 semi circles and one quarter circle, findjng the area of the weird blue shape above is easy, but the oval like shape is harder
Well, it took me like 10 minutes to figure it out
Just move the x to the z and you can get a triangle in the quarter of a circle so (pi4^2/4) - (4x4÷2) = 4pi-8
I am an Indian and I was able to solve it. But it took me around 5 min and pen and paper
Imagine cutting the bottom white shape with a vertical line starting at the top corner, the two pieces recombine themseves in a. Square of side 2, there are two of such white shapes so the area of the blue shape is just the area of the big quarter circle minus the area of two squares:( pi×4^2)/4-2×2^2=4pi-8
No kidding, I remember doing this in grade 6.
this is such a good math problem
Find the blue area? It’s right here. I could do that in Pre-K, I knew all my colors.
I sloved it by my mouth.
This is easy. Us singaporeans also learn it in primary school but the method we use is simpler😊
I was eating sand and throwing rocks in primary school.
Tangram?
2x + 2y + 2z = 4π...(1.1)
2x + 2z = 4π - 2y....(1.2)
But : y = (4) .... (2)
4π - 2•(4) = 4π - 8 = 4(π - 2)
Dividing them into smaller squares and then we can find the area
Or just move the x areas to fit z. Then the figure will become as a triangle and a segment of 90° together making a quarter circle.
Then area of blue area= area of segment = area of quarter circle - area of triangle.
As it is a quarter circle, the angle of sector is 90° and hence you can simply use the formula 1//2×b×h for calculating the area of the triangle.
Solving:
πr²/4 - 1/2 × b × h
3.14 × 4 × 4/4 - 1/2 × 4 × 4
12.56-8
4.56
#MATHS IS FUN
Using integration here would be even harder than using the simpler method
1/4Circles Substractions!
The real question here is not to find y
but to find why a student would every need to do a problem like this.
Split the lower part diagonaly and you'll see that they each flip perfectly into the top and bottom parts, you're left with a 1/4 circle minus the right angle triangle
so 4pi-8
IMHO would be much easier if you first substract the A of the two half circles from the A of the big quater circle to get (2z) and then add deal with (the two) x, no?
move the x to be next to the z boom its a quarter circle and a triangle
That move is what is expected to be done by a primary school student? A visual idea instead of more analytical concepts...?
Same answer the difference is i used square instead of triangle
Task: Find blue area.
China primary school: Does highly complex math procedure to calculate surface of the blue area.
US primary school: There it is! 👉
What do I think is highly problematic in this task is that it isn't worded correctly.
We can only conclude that two "x" areas are equal if the angle between that line and bottom side (where 4 is written) is 45°.
Since this isn't defined, one can only assume two "x" areas are equal.
In math, tasks need to be properly worded and leave nothing to assumption.
This is why they have excellent computer hackers.
I did not use any triangles instead I drew perpendicular radius from semi circles to get the region in semi circle not in x region that is separated by the the y region with the perpendicular radius, can found out by visualizing as the semicircle trapped in a rectangle leaving that region between the perimeter of rectangle and semicircle which can be solved just by subtracting the area of rectangle by the area of semicircle
I found that blue area after watching this video for 1 second saved 2 minutes and 48 seconds
answer=4pi-8
calculated in brain in a minute
Having to learn the formulas by heart will not foster any conceptual thinking nor creativity. You encounter that pattern later on (when they’re much older) where students are asked to solve complex problems that are outside of any teaching books that require systems thinking. I fear most educational systems still rely on old teaching methods
whats the proof, that the triangle has right angle? Also, why can we be sure that the line splits the quarter curcle in half? (I know these things are obvious on first sight, but I cant find a proof why it MUST be like that)
Ok, it's easy but looks difficult. Your method is unnecessarily complicated. You can flip the parts around to make the area a quadrant less a triangle which is much simpler. The quadrant is Pi x R x R/4 which is 4Pi. The triangle is 4 x 4/2=8. So the blue area is 4Pi-8. Done in two lines.
this is actually 6th grade math in china; i did this problem some time back
I know just by rotating x we can get the answer by ar(quadrant) -ar(triangle) but i have done a different method. I divided the semicircle into squares such that there would be 4 squares in the quadrant then i took a big square of side 4 which is bigger than quadrant the only problem i had was thr square which has both x inside so i made a circle with it and a square outside a circle and subtracting the area and divide into two parts and hence adding all these and subtracting from the quadrant area i got the answer as 4π-8sq unit!
Don’t even need algebra to solve it
Oh thats nothing, im typing from womb and studying gamma rays here
It's much simpler than that.
Each white part surface is 4, you don't need to calculate anything to see that. Just take one white part, cut it in two pieces (that fit into a 2x2 square), one concave, the other convex. Stick the two part on their round section, you get a 2x2 square.
Then the final result 4pi-8 is straightforward. No equation.
Still the title was clickbait. There is no way it can be asked to primary school kid, unless this is a kind of contest for child prodigy.
принципиально не вижу проблем. по сути, знание формул по кругу и проблема решена.
сплошные половинки и четвертинки.
Tсть даже вариант "решения" через координаты x^2+y^2=r^2, через вычисление площади треугольника через шаг dX.
- используем компьютер, считать очень много.
это как если бы была теория о соотношении площадей по приведённому примеру, и её надо было проверить таким извращённым способом. :)
Please state the source, that this is a problem the give in chinese primary schools.
How would we do integrating?
Too easy😂
(From Indian perspective)