How to solve the three-circle problem from the 2022 GCSE math exam
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- Опубліковано 1 жов 2024
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Here's the last question from the 2022 GCSE maths paper that made the news. We have three circles as shown and each radius is 4 cm. We have to find the area of the shaded region in the middle. I made a horrible mistake last time when I said the area of a circular sector is r*theta. The correct formula should be A=1/2*r^2*theta and theta has to be in radians.
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and you'll get a lot of views as well
If you didn't make (and show) your mistakes, people would respect you less. Well done for showing not only your fault, but the solution.
Bruh so many people here saying how clever they are and that they have no idea how this is "notorious". Its a GCSE question done by all 16 yo a lot of whom have no interest in maths, just because youre doing complex analasis or linear algebra at college or uni doesnt mean that a 16 yo kid cant struggle without you saying how much better you are, why cant we appreciate that some people struggled and be happy that this vid might help
they can't get the appreciation they want from their teachers or parents, so they resort to youtube comments to boast about their math skills
i too think the question isn't hard at all tho, all you need is practical knowledge
I am 16, turning 17, and I have never heard of this shit. But that is because our school system here is dog crap. I would have failed this very hard.
I had this question during my gcse, was the only one in my class who had got it right on the test and ended up with 235/240 marks out of all 3 papers
yall are learning linear algebra at college or uni?????? wtf?
@@activatewindows7415pretty standard for STEM majors
You wouldn’t be expected to know about radians at GCSE. The area of the sector is one sixth of the area of a circle.
That looks like a solvable question. You don't need to make a fancy construction once you realize the broken arches can be made from by adding 2 circle sectors and subtracting the extra triangle.
Seeing the previous problem (13) with proving the semicircles grown on the right angle triangle reminded me of the last problem in my maths textbook for circular arcs (romanian school). We had to find our the area of the lunules of Hippocrates. You get shaded area of arches drawn between semicircles drawn on cathetes and semicircle of hypotenuse passing by the right angle point, which turns out to be exactly the area of the right triangle.
Seeing the similarity of previous problem and having this come up leads one to just consider addition and subtraction of surface areas, knowing this exam is precalculus.
You are expected to know
@@anghme28ang11nope
That is a UK problem then.
Because average Italian, French and German 16yo students know about trigonometry and radians.
GCSE are joke exams. 💅🏻
Your second sentence is weird.
Yes the area of THIS sector is a sixth, but you have to show that it's a sixth in the working out, you can't say "cause I see it's a sixth" 😅
Great video. The only comment I will make as someone who teaches maths here in the UK, is radians are not taught until A level (beyond GCSE). GCSE students are taught area of a sector by considering it a fraction of the area of the whole circle using degrees: (theta/360) multiplied by pi.radius^2
That is a UK problem then.
Because average Italian, French and German 16yo students know about trigonometry and radians.
GCSE's are very weak exams.
💅🏻💅🏻💅🏻💅🏻💅🏻
Continental kids are better that Brits up until uni level.
Then in uni, with the relationship you have with the work environment, UK is the very best.
So yeah... if you have a kid in England, make him study abroad from 11 to 18 and when he comes back he can kick arse to his peers 😂
@@MartiniComedian you sound like an incredibly grating person to be around, i hope only the worst for you, much love from bob 10
@@MartiniComedianwhat ? Different countries learn different syllables. Other countries might learn about radians sooner in the uk but the uk would learn other things sooner as well. Weird comment.
@@MartiniComedian mate people just have different curriculums. We were taught trigonometry at 13 years old. And while one GCSE is not that bad, the fact that you have to do 10 subjects, with 2 papers for each, which will go on your job applications for the rest of your life, they are fucking brutal.
"Great video. The only comment I will make as someone who teaches maths here in the UK, is radians are not taught until A level (beyond GCSE)."
I didn't see the point of using radians, I'd not have done it that way, but I have to say the 1968 SCE "O" grade paper included radians, but even that was beyond GCSE... I'd have calculated the area of one 60 degree sector, (1/6th of a circle's area) then the area of an equilateral triangle of side 4, subtract that from the sector area, giving the segment area, and subtract 4 x one segment's area from 2 x a sector's area, we have the area of the shaded portion. But I have to express that in terms of pi? I'd have divided my answer by pi, giving a neat answer (I think) And I'd hope not to be penalised for not following a prescribed path...
In my day, pi was expressed as 22/7 (no calculators in 1968...) and circle radii/diameters expressed as multiples of 7... (they weren't that cruel). Maybe we were permitted/obliged to use slide rules and log tables?
I didn't advance much beyond "O" grade maths as a discrete subject, only as part of machine shop and sheet metal calculations... (where everything's pared down to its simplest, since complexity brings with it opportunities for error)
GCSE maths will only be in degrees no angles in radians. Radians are only introduced to maths students doing A level in the UK.
Yeah, I was able to understand this because I was in Year 12 when the question came out, but then I realised I was using A-level formulas to do it, idk how the kids in the year below me were coping with it, most of whom weren't even going to continue maths to A-level!
They use an alternate formula, pi*r^2*(theta/360). It simplifies to the same formula but is a lot more digestible for them.
This was in the non-calculator exam and it brought back some funny memories of the exam, it's nice you're covering it nice video!
Seriously?
@@lawdohio Yes, I sat my GCSEs and now I’m doing A Level Maths
@@sampan04 Brutal. It's not that bad but under exam conditions... I think anyone could justifiably blank out.
@@lawdohio yeah I think I got only 2 marks in that question, I was pretty exhausted as it was the last question on the exam but I did learn a lot from that experience
It was so weird lol everyone was so confused after the exam.
For the people in the comments ! Yes this question isn’t incredibly hard to solve if you know your stuff, but remember that most of the students that did this question were 16 year olds that dgaf with probably around 5~10 minutes left if lucky 😭
And as such this is probably the differentiator question for those that get absolutely top grades vs just very high grades. You're not going to fail GCSE because you can't do the one hardest question on the paper.
An alternative approach is to find a segment of either of the outer circles (radius 4cm, angle of 120 degrees (since radians aren't technically in the GCSE syllabus)) and then multiply that by 4 and subtract from 16pi.
That was my approach yeah
The only thing that gave me a pause was proving that the intersection points (vertically) marked 1/3 of the circumferences. Because of the radii of the center and one side circle, you wind up with triangles of identical side lengths, which implies 60-degree angles, etc.
@@josepherhardt164 saying in the q A, B and C have the same r, so its kind of already implied and ofc easy to prove... its also easy prove CC A-C is a straight line... after that its becomes rather simple
Shaded area = r²(2π/3) - 4[ ( r²π/6) - ( r(r√3)/4 ) ]
Clarifications
sectors containing shaded regions = 2 * r²(π/3) = r²(2π/3)
sector segment = ( r²π/3)/2 = ( r²π/6)
triangle = ( r( (r/2)√3)/2 ) = ( r(r√3)/4 ) [ Base=r, Height=(r/2)√3 ]
@@josepherhardt164 I confess I over complicated that part when I did it in my head. I did inverse cos(1/2) (requiring me to further use the result that triangles from the diameter of a circle to the circumference are right-angled) instead of just observing the perfectly obvious equilateral triangle.
Yeah, radians superflous as the intersections are all based on perfect hexagons thus equilateral triangle based, and thus the circle is cut in sixths (or thirds, depending which approach you go)
The "rule of thumb" I give my students is spend on average 1 minute per mark. This is based on a 90 minute exam, for 80 marks, with 10 minutes for checking work.
Of course there are some 1 mark questions at the start of the paper that can be answered in seconds, and I think you really do need to work quickly on those basic questions to build up a time cushion because in my view some of these later questions can be quite involved, and will take longer (in minutes) than the number of marks being awarded.
This was a video of over 8 minutes and yes, there was some time taken for explanation etc but as a calculus professor BPRP knows exactly where to start and which direction to take. I think most 15-16 year olds would in any event take longer than 5 minutes to answer it, especially given the number of steps and the need to methodically set out working.
I do think that's a bit of a shame as it over-rewards simple questions and under-rewards the difficult ones. If I were designing the paper I would increase the number of marks and keep the very basic questions that can be answered in 30 seconds or less at 1 mark, whilst weighting the difficult questions (normally up to 5/6 marks) more and giving them say 10 marks.
Exam boards should be able to work out what is the required amount of time to spend on a question, and allocate marks accordingly, but until they do that, I'll keep advising (a) pick up the easy marks quickly (b) don't dwell on a question too long, move on to the next and (c) come back later if you have time". That's general exam strategy, not specific to maths, of course.
half the people never get to the end of the paper, there is no point weighting questions more if people aren't going to end up there, thats just lowering the threshold and making it harder to classify into students that at different grades.
All it does is makes it better for students that are good at pattern recognition and have done many past papers to get the answer right than for students who take their time to think through a solution.
@@rbanerjee605 I'd have to see some evidence for the claim that "half the people never get to the end of the paper" but leaving that aside, my point is that all marks should be worth the same.
If the problem is (as you state) that too many students are not getting to the end of the paper, then that is a solvable problem - you either (i) have fewer questions or (ii) increase the time allowed to complete the paper or (iii) mix up the order of the questions more between the "simple" and "complex" ones.
If, say, the order of the questions were mixed up, (some) more difficult questions would come before (some) easier questions, mitigating the issue somewhat. Assuming there will be a number of students who do not read ahead, and who will simply deal with questions sequentially, they may miss out on some simple marks altogether because they are further in, whilst possibly struggling with the earlier, harder questions.
Therefore, as long as the structure of the paper is such that you get basic, 1 mark questions at the beginning, with progressively harder questions, I think my argument holds, which is that a reasonably good measure of the difficulty of a question and the number of marks it should be awarded is the average time it should take to answer it.
Otherwise, not all marks are created equal and I can't see any logic for that approach.
@@tanelkagan The purpose of examinations is to distinguish different skill levels. Adding some disproportionately difficult questions helps distinguish the highest skill levels. If they're kept to the end of the exam then they won't impact less-skilled students who will attempt the questions in order.
@@4hodmt Of course, I don't think it's ever been disputed that exams are designed to distinguish between students of different levels.
Mixing the order of questions up was not something I was pushing - I was merely exploring the possible implications of such an approach given the alleged problem of "half of students not finishing the paper".
I don't believe that anyone sits down and specifically tries to design an exam paper so that half the students will not finish it. If we are assessing different skill levels, it is far better and far more meaningful to do that based on questions that *are* answered, rather than questions that are not.
This goes back to my original point - if the order of questions is to be kept as it is, we should still reward questions as finely as possible based on complexity, with the time taken to answer the question being a measure of that complexity.
In simple terms, it should take on average 5 times as long to answer a 5 mark question than it does to answer a 1 mark question. That's all. Where you have 1 mark questions that you can answer in 20 seconds and 5 mark questions that take 8-10 minutes, something isn't quite right and that's why I would prefer to see better time/mark weighting. I can't see how anyone could logically argue against that.
Really hoping for questions on Thursday to be beauties like that. I would run out the exam hall in glee if it was simply just trig lol
I solved it using integration 💀.
Clearly the application of this problem is to calculate how much fabric is needed to make a thong. 👙☺
bruhh
I was gonna say integration but remembered it's GCSE
I was feeling so confident going into this exam, i finished the rest of the paper with half an hour left and i thought it was going so well, and then the last question hit me 😭it took me all of that remaining half hour to solve this!!
It's not that hard, it's just a sector of a circle = (1/3)pi×4^2
MINUS
Area of triangle= (1/2)absinC = (1/2)×4^2×sin120 = 8×sqrt(3)/2 = 4sqrt(3)
Then just a full circle minus 4 of those D shapes, easy.
i got a whooping 1/5 marks on this question while doing it as practice for my gcses this year lol
good luck!
Don't dismiss a single mark. They all count. Do as much as you can on any question.
@@ellentronicmistress4969 yea a single mark can be a big difference and thankfully I learnt that before the real thing, thanks
The worst thing is my math teacher put a somewhat similar question for me on the board, just because I do calculus in my free time, and judging by his face, he wanted me to find it's area without a limit. Solved it like you did. After class he told me, that he wanted it to be somehow get solved with the golden ration, like wtf. anyway, got an A :)
LOL. Your teacher has conflated sqrt(3) with sqrt(5).
How do you solve an area with a limit?
Limits apply to functions with a range of values that converge to a number. This area function simplifies to A(r) = (r^2)[ √3 - (pi)/3 ]. Thus, lim(r-->n) A(r) for some value "n" converges to an area A(n).
If you want an area equal to the golden ratio, then solve for "r" in, (phi) = (r^2)√3 - (pi*r^2)/3 ----> (phi) = (r^2)[ √3 - (pi)/3 ] ----> r = √ [ (phi) / (√3 - (pi)/3) ] = 1.537074995 .
If you want a radius equal to the golden ratio, then substitute "r =phi" into A(phi) = 1.792969103
I looked at this paper as a final year engineering student and it stumped me for a few minutes too
I got
|(-16π - 48 √3) ÷ 3|
without using the formula.
I divided B into 6 equal pieces, then made a triangle for every piece.
I calculated the area of the triangle (At).
Then I calculated the area of the small thing left,
As = (⅙ circle B - At).
Finally, it will be
Shaded Area = 16π - ((As ×8) + At×4).
I simplified and got my answer.
I liked the video however theres one thing to point out.
I sat this exact paper and this was the one question I couldn's solve. I still have nightmares.
Anyway, the method you used utilised radians, which is very convenient for finding areas. But in our GCSE specification we do not learn about radians so I don't think this method is fair for GCSE students.
I mean, it's still pretty simple. A 60° sector is just 1/6 the area of the circle.
This pretty kaizo for gcses
Asian students: Notorious? Do you mean *not serious?*
can't relate more to anything other than this😔😭
I know it's a meme but seriously, this question seems so basic. It doesn't require anything above middle school math and the geometry is quite simple.
@@pkuvincentsu An exam for 16 year olds asking questions meant for 16 year olds? Who would've guessed...
@@pkuvincentsucan confirm, something like this was in my middle school entrance exam
@@hammer3721 So... why is this question "notorious" then?
I did it slightly differently with a mix of geometry and basic algebra: if you draw some additional circles of radius 4 centered on intersections, you can partition the central circles into 6... let's call them concave triangles (CT) and 12 ... almonds (A). The grey area is then equal to 2CT + 2A. Since the circles have radius 4, we know 6CT + 12A = 16Pi and we also know that the area of an equilateral triangle of side 4 can be described as follow CT + 3A/2 = 4 Sqrt(3). From there you manipulate these two equations to reach 2CT + 2A = 16 Sqrt(3) - 16 Pi / 3.
I just did integration, I converted it into geometric coordinates considering B as origin, found each coordinate, found equations of the circles, found points of intersection of the circles the integrate from 0 to 2 for 1 part,, then 16pi-8(segment area) = ans, should mention each segment is that area from 0-->2 that's part of the A and C circles that are 8 in total
Pretty interesting way to do it damn, but yh this is done by 16 year olds in the UK 😭, 80% which dgaf
samee I too did it with integration
but tbh its a lengthy way.. (especially if you dont know remember the formula)
Not sure you would get the marks as integration isn't on the GCSE syllabus.
Every single one of these geometric problems would be a cakewalk with integration. I think the students doing this either weren't taught integrals yet, or aren't allowed to use them.
Integrate to find area under curve of root(16-x²) from 2 to 4, let this be = k
Therefore, the required answer is 16pi - 8k
And k comes out to be = 8pi/3 - 2(root3)
Answer : 16pi - 64pi/3 + 16(root 3)
= 16(root 3) - 16pi/3
Value of k :
Put x = 4 sin y
dx = 4 cos y dy
When x = 2 then y = pi/6
When x = 4 then y = pi/2
Therefore, now we have to integrate 16cos²y from pi/6 to pi/2
Put cos²y = 1/2 + cos 2y/2
Therefore after integrate it will be 16(y/2 + sin 2y/4)
Put the limits
16(pi/6 - (root3)/8)
= 8pi/3 - 2(root3)
Hope u like this 😊😊
It’s a nice idea but sadly would get no marks in the exam as integration is not on the curriculum for maths GCSEs 😢
@@Daniel-yc2ur Ohh
Remember these are 15 year olds taking this exam
The mark scheme for this question did give a mark for the correct answer (or equivalent), so they'd likely get 1 of the 5.
yep i’m not doing all that thank god i know integrals
A less cumbersome method:
Call the upper intersection of the circles A and B, Y, and that of B and C, X.
BX=BC=XC=4. BCX is therefore equilateral, so 6 such triangles will form a hexagon in Circle C. Call each triangle area A(T)
Equally, Circle C comprises 6 pie slices of equal size - the triangles plus a curved portion. Call each slice area A(P). It's just 1/6 of the area of the circle.
Next, the part in each slice but not in each triangle is a sliver. Call the sliver area A(S)
But also BXY is another congruent triangle, and half the area we want, with 2 slivers extra.
So for the whole of the desired area, we need 2A(P)- 4(A(P)-A(T)) = 4A(T)-2A(P)
A(P) = 1/6(pi*4^2)
A(T) = 1/2(4*(2sqrt(3))
4A(T)-2A(P) = 2(4*(2sqrt(3))-1/3(pi*4^2) = 16sqrt(3)-16pi/3
lol wtf? how was this notorious? this is pretty basic geometry and by GCSE level it should be no problem. Idk how the British system works but it should be in any country
For the area of equilateral triangle, wouldn't it be easier if we just use "Area = 1/2 ab sin C"?
This one was easy thanks to your explanation
Only a real man can admit mistakes in public
Apparently the term for the shape bound by a chord and the circumference is a *circular segment.*
explain how it is bounded by a chord
@@savitatawade2403 It is bounded by a chord *and the circumference.* In other words, that third shape he found the area for.
@@ZipplyZane oh I thought you were referring to the overall area
I remember having this in my paper. It was horrible but managed to get the 5 marks. There’s another horrible question in 2023 papers near end of paper 2 I think but might not be avliable yet.
what was the question again? i remember everyone near me getting paper 1s last question wrong
@@DotDotEight it was the hexagon one
I'd call it fiddly and tedious, but horrible? It's about picking the problem to pieces then assembling the pieces.
@@3TAN12Eoh I remember that one when I sat the paper. But I'm so grateful for 2022's circle question because they were pretty similar so I understood what to do
@@3TAN12E oh that one? i coulda sworn i did something similar in class but maybe im wrong seeing as i sat this same paper. i remember doing this question but i forgot if i got to the correct conclusion
this is a hard problem? WHAT???
Yes😅
Hey dont said i'm dumb
If you have a function f(x)=x+inf. What would be its derivative? On one hand infinity is like a constant so its derivative is 0 and that of x is 1 so f'(x)=1. But on the other hand the function is always inf. and, again, infinity is like a constant so f'(x)=0. So is there a definite answer?
"inf" isn't a number so you can't really use it in equations like that - the only time bprp uses it is to represent limits like when x -> inf of 0.5^x: you can sort of write that as 0.5^inf but that's only really a representation - you can't actually take 0.5 to the power of infinity. Your question is a bit like asking what inf/inf equals: on the one hand infinity is infinity no matter how much you divide it but on the other hand inf*1 = inf, divide by inf on both sides you get 1=inf/inf so inf/inf must be 1 but then you'd also get inf/inf = 2 with the equation inf*2=inf. In short, equations with infinity don't make sense because infinity isn't a number
You could have gone the calculus way by integrating (r^2 - x^2)^{1/2} - r/2 from x = - r*sqrt(3)/2 and + r*sqrt(3)/2. After that you just need to substract 4 times the result to the area of the circle. Not sure if it is faster, though.
Well you see, this was a non-calculator exam and integration is not taught at GCSE level so you cannot do that as you won’t get marks for it
@@Colea1010 They don't give you the mark if you use a technique that they didn't teach you directly? I didn't know that. Also, you don't need a calculator.
@@jeremielhomme8572 yeah because you get a mark for the correct answer, but most the marks come from working out. They’ll have some methods for working out listed and if your method isn’t there - like if you use integration, then you don’t get the marks for it. It’s happened to me a few times doing past papers - for example 1 Q asked 27^x = 3, and doing log_27(3) to get the answer wouldn’t award more than 1 mark (3 mark question btw), and you instead had to recognize that cbrt(27) = 3 and cbrt(27) = 27^(1/3) and get x = 1/3 that way
@@Colea1010 ah ok. Thank you for the context. I am not familiar with that exam. It seems pretty hard!
@@Colea1010 That's not true. You get marks for any mathematically valid method (unless they explicitly state a method, which they don't here.) The examiner may not understand it, but they would escalate it to a chief examiner as it's not on the mark scheme.
Since 1 circle is made up of 6 equilateral triangles and 6 minor segments:
Let O = area of a circle = πr²
let ∇ = area of equilateral triangle = ¼r²√3
let ⌓ = area of minor segment =?
let ⋎ = shape internal to area of ∇ such that, ∇= ⋎ +2⌓
hence we can form a system of simultaneous equations;
A = area required = 2⋎ + 2⌓
O = 6∇ + 6⌓
∇ = ⋎ +2⌓
Solving for unknown ⌓ by seeing that
2⌓ = ⅓O -2∇
2⋎ = 6∇ - ⅔O
adding these two eqn. gives solution,
A = (⅓O -2∇) + (6∇ - ⅔O)
= 4∇ - ⅓O
= 4(¼r²√3) -⅓(πr²)
sub in r= 4cm
gives area A = 16(√3 - ⅓π) cm² ㋡㋡
Finally some real math notation
@@TheSourovAqib I struggle with "real math notation" But I know how to calculate the area of a circle, and I know that, as a proportion of the circle that circumscribes it, a regular hexagon has an area cos 60 / 1 of the circle's area... from that, I can calculate the six segments' total area, and thence the area of four of them, the area of two sectors, and subtract the smaller figure from the larger. that gives the area of the shaded portion,
It's not expressed in terms of pi, but, divide by pi, you get there... in a rough and ready way.
@@robertlawson8572 indeed
@@TheSourovAqib A somewhat brief response... Having worked in engineering for most of my life, I found practical maths needed to be cut down to its bare bones to be performed quickly and (sufficiently) accurately on the shop floor. Esoteric stuff doesn't fly well in the workshop.
The time would be better spent checking your other answers than solving this for a measly 5 marks.
IMO, your method is unnecessarily complicated, simply because it uses radians instead of realizing the sector is 60°, which is 1/6 of a circle. You should just see that the sector is πr^2/6 without having to do conversions between degrees and radians.
I personally prefer using the method where you see the top grey area as a triangle minus a bulge, and a bulge as a sector minus a triangle. This gives youthe result that the top area is two triangles minus a sector. Multiply that by two, and you get the total grey area.
This is a classic example of why Pi is terrible and we should use Tau. Asector = 1/2*pi*r*theta.... yuck.... where in with tau its obvious that Asector = tau*r*theta .... clearly theta is radians around the circle so its a portion of a circle.
edexcel really just decided nah screw the 2005-06 kids (this was the only question i struggled on)
I guess we can solve it easily using the areas of ellipses which is πab
They brought this question back this this years a level paper but much easier
Maybe a little easier: Let's call the equilateral triangle area T, and the little crescent area C. The top shaded area is just a copy of the equilateral triangle with 2 crescents deleted from the bottom and 1 crescent added to the top. So its area is T - 2C + C or simply, T - C. The video shows that T = 4√3, and C = 8π/3 - 4√3. So the top shaded area is T - C = 4√3 - (8π/3 - 4√3) which simplifies to 8√3 - 8π/3. Double that to get both shaded areas and we get 16√3 - 16π/3.
Also, you don't need to know about radians to solve the area of the pie-shaped piece. It's simply 1/6 of a full circle, so its area is πr²/6 which is 16π/6 or 8π/3.
Absolutely no radians needed! It's a piece of a perfect hexagon, either you are looking at the whole crescent and thus 1/3 of it, or you're looking at the half crescent in which case it's a 1/6th of it, either way, radians superflous
ermmm i dont remember GCSES being like this lol. although once you notice that the outer circles share an intersection with the centre circle, youre basically done.
-8pi/3 - 8pi/3 is not 0, it is -16pi/3. What kind of a chancer is this guy.
You can make it easier if you use the formula for the area of a circular segment.
Even as a grade 9 maths student, if I got this, I would be very unhappy.
I believe the first time you tried solving this problem, you used the formula r multiplied by theta, which is used to find the arc length of a certain section of the circle.
It may seem weird but I really wish that this question came up on my maths paper, it looks really cool for me and i think it’s pretty easy for me.
brudda this question was so easy i got full marks you mfs just dont look at circles properly
Can't you just solve this easily with integrals? Or are you not allowed to use them?
The GSCE is similar to a GED in the U.S., and it would be crazy if this appeared on the GED.
This was the last question on the higher paper, so only the candidates who were getting the highest of grades would be expected to get it right.
Not quite, a student will take GCSEs when they are 15 or 16.
blue pen yay
I was able to solve this as a 10th grader CBSE student in India without radians
The point here is to not freak out and think you have to subtract stuff from a square. My approach: Complete the red circle, draw the vertical chord lines where the circles intersect, and then draw the radii from the three centers to the intersection points. After that, it becomes clear.
we used to do questions like this in 10th grade in india cbse
thats the same age as this question was built for
@@gabriql yeah, so i don't see why it's considered hard; it's quite a basic question for 16 year olds
@@dingus42 it really isnt that paper is for EVERY 16 year old in the country, even if they dont choose it; its not like only for gifted students. I don't think you understand also that they had about 5 mins for that length question, and also have around 9 other subjects to revise for other than maths. It is an easy question for someone in hindsight maybe, but dont say its easy
@@gabriql idk, it is genuinely very very basic geometry, literally just area of a circle and area of a triangle are needed. concepts which smart upper primary-lower secondary school kids can solve quite easily. Even if you use trig it's quite simple trig, so by 16 i would not say it's unfair or particularly hard at all.
ahaha i remember turning the page, seeing that monstrosity, and turning back to the rest of the paper immediately
think only 1 person in my year managed to figure out how to solve the question in the exam although they didn't have time to finish it
I believe the area is wrong. You missed the small wedge in the triangle.i will post a solution to it tomorrow.Great job.😊
No, the result is correct.
@@mcichael9661 Sorry. THE RESULT IS CORRECT. I HAD PROvEN IT USING CALCULUS .IT IS UP IN MY CHANNEL. THANK YOU.
I got the same result as the video using a different technique. The answer given is correct.
what if i use calculus
i know the area of R1 will become the top circle minus the bottom circle
red digital pen blue digital pen ...... not as catchy lol.
lacrimosa must have started playing in these students heads once they turned the page to this horror😂😂
I can't wait for that day many years in my future when I'll be sifting through the fruit section of a mostly empty supermarket late one evening and suddenly this problem appears out of a portal from another dimension and demands I solve it... Thanks to this video, I'll be ready.
lol i sat this one, spent 20 minutes on that question, got halfway through and ran out of time
wait isn’t this easy? This is just year 1 math no?
no
Not at all. It's fairly difficult geometry. At least at, say, year 9 level. But in all honesty, GCSE difficulty.
@@cacalightx area of circle - 4 x area of a convex triangle (1/6 of a circle - an equilateral triangle with length r + 1/6 of a circle) Pythagorean theorem is used, but apart from that, it should be year 1, no?
@@I_like_smashburgers I agree, it's extremely easy to solve. The result is not "nice looking" but that's about it.
You have to bear in mind that this was in a GCSE paper which is compulsary for all 16 yo in the uk no matter your interest or skill level in maths, if this was a level (which is done by 18 year olds who have chosen maths to study) then even if the question was much harder it would be more reasonable. The reason why this made a lot of people upset was that a lot of people couldnt care less about maths and just needed to pass their GCSE to get into an unrelated a level. When you get to the a level stage (2 years before college and one year after GCSE) you get a lot more people who actually like maths and thefore probably wouldnt mind that a question that equivically hard would be there
Nice one!
But why don't you solve all equations for r as a variable, rather than a constant 4?
So, at the end we can have a general type also, you never know, may be a nice formula will emerge...
Thank you!
This was my gcse paper and I was one of two people in school who got the same answer 😎 peaked that day
Just a doubt can't we use calculus.
I am here to correct my comment. I said that i will solve for the shaded area using calculus. I did it and get the same result. Check my three parts calculus solution for the saded area. This is a hand written solution.Remember that i am Archimedes. We do everything by hand.
I did my gcse in maths more than a decade ago. They didn't teach us about radians, sectors or segments. Things have changed a bit apparently.
radians aren't in gcse and aren't needed to answer the question, it will also work in degrees.
We still don’t do radians but the rest is included 😅
So it would have been pi*r^2ø/360
@@powerhouseplays that's good. Hopefully theyll continue to expand the topics they teach
To solve this question you don't need to use radians and they still aren't at GCSE level, sectors and segments have been part of Maths in the UK since forever, and you actually do them in geometry in year 7 at age 11/12. In this question they wanted students to notice the equilateral triangle in the segments and go from there to solve the problem, which can be done by knowing the basics of circles and triangles.
Hey, can you solve imo 2007 p6, its very intuitive
yeap i gave igcse exam on that year and that was the paper got memories back
thank heavens I did iGCSE 😭🙏
I was part of the year this question was on 😭
I got this wrong in my GCSE’s and still got an A*
They usually put a really difficult question on to try and stop people getting top grades
Wow
My advice if you get a question like this is at least have a crack at it and show working. Even if you don't get the solution, doing this can still net you a good portion of the marks. Also consider if your time is better spent trying to get the whole solution to this or doing what you can on it efficiently and taking a second look at any other (perhaps easier) questions where you can still improve your mark.
I'm a physicist by training, 1st class Masters from a uni generally considered in the top 10 globally, and the solution wasn't immediately obvious to me - so if you're one of those who didn't get the answer under time constrained exam conditions, please don't feel bad about it. This is the hardest GCSE question I've seen. Of course taken step by step the maths is not difficult, but the devil is in figuring how to break it down. These kind of questions invariably boil down to figuring out that you can get the area of an odd shape by subtracting the area of two simple shapes.
08:00 i like statistics, and numerical analysis.
I realise you have never did any mechanics question.
How about this:
Area of the Square suroundin middle circle is 4R^2
Eliminate area covered by left and right half-circles = subtract pi R^2
Now you have to get rid of the wavy bit along the top. This is a curve x^2 + y^2 = R^2 that goes from 0 to R/2 , four times across the top, and four times underneath. You need to subtract this too.
The area of this is 8 { area under the curve from x=0 to x=R/2 ; curve is y^2 = R^2 - x2...
Yeah we get to do some integration
I = integral(0 to R/2) of sqrt(R^2-x2) dx
The overall result is 4R^2 - pi(R^2) - 8 I
The integral I comes out to be R^2/2 ( arcsin(X/R) + (X/R) sqrt(1-(X/R)^2) )
Which evaluates to zero at x=0 and to
I = R^2 /2 arcsin( 1/2) +(1/2)sqrt(1-1/4)
I = R^2 /2 arcsin( 1/2) +(1/4)sqrt(3)
arcsin(1/2) is pi / 6
4R^2 - piR^2 - 8 I = ... = R^2 ( 4 - sqrt(3) - Pi / 3)
Which I think is what you got.
A task in my maths book when I was 15 was a lot like this. It was my favorite one I had ever done at that point and it really further solidified my love to maths. I think I may still remember the exact page and task number of it.
Alt way: Take B as origin and convert the whole system into coordinates and use integration....
Glad I wasn't the only one
Can You make a video on Gudermannian functionn Calculus? And also Diffrential Equations of 2nd order?
I should’ve got this question right. I scribbled out my answer because it looked dumb and did some stupid assumption that gave me a “better looking” but *incorrect* answer, still got a grade 9 overall tho (highest grade possible)
I am from Ethiopia I like your teaching videos and I masterd at mathematics especially calculus and I have a dream to join mathematics or physics but I joined medical school because of our country give minor attention to maths and physics field
i dont see how this is notorious but ok
That's probably because you lack the context, but ok
This question is for 15/16 year olds, most of whom don’t have any interest in maths
great that you got it right but its a GCSE maths question and the techniques that you used i dont even think theyd count it as a marking point. thats like doing integration and differentiation in a gcse higher maths question. not necessarily incorrect but doesnt follow the spec
am i the only one that thinks it's not that bad...?
yup, solved it in abt 2 mins before watching the vid no idea why some people trying to use integrals the solution is so obvious
This is my least favorite type of problem. Each time I see a problem where I don’t even need to think to find a way to do it and then it’s all boring computation… I swear they just have to put those types of questions in every single test. If there were some clever way of doing it fast I won’t be complaining but there often isn’t one that is accessible to people who don’t know absurdly complicated active research stuff.
4 * integral sqrt(16-x^2) - sqrt(16 - (x-4)^2) from 0 to 2 ≈ 10.957652, which is the same as BPRP's "exact" answer
I dont think they do integrals in gcse...
gcse is done by15/16 year olds, no integrals in gcse mathematics
I REMEMBER THIS IN 2022 IT DESTROYED MEEEE 😭😭😭 i ended up just putting 16pi as my answer 😅
I did this too, now sitting A Level maths and this still looks tough lol
@@SoupMagoosh same here! - gl with exams it’s In less than 2 weeks 😭😭😭
By some fluke got this in about 12 minutes first time. I worked out the cord length (4 * sqrt(3)) and the angle subtended (120 degrees). Then worked out the segment area with that cord length wand subtended angle. That is the sector area - area triangle of height 2 and base 4 * sqrt(3). There are obviously 4 of those sectors where the two outer circles intersect the first one so just subtracted 4 of those segments from the area of a circle with radius 4.
In any event, a tough question and a lot of time for just 5 points.
All are talking about GCSE What is GCSE?
Please try the 2016 JEE Advance paper. It's a tough pill you will like it
As a check, you can see by inspection that the shaded area is a little less than 1/3 of the area of one of the circles, i.e,
Here is a different way to do it: Let AB = 1, then let D be the upper point of intersection between circles A and B. Then find angle, E, between segments AB and BD. Drop vertically down from D, point F, at intersection of segment AB. Point F must be mid-point of AB. Thus, from right-triangle, Z, formed by FB, BD, DF, angle E can be solved by arccos(BF / BD) = arccos(1/2) = pi/3. Area of corresponding sector is then, Y = pi/3 * 1/2 * r^2. Radius is same as segment AB, so sector area becomes, Y = pi / 6. Area of triangle Z is 1/2 * BF * DF. Because Z is equilateral, it height, segment DF, is srq(3)/2. Area of Z is then 1/2 * 1/2 * sqr(3)/2 = sqr(3)/8. The area of one full circle, W, is pi*r^2. Because r = 1, W = pi. Then, to get the area in question, V, that's the circle, W's, area minus 8 slivers; each sliver is sector, Y, minus triangle, Z. Or, W = V - 8 * (Y - Z) = pi - 8 * ( pi/6 - sqr(3)/8 ) = pi - (8/6)*pi + sqr(3) = sqr(3) - (2/6)pi. To get the final answer, V, requires scaling AB back up to 4, which results in 16 times the original calculated area, or 16 * V = 16*sqr(3) - (16/3)*pi. I actually paused the video this time! :D It's basically the same way blackPenRedPen solved it, but using the full circle, instead of a quarter-circle.
I was one of the ones who did this paper
I'm doing A-level maths now and at first glance I still didnt know how to solve it lol
Alternatively you could draw the 6 small equilateral triangle created by the intersections of the circle and conisder the top, bottem and both on the right ones. After that move the gray circular segment such that the gray areas look like a equiliteral triangle with 1 circular segment missing and move the white area of the top and bottom equiliteral triangle to the right side of the white triangles on the right. Then the gray area is:
A_gray = 4*A_triangle - 1/3*A_circle = 4*(1/2*4*sqrt(3)/2*4) - 1/3*(PI*4^2) = 16sqrt(3) - 16PI/3
Solved it faster than the duration of the video and correctly, so I call it a win.
Notorious…? I tried it and it wasn’t hard at all… You can do it without radians, too.
GCSE students haven't met radians yet...