That’s actually mad first person that solved it in the exam that I know of, it is not hard after knowing that it wants you to use sectors but in the exam I would have never worked it out. The worst part is that is spent 10 minutes thinking that I could work it out so wasted time that i could have used to check my other answers which had 5 marks of silly mistakes
Without knowing the formulas for the equilateral triangle and the circle segment, but knowing pythagoras and area of circle, you can still get there. You can half the triangle to a Pythagorean triangle with hypotenuse 4 and short side 2. And you can easily see that 6 segments fit in a circle, so area of the circle, minus 6x surface of triangle, div by 6 is the circle segment. I think that showing how to solve with minimal formula knowledge is often best (you can still show the formulas for the more advanced viewers).
My solution was like area of circle minus area of triangle to find the small leaf parts then use it to find the shaded area. I found triangle area using cosine rule as I know the angles as the triangles are equilateral
Another way to do it is to recognize that each area that you’re not trying to solve for (I’ll call that the negative area) is symmetrical across the axis of the intersection points (DE and FG). Based on the triangles drawn out, the angle creating that segment is 120° because it’s two equilateral triangles. You can then solve for one sector, then multiply it by 4. At that point, you just subtract this negative area from the area of the circle (16pi), and you have the positive area
another variant... also using the sectors that has the blue areas minus the 4 sector segments OS = occupied sectors SS = the sector segnemt (with he green triangle) SA = Sector arc area OS = (120/360 * πR^2) = 16π/3 SS = (4^2/2*sin(60)) [short version of finding the isosceles triangle area] SA = (60/360) * πR^2) = Area = OS - 4(SA-SS) -> Aea = (16π/3) - 4( (16π/6) - (sqrt(48) ) Note: the green triangle are is bh/2 = 4(sqrt(12))/2 = 2sqrt(12) = sqrt(4 *12) = sqrt(48) = 4^2/2*Sin(60),
I solved it by putting the circles in a grid, while ignoring the left most circle. This gives you two circles with the formulae: x² + y² = 4² and (x - 4)² + y² = 4². Substitute one in another and you get x = 2, which is the x-cordinate of the intersection. Now use integration of the middle circle with 2 and 4 as limits (or use the other circle with 0 and 2 as limits, but this formula is easier to integrate) this comes out to be 2⅔π - 2√3. Multiply this by 8 because there are 8 of these in the middle circle and then get the surface of the circle, which is 4²π, and subtract 8 × (2⅔π - 2√3) from that and you get -16/3π + 16√3
I solved this question in a similar way but instead turned the circles into semicircles. Then I integrated both of the semi-circles from 0-2. I then took the area of the right most semi-circle and subtracted it from the area of the middle semi-circle. This gave me the area of the shaded part in the first quadrant, so I multiplied it by 4 to get the answer
I started by finding the area of the circular segment, then just subtracted two of those from two sectors. I also kept it in terms of r until the end for easier bookkeeping.
Also I got the answer but firstly I got a wrong answer actually I took a long way firstly I calculated the coordinate of point F and G then I calculated the slope of line BF and BG then I found the angle between both of the line segments ( tanx = m1 - m2 / 1+ m1 × m2 ) after calculating jt I use the unitary method to calculate the area which will be off BFCG I just did a mistake in taking the angle between the two line segment it came out to be tan theta equal to |√3| and I took theta as 60° but it should be 120° to get the correct answer. If you want I can tell the complete method
It's basically two equilateral triangles, each with one circular segment added and two dropped. So we have to subtract one circular segment from each of the two equilateral triangles: Area of equilateral triangle = r²√3/4 Area of circular sector = r²π/6 Area of circular segment = r²π/6 - r²√3/4 = r²(π/6 - √3/4) A = 2 * (r²√3/4 - r²(π/6 - √3/4)) = 2r²(√3/4 - π/6 + √3/4) = 2r²(2√3/4 - π/6) = 2r²(√3/2 - π/6) = r²(√3 - π/3) A = 4²(√3 - π/3) = 16(√3 - π/3) = 10.958
I feel great again. It took me more than several minutes to look at it and tried to work out the solution without watching your answer. In the end I could solve it and am so happy. Keep up the great work please. I used to love geometry as a kid and I go back to my childhood thanks to you:)
Just imagine the centre of the middle circle as (0,0). Then find all the coordinates required and then use integration to find the area under the curve
Omg. This is a UA-cam channel and I'm not in GCSE - whatever that is. I'm here because I think math is fun. My first thought when I saw this problem was that it's an integration problem. Can you just let those of us who want to talk about a calculus approach talk about the calculus way to do it? Gah
I know a few others may have already stated this solution or a similar solution, but I want to type it out here anyway: using the diagram at 2:17 as a reference (note: [ABCD] means the area of shape ABCD): Connect D to E and F to G instead of what was shown in the video. This creates 4 "caps" which are congruent by symmetry which can be solved for quite easily. To find the area of one "cap": we will subtract triangle ADE from circular sector ADBE; for the same reason the equilateral triangles can be drawn in the video, angle DAE is 120 degrees, so the [ADBE] = 1/3(circle A) = 16pi/3 next, we find [ADE], which can be done in a multitiude of ways like [triangle] = ab sin(c)/2, special right triangles, etc; so [ADE] = 4sqrt(3) from this, the area of a "cap" (such as DBE) is 16pi/3 - 4sqrt(3) lastly, we subtract 4 of these from one circle, so we get 16pi - 4(16pi/3 - 4sqrt(3)) = 16sqrt(3) - 16pi/3, just as the video said.
Ooh, which method did you use? I found the area of the region bounded by line(DE) and arc(DBE) by finding that points D, E, and a third one (I named it H) can form an equilateral triangle in circle(A). By subtracting the area of circle(A) with triangle(HDE), you will find three times the area of the region bounded by line(DE) and arc(DBE). Dividing that value by three and multiplying by four gives the area of the unshaded white regions of circle(B). Subtract the area of circle(B) with that value and you'll get the area of the blue regions.
@@mouselmao interesting. I found that, since the circles are the same and the centers are exactly 4 units apart and on a straight line, you can divide them by lines DE and FG. You will now have 4 areas inside of circle B that are each identical circle segments. For the formula you only need the corresponding angle, which you calculate with the height of the segment (r/2 = 2). Then you just substract that from the area of circle B. Tbh I needed to look up the formula for circle segments though to come to this idea
@@mouselmao I don't know what all methods exist, but I used calculus. That might sound overly complicated, but the concept is actually pretty simple. I don't know what the GCSE allows, but this method should be within the abilities of a 16 or 17 year-old if they ever took calculus. ___ The equation for circle A is y = sqrt(-x^2 + 16) The equation for circle B is y = sqrt(-x^2 + 8x) We don't need the equation for circle C, but if you're curious it's y = sqrt(-x^2 + 16x - 48) These equations are taken from the general equation of a circle (x^2 + y^2 = R^2) and translating them to the right along the x-axis. "sqrt" means "square root". We will find the area by integration. For that we need limits. By simple inspection, it's obvious that the x-coordinate of points D and E is exactly halfway between points A and B. Point A lies at x=0. Point B lies at x=4. This means that D and E lie at x=2. To find the area of the blue regions, we'll need to find the area of circle B and subtract the areas of A and C. The area of circle B is twice the integral of sqrt(-x^2 + 8x) over [0,8], or 2 ∫ sqrt(-x^2 + 8x)dx over [0,8] = 16π. We could have also used πr^2, but I did the integral version just to show that calculus is easily applicable to this situation. I said before that we don't actually need the equation for circle C. This is because the areas we need to subtract from B are all equal to each other, and A has a slightly prettier equation. So to find the area of the two shaded regions, we take 16π - 8 ∫ sqrt(-x^2 + 16)dx over [2,4] and that gets us the correct answer 16sqrt(3) - 16π/3. The only drawback to this method is that you'd need to consult an integral table, which is generally provided during calculus exams. Aside from that, I'd argue that this method is just as clean-if not cleaner-than using geometry to analyze the problem. It just doesn't look like it because UA-cam comments aren't exactly friendly to math notation.
@@LordAmerican Ooh, that's so cool! I never thought of modeling the circles as relations and finding that area by using integration. That's a really neat method
@@3iknet327 Yeah I did this method too. Basically circle B minus 4*the circle segments (I too had to look up the equation).The angle stumped me for a bit until I realised that the arc DBE was made up of 2 radia. Had to look up the sine for it, but yeah glad that I got there in the end :¬)
Managed to figure it out during the exam but I'm incredibly happy to see this question show up here. Props to Edexcel - it's a really well made and well crafted question that's not incredibly overcomplicated or wordy, just heavily problem solving. Was fun to find out in the exam.
@@tobysuren I’m not sure why they didn’t comprehend after you explained, but “giving an exam” is a pretty common phrase here in India which just means taking an exam.
Another useful strategy when dealing with problems like this is "if you have symmetries, use them" - I personally used symmetries to divide the thing to be solved into 4 pieces that I called X, then I defined a piece Y, which is half of the intersection of two circles bit, and X and Y together form a quarter of the central circle. That's a lot easier to draw and keep track of. Just don't forget that your final answer is 4X at the end. Pretty symmetries usually lead to pretty solutions.
It is a symmetric question and not of calculation. The intersection area of circles is 1/3*2 leaving the shaded area 1/3 of any circle. 1/3 π 4*4 = 16π/3
I think you can do this in a slightly less tedious way. In the drawing at 2:40 you can draw a line from D to E. You now have a segment of the 1st circle defined by the chord DE. You can then use the formula for the area of a segment of a circle: A=r²(α*π/360-sin α/2). Triangle ABD is an equilateral triangle so angle BAD=60. Angle DAE=2*BAD=120=α. So A=r²(π/3-√3/4). You have 4 of these segments with a total area of 4r²(π/3-√3/4) which with r=4 gives 64π/3-16√3. Subtract this from the area of the middle circle (16π) to get 16π-(64π/3-16√3)=16√3-16π/3≈10.96
I'm guessing they didn't learn that formula yet. Because by just knowing that the formula exist, you really shouldn't need more than a couple of seconds to figure out the strategy.
This was the final question on the British GCSE Mathematics rexam on May 20th. The students always complain about the final question and it usually a tough geometry one.
Yep, I remember everyone online and real life talking about it lmfao. I did my GCSEs this year and imo the last question of paper 3 was even more challenging than this. 😭😭
i got this in my gcse, one of my friends actually got it which is mad. also, the paper (specifically this question actually) was leaked on a discord server about 3 hours before the exam was sat but only a few people saw it.
No need for the equation at the end, since the blue "triangle" is equal to green+2orange-1orange (all multiplied by 2 since you need 2 of them, so two times green+orange) . The final result is the same, just simpler at that point.
Although i thought it may have been overkill, I made a graph with a circle in coordinates (-2,0), integrated between 0 and 2, then multiplied by 8(simetries) to get the white inside area. Finally, I subtracted the circle area from that
as my physics teacher once said, integration is the uncivilized way - all too often there is a clever trick that lets you skip the integral. My favorite such flex is skipping most flux calculations because there is either a neat symmetry or the shape is enclosed or something else.
This problem is easily solved by integrating: Let's take a circle at the center, radius 4 and center at point B. Let's find the arc of this circle in the first coordinate quarter, as y = sqrt(16 - x^2). We will integrate this arc over dx from 2 to 4. I will not describe the course of the solution, it will turn out: (8*pi/3 - 2*sqrt(3)) The figure shows that the circle in the center is divided by two other circles into 8 such areas, respectively, their total area is equal to: (64*pi/3 - 16*sqrt(3)) The area of the circle in the center is: pi*r^2 = 16*pi It remains only to subtract the 8 areas found earlier from the area of the circle, and we will get a ready-made formula for solving this problem: S = 16*pi- 64*pi/3 + 16*sqrt(3) = 16(sqrt(3) - pi/3) Approximately this number is equal to 10.96. Problem solved.
this is one of the few ones in here I could actually do! I just split the parts where both circles overlap in half by drawing a line, then found the area of one of the segments, multiplied it by 4, then subtracted that from the total area of the circle
This is what I ended up doing, for the most part, though I think I went around my elbow to reach my nose in actually determine that arc area. I'm just glad I ended up with the right answer. I was certain I'd messed something up when those root 3's hung around.
For me, If I am willing to remember any circle formulas, I would just use area of the circle - 4 equal segments with a height of r/2 Or without knowing formulas take an area of a sector - the area of isosceles triangle, the high of which is r, and the base is a double the hight in 3 side r triangle, basically the longer version of your explanation
I used 8(16pi/6)-4(4pi/3). This effectively calculates eight arcs and then subtracts the four triangles which were double counted within those arcs. It makes the calculation just a tiny bit neater and easier to manage rather than subtracting eight triangles and then adding four of them back in.
Plus you avoid all those nasty irrationals and get to simplify the answer down to terms of Pi only, which the question explicitly asks for. I’m guessing this is probably the solution in the marking guide, and therefore the solution that gets you full marks.
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It's simple. From the circle you should subtract 4 equal "circle segments" en.wikipedia.org/wiki/Circular_segment. The student does not need to know the formula for the area of a sector of a circle, it is easy to derive. The segment of the circle we are interested in is the difference between 1/3 (not 1/6 my mistake) of the circle in the radius R and an equilateral triangle with the side R.
Exactly how I did it. I’d forgotten the exact formula for the area of a sector but I thought it was pretty intuitive to take the area of one of those sectors as a third of the area of one circle.
Won't bother watching the video since your comment gives the same solution I had :) So it's circle - 4 (1/3 circle - triangle) = 4 triangle + 1/3 circle = sqrt(3) + 4/3 pi? Maybe 50% chance of getting something wrong but still more fun to solve these without writing anything down.
I used almost the same, but used 8 areas ... which are 1/6 (not 1/3) of the circle minus a square angled triangle (I don't remember the formula of an equilateral by mind :))
It's an interesting solution. I thought of a very different one though. We can break the blue area into 4 equal parts, due to symmetry, and find one part's area as a difference between two finite integrals.
@@cprox1000 construct a semicircle of radius 4 and integrate with the limits from 2 to 4. Then multiply by 8. Finally subtract that to the area of the circle.
I went from having a hard time solving the easiest of questions to easily solving more complex ones by watching your videos in less than a month. Thanks a lot!
People studying for Colégio Naval and Epcar watching this: easy However, it's a good question! Awesome solution too! *For people who don't know what's CN and Epcar: both are military schools for students between 15 and 18 years old in Brazil.
You could just use the 4 equal segments that are 120 degrees. It would have been way waaay faster. S = (full circle area) - (segment) * 4 S = pi*R^2 - ( (2/3 * pi - sin(120)) * (4^2)/2 ) * 4
did this one pretty quick in my head. ignoring r for a bit for better intuition, triangles have area √3/4, and wedges of 1/6 of a circle are π/6. intuition confirms both of these are close to 0.5, so far so good. then taking differences until the result matches the shapes in the diagram (wedge + two triangles on the sides - two wedges cutting into it) gives the answer up to a scale factor. putting the r^2 in gives the answer you found
you can also construct a chord that passes through the intersection of two circles, you'll then see a triangle that has an angle of 120 deg., its area is also the same as the triangle in his solution (4*sqrt(3)). On the other side of the chord is a circular segment, which has an area of 16*(pi/3)-(4*sqrt(3)), to determine this value, calculate the area of the 120 deg sector and subtract it with the triangle's area. All overlapping regions make 4 of these circular segments, thus, deducting the total area of those circular segments (4(16*(pi/3)-(4*sqrt(3))) from the area of 1 circle (16*pi) will give us the wrong answer.
@@fredgoodyer4907 I got the same answer. But it was difficult for me to explain in plain text, so some (or all of it) may be misunderstood. In any case, there is not just one solution to this problem.
I also used this method and I think it is more elegant than the one presented in the video. I also think you do yourself a disservice; I think it is incredibly well explained for such an inherently diagrammatic problem, and that you did a great job 😊 It just confused me that you ended the comment “wrong answer” 😅
You can also do it with the circle function r^2 = x^2 + y^2 and integrating. Do this for circle A and circle B, integrating from 0 to 2, which gives you: 4 integral_0^2 (-sqrt(16 - (-4 + x)^2) + sqrt(16 - x^2)) dx = 4 (4 sqrt(3) - (4 π)/3)≈10.958
You can also solve this problem using integrals, ((integral from 0 to 2 of sqrt(16-x^2)) - (integral from 0 to 2 of sqrt(16-(x-4)^2))) * 4 The first integral is the area of one quarter of the circle with origin (0,0), the second integral is the area of the one quarter of the right circle. So if i subtract this areas, i get a quarter of the desired region. Then just multiply by 4 and get the total area Edit: the region of integration is from 0 to 2 because in x=2 is the intercept of both circles
@@ratpoisonedcupcake2827 you don't have to be a part of that bright brunch to solve a problem like this. You lack experience. Keep on learning math and asking questions this way and you would join that bright brunch in no time.
When I was 16 we had a hardass of a math teacher who would've escorted us off the school if we couldn't solve this problem. Okay maybe that's slightly exagerrated, but the point is that the problem isn't even that hard when you know a few tricks. For example, you can show, that the triangles you get are equilateral and they always have angles of 60° and sin(60°) = sqrt(3)/2. If you know that beforehand, then the problem is very easy.
I think there's an easier way. If you run a line from D to E, it gives you a segment (chord) with a radius of 4 with a height of 2. (I don't remember the formula), but just calculate the area of that segment, multiply by 4 and subtract the total area of the circle. The remainder should be the correct answer.
I did this. This is by far the most straight forward answer. Area of segment is As = 0.5(θ-Sinθ)r^2 4 segments is 4As = 2(θ-Sinθ)r^2 Area of a circle is πr^2 Then the area of two curvy sections is A = πr^2 - 2(θ-Sinθ)r^2 θ is 1/3 a full rotation, since there are three circles equally spaced; θ is thus 2π/3 The formula is finally A = πr^2 - 2(θ-Sinθ)r^2 = 16π- 32(2π/3-sqrt[3]/2) = -16π + 16sqrt[3] = 10.958
My answer before watching the entire video: Since the radius of the circle never changes no matter what point on the circle’s edge is measured from that point to the circle’s centre. Therefore points BFC, BCG, BEA, BAD all form a equilateral triangle with all sides equal to the radius. From this we can take one of these triangles like BFC and used pythagorus theorem to find the length of the median. c^2 = a^2 + b^2 4^2 = 2^2 + b^2 Sqrt(16-4)=b b=3.4641 (4 d.p.) Next we’re going to draw a line from B to the top of the circle and to the bottom, a line between D and F and a straight line across the highest point of the circles, this will create a new triangle that shared an edge of BF. We can determine the angle of B in this new triangle by using the inverse of Sine as: Sine^-1(2/4) = 30. By deduction angle F is determined to be 60. Next we’re going to calculate the distance between point F and the top line: Radius - Median line of BFC 4 - 3.4641 = 0.5359 Next We calculate half of the area of the empty spaces near point F by calculating a 30 degree portion of the circle and minus it from a rectangle made from the B to top line, the line going across the top of the circles till just above the F point and from the top line down to F and a line coming perpendicular from B to F: (2 x 0.5359) - (((30/360) x pi x 4^2) - (2 x 3.4641 / 2) = 0.3471(4 d.p.) Finally to calculate the Area we draw a vertical line from C to intersect with the line going across the top of the circles and a vertical line from C to intersect with the bottom of the circles line. This will form a rectangle going from B to top line to C to bottom line which we will calculate the area of and minus all the empty spaces: 2 x ((4 x 8) - ((1/2 x pi x 4^2) + (4 x 0.3471)) = 10.9577 4 d.p.) Edit: Ayee I got it right but it’s interesting that I worked out the top and bottom diverts at DFEG between the circle rather than the intersecting area of the circle
So this is the way I solved it using middleschool knowledge, similar to the one in the video: 1. I started by calculating the green triangle because it's just an equalateral triangle 2. Then, I filled one circle with 6 equalateral triangles, forming a regular hexagon (if we think about it, the area of a regular hexagon is just 6*equalateral triangle area) 3. Now we can see that Circle Area = Hexagon Area + 6 Orange Parts => Orange Part = (Circle Area - Hexagon Area)/6 4. We can find the area of an "oval" shape by adding 2 equalateral triangles with 4 orange parts 5. Using: Circle Area - "Oval" shape Area = Shaded Area we get the area of the shaded area.
I got EXACTLY that answer! I simply taped together 50 sheets of graph paper (5x10), taped them to the concrete floor of my garage, drew the circles on the paper (used my empty garbage can as a template), and counted the blocks. Once I did that, I used a simple ratio of the garbage can radius vs. the example. My wife yelled at me because I was supposed to be cleaning the garage. My daughter got in trouble for not turning in her science report because I used all the graph paper. I probably could have figured the math 30 years ago when I got out of school, but, as a practicing engineer, I mostly just fight with contractors about payment issues and contract disputes.
Draw lines DE and FG. You then have 4 circular segments of 120°. To find the area of the 4 triangles to subtract, it's simply two of the equilateral triangles cut in half vertically, so the same area as an equilateral triangle. Different approach, same key concepts. Circle B minus circles segments found using equilateral triangles (except my circle segments incorporate the area of said triangles, so no need to subtract them again in the final equation) Circle segment areas: 4(Pi*r^2*120°/360°) - 4(4*sqrt3) =64pi/3 - 16sqrt3 Final solution: 16pi - 64pi/3 +16sqrt3 =16sqrt3 - 16pi/3 = 10.958
Just find the segment area cut by left most circle(or the right one doesn't matter), angle DAE is 120° which can be found by very basic trigonometry, then subtract that area 4 times from the area of 1 circle (Segments DAE, EBD, FBG and GCF all are exactly the same) PS: I am 16 and love your channel it provides very interesting problems for me to scratch my head around
how would this work? wouldnt be the orange segments there, like if i subtract the green diamonds, i still have the orange an blue area, how do i get rid of the orange then?
I did it differently. I calculated the area of the equilateral triangle and subtracted it from the area of the sector to give the area of the segment. Then it was simply 2 (area of the triangle - 2 segments + 1 segment), or 2(area of triangle - area of segment.) or 2(√12*2 - (∏*16/6 - √12/4)) = 10.9576521 square units.
I got this without needing to know area of equilaterals. Just used third sectors minus [length times width area right triangles] to find third circular segments. Subtract 4 from a circle and you have the answer.
I solved it by calculating the area of a circle sector with the angle of 120deg (got the angle from the equilateral triangles), then subtracted the inner isosceles triangle (sides of r, r, and a circle chord of 120 deg) from that, then multiplied the result by 4 and subtracted that from the area of the circle.
I first subtracted the area of a 120-degree-30-degree-30-degree triangle (0.5*2*4*sqrt(3)) from a third of an area of a circle with a radius of 4(16pi) for a value of 16pi - 0.5*2*4*sqrt(3) for the area of the part of the 120-degree sector that is not part of the area of the triangle mentioned earlier in this sentence. I then multiplied this value by 4 since there were four such areas that had to be eventually subtracted from the area of a circle with a radius of 4 to get the value of 64pi - 2*2*4*sqrt(3). After subtracting this value from the area of a circle with a radius of 4(16pi), I get 16*sqrt(3) - (16/3)pi or 10.958 approx.
I live in Calgary, and I learned how to do this stuff using similar strategy but rather direct equations for the “circular segment”. It was moved from 11th to 9th grade math the year I took math. I found it the part of the course, along with my friends. Had a great teacher, which may have played a part. However I remember nothing I learned in school so I don’t remember equations
I find it easier to consider that the piece of the circle described by A-D-E is a third of the entire left circle. Hence it has a size of r^2*pi / 3. From this area we have to substract the triangle given by A-D-E with area r^2 * sqrt(3)/4. The area of the remaining arc has then to be multiplied by 4 and substracted from the total area of the central circle. The result is also r^2 * (sqrt(3) - pi/3).
@@leif1075 yep i didn't knew one formula , and actually that formula is in class 11 maths in cbse but i am in 10th because i got admitted in school at 4 year instead of 3
@@I_guess_its_over dude I also study in 10th grade but if you have studied both circles chapters in 10th ncert you might be able to solve those questions.
@@kshitijsingh2412 it's start of 10th class for me , right now summer vacations going on , but my favourite is maths so i have already learnt those chapters but not practiced enough to solve this
I'd like to add my solution. After you'd shown 3:28 it became clear to me, that since we deal with circles and radius fits 6 times around that, that one could imagine an orange section between D and F pointing upward. Because BDF is equilateral as well this section must be the same as the other orange sections shown already. If we subtract that area from the Blue, one of the Orange sections can be filled by it to adjust to an equilateral side of a triangle, I'll take OrangeBF. So one Blue section must be BDF - OrangeBD. Multiply by 2 because of the bottom one, voila. I like your solution better, because it is very generic. Make a formula, solve. Whenever I encounter geometry, though, I can't help build legos, meaning stiching pieces of area and volume up out of something and it does not always work out if you don't find the right pieces to build from. But then, no more math tests for me in the forseeable future and building legos is arguably more fun. :D
There are many possible solution methods, particularly if calculus is allowed as a tool to aid in the solution. It is a nice problem. The method shown in the video is very nice.
my solution was to take the integral of a half circle with the and 2 to 4 the i took the area of the circle and subtracted 8 times the integral. a different method but the same idea
@@evandrofilipe1526 They're not expected to know calculus for GCSE, and as such it wouldn't be in the mark scheme. But it would be utterly ridiculous if an entirely correct and valid solution, with all working clearly shown, were to be penalised just because the student used methods which are outside the syllabus. If it is the case that no marks would be given, then there's something seriously wrong with the exam system and the philosophy of mathematics education.
I actually managed to figure this out during the exam! Unfortunately I didng have enough time to fully write down the method but I reckon I got 2 or 3 Mark's for it
I've done tons of hexagons using this manner of overlaid circles, so it was the first thing that occurred to me. Knowing it makes equilateral triangles and congruent segments in the circles, I noted that the two regions we want the total area of are each 1 Triangle - 2 Segments + 1 Segment = 1 Triangle - 1 Segment And 1 Triangle + 1 Segment = Circle / 6 (since they're sectors defined by 60 degree angles), so (Circle / 6) - 1 Triangle = 1 Segment Dividing one of the triangles in half gives a 3-60-90 triangle, and I remember the side length ratios for those, so calculating triangle area was easy. Then I just plugged in the values to the 1T - 1S, multiplied by 2, and had it. This was absolutely one of the easier ones I've seen on this channel, for me, definitely thanks to my familiarity with the layout. Fun problem!
If you draw a line halfway from A to B, you can integrate the area underneath the curve, then multiply by eight, then subtract that value from the area of one circle. Nice to see a solution that doesn't require calculus though.
If you know the area of a Hexagon you can solve the problem even faster through simplification: AH = Area of Hexagon, AC = Area of Circle, x= shaded region, s = side length of hexagon, 1st: Draw an inscribed Hexagon with vertices on the intersections and centers of the circles, as well as the triangles that make up the hexagon. Observation 1: The 6 arc sections are equal to the AC - AH. (6 arc = AC - AH) Observation 2: The shaded region is 1/3rd of the AH - 2 arc sections. (x = 1/3 AH - 2 arc) 2nd: Combining these 2 observations by substitution it is clear that x = 1/3 AH - 1/3(AC -AH) 3rd: Simplify the equation to get x = 1/3 ( 2AH - AC) The area of a hexagon is 1/2 (3root3)s^2 and the area of a circle is pi r^2 (note that r=s here) 4th: Substitute the areas in and simplify to get the equation 1/3 r^2 (3root3 - pi) Finally: Calculate for r=4 and you get 10.958!
I did it similarly, except I calculated the excluded area based on 2 overlapping sectors minus the equilateral triangle of their intersection (times 4, of course). Nice problem that took a little thought (but thankfully no calculus!). You can also simplify things by making r=1 initially and then scaling by r^2 at the end.
A simpler way to solve this problem might be using the following approach: In the leftmost circle - Area of the sector enclosed by line segments AB and AD = 1/6 of the total area of the circle = 1/6 * pi * r^2 (because angle DAB is 60 degrees) Similarly, area of the sector on the middle circle enclosed by line segments BD and BA = 1/6 * pi * r^2 if we add both and remove area of the equilateral triangle ABD then we get the area of the half of the intersection of Leftmost and the middle circles. Let's denote this area by 'X' Therefore, the area of the coloured figure = pi* r^2 - 4X
Or, if you're feeling adventurous, simply take the equations of middle and RH circle: x^2 +y^2 =r^; (x-r)^2 + y^2 = r^2 respectively. Calculate where they intersect; x = r/2 and perform a double integral (∬) with x between zero and r/2 and y between [r^2 - (x-r)^2] and [r^2 - x^2]. After a bit of trig substitution you'll get 4(√3 - π/3); this is for one quarter of the area, of course, so the final area will be 16(√3 - π/3) QED😁. A bit above GCSE (it's first year degree maths, well it was on my degree) but fun if you bored and have a few minutes to fill.
me and a few friends managed to figure this out after the exam, but in the actual thing i couldnt. not because i had tried and failed, i just didnt have enough time. i guess i choked in some other questions under the stress and the exam was just a big time crunch for me. i was literally speedrunning questions as the examiners were saying there was a few minutes left lol
I paused the video on 1:27 and the answer is pretty obvious if you know integrals. Just calculate the area between points ADB, divided by two (vertically in half) by counting a defined integral from -1 to 0 from the function, that represents the circle with the center in point A. The function of the circle with a center in zero coordinates is x^2+y^2=16, so if we change x to (x+4), the circle will move by 4 points to the left. The result function is (x+4)^2+y^2=16. Calculate the integral from -1 to 0, it will give us the area between the X axis and the function and then just take 8 of these parts from the full circle and you will get the highlighted zones. And that's it - the answer in 2 steps. This is pretty straightforward approach if you need to calculate difficult areas, which can be limited by functions. English is not my native language, so I'm sure, that I call some things the wrong way, but I hope you got the idea. I watched the video till the end and found much easier solution, when you even don't need any calculations with pi, except for the whole circle area. In the beginning, when you line up the points, also connect DF and EG. Now the whole circle area consists of 6 equal triangles and 6 equal circular segments on the outer sides of these 6 triangles. The area of each triangle can be obviously easily calculated and we can calculate the area of the circular segment by just taking out the area of 6 triangles from the area of the whole circle and dividing the result by 6 ((circleS-6*triangleS)/6). The highlighted blue area consists of 2 equal zones (triangle+segment-2segments) or, if to simplify 2*(triangle-segment). And that's it. Again the answer in 2 steps, but with much less calculations.
I did this as a mental exercise to see if I could relate the blue area to the basic shapes present. I recognized that if I draw an inscribed hexagon on the central circle, then I can describe the blue area as: Let A=full circle area, H=Area of Inscribed hexagon, B = Blue area to be found Then: B = A - 2/3A - 2/3 (A-H)
Cool i figured out an identical strategy. Special Area = Circle Area - 4*Chord Area, where the chord is from E to D. I knew there was some formula on wikipedia for chord area but i was too lazy to look it up lol, and i knew i cud find the length of E to D too since that line is part of a iscosc triangle with 2 radii legs. Im happy with figuring out the strategy of attack.
I approached it a little bit differently. I saw that if you connect the chords, BDF and BEG are equilateral triangles of side length 4. By symmetry all of the segments are the same (16pi/6 sector - 4 √3 equilateral triangle). The blue areas are one equilateral triangle minus two segments plus one segment, so (4√3 - (8pi/3-4√3)) = 8√3-8pi/3. The total for two of those blue areas is therefore 16√3-16pi/3.
This was also fun to do with integration, half circles, trig sub, and double angle identity. Would reccomend seeing how far you can get without writing anything down. You definitely have the better solution than integration, though.
idk about that tbh... what I first thought of is simply taking half circles, getting the intersects between them, which would be simply setting them equal: sqrt(1-x^2) = sqrt(1-(x-1)^2 which is fairly simple so solve, or you just make the educated guess that its 1/2 (which it is if you look at it so yeah). Then you take the integral of both equations between 0 and said point (here called F) and then take it times 4. The integral is pretty hard ig, tho you have hard borders and most calculators can do it. idk if those are allowed? If they are, way easier than above solution.
@@pottedrosepetal6906 You're on the right track, but you've missed that r=4, which changes the equation for the half-circles and the intersection point. And yeah, you could solve the integrals by looking them up or using a calculator, but I think it's more fun as a mental math challenge.
I solved a similar problem in the past. It was something along the lines of 'how close do you have to bring two similar circles so that the area of their non-overlapping regions and the area of the overlapping region are equal'
This basically what my initial thought was, but I haven't taken or done any kind of trig/calc in like 8 years so as far as I got was: π(4)^2 - 8(integral of the arc from the point of intersection to the midpoint of the inner circle from 0 to x) I tried to figure out what that x would be making a right triangle with a 2 unit side, but I don't remember enough trig to use that. Also, is there any way to really figure out the integral?
Easy. The areas to keep track of are as follows: the circular sector enclosed by F and C, the segment created by cutting it with its chord and the area of the circle itself. Which of course is 16×pi, as we know. The area of a sector is r^2/2×theta. Theta in this case is pi/3. The area of the sector is therefore 16×pi/6, or 8pi/3. Now the area of the segment. That will be equal to the sector minus the big triangle formed by the 2 points on the circumference and the centre of the circle. Since it's an equilateral triangle, the usual formula devolves to (r/2)^2×sqrt(3). r being 4 here leads to the inner triangle having an area of 4sqrt(3). Subtracting this from the sector gets (8pi - 12sqrt(3))/3. This is the area of the segment. Now to build the rest. Adding the area of the segment to that of the sector will get the area of 1/2 of one of the wedges. Multiplying that by 4 will get the area of both and subtracting that from 16pi will result in the total shaded area. Sector + segment = 8pi/3 + (8pi - 12sqrt(3))/3 = (16pi - 12sqrt(3))/3. 4×(sector + segment) = (64pi - 48sqrt(3))/3. Shaded area = circle - 4×(sector + segment) = 16pi - (64pi - 48sqrt(3))/3 = (48pi - 64pi + 48sqrt(3))/3 = 16sqrt(3) - 16pi/3. This is equal to 10.958 (3 s.f.)
I did my maths GCSE this year as a year 10. It was a pretty difficult question indeed, but luckily here in Britain we have similar questions like this in our national maths challenges (Intermediate and senior maths challenges especially), so students who did well in those would have a good time doing these sorts of questions. To my disappointment, I would have lost a single mark, as I had put the wrong denominator (6) in my answer, however had the correct answer somewhere in my working. Lesson learned for paper 2 and 3: check your answers!!
@@generaltoasty3990 I did the maths challenges, they are indeed a very good way to revise and help you prepare you for any challenging question such as this.
Hey, fellow Y10 here, I thought the question was quite interesting- The question actually felt quite unique from the usual past papers- Even though the final questions usually differed on past papers, I was expecting a problem solving vector question or a disguised trigonometry question. Tough to hear about that lost mark, I get that a lot- 😅🤣 How did you manage to take your Gcses a year early? For me, there was quite a long process that had to be done of tests and arrangements in our school trust, so I'm curious to hear how it worked for you
@@tricksterdim hello there fellow year 10, great to see you here. I go to a grammar school that only has sets for maths, which is (probably) unusual. Of these sets, set 1 will do 2 sets of mock exams to test the students' capability and their potential to get a grade 9 in the GCSE. If they do not get a grade 9 in the second set of mock exams, then they will not do the GCSE early, otherwise they will, which seems to be a pretty fair and efficient way of enrolling for early GCSE exams. I'm sure this may come as a suprise to you, but I was not aware that other schools had less linear and more extensive methods to enroll students for an early GCSE, so I thank you for making me aware of this! Anyways, best of luck in papers 2 and 3!
@@generaltoasty3990 Wow! I didn't know about this type of method! I congratulate you for your move! What an interesting and peculiar process- I've heard that Public Schools don't really do the year transfers, as they generally have different priorities for students to meet- because of this, my transfer was a very messy process, especially as I was the only one in the trust who had been considered for a transfer. Mine was a bit of a peculiar scenario where I was randomly asked to do some gcse practice tests a year ago after they initially considered the year transfer when they saw the hours of extra work I had been doing to see if it should happen. I got 9s on all of the tests, and previously I was working in private facilities within a public school anyway- I loathed the public classes-, so they went forth with it a few months after the tests. Great to hear there are more bright students out there who actually are able to do these things- In public schools especially it's very hard to find truly dedicated students, so I congratulate you on that! Good luck for the next papers!
I tried doing this elegantly... but didn't come up with your amazing solution. I threw in the towel, and just did an integral. The integral ranged from 2 to 4 (since we know both those points on the x axis) bounded on the upper part by the offset circle (middle circle) and bounded on the lower part by the origin circle (leftmost circle). This gives 1/2 the area of the top blue region. Multiply by 4 and there's the brute force answer.
I solved the problem by drawing a line from E to D. Then i calculated the area of the segment EDB. I was able to do that because i knew from geometry that the angle DAE is 120°. That means that ADE must be 30°. I then calculated the area of the triangle EAD and subtracted it from the sector. Finally i subtracted 4 segments from the area of a full circle to get the shaded area.
Seems easier to me to note that after constructing lines DE and FG, we have four circle segments with a theta of 120 degrees (due to the equilateral triangles). These have an area of (pi/3 - SQRT(3)/4)r^2. The area DFB+BEG is then the area of circle B, pi*r^2, minus four time the sector area, or (pi - 4*pi/3 - SQRT(3)) * r^2. Multiply out and re-factorise for pi and you have your answer.
I personally thought it through in a slightly different manner. I’m not entirely sure if it’s any easier or more difficult given I didn’t crunch the numbers, just thought out the process, but here’s how I did it. I took the 60° circle segment minus the equilateral triangle that (for whatever reason) I chose to take from BDF. Keep in mind, I almost always take a backroadsy kind of path for these kinds of equations, but I took that orange segment and subtracted it once from the equilateral triangle since the other subtraction is replaced by the top curve. That gives you the area of a single blue segment, so all that’s left to do is just multiply by 2 and you’ve got your answer. It was fun to think it out first and then see a different method to show that there’s multiple ways to do almost any problem out there if not all.
I did it two ways: the first was subtracting the four 120 °segments from the whole circle. Then I came up with your method as a check. I really think this method is the most elegant! The way I wrote it down was A/2 = 2At - As where At is area of the triangle and As is the area of the sector. (You can make a nice drawing of this.) Then just plug in the formulas for At and As and simplify.
Fun puzzle, straightforward solution. Only substituting r=4 in the end rather than the beginning would have been a bit cleaner and you could use more geometrical moving around and do with less calculations
Another approach: I calculated the area of the equilateral triangle CDE, inscribed in the circle with center B, to deduce the area of the arc DAE. The area of the circle deducted from 4 times the area of this DAE arc gives us the area sought. Area of the triangle CDE. I draw the vertices CD, DE and EC. I draw the intersection of DE and AC in H. I draw the triangle inscribed in the semicircle BDE. I calculate the length DH = r * sin60. I calculate the base DE = DH * 2 and the height CH = r * 3/2, hence the area of the triangle CDE, by the calculation DE * CH / 2. Area of arc DAE = (area of circle - area of triangle) / 3. Area sought = area of the circle - 4 times the area of the arc. Reduced formula = r^2 * (2 * sin60 - pi/3) (traduction Google Chrome)
Can someone explain how he went from 128pi/6 to 32pi/6 and how he solved for x in the end ( so how he went from the second to last equation to the final equation where x=16sqr3-16pi/3
On the right side of the equation it's 128pi/6. We need to have the left side represented as a term over 6. So we can represent 16pi as 96pi/6 so that these terms are compatible. Subtract 96pi/6 from both sides, that's 0 = x + 16sqrt3 + 32pi/6, as shown in the video.
2 circles first instead of 3. We must find the area (total area) between them. Area of Triangle ADE and the area between line DE and circle " I) ". We find the total area. The angle A is 120 degrees so the area ( name it:A) is the 1/3 of the circle area = 1/3 16pi. Find the area of the ADE triangle (name it: B)- is easy to do. So the area "I)" is A-B. For the 2 circles is 2 x (A-B). For 3 circles is 4 x (A-B). So the answer is 16pi-4 (A-B). I came in the same result as video shown!
This answer is easy if you understand Arcs and Chords and among these this is one of the easiest. So what we start with is actually the definition of the inscribed Hexagon, which is one of the geometric givens. But in this case having defined the hexagon we are going to ignore 2 of the vertices leaving a rectangle of sides r and r* SQRT(3). Next we are going to ignore sides r. 360/6 = 60° but DE is both a chord and an arc of 2 * 60° = 120° which is 1/3 of a circle. So area between the Arc and the origin is pi/3 the area between the chord and center for either a side of a equilateral inscribed hexagon or triangle is the same, 1/4 SQRT(3) So pi/3 - 1/4 SQRT(3) gives the area between the Arc and the chord. The area arced inside the chord is equal to the outside since they intersect each other at the same displacement from the horizon. Thus total overlapping area is 4/3pi - SQRT (3) and the area we are interested in is that area subtracted from pi So Blue Area = pi - 4/3pi + SQRT(3) = SQRT(3) - pi/3 Pi/3 = 1.047 SQRT (3) =.866025 x 2 = 1.732050 1.732-1.047 = 0.685 So now we have to rescale for r 0.685 * r^2 = 0.685 * 16 = 6.85 + 4.110 = 10.96
I did the question in the exam and honestly its like the best question they could have ever asked it doesnt require some "fancy" math you need to learn at gcse instead it requires insight. I hope edexcel makes more of these problems in their exams in the future!
Did this by calculating sector BDAE - triangle DAE (easy as DAB is equilateral), which gives the area of the segment DAE. The required area is simply the area of the centre circle minus 4 x (area of the segment DAE). I think this is simpler arithmetic and therefore a bit safer.
my calculus oriented brain jumped straight to calculus. I found out the area of one quarter of the area to be found by plotting two functions, then multiplied my result by 4. Needless to say, I actually got it right :)
I did it a different way and still got the same answer. I noticed that BDF formed an equilateral triangle with sides R. I then noticed that the segment formed by chord DF was identical to the segements formed by chord DB and BF, Therefore 1 segment could be ignored. So the answer to the question was 2(Area of Triangle - Area of Segment). Area of triangle = (Base x height) / 2 = (4 x 2 x Sqrt(3)) = 4 x Sqrt(3) Area of Sector = r^2 x (angle in radians) / 2 = 16 x (Pi/3) / 2 = 16Pi/6 = 8Pi/3 Area of Segment = Area of Sector - Area of Triangle = 8Pi/3 - 4Sqrt(3) Area of Blue areas = 2 x (Area of Triangle - Area of Segment) = 2 (4Sqrt(3) - (8Pi/3 - 4Sqrt(3)) = 2(8Sqrt(3) - 8pi/3) = 16(Sqrt(3) - Pi/3) 16(Sqrt(3) - Pi/3) is the answer. Multiplying by 3/3 gets you 16(3Sqrt(3) - Pi) / 3, which was the answer in the video.
i did this exam, and no one in my school that i spoke to managed to do it. there were videos of teachers struggling to do it. i don't think the average 16 year old in the uk could solve this
You needn't feel bad about this. I believe most students can't do it because they are not trained to handle this sort of questions, let alone in an exam situation. It's more about the training than the talent. I showed this to a year 9 girl and she gave me a 4-part full solution. However, she is trained to do these complex multiple step problems for a couple of years. If you were trained the same way, you could also do it.
Too much learning by rote! They need to explain how to solve problems by breaking them into simpler parts. A maths teacher doesn't need to be a genius at maths, it is more important to know how to explain things to students.
Smart solution. I was calculating the triangle BDE to get the area of the part of the circle that is determined by DEA. Then I subtract this area 4 times from the area of the circle.
People in the same age range of 10th grade (year 11) did the test. It is pretty doable looking back on it, definitely not the same in test conditions and no source of guidance
Before I even hear the answer I took a stab at it, wrote out equations for each circle, solved for y. Used two of the equations, used both positive equations and set equal to find intersection. Point at -2, mid way between the two centres duh. Y is 2sqrt(3). Arc BD is 1/6 of circle. Area of circle is 16pi. Area of slice of circle A with arc BD is 8pi/3 same for slice of circle B with arc DF. Triangle BDF equilateral has area 4sqrt(3). Take the triangle away from the slice to get area of the edge to be 8pi/3 - 4sqrt(3) now I have everything I need to find the area of the shaded parts. The area of the blue shaded area enclosed by BDF is found by taking the area of the slice and subtracting the two side slivers, so 8pi/3 -2(8pi/3 - 4sqrt(3)). Just times that by two to get the area of both of them. Final answer is 16(-pi/3 +sqrt(3)). Thank you for coming to my TED talk.
Damn right on the money, quite proud I got that like right away. Only part we did differently was me being smooth brain finding the intersection of the two circles by solving lmao.
I try to use integral By int dx : 4 int[2,4] of sqrt(16-(x-4)^2) - sqrt (16 -x^2) dx By int dy : 4 (int[0,sqrt12] of 4 - sqrt (16-y^2) dy + int[sqrt12,4] of 4 - (4 - sqrt(16-y^2)) dy): Got the same result. 16✓3 - 16π/3
Unfortunately for them, don't learn integration at gcse level, shame tbh, integration and differentiation remove the need to know 90% of these very specific methods to solving problems yet no one seems to learn even the basics until a levels in the uk
Or you could do the Integral approach: Take the integral from -2 to 2 of the circle equation; y = sqrt(4^2 - x^2 ) [sqrt(4^2 - x^2 )] lower = -2, upper = 2 and subtract 2 times the integral from 2 to 4. This will result in 5.478826. Then just remember to multiply by 2 because we only looked at the upper half of the circles and you have 10.958
Thanks!
let’s go lol
11/2 = 5,5
He just made all that up, trickery.
@@whaynelongjhonsondanglesmo986 🤣🤣🤣
@@dimitararabadzhiev ?
i actually managed to figure out how to solve this during the exam because of you! i remembered a video about a lens and that got me the method
That’s actually mad first person that solved it in the exam that I know of, it is not hard after knowing that it wants you to use sectors but in the exam I would have never worked it out. The worst part is that is spent 10 minutes thinking that I could work it out so wasted time that i could have used to check my other answers which had 5 marks of silly mistakes
@@idontknowwhattonamethis75h82 you can solve it with calculus using integrals
@@acuriousmind6217 Dont do calculus at GCSE
@@acuriousmind6217 lmao you go back to your gcses and do calculus
Yeah no this question’s getting way too much credit honestly, it wasn’t meant to be easy but it definitely wasn’t impossible
Without knowing the formulas for the equilateral triangle and the circle segment, but knowing pythagoras and area of circle, you can still get there. You can half the triangle to a Pythagorean triangle with hypotenuse 4 and short side 2. And you can easily see that 6 segments fit in a circle, so area of the circle, minus 6x surface of triangle, div by 6 is the circle segment. I think that showing how to solve with minimal formula knowledge is often best (you can still show the formulas for the more advanced viewers).
Nice approach!
I solved this so much more differently.... wow. My solution was super long took up whole foolscap
This is like what I was thinking.
Yeah, I used Pythagoras to solve the triangles area.
My solution was like area of circle minus area of triangle to find the small leaf parts then use it to find the shaded area. I found triangle area using cosine rule as I know the angles as the triangles are equilateral
Another way to do it is to recognize that each area that you’re not trying to solve for (I’ll call that the negative area) is symmetrical across the axis of the intersection points (DE and FG). Based on the triangles drawn out, the angle creating that segment is 120° because it’s two equilateral triangles. You can then solve for one sector, then multiply it by 4. At that point, you just subtract this negative area from the area of the circle (16pi), and you have the positive area
That was exactly my method. Essentially 16π -4(larger segment area). Got the same result expressed differently x=8√12-16π\3
yep I'm 12 and this is exactly how i worked it out. Way easier than the video
another variant...
also using the sectors that has the blue areas minus the 4 sector segments
OS = occupied sectors
SS = the sector segnemt (with he green triangle)
SA = Sector arc area
OS = (120/360 * πR^2) = 16π/3
SS = (4^2/2*sin(60)) [short version of finding the isosceles triangle area]
SA = (60/360) * πR^2) =
Area = OS - 4(SA-SS) ->
Aea = (16π/3) - 4( (16π/6) - (sqrt(48) )
Note: the green triangle are is bh/2 =
4(sqrt(12))/2 = 2sqrt(12) = sqrt(4 *12) = sqrt(48) = 4^2/2*Sin(60),
what i cant figur out, is why this was pussling?... its litterally takes 5 minuits to solve... and is quite easy...
@@hana-ov1ju wow you're so smart 🤯
I solved it by putting the circles in a grid, while ignoring the left most circle. This gives you two circles with the formulae: x² + y² = 4² and (x - 4)² + y² = 4². Substitute one in another and you get x = 2, which is the x-cordinate of the intersection. Now use integration of the middle circle with 2 and 4 as limits (or use the other circle with 0 and 2 as limits, but this formula is easier to integrate) this comes out to be 2⅔π - 2√3. Multiply this by 8 because there are 8 of these in the middle circle and then get the surface of the circle, which is 4²π, and subtract 8 × (2⅔π - 2√3) from that and you get -16/3π + 16√3
I am a fan of mathematics and studying in class 9th
And I love this comment for some reason
I solved this question in a similar way but instead turned the circles into semicircles. Then I integrated both of the semi-circles from 0-2. I then took the area of the right most semi-circle and subtracted it from the area of the middle semi-circle. This gave me the area of the shaded part in the first quadrant, so I multiplied it by 4 to get the answer
I started by finding the area of the circular segment, then just subtracted two of those from two sectors. I also kept it in terms of r until the end for easier bookkeeping.
Same here sir 👋
Same thing here 👍🏼
Yeah this is the correct and most simple way
Same here
Same here
First time I actually solved one of your puzzles without seeing the answer first; thanks for all the videos, man, you rock.
same bro
Me too !
Also I got the answer but firstly I got a wrong answer actually I took a long way firstly I calculated the coordinate of point F and G then I calculated the slope of line BF and BG then I found the angle between both of the line segments ( tanx = m1 - m2 / 1+ m1 × m2 ) after calculating jt I use the unitary method to calculate the area which will be off BFCG I just did a mistake in taking the angle between the two line segment it came out to be tan theta equal to |√3| and I took theta as 60° but it should be 120° to get the correct answer.
If you want I can tell the complete method
It's basically two equilateral triangles, each with one circular segment added and two dropped. So we have to subtract one circular segment from each of the two equilateral triangles:
Area of equilateral triangle = r²√3/4
Area of circular sector = r²π/6
Area of circular segment = r²π/6 - r²√3/4 = r²(π/6 - √3/4)
A = 2 * (r²√3/4 - r²(π/6 - √3/4)) = 2r²(√3/4 - π/6 + √3/4) = 2r²(2√3/4 - π/6) = 2r²(√3/2 - π/6) = r²(√3 - π/3)
A = 4²(√3 - π/3) = 16(√3 - π/3) = 10.958
I feel great again. It took me more than several minutes to look at it and tried to work out the solution without watching your answer. In the end I could solve it and am so happy. Keep up the great work please. I used to love geometry as a kid and I go back to my childhood thanks to you:)
Just imagine the centre of the middle circle as (0,0). Then find all the coordinates required and then use integration to find the area under the curve
They wouldnt learn that at that age
Integration is beyond GCSE level. This was in a GCSE paper
Good idea but this is GCSE
Omg. This is a UA-cam channel and I'm not in GCSE - whatever that is. I'm here because I think math is fun. My first thought when I saw this problem was that it's an integration problem. Can you just let those of us who want to talk about a calculus approach talk about the calculus way to do it? Gah
@@thomcarls2782 yeah sure
I know a few others may have already stated this solution or a similar solution, but I want to type it out here anyway:
using the diagram at 2:17 as a reference (note: [ABCD] means the area of shape ABCD):
Connect D to E and F to G instead of what was shown in the video. This creates 4 "caps" which are congruent by symmetry which can be solved for quite easily.
To find the area of one "cap": we will subtract triangle ADE from circular sector ADBE; for the same reason the equilateral triangles can be drawn in the video, angle DAE is 120 degrees, so the [ADBE] = 1/3(circle A) = 16pi/3
next, we find [ADE], which can be done in a multitiude of ways like [triangle] = ab sin(c)/2, special right triangles, etc; so [ADE] = 4sqrt(3)
from this, the area of a "cap" (such as DBE) is 16pi/3 - 4sqrt(3)
lastly, we subtract 4 of these from one circle, so we get 16pi - 4(16pi/3 - 4sqrt(3)) = 16sqrt(3) - 16pi/3, just as the video said.
That feeling when you used a completely different approach and still got the very same result
Ooh, which method did you use?
I found the area of the region bounded by line(DE) and arc(DBE) by finding that points D, E, and a third one (I named it H) can form an equilateral triangle in circle(A). By subtracting the area of circle(A) with triangle(HDE), you will find three times the area of the region bounded by line(DE) and arc(DBE). Dividing that value by three and multiplying by four gives the area of the unshaded white regions of circle(B). Subtract the area of circle(B) with that value and you'll get the area of the blue regions.
@@mouselmao interesting. I found that, since the circles are the same and the centers are exactly 4 units apart and on a straight line, you can divide them by lines DE and FG. You will now have 4 areas inside of circle B that are each identical circle segments. For the formula you only need the corresponding angle, which you calculate with the height of the segment (r/2 = 2). Then you just substract that from the area of circle B. Tbh I needed to look up the formula for circle segments though to come to this idea
@@mouselmao I don't know what all methods exist, but I used calculus. That might sound overly complicated, but the concept is actually pretty simple. I don't know what the GCSE allows, but this method should be within the abilities of a 16 or 17 year-old if they ever took calculus.
___
The equation for circle A is y = sqrt(-x^2 + 16)
The equation for circle B is y = sqrt(-x^2 + 8x)
We don't need the equation for circle C, but if you're curious it's y = sqrt(-x^2 + 16x - 48)
These equations are taken from the general equation of a circle (x^2 + y^2 = R^2) and translating them to the right along the x-axis. "sqrt" means "square root".
We will find the area by integration. For that we need limits. By simple inspection, it's obvious that the x-coordinate of points D and E is exactly halfway between points A and B.
Point A lies at x=0.
Point B lies at x=4.
This means that D and E lie at x=2.
To find the area of the blue regions, we'll need to find the area of circle B and subtract the areas of A and C.
The area of circle B is twice the integral of sqrt(-x^2 + 8x) over [0,8], or 2 ∫ sqrt(-x^2 + 8x)dx over [0,8] = 16π. We could have also used πr^2, but I did the integral version just to show that calculus is easily applicable to this situation.
I said before that we don't actually need the equation for circle C. This is because the areas we need to subtract from B are all equal to each other, and A has a slightly prettier equation.
So to find the area of the two shaded regions, we take 16π - 8 ∫ sqrt(-x^2 + 16)dx over [2,4] and that gets us the correct answer 16sqrt(3) - 16π/3.
The only drawback to this method is that you'd need to consult an integral table, which is generally provided during calculus exams. Aside from that, I'd argue that this method is just as clean-if not cleaner-than using geometry to analyze the problem. It just doesn't look like it because UA-cam comments aren't exactly friendly to math notation.
@@LordAmerican Ooh, that's so cool! I never thought of modeling the circles as relations and finding that area by using integration. That's a really neat method
@@3iknet327 Yeah I did this method too. Basically circle B minus 4*the circle segments (I too had to look up the equation).The angle stumped me for a bit until I realised that the arc DBE was made up of 2 radia. Had to look up the sine for it, but yeah glad that I got there in the end :¬)
Managed to figure it out during the exam but I'm incredibly happy to see this question show up here. Props to Edexcel - it's a really well made and well crafted question that's not incredibly overcomplicated or wordy, just heavily problem solving. Was fun to find out in the exam.
@All Info in One ?
@All Info in One I did the exam, if that's what you mean.
@@tobysuren No but did you give it
@@elysian7817 ? not completely understanding what you mean.
@@tobysuren I’m not sure why they didn’t comprehend after you explained, but “giving an exam” is a pretty common phrase here in India which just means taking an exam.
Another useful strategy when dealing with problems like this is "if you have symmetries, use them" - I personally used symmetries to divide the thing to be solved into 4 pieces that I called X, then I defined a piece Y, which is half of the intersection of two circles bit, and X and Y together form a quarter of the central circle. That's a lot easier to draw and keep track of. Just don't forget that your final answer is 4X at the end.
Pretty symmetries usually lead to pretty solutions.
It is a symmetric question and not of calculation. The intersection area of circles is 1/3*2 leaving the shaded area 1/3 of any circle. 1/3 π 4*4 = 16π/3
Same. Way easier for me to do it that way
@@G0elAyush the answer isn’t 16pi/3 though?
explanation and visual was beyond perfect. truly appreciate the effort Presh 👏👏👏
I think you can do this in a slightly less tedious way. In the drawing at 2:40 you can draw a line from D to E. You now have a segment of the 1st circle defined by the chord DE. You can then use the formula for the area of a segment of a circle: A=r²(α*π/360-sin α/2). Triangle ABD is an equilateral triangle so angle BAD=60. Angle DAE=2*BAD=120=α. So A=r²(π/3-√3/4). You have 4 of these segments with a total area of 4r²(π/3-√3/4) which with r=4 gives 64π/3-16√3. Subtract this from the area of the middle circle (16π) to get 16π-(64π/3-16√3)=16√3-16π/3≈10.96
I'm guessing they didn't learn that formula yet. Because by just knowing that the formula exist, you really shouldn't need more than a couple of seconds to figure out the strategy.
@@MrMattie725 Maybe but Presh uses it too, at 4:04.
This was the final question on the British GCSE Mathematics rexam on May 20th. The students always complain about the final question and it usually a tough geometry one.
Yep, I remember everyone online and real life talking about it lmfao. I did my GCSEs this year and imo the last question of paper 3 was even more challenging than this. 😭😭
using integration for this one is just being lazy tbh
i got this in my gcse, one of my friends actually got it which is mad.
also, the paper (specifically this question actually) was leaked on a discord server about 3 hours before the exam was sat but only a few people saw it.
No need for the equation at the end, since the blue "triangle" is equal to green+2orange-1orange (all multiplied by 2 since you need 2 of them, so two times green+orange) . The final result is the same, just simpler at that point.
Although i thought it may have been overkill, I made a graph with a circle in coordinates (-2,0), integrated between 0 and 2, then multiplied by 8(simetries) to get the white inside area. Finally, I subtracted the circle area from that
british gcse students do not know how to integrate
as my physics teacher once said, integration is the uncivilized way - all too often there is a clever trick that lets you skip the integral. My favorite such flex is skipping most flux calculations because there is either a neat symmetry or the shape is enclosed or something else.
Many people would consider that inelegant
@@rv706 ain't us engineers. If it's quicker, we use it.
I also did an integral INT ( sqrt(16 - x²) - sqrt(16 - (x-4)²) | [0 . . 2]
IT took 15 min, but I still got it! :)
This problem is easily solved by integrating:
Let's take a circle at the center, radius 4 and center at point B. Let's find the arc of this circle in the first coordinate quarter, as y = sqrt(16 - x^2).
We will integrate this arc over dx from 2 to 4. I will not describe the course of the solution, it will turn out:
(8*pi/3 - 2*sqrt(3))
The figure shows that the circle in the center is divided by two other circles into 8 such areas, respectively, their total area is equal to:
(64*pi/3 - 16*sqrt(3))
The area of the circle in the center is:
pi*r^2 = 16*pi
It remains only to subtract the 8 areas found earlier from the area of the circle, and we will get a ready-made formula for solving this problem:
S = 16*pi- 64*pi/3 + 16*sqrt(3) = 16(sqrt(3) - pi/3)
Approximately this number is equal to 10.96.
Problem solved.
this is one of the few ones in here I could actually do! I just split the parts where both circles overlap in half by drawing a line, then found the area of one of the segments, multiplied it by 4, then subtracted that from the total area of the circle
This is the simplest method.
Nice approach!
This is what I ended up doing, for the most part, though I think I went around my elbow to reach my nose in actually determine that arc area. I'm just glad I ended up with the right answer. I was certain I'd messed something up when those root 3's hung around.
how did u solve for the area of the segments?
@@luciferbroke7875 using this formula: (½) × r^2 × [(π/180) θ - sin θ]
radius is whatever it said in the video and the angle is 120
For me, If I am willing to remember any circle formulas, I would just use area of the circle - 4 equal segments with a height of r/2
Or without knowing formulas take an area of a sector - the area of isosceles triangle, the high of which is r, and the base is a double the hight in 3 side r triangle, basically the longer version of your explanation
I used 8(16pi/6)-4(4pi/3). This effectively calculates eight arcs and then subtracts the four triangles which were double counted within those arcs. It makes the calculation just a tiny bit neater and easier to manage rather than subtracting eight triangles and then adding four of them back in.
I found the same. That makes it a little bit easier for this case.
Yeah, I did the same. Seemed a bit cleaner somehow.
Plus you avoid all those nasty irrationals and get to simplify the answer down to terms of Pi only, which the question explicitly asks for.
I’m guessing this is probably the solution in the marking guide, and therefore the solution that gets you full marks.
💜💜🇮🇳💜💜I am from India and preparing for SSC CGL examination, its one of the largest competitive examinations in the world, a total of 2.1 million candidates apply for this every year for getting employment in top government organisations. Your videos are really helpful for me to improve my maths knowledge in Geometry section. 💜💜💜💜🇮🇳💜💜💜
Imagine calculus doesn't exist..... people lived like that.
I also thought of Application of integrals, by taking middle circle's centre as origin, and we can easily solve that
It's simple. From the circle you should subtract 4 equal "circle segments" en.wikipedia.org/wiki/Circular_segment. The student does not need to know the formula for the area of a sector of a circle, it is easy to derive. The segment of the circle we are interested in is the difference between 1/3 (not 1/6 my mistake) of the circle in the radius R and an equilateral triangle with the side R.
Exactly how I did it. I’d forgotten the exact formula for the area of a sector but I thought it was pretty intuitive to take the area of one of those sectors as a third of the area of one circle.
@@thethinkinlad We made mistake, it is 1/3 of circle, so it is isosceles triangle triangel. But idea is OK
Won't bother watching the video since your comment gives the same solution I had :) So it's circle - 4 (1/3 circle - triangle) = 4 triangle + 1/3 circle = sqrt(3) + 4/3 pi? Maybe 50% chance of getting something wrong but still more fun to solve these without writing anything down.
The height of the isosceles triangle is cos(60°) and its base is 2 sin(60°).
I used almost the same, but used 8 areas ... which are 1/6 (not 1/3) of the circle minus a square angled triangle (I don't remember the formula of an equilateral by mind :))
I missed this channel all these years but I’m subbed. Please don’t disappear.
It's an interesting solution. I thought of a very different one though. We can break the blue area into 4 equal parts, due to symmetry, and find one part's area as a difference between two finite integrals.
I wish I knew calculus. . .
I thought of using an integral but I have no idea how I'd set up the bounds
@@cprox1000 The circles are equal, so they're clearly intersect at x=r/2.
@@cprox1000 construct a semicircle of radius 4 and integrate with the limits from 2 to 4. Then multiply by 8. Finally subtract that to the area of the circle.
@@MultiWilliam15 exactly what I would do
I went from having a hard time solving the easiest of questions to easily solving more complex ones by watching your videos in less than a month. Thanks a lot!
People studying for Colégio Naval and Epcar watching this: easy
However, it's a good question! Awesome solution too!
*For people who don't know what's CN and Epcar: both are military schools for students between 15 and 18 years old in Brazil.
You could just use the 4 equal segments that are 120 degrees.
It would have been way waaay faster.
S = (full circle area) - (segment) * 4
S = pi*R^2 - ( (2/3 * pi - sin(120)) * (4^2)/2 ) * 4
This was in a gcse exam, this method wasn’t taught in the course
did this one pretty quick in my head. ignoring r for a bit for better intuition, triangles have area √3/4, and wedges of 1/6 of a circle are π/6. intuition confirms both of these are close to 0.5, so far so good. then taking differences until the result matches the shapes in the diagram (wedge + two triangles on the sides - two wedges cutting into it) gives the answer up to a scale factor. putting the r^2 in gives the answer you found
you can also construct a chord that passes through the intersection of two circles, you'll then see a triangle that has an angle of 120 deg., its area is also the same as the triangle in his solution (4*sqrt(3)). On the other side of the chord is a circular segment, which has an area of 16*(pi/3)-(4*sqrt(3)), to determine this value, calculate the area of the 120 deg sector and subtract it with the triangle's area. All overlapping regions make 4 of these circular segments, thus, deducting the total area of those circular segments (4(16*(pi/3)-(4*sqrt(3))) from the area of 1 circle (16*pi) will give us the wrong answer.
Is this the wrong answer???
@@fredgoodyer4907 I got the same answer. But it was difficult for me to explain in plain text, so some (or all of it) may be misunderstood. In any case, there is not just one solution to this problem.
I also used this method and I think it is more elegant than the one presented in the video. I also think you do yourself a disservice; I think it is incredibly well explained for such an inherently diagrammatic problem, and that you did a great job 😊 It just confused me that you ended the comment “wrong answer” 😅
@@fredgoodyer4907 AH, this was months ago so i forgot that i put a joke at the end lol
@@billieticklish Haha wow that went so far over my head 😆
You can also do it with the circle function r^2 = x^2 + y^2 and integrating. Do this for circle A and circle B, integrating from 0 to 2, which gives you:
4 integral_0^2 (-sqrt(16 - (-4 + x)^2) + sqrt(16 - x^2)) dx = 4 (4 sqrt(3) - (4 π)/3)≈10.958
you can also do it with any CAD program..... 10.957652
You can also solve this problem using integrals, ((integral from 0 to 2 of sqrt(16-x^2)) - (integral from 0 to 2 of sqrt(16-(x-4)^2))) * 4
The first integral is the area of one quarter of the circle with origin (0,0), the second integral is the area of the one quarter of the right circle. So if i subtract this areas, i get a quarter of the desired region. Then just multiply by 4 and get the total area
Edit: the region of integration is from 0 to 2 because in x=2 is the intercept of both circles
great duh
I basically did it the same way since I don't have a good bag of trig tricks.
@@JaimeWarlock but this time going with basic logic is easier....
I think when you put the 'x' in the equation to find the value of x; shouldn't there be 16π = 2x + 4(4√3) and so on...?
I believe that is correct and that the accurate answer is half of what he concluded
@@charleswagner4179 No, the answer given at 6:00 is absolutely correct
He took the area of both shapes combined as x so he is correct
"Give your answer in terms of PI"....
Approximately 10.958, "and thats the answer!" CAN YOU PLEASE EXPLAIN TO ME HOW 10.958 IS IN TERMS OF PI???????
Divide π by π and then multiply by ≈ 10.958
As a secondary maths teacher, this really wasn't that bad for a level 9 question. They're in there to challenge the brightest of the bunch
Unfortunately I am not part of that bright bunch so could you please help explain to me why he didnt take the blue area as 2x instead of x?
@@ratpoisonedcupcake2827 because the task was to determine the total area (x) of both sectors
@@ratpoisonedcupcake2827 you don't have to be a part of that bright brunch to solve a problem like this. You lack experience. Keep on learning math and asking questions this way and you would join that bright brunch in no time.
@@ratpoisonedcupcake2827 he could have, and then at the end give 2x as the final answer instead. Doesn't make a difference indeed 😉
When I was 16 we had a hardass of a math teacher who would've escorted us off the school if we couldn't solve this problem. Okay maybe that's slightly exagerrated, but the point is that the problem isn't even that hard when you know a few tricks. For example, you can show, that the triangles you get are equilateral and they always have angles of 60° and sin(60°) = sqrt(3)/2. If you know that beforehand, then the problem is very easy.
I think there's an easier way. If you run a line from D to E, it gives you a segment (chord) with a radius of 4 with a height of 2. (I don't remember the formula), but just calculate the area of that segment, multiply by 4 and subtract the total area of the circle. The remainder should be the correct answer.
I did this. This is by far the most straight forward answer. Area of segment is
As = 0.5(θ-Sinθ)r^2
4 segments is 4As = 2(θ-Sinθ)r^2
Area of a circle is
πr^2
Then the area of two curvy sections is
A = πr^2 - 2(θ-Sinθ)r^2
θ is 1/3 a full rotation, since there are three circles equally spaced; θ is thus 2π/3
The formula is finally
A = πr^2 - 2(θ-Sinθ)r^2
= 16π- 32(2π/3-sqrt[3]/2)
= -16π + 16sqrt[3]
= 10.958
@@ND-im1wn
😃👍
My answer before watching the entire video:
Since the radius of the circle never changes no matter what point on the circle’s edge is measured from that point to the circle’s centre.
Therefore points BFC, BCG, BEA, BAD all form a equilateral triangle with all sides equal to the radius.
From this we can take one of these triangles like BFC and used pythagorus theorem to find the length of the median.
c^2 = a^2 + b^2
4^2 = 2^2 + b^2
Sqrt(16-4)=b
b=3.4641 (4 d.p.)
Next we’re going to draw a line from B to the top of the circle and to the bottom, a line between D and F and a straight line across the highest point of the circles, this will create a new triangle that shared an edge of BF.
We can determine the angle of B in this new triangle by using the inverse of Sine as:
Sine^-1(2/4) = 30.
By deduction angle F is determined to be 60.
Next we’re going to calculate the distance between point F and the top line:
Radius - Median line of BFC
4 - 3.4641 = 0.5359
Next We calculate half of the area of the empty spaces near point F by calculating a 30 degree portion of the circle and minus it from a rectangle made from the B to top line, the line going across the top of the circles till just above the F point and from the top line down to F and a line coming perpendicular from B to F:
(2 x 0.5359) - (((30/360) x pi x 4^2) - (2 x 3.4641 / 2) = 0.3471(4 d.p.)
Finally to calculate the Area we draw a vertical line from C to intersect with the line going across the top of the circles and a vertical line from C to intersect with the bottom of the circles line. This will form a rectangle going from B to top line to C to bottom line which we will calculate the area of and minus all the empty spaces:
2 x ((4 x 8) - ((1/2 x pi x 4^2) + (4 x 0.3471)) = 10.9577 4 d.p.)
Edit: Ayee I got it right but it’s interesting that I worked out the top and bottom diverts at DFEG between the circle rather than the intersecting area of the circle
So this is the way I solved it using middleschool knowledge, similar to the one in the video:
1. I started by calculating the green triangle because it's just an equalateral triangle
2. Then, I filled one circle with 6 equalateral triangles, forming a regular hexagon (if we think about it, the area of a regular hexagon is just 6*equalateral triangle area)
3. Now we can see that
Circle Area = Hexagon Area + 6 Orange Parts =>
Orange Part = (Circle Area - Hexagon Area)/6
4. We can find the area of an "oval" shape by adding 2 equalateral triangles with 4 orange parts
5. Using:
Circle Area - "Oval" shape Area = Shaded Area
we get the area of the shaded area.
absolute genius omg
You are the only person I’ve seen that has done the same method I did. Good job!
Same solution I used; nice.
I got EXACTLY that answer! I simply taped together 50 sheets of graph paper (5x10), taped them to the concrete floor of my garage, drew the circles on the paper (used my empty garbage can as a template), and counted the blocks. Once I did that, I used a simple ratio of the garbage can radius vs. the example. My wife yelled at me because I was supposed to be cleaning the garage. My daughter got in trouble for not turning in her science report because I used all the graph paper. I probably could have figured the math 30 years ago when I got out of school, but, as a practicing engineer, I mostly just fight with contractors about payment issues and contract disputes.
Damn.... I focused on one of the parts and got it, but then of course realised the question was to solve for two of them..... part marks?? :)
Draw lines DE and FG. You then have 4 circular segments of 120°. To find the area of the 4 triangles to subtract, it's simply two of the equilateral triangles cut in half vertically, so the same area as an equilateral triangle.
Different approach, same key concepts. Circle B minus circles segments found using equilateral triangles (except my circle segments incorporate the area of said triangles, so no need to subtract them again in the final equation)
Circle segment areas:
4(Pi*r^2*120°/360°) - 4(4*sqrt3)
=64pi/3 - 16sqrt3
Final solution:
16pi - 64pi/3 +16sqrt3
=16sqrt3 - 16pi/3 = 10.958
Just find the segment area cut by left most circle(or the right one doesn't matter), angle DAE is 120° which can be found by very basic trigonometry, then subtract that area 4 times from the area of 1 circle
(Segments DAE, EBD, FBG and GCF all are exactly the same)
PS: I am 16 and love your channel it provides very interesting problems for me to scratch my head around
I did the same way
Same
Wait are you sure you aren’t missing some pieces?
@@Panin2001oca no
how would this work? wouldnt be the orange segments there, like if i subtract the green diamonds, i still have the orange an blue area, how do i get rid of the orange then?
I did it differently. I calculated the area of the equilateral triangle and subtracted it from the area of the sector to give the area of the segment. Then it was simply 2 (area of the triangle - 2 segments + 1 segment), or 2(area of triangle - area of segment.) or 2(√12*2 - (∏*16/6 - √12/4)) = 10.9576521 square units.
I got this without needing to know area of equilaterals. Just used third sectors minus [length times width area right triangles] to find third circular segments. Subtract 4 from a circle and you have the answer.
You are too intelligent to solve this problem
That's the same thing I did during the exam
Nice
AE=R=4
S1=((AE)^2 - (AB/2)^2)*AB/2*p - ellipse
S2=pR^2=16p - circle
S3=16*p - 4*sqrt(12)*p - (1 Circle - 2 ellipses) - Answer
Feel like it’s very important to remember that in the exam a question like this is supposed to take about six minutes !
Thanks, that was what I was wondering.
The pressure comes from those seconds ticking down !
So the video does it perfectly in time 😂
I obtained the same answer by using calculus instead of geometry. A very fascinating problem. thank you.
Same here, but i found it being less efficient taking around 10 minutes.
I'll cough it up to my being rusty at finding areas under 2d curves.
I solved it by calculating the area of a circle sector with the angle of 120deg (got the angle from the equilateral triangles), then subtracted the inner isosceles triangle (sides of r, r, and a circle chord of 120 deg) from that, then multiplied the result by 4 and subtracted that from the area of the circle.
same
I first subtracted the area of a 120-degree-30-degree-30-degree triangle (0.5*2*4*sqrt(3)) from a third of an area of a circle with a radius of 4(16pi) for a value of 16pi - 0.5*2*4*sqrt(3) for the area of the part of the 120-degree sector that is not part of the area of the triangle mentioned earlier in this sentence. I then multiplied this value by 4 since there were four such areas that had to be eventually subtracted from the area of a circle with a radius of 4 to get the value of 64pi - 2*2*4*sqrt(3). After subtracting this value from the area of a circle with a radius of 4(16pi), I get 16*sqrt(3) - (16/3)pi or 10.958 approx.
I live in Calgary, and I learned how to do this stuff using similar strategy but rather direct equations for the “circular segment”. It was moved from 11th to 9th grade math the year I took math. I found it the part of the course, along with my friends. Had a great teacher, which may have played a part. However I remember nothing I learned in school so I don’t remember equations
I find it easier to consider that the piece of the circle described by A-D-E is a third of the entire left circle. Hence it has a size of r^2*pi / 3. From this area we have to substract the triangle given by A-D-E with area r^2 * sqrt(3)/4. The area of the remaining arc has then to be multiplied by 4 and substracted from the total area of the central circle. The result is also r^2 * (sqrt(3) - pi/3).
"16-years old across the country stumped"
Me being a 16 year old : "Finally! I will be able to solve something"
Question:
Me:
What does that mean? You couldn't solve it?
@@leif1075 yep i didn't knew one formula , and actually that formula is in class 11 maths in cbse but i am in 10th because i got admitted in school at 4 year instead of 3
@@I_guess_its_over dude I also study in 10th grade but if you have studied both circles chapters in 10th ncert you might be able to solve those questions.
@@kshitijsingh2412 it's start of 10th class for me , right now summer vacations going on , but my favourite is maths so i have already learnt those chapters but not practiced enough to solve this
@@I_guess_its_over what formula I'm curious?
I'd like to add my solution.
After you'd shown 3:28 it became clear to me, that since we deal with circles and radius fits 6 times around that, that one could imagine an orange section between D and F pointing upward. Because BDF is equilateral as well this section must be the same as the other orange sections shown already. If we subtract that area from the Blue, one of the Orange sections can be filled by it to adjust to an equilateral side of a triangle, I'll take OrangeBF.
So one Blue section must be BDF - OrangeBD. Multiply by 2 because of the bottom one, voila.
I like your solution better, because it is very generic. Make a formula, solve.
Whenever I encounter geometry, though, I can't help build legos, meaning stiching pieces of area and volume up out of something and it does not always work out if you don't find the right pieces to build from. But then, no more math tests for me in the forseeable future and building legos is arguably more fun. :D
Your solution is better, since it is much simpler.
Make a video on the Collatz conjecture if you haven’t already
He will make one if he can prove it.
There are many possible solution methods, particularly if calculus is allowed as a tool to aid in the solution. It is a nice problem. The method shown in the video is very nice.
my solution was to take the integral of a half circle with the and 2 to 4 the i took the area of the circle and subtracted 8 times the integral. a different method but the same idea
Calculus wasn't allowed so you'd get no marks unfortunately
@@evandrofilipe1526 They're not expected to know calculus for GCSE, and as such it wouldn't be in the mark scheme. But it would be utterly ridiculous if an entirely correct and valid solution, with all working clearly shown, were to be penalised just because the student used methods which are outside the syllabus.
If it is the case that no marks would be given, then there's something seriously wrong with the exam system and the philosophy of mathematics education.
@@davidrobins1021 yes I agree that they can teach it in whatever way they want but if your solution is correct it should be marked as such.
one thing to note with this method is that this was on a non-calculator paper, so any functions used would have to be done by hand
I actually managed to figure this out during the exam! Unfortunately I didng have enough time to fully write down the method but I reckon I got 2 or 3 Mark's for it
I've done tons of hexagons using this manner of overlaid circles, so it was the first thing that occurred to me.
Knowing it makes equilateral triangles and congruent segments in the circles, I noted that the two regions we want the total area of are each
1 Triangle - 2 Segments + 1 Segment = 1 Triangle - 1 Segment
And 1 Triangle + 1 Segment = Circle / 6 (since they're sectors defined by 60 degree angles), so (Circle / 6) - 1 Triangle = 1 Segment
Dividing one of the triangles in half gives a 3-60-90 triangle, and I remember the side length ratios for those, so calculating triangle area was easy.
Then I just plugged in the values to the 1T - 1S, multiplied by 2, and had it.
This was absolutely one of the easier ones I've seen on this channel, for me, definitely thanks to my familiarity with the layout. Fun problem!
If you draw a line halfway from A to B, you can integrate the area underneath the curve, then multiply by eight, then subtract that value from the area of one circle. Nice to see a solution that doesn't require calculus though.
Haha, did the same
Why do you calculate the formula of a sector, if you can just use the result?
If you know the area of a Hexagon you can solve the problem even faster through simplification:
AH = Area of Hexagon, AC = Area of Circle, x= shaded region, s = side length of hexagon,
1st: Draw an inscribed Hexagon with vertices on the intersections and centers of the circles, as well as the triangles that make up the hexagon.
Observation 1: The 6 arc sections are equal to the AC - AH. (6 arc = AC - AH)
Observation 2: The shaded region is 1/3rd of the AH - 2 arc sections. (x = 1/3 AH - 2 arc)
2nd: Combining these 2 observations by substitution it is clear that x = 1/3 AH - 1/3(AC -AH)
3rd: Simplify the equation to get x = 1/3 ( 2AH - AC)
The area of a hexagon is 1/2 (3root3)s^2 and the area of a circle is pi r^2 (note that r=s here)
4th: Substitute the areas in and simplify to get the equation 1/3 r^2 (3root3 - pi)
Finally: Calculate for r=4 and you get 10.958!
I did it similarly, except I calculated the excluded area based on 2 overlapping sectors minus the equilateral triangle of their intersection (times 4, of course). Nice problem that took a little thought (but thankfully no calculus!). You can also simplify things by making r=1 initially and then scaling by r^2 at the end.
i used a calculus approach for this problem, but i didn't solve the integral by hand, so it was fairly simple.
A simpler way to solve this problem might be using the following approach:
In the leftmost circle - Area of the sector enclosed by line segments AB and AD = 1/6 of the total area of the circle = 1/6 * pi * r^2 (because angle DAB is 60 degrees)
Similarly, area of the sector on the middle circle enclosed by line segments BD and BA = 1/6 * pi * r^2
if we add both and remove area of the equilateral triangle ABD then we get the area of the half of the intersection of Leftmost and the middle circles. Let's denote this area by 'X'
Therefore, the area of the coloured figure = pi* r^2 - 4X
I was so satisfied when I got to a sensible answer at the end of this.
Or, if you're feeling adventurous, simply take the equations of middle and RH circle: x^2 +y^2 =r^; (x-r)^2 + y^2 = r^2 respectively. Calculate where they intersect; x = r/2 and perform a double integral (∬) with x between zero and r/2 and y between [r^2 - (x-r)^2] and [r^2 - x^2]. After a bit of trig substitution you'll get 4(√3 - π/3); this is for one quarter of the area, of course, so the final area will be 16(√3 - π/3) QED😁. A bit above GCSE (it's first year degree maths, well it was on my degree) but fun if you bored and have a few minutes to fill.
me and a few friends managed to figure this out after the exam, but in the actual thing i couldnt. not because i had tried and failed, i just didnt have enough time. i guess i choked in some other questions under the stress and the exam was just a big time crunch for me. i was literally speedrunning questions as the examiners were saying there was a few minutes left lol
Same here lol, the fire alarm went off in the middle of our exam and I didn't even attempt it in the end 😭
I paused the video on 1:27 and the answer is pretty obvious if you know integrals. Just calculate the area between points ADB, divided by two (vertically in half) by counting a defined integral from -1 to 0 from the function, that represents the circle with the center in point A. The function of the circle with a center in zero coordinates is x^2+y^2=16, so if we change x to (x+4), the circle will move by 4 points to the left. The result function is (x+4)^2+y^2=16. Calculate the integral from -1 to 0, it will give us the area between the X axis and the function and then just take 8 of these parts from the full circle and you will get the highlighted zones. And that's it - the answer in 2 steps. This is pretty straightforward approach if you need to calculate difficult areas, which can be limited by functions.
English is not my native language, so I'm sure, that I call some things the wrong way, but I hope you got the idea.
I watched the video till the end and found much easier solution, when you even don't need any calculations with pi, except for the whole circle area. In the beginning, when you line up the points, also connect DF and EG. Now the whole circle area consists of 6 equal triangles and 6 equal circular segments on the outer sides of these 6 triangles. The area of each triangle can be obviously easily calculated and we can calculate the area of the circular segment by just taking out the area of 6 triangles from the area of the whole circle and dividing the result by 6 ((circleS-6*triangleS)/6). The highlighted blue area consists of 2 equal zones (triangle+segment-2segments) or, if to simplify 2*(triangle-segment). And that's it. Again the answer in 2 steps, but with much less calculations.
My teacher asked me only one part of the shaded area and I just considered BDE as a sector and solved as a sector area. The radius was 5 cm.
Are you Myanmar? SAME! May I know, which courses are you attending?
I did this as a mental exercise to see if I could relate the blue area to the basic shapes present.
I recognized that if I draw an inscribed hexagon on the central circle, then I can describe the blue area as:
Let A=full circle area, H=Area of Inscribed hexagon, B = Blue area to be found
Then: B = A - 2/3A - 2/3 (A-H)
I’d like to see the data on this question to see how many people got it right.
Cool i figured out an identical strategy. Special Area = Circle Area - 4*Chord Area, where the chord is from E to D. I knew there was some formula on wikipedia for chord area but i was too lazy to look it up lol, and i knew i cud find the length of E to D too since that line is part of a iscosc triangle with 2 radii legs. Im happy with figuring out the strategy of attack.
Took two minutes. I had actually solved this more than ten years ago. Nice to see I can still do it.
I approached it a little bit differently. I saw that if you connect the chords, BDF and BEG are equilateral triangles of side length 4. By symmetry all of the segments are the same (16pi/6 sector - 4 √3 equilateral triangle). The blue areas are one equilateral triangle minus two segments plus one segment, so (4√3 - (8pi/3-4√3)) = 8√3-8pi/3. The total for two of those blue areas is therefore 16√3-16pi/3.
Yup this is the way I went about it too
This was also fun to do with integration, half circles, trig sub, and double angle identity. Would reccomend seeing how far you can get without writing anything down. You definitely have the better solution than integration, though.
idk about that tbh... what I first thought of is simply taking half circles, getting the intersects between them, which would be simply setting them equal: sqrt(1-x^2) = sqrt(1-(x-1)^2 which is fairly simple so solve, or you just make the educated guess that its 1/2 (which it is if you look at it so yeah). Then you take the integral of both equations between 0 and said point (here called F) and then take it times 4. The integral is pretty hard ig, tho you have hard borders and most calculators can do it. idk if those are allowed? If they are, way easier than above solution.
@@pottedrosepetal6906 You're on the right track, but you've missed that r=4, which changes the equation for the half-circles and the intersection point.
And yeah, you could solve the integrals by looking them up or using a calculator, but I think it's more fun as a mental math challenge.
I solved a similar problem in the past. It was something along the lines of 'how close do you have to bring two similar circles so that the area of their non-overlapping regions and the area of the overlapping region are equal'
This basically what my initial thought was, but I haven't taken or done any kind of trig/calc in like 8 years so as far as I got was:
π(4)^2 - 8(integral of the arc from the point of intersection to the midpoint of the inner circle from 0 to x)
I tried to figure out what that x would be making a right triangle with a 2 unit side, but I don't remember enough trig to use that. Also, is there any way to really figure out the integral?
@@yellowlabb sure there is, but you need nasty substitution. Its not pretty.
Easy. The areas to keep track of are as follows: the circular sector enclosed by F and C, the segment created by cutting it with its chord and the area of the circle itself. Which of course is 16×pi, as we know.
The area of a sector is r^2/2×theta. Theta in this case is pi/3. The area of the sector is therefore 16×pi/6, or 8pi/3.
Now the area of the segment. That will be equal to the sector minus the big triangle formed by the 2 points on the circumference and the centre of the circle. Since it's an equilateral triangle, the usual formula devolves to (r/2)^2×sqrt(3). r being 4 here leads to the inner triangle having an area of 4sqrt(3). Subtracting this from the sector gets (8pi - 12sqrt(3))/3. This is the area of the segment.
Now to build the rest. Adding the area of the segment to that of the sector will get the area of 1/2 of one of the wedges. Multiplying that by 4 will get the area of both and subtracting that from 16pi will result in the total shaded area.
Sector + segment = 8pi/3 + (8pi - 12sqrt(3))/3 = (16pi - 12sqrt(3))/3.
4×(sector + segment) = (64pi - 48sqrt(3))/3.
Shaded area = circle - 4×(sector + segment) = 16pi - (64pi - 48sqrt(3))/3 = (48pi - 64pi + 48sqrt(3))/3 = 16sqrt(3) - 16pi/3. This is equal to 10.958 (3 s.f.)
I did my maths GCSE this year as a year 10.
It was a pretty difficult question indeed, but luckily here in Britain we have similar questions like this in our national maths challenges (Intermediate and senior maths challenges especially), so students who did well in those would have a good time doing these sorts of questions.
To my disappointment, I would have lost a single mark, as I had put the wrong denominator (6) in my answer, however had the correct answer somewhere in my working. Lesson learned for paper 2 and 3: check your answers!!
Little note for anyone reading this: SMC (senior maths challenge) questions are an underrated source of revision!!!
@@generaltoasty3990 I did the maths challenges, they are indeed a very good way to revise and help you prepare you for any challenging question such as this.
Hey, fellow Y10 here, I thought the question was quite interesting- The question actually felt quite unique from the usual past papers- Even though the final questions usually differed on past papers, I was expecting a problem solving vector question or a disguised trigonometry question. Tough to hear about that lost mark, I get that a lot- 😅🤣
How did you manage to take your Gcses a year early? For me, there was quite a long process that had to be done of tests and arrangements in our school trust, so I'm curious to hear how it worked for you
@@tricksterdim hello there fellow year 10, great to see you here.
I go to a grammar school that only has sets for maths, which is (probably) unusual. Of these sets, set 1 will do 2 sets of mock exams to test the students' capability and their potential to get a grade 9 in the GCSE. If they do not get a grade 9 in the second set of mock exams, then they will not do the GCSE early, otherwise they will, which seems to be a pretty fair and efficient way of enrolling for early GCSE exams.
I'm sure this may come as a suprise to you, but I was not aware that other schools had less linear and more extensive methods to enroll students for an early GCSE, so I thank you for making me aware of this!
Anyways, best of luck in papers 2 and 3!
@@generaltoasty3990 Wow! I didn't know about this type of method! I congratulate you for your move! What an interesting and peculiar process-
I've heard that Public Schools don't really do the year transfers, as they generally have different priorities for students to meet- because of this, my transfer was a very messy process, especially as I was the only one in the trust who had been considered for a transfer. Mine was a bit of a peculiar scenario where I was randomly asked to do some gcse practice tests a year ago after they initially considered the year transfer when they saw the hours of extra work I had been doing to see if it should happen. I got 9s on all of the tests, and previously I was working in private facilities within a public school anyway- I loathed the public classes-, so they went forth with it a few months after the tests.
Great to hear there are more bright students out there who actually are able to do these things- In public schools especially it's very hard to find truly dedicated students, so I congratulate you on that!
Good luck for the next papers!
I tried doing this elegantly... but didn't come up with your amazing solution. I threw in the towel, and just did an integral. The integral ranged from 2 to 4 (since we know both those points on the x axis) bounded on the upper part by the offset circle (middle circle) and bounded on the lower part by the origin circle (leftmost circle). This gives 1/2 the area of the top blue region. Multiply by 4 and there's the brute force answer.
I solved the problem by drawing a line from E to D. Then i calculated the area of the segment EDB. I was able to do that because i knew from geometry that the angle DAE is 120°. That means that ADE must be 30°. I then calculated the area of the triangle EAD and subtracted it from the sector. Finally i subtracted 4 segments from the area of a full circle to get the shaded area.
Me too, much easier this way
Seems easier to me to note that after constructing lines DE and FG, we have four circle segments with a theta of 120 degrees (due to the equilateral triangles). These have an area of (pi/3 - SQRT(3)/4)r^2. The area DFB+BEG is then the area of circle B, pi*r^2, minus four time the sector area, or (pi - 4*pi/3 - SQRT(3)) * r^2. Multiply out and re-factorise for pi and you have your answer.
"I thank Joe for this suggestion"
"Joe who?"
JOE MAMA
I personally thought it through in a slightly different manner. I’m not entirely sure if it’s any easier or more difficult given I didn’t crunch the numbers, just thought out the process, but here’s how I did it. I took the 60° circle segment minus the equilateral triangle that (for whatever reason) I chose to take from BDF. Keep in mind, I almost always take a backroadsy kind of path for these kinds of equations, but I took that orange segment and subtracted it once from the equilateral triangle since the other subtraction is replaced by the top curve. That gives you the area of a single blue segment, so all that’s left to do is just multiply by 2 and you’ve got your answer. It was fun to think it out first and then see a different method to show that there’s multiple ways to do almost any problem out there if not all.
I did it two ways: the first was subtracting the four 120 °segments from the whole circle. Then I came up with your method as a check. I really think this method is the most elegant! The way I wrote it down was A/2 = 2At - As where At is area of the triangle and As is the area of the sector. (You can make a nice drawing of this.) Then just plug in the formulas for At and As and simplify.
Fun puzzle, straightforward solution. Only substituting r=4 in the end rather than the beginning would have been a bit cleaner and you could use more geometrical moving around and do with less calculations
Another approach:
I calculated the area of the equilateral triangle CDE, inscribed in the circle with center B, to deduce the area of the arc DAE. The area of the circle deducted from 4 times the area of this DAE arc gives us the area sought.
Area of the triangle CDE. I draw the vertices CD, DE and EC. I draw the intersection of DE and AC in H. I draw the triangle inscribed in the semicircle BDE. I calculate the length DH = r * sin60. I calculate the base DE = DH * 2 and the height CH = r * 3/2, hence the area of the triangle CDE, by the calculation DE * CH / 2.
Area of arc DAE = (area of circle - area of triangle) / 3.
Area sought = area of the circle - 4 times the area of the arc.
Reduced formula = r^2 * (2 * sin60 - pi/3)
(traduction Google Chrome)
Can someone explain how he went from 128pi/6 to 32pi/6 and how he solved for x in the end ( so how he went from the second to last equation to the final equation where x=16sqr3-16pi/3
On the right side of the equation it's 128pi/6. We need to have the left side represented as a term over 6. So we can represent 16pi as 96pi/6 so that these terms are compatible. Subtract 96pi/6 from both sides, that's 0 = x + 16sqrt3 + 32pi/6, as shown in the video.
2 circles first instead of 3. We must find the area (total area) between them. Area of Triangle ADE and the area between line DE and circle " I) ". We find the total area. The angle A is 120 degrees so the area ( name it:A) is the 1/3 of the circle area = 1/3 16pi. Find the area of the ADE triangle (name it: B)- is easy to do. So the area "I)" is A-B. For the 2 circles is 2 x (A-B). For 3 circles is 4 x (A-B). So the answer is 16pi-4 (A-B). I came in the same result as video shown!
Bro that was ez
This answer is easy if you understand Arcs and Chords and among these this is one of the easiest.
So what we start with is actually the definition of the inscribed Hexagon, which is one of the geometric givens. But in this case having defined the hexagon we are going to ignore 2 of the vertices leaving a rectangle of sides r and r* SQRT(3). Next we are going to ignore sides r.
360/6 = 60° but DE is both a chord and an arc of 2 * 60° = 120° which is 1/3 of a circle.
So area between the Arc and the origin is pi/3 the area between the chord and center for either a side of a equilateral inscribed hexagon or triangle is the same,
1/4 SQRT(3) So pi/3 - 1/4 SQRT(3) gives the area between the Arc and the chord.
The area arced inside the chord is equal to the outside since they intersect each other at the same displacement from the horizon.
Thus total overlapping area is 4/3pi - SQRT (3) and the area we are interested in is that area subtracted from pi
So Blue Area = pi - 4/3pi + SQRT(3) = SQRT(3) - pi/3
Pi/3 = 1.047
SQRT (3) =.866025 x 2 = 1.732050
1.732-1.047 = 0.685
So now we have to rescale for r
0.685 * r^2 = 0.685 * 16 = 6.85 + 4.110 = 10.96
I did the question in the exam and honestly its like the best question they could have ever asked it doesnt require some "fancy" math you need to learn at gcse instead it requires insight. I hope edexcel makes more of these problems in their exams in the future!
Did this by calculating sector BDAE - triangle DAE (easy as DAB is equilateral), which gives the area of the segment DAE. The required area is simply the area of the centre circle minus 4 x (area of the segment DAE). I think this is simpler arithmetic and therefore a bit safer.
my calculus oriented brain jumped straight to calculus. I found out the area of one quarter of the area to be found by plotting two functions, then multiplied my result by 4. Needless to say, I actually got it right :)
I did it a different way and still got the same answer.
I noticed that BDF formed an equilateral triangle with sides R. I then noticed that the segment formed by chord DF was identical to the segements formed by chord DB and BF, Therefore 1 segment could be ignored. So the answer to the question was 2(Area of Triangle - Area of Segment).
Area of triangle = (Base x height) / 2 = (4 x 2 x Sqrt(3)) = 4 x Sqrt(3)
Area of Sector = r^2 x (angle in radians) / 2 = 16 x (Pi/3) / 2 = 16Pi/6 = 8Pi/3
Area of Segment = Area of Sector - Area of Triangle = 8Pi/3 - 4Sqrt(3)
Area of Blue areas = 2 x (Area of Triangle - Area of Segment) = 2 (4Sqrt(3) - (8Pi/3 - 4Sqrt(3)) = 2(8Sqrt(3) - 8pi/3) = 16(Sqrt(3) - Pi/3)
16(Sqrt(3) - Pi/3) is the answer.
Multiplying by 3/3 gets you
16(3Sqrt(3) - Pi) / 3, which was the answer in the video.
i did this exam, and no one in my school that i spoke to managed to do it. there were videos of teachers struggling to do it. i don't think the average 16 year old in the uk could solve this
You needn't feel bad about this. I believe most students can't do it because they are not trained to handle this sort of questions, let alone in an exam situation. It's more about the training than the talent. I showed this to a year 9 girl and she gave me a 4-part full solution. However, she is trained to do these complex multiple step problems for a couple of years. If you were trained the same way, you could also do it.
Yeah but a average 12 year old like me from India could. I think it is not hard enough for teachers to *struggle* with it
Too much learning by rote! They need to explain how to solve problems by breaking them into simpler parts. A maths teacher doesn't need to be a genius at maths, it is more important to know how to explain things to students.
I hope they were music teachers that couldn't do this....
@@geoninja8971 yeah dude
Smart solution. I was calculating the triangle BDE to get the area of the part of the circle that is determined by DEA. Then I subtract this area 4 times from the area of the circle.
It was an amazing questions and was pretty doable even for a 10th grade students like me! I did it under 14 min
Thank God you didn't say you are 2nd grade student in India and solved it under 2 minutes. 😂
Welp i have passed 10th too but couldn't do that because our Surface Area and Volume was completely omitted
@@Shreysoldier Cbse always there to omit important chapters that are required for higher classes
People in the same age range of 10th grade (year 11) did the test. It is pretty doable looking back on it, definitely not the same in test conditions and no source of guidance
It was a 3 mark question so 14 minutes is way too long in an exam.
Before I even hear the answer I took a stab at it, wrote out equations for each circle, solved for y. Used two of the equations, used both positive equations and set equal to find intersection. Point at -2, mid way between the two centres duh. Y is 2sqrt(3). Arc BD is 1/6 of circle. Area of circle is 16pi. Area of slice of circle A with arc BD is 8pi/3 same for slice of circle B with arc DF. Triangle BDF equilateral has area 4sqrt(3). Take the triangle away from the slice to get area of the edge to be 8pi/3 - 4sqrt(3) now I have everything I need to find the area of the shaded parts. The area of the blue shaded area enclosed by BDF is found by taking the area of the slice and subtracting the two side slivers, so 8pi/3 -2(8pi/3 - 4sqrt(3)). Just times that by two to get the area of both of them. Final answer is 16(-pi/3 +sqrt(3)). Thank you for coming to my TED talk.
Damn right on the money, quite proud I got that like right away. Only part we did differently was me being smooth brain finding the intersection of the two circles by solving lmao.
I try to use integral
By int dx :
4 int[2,4] of sqrt(16-(x-4)^2) - sqrt (16
-x^2) dx
By int dy :
4 (int[0,sqrt12] of 4 - sqrt (16-y^2) dy + int[sqrt12,4] of 4 - (4 - sqrt(16-y^2)) dy):
Got the same result. 16✓3 - 16π/3
Unfortunately for them, don't learn integration at gcse level, shame tbh, integration and differentiation remove the need to know 90% of these very specific methods to solving problems yet no one seems to learn even the basics until a levels in the uk
I could not have done that at age 16. I could have done it at age 18 though. At 16, I'd have used his approach.
Or you could do the Integral approach: Take the integral from -2 to 2 of the circle equation; y = sqrt(4^2 - x^2 )
[sqrt(4^2 - x^2 )] lower = -2, upper = 2 and subtract 2 times the integral from 2 to 4. This will result in 5.478826. Then just remember to multiply by 2 because we only looked at the upper half of the circles and you have 10.958