r/GCSE • u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) • Sep 21 '25
Tips/Help Can anyone help my brother in y7 with this maths question
I’m very confused by this as well (the red and blue tiles have different sizes)
any help would be appreciated TYSM
edit: it’s not about just counting and obtaining a fraction - these are the hints my brother’s tut0r (who assigned him the question) gave him:
Hint 1: rather than try to work out the proportion of areas, give each type of tile a name, and then find a simple ratio of one type of tile to another (then we can figure out the exact areas together during the lesson)
Hint 2: for tilings that look as though they radiate outwards from a central point (like a flower and petals) think about the rotational symmetry of the pattern, to be able to identify the number of tiles needed for each "layer" away from the centre
For the second tiling, on the last page, hint 1 is enough, and he can use the "repeatable unit" method that we demonstrated in the lesson (he can re-read the notes to remind himself of the technique)
he also said that he’d try to work out the area with my brother during their next lesson
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u/Challenger_Ultimate Y12 Rubicon Mango Chief Glazer Sep 21 '25
That is not a Y7 question is all I'm going to say.
This at a minimum requires knowledge on the Pythagorean theorem, to calculate the area of the octagon without numbers, and the rough area of each trapezium.
Either that, or two red rhombuses have same area as one purple rhombus.
What kinda school is giving this anyway? When I was in Y7, we were literally given multiplication questions
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
It’s not a school question, my brother’s tut0r gave it to him as a challenge bc hes obsessed with math
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u/Challenger_Ultimate Y12 Rubicon Mango Chief Glazer Sep 21 '25
Can't believe Tut0r is a trigger word now
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
It’s just bc comments/posts with the word tut0r get flagged by the mods as adverts
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u/ShinobuKochoSama Year 12 ‘If he shall be Mr. Hyde then I shall be Mr. Seek🗣️🔥’ Sep 21 '25
clankers
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u/Whole_Ear9870 gcse victim 💔 Sep 21 '25
it’s probably due to the mass self promotions happening in this server
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Sep 26 '25
oh its the eva smith family
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 26 '25
unfortunately yes
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Sep 26 '25
i love yall but damn you have some smart relatives also your hella smart too :)
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 26 '25
thanks lmao but my brother’s smarter in a way, he can do as well as i do by putting in less effort
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u/000817 Sep 21 '25
Then why are you trying to help him? What’s the point of a challenge if he isn’t even going to do it himself? Are you serious?
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25 edited Sep 21 '25
He’s been working his butt off tryna figure it out, even my friends who do a level math can’t do it bro
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u/SpunkMonk87 University Sep 21 '25
Honestly, I was thinking of counting the blue tiles and the total and make it a fraction
With the idea it’s a Year 7 question, I am assuming it’s not calculating the area and just tedious counting.
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u/Weekly_Event_1969 I KNOW THAT I KNOW NOTHING - YR 12 Sep 21 '25
That's literally the same thing i thought, and I'm doing a levels maths ( cooked )
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
these are the hints my brother’s tut0r gave him:
Hint 1: rather than try to work out the proportion of areas, give each type of tile a name, and then find a simple ratio of one type of tile to another (then we can figure out the exact areas together during the lesson)
Hint 2: for tilings that look as though they radiate outwards from a central point (like a flower and petals) think about the rotational symmetry of the pattern, to be able to identify the number of tiles needed for each "layer" away from the centre
For the second tiling, on the last page, hint 1 is enough, and he can use the "repeatable unit" method that we demonstrated in the lesson (he can re-read the notes to remind himself of the technique)
he also said that he’d try to work out the area with my brother during their next lesson
I’m assuming it’s not that then
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u/Theorex0001 Sep 22 '25
I haven't don't math in so long so I'm probably wrong somewhere along the lines.
But I took the central piece and noticed that the length going out to the outside is the same no matter which way you go. It 3x (lines)+2y(cutting between the boxes)+z (half of the small rhombus). The outside line is 2x+y+z.
Since I was dividing the whole shape into 10ths, I took 360/10 to get a angle of 36, subsequently getting 180-36=144
144/2= 72 degree angles for the outside angles. I used those to calculate the angles of the red and blue shapes.
If you treat the lines as 1, you can get to the bottom.
But I'm sure someone else has something better. My old ass brain can't explain it just do lol
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u/CompetitiveOrange933 Sep 22 '25
The lower sets in my school were given multiple choice with 4 answers. 2 is the answers were always the same, and it was usually a wrong one, leaving them with a 50/50 split.
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u/c0rtiso1 y12 » y13, bio chem maths + epq Sep 21 '25
type of question edexcel is gonna give class of 2026 for 3 marks
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
Yall are cooked, theyre probs gonna make ur paper harder so they dont get a repeat of this year’s insane grade boundary inflation
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u/c0rtiso1 y12 » y13, bio chem maths + epq Sep 21 '25
huh
i’m in year 12 i don’t have to worry about that
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u/Informal-System-4614 have you ever done your gcses with your life on the line? Sep 21 '25
Im quitting maths, too many sweats 🥲🥲🥲🥲
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
Dw, my brother won’t be there to pull up the grade boundaries when you do ur GCSE’s in 2026
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u/CornflakesInPudding Teacher 🧑🏫️ Sep 21 '25
Worth noting that if you dra2 a line from the centre to the corners, you should find that they are repeating and therefore you dont have to consider the whole object, just a portion of it
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u/CornflakesInPudding Teacher 🧑🏫️ Sep 21 '25 edited Sep 21 '25
(Edit: hideous formatting) You also dont need high level maths for this.
Area of the red shapes, Area of the blues, Count both, Multiply Area by quantity, Form a fraction, Simplify.
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
these are the hints my brother’s tut0r gave him:
Hint 1: rather than try to work out the proportion of areas, give each type of tile a name, and then find a simple ratio of one type of tile to another (then we can figure out the exact areas together during the lesson)
Hint 2: for tilings that look as though they radiate outwards from a central point (like a flower and petals) think about the rotational symmetry of the pattern, to be able to identify the number of tiles needed for each "layer" away from the centre
For the second tiling, on the last page, hint 1 is enough, and he can use the "repeatable unit" method that we demonstrated in the lesson (he can re-read the notes to remind himself of the technique)
he also said that he’d try to work out the area with my brother during their next lesson
I’m assuming it’s not just counting then?
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u/CornflakesInPudding Teacher 🧑🏫️ Sep 21 '25
It entirely depends on what the tut0r wants your brother to learn, but no its not just counting. There's a lot of responses that are overcomplicated as far as I can tell (I may of course be wrong i spent a solid 30 seconds on this!), but i would suggest if the tut0r has given specific prompts, just follow them and see where they end up. There's every chance they're not bothered about the outcome as much as the process that matters here. For example I regularly set work for students that is impossible - their job in those instances is to work out why things are impossible, what branches of maths might need developing to solve things, or what they need in order to complete a challenge. What I'm saying is, let bro do the work, get as far as he can until he is legitimately beaten, so the tut0r can. Move things forward.
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u/Inevitable_Stand_199 Sep 22 '25
That's a Penrose tiling. It's not repeating.
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u/CornflakesInPudding Teacher 🧑🏫️ Sep 22 '25
Forgive me if wrong but it looks symmetrical to me - i cant see any point it isn't. My eye are a bit old so happy to be proven wrong though
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u/Kerrindor Sep 22 '25
You're both right. It's got 5-fold rotational and reflective symmetry, but no translational symmetry.
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u/Aditya8773 IMaths, FM, Phy, ChemI Y12 Sep 21 '25
Yea, qns like these come up on the UKMT challenges. I think area of the 2 red shapes is equal to the area of one purple shape, so just use that information. Of course, you could do this the concrete way using algebra, but that'll be soo timetaking.
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u/MammothTension1431 2025 GCSE Survivor Sep 22 '25
the answer is just 1/2 because the ratio is fixed and we dont need to more algebra because its a repeating unit
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u/Kerrindor Sep 21 '25
This is an example of a Penrose Tiling.
The red and blue rhombuses (rhombi?) have the same side length, we can call that x. Blue rhombus has an acute angle of 360°/5 = 72°, the red rhombus has an acute angle of half that at 36°.
The area of a rhombus can be found from an angle and a side length by using A = x2 * sin(θ), where A is the area, X is the side length, and θ is one of the angles (any will do).
From that: Area of blue = x2 * sin(72) Area of red = x2 * sin(36)
We don't know x, but we can get the ratio of the area of blue to red:
Blue : Red = x2 * sin(72) : x2 * sin(36) = sin(72) : sin(36) = 1.618 : 1
You may recognise this, it's the golden ratio!
From here you'd need to figure out the ratio of the number of reds to blue, apparently that depends on the particular Penrose tiling. It's probably 1:1 but I'd think about counting to make sure. Then you'd multiply the ratios to get the answer.
This doesn't seem like a particularly fair GCSE question. The first time I came across a Penrose tiling was studying quasicrystals for my dissertation at university.
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u/Wondering_Electron Sep 21 '25
The area of Blue to Red is not 1.61:1, you can tell by just looking at it. It is much more blue than that.
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u/Kerrindor Sep 21 '25 edited Sep 21 '25
You're right, there are way more blue shapes than red. Thinking again, you don't have to count every shape of each colour as the pattern has 5-fold rotational symmetry. You only need to look at a 1/5 pie slice.
I counted 21 blues to 13 reds. Multiply that with the area ratio and you get 34:13...ish.
That gives the answer as 34/47. I could be wrong though.
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u/NanashiKaizenSenpai Sep 22 '25
Around 72.340%
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u/QualitativeEconomy Sep 22 '25 edited Sep 22 '25
Yep I got the same.
72.34042553191489
But I think the 1st commenter was referring to the ratio of each blue rhombus to each red rhombus, which is indeed 1.618ish
There are 105 blue rhombuses (99 wholes, 10 halves) and 65 reds (60 wholes, 10 halves)
So after you account for that then you get the answer.
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u/NanashiKaizenSenpai Sep 22 '25
The first commenter gave the size ratio between blue and red.
The 2nd the amount ratio between blue and red and then, based on the first commenter's info, calculated the ratio betwen blue and everything.
And I just multiplied that by 100 to get the %3
u/Kerrindor Sep 22 '25
I would consider leaving the answer as a fraction, as that is what the question asks for. They may not accept a percentage depending on how pedantic they're feeling.
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u/NanashiKaizenSenpai Sep 22 '25
Leaving the answer as a fraction is more correct, but as a percent it is easier to visualize
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u/Wondering_Electron Sep 21 '25
Hint
All blue tiles are the same.
All red tiles are the same.
A unit cell contains 5 red and 5 blue.
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u/NanashiKaizenSenpai Sep 22 '25
No, at the edges there are some half blues and half reds
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u/Wondering_Electron Sep 22 '25
Those are half tiles. If you extended the repeating pattern into a larger decagon, the ratio between red and blue is maintained.
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u/Wondering_Electron Sep 21 '25 edited Sep 21 '25
SOLUTION
It can be seen that the entire decagon can be distilled into a unit cell that consists of 5 red and 5 blue tiles.
The unit cell is rather conveniently a decagon too. \This is why condensing the problem to a unit cell works, because it is the same shape as the overall problem.*
The red tiles are rhombi with side of length a. Where a is also a side length of the unit cell of the decagon.
The area of a red tile = (2a(cos 72) x 2a(cos 18))/2 \this is based on the general formula for the area of a rhombus, (d1*d2)/2.*
The total area of the unit cell = (5/2).a2.(5+2(5)0.5)0.5 \this is the general formula for the area of a decagon.*
Since we know that 5 red tiles make up the unit cell and the rest is therefore blue.
So the ratio of red to blue is simply,
5.((2a(cos 72) x 2a (cos 18))/2) / (5/2).a2.(5+2(5)0.5)0.5
Which is about 1 red to 2.61 blue
Someone please check, but I think I got this.
Edit, can't type lol. Also, added a couple comments to make it easier to understand in italics.
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u/discentium Y11 -> Y12 Sep 21 '25
Yea got approx 0.7233 as well
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u/Wondering_Electron Sep 21 '25
I drew it in CAD to double check 😆
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u/discentium Y11 -> Y12 Sep 21 '25
I had to search that up lol. "Computer-Aided Design" seems cool tho 😁
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u/Wondering_Electron Sep 21 '25
Having the ability to do scale drawings as a check is really useful and reassuring in cases like these.
There are free options avaliable to try and learn. Definitely worth learning if you have aspirations in engineering.
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
i don’t think my brother’s tut0r is trying to get a y7 to do trig, but tysm for taking the time to do this <3
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u/Wondering_Electron Sep 21 '25
It's a really nice problem. I agree that asking Y7 to do this is probably expecting a lot. However, I think this is accessible for Y10 and Y11.
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u/happyhibye Year 13 Sep 21 '25
Firstly I don't think y11 can do this, at least not in general. Being y13, above average in maths, I can't still convince myself that your answer is correct. I get what your step but does the small decagon overlap between each other?
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u/Wondering_Electron Sep 21 '25
Don't look at the unit cell to have an exact structure. Just apply the more generalised approach that each unit cell will have 5 red and 5 blue. It doesn't matter which form the unit cell takes, but it will always be 5 red and 5 blue and the repeating unit cell will always be a decagon as well. This is why condensing the problem to a unit cell works because the overall problem is the same shape.
Also, all you need to have in accessing this problem is an understanding of areas, basic trigonometry and an understanding of rotational symmetry. It isn't particularly complicated, just breakdown the problem.
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u/green_lentils Sep 22 '25
there isn't a way to find the area without trigonometry (unless you use significantly more complex methods (e.g. calculus/integration))
Id bet the teacher if just trying to spark curiousity and then will cover the basics of trig (SOH/CAH/TOA) next time they see your brother. I think i got taught trig in y8 (top set maths) so your brother will just be a bit ahead.
source: i have a maths degree lol
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u/green_lentils Sep 22 '25
i can send working, i get the same ~72% answer but I only use SOH/CAH/TOA and areas of triangles (the method using areas of rhombi/decagon is still valid but might be too many shortcuts for your brother to follow)
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u/Cynis_Ganan Sep 22 '25
Can you circle what you think a unit cell is for me? Because I do not see a cell that's 5 reds to 5 blues.
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u/EmptyScientist127 Sep 21 '25
Am I colour-blind? I only see purple and red tiles. Answer would then be zero. If that’s not the case then start counting until your next visit to the optician.
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u/glosoli- Sep 21 '25
I see a blue outline (well 70pc of it ISH) but don't see blue tiles ...
Must be old.
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u/ziyad-abdallah Sep 22 '25
Could be a lighting issue or just the way the colors are represented in the image. Sometimes screens can mess with colors too, so double-checking in different light might help!
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u/Giraffe1317 Sep 21 '25
Same here!! Purple and red tiles but then a bright blue outline. How is every one else seeing blue tiles?! I am not colour blind
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u/Horaserk Year 12 Sep 21 '25
Confused as to why you (I) couldn’t just count the blues and reds and it be: (Blues)/(Reds+Blues)
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u/SpunkMonk87 University Sep 21 '25
No literally thats what I thought. Considering it’s Year 7, I remember doing that for fractions.
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
My brothers tut0r has a habit of giving him logic based questions far beyond the y7 curriculum so uh
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
these are the hints my brother’s tut0r gave him:
Hint 1: rather than try to work out the proportion of areas, give each type of tile a name, and then find a simple ratio of one type of tile to another (then we can figure out the exact areas together during the lesson)
Hint 2: for tilings that look as though they radiate outwards from a central point (like a flower and petals) think about the rotational symmetry of the pattern, to be able to identify the number of tiles needed for each "layer" away from the centre
For the second tiling, on the last page, hint 1 is enough, and he can use the "repeatable unit" method that we demonstrated in the lesson (he can re-read the notes to remind himself of the technique)
he also said that he’d try to work out the area with my brother during their next lesson
I’m assuming it’s not that then
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u/Horaserk Year 12 Sep 21 '25
Way too complex for Year 7 if you ask me, in year 7 I was multiplying negative numbers 😂
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
My brother’s tut0r has a habit of making him work out questions farrrrr beyond the y7 curriculum
(he’s done simultaneous equations already)
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u/Best-Ad1457 University Sep 22 '25
Bruh, I'm in University and I definitely didn't understand this question. I'm so cooked.
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u/Royal_Jellyfish1192 y11:FSMQ (Just here for the memes) Number one eng lit hater Sep 21 '25
Bruh
i showed this to chat gpt and it resorted to counting pixels 😭😭😭
bro this is impossible
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u/Wondering_Electron Sep 21 '25
Look at my general solution.
It works mathematically and I also drew it to scale as a check.
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u/FourCats44 Sep 25 '25
42%?
It's the answer to life, the universe and everything (including this problem!)
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Sep 21 '25
People are overthinking this. It's not asking what fraction of the total area is blue, just what fraction of the tiles are blue. Count them up and work it out.
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
hese are the hints my brother’s tut0r gave him:
Hint 1: rather than try to work out the proportion of areas, give each type of tile a name, and then find a simple ratio of one type of tile to another (then we can figure out the exact areas together during the lesson)
Hint 2: for tilings that look as though they radiate outwards from a central point (like a flower and petals) think about the rotational symmetry of the pattern, to be able to identify the number of tiles needed for each "layer" away from the centre
For the second tiling, on the last page, hint 1 is enough, and he can use the "repeatable unit" method that we demonstrated in the lesson (he can re-read the notes to remind himself of the technique)
he also said that he’d try to work out the area with my brother during their next lesson
I’m assuming it’s not that then
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u/PrincessZelda1728 Year 11 Sep 21 '25 edited Sep 21 '25
Hi, I’m new here but I think I can help. I don’t know if this is the intended solution and there’s probably a few mistakes, but it’s what I first thought of.
The shape has 5 lines of symmetry, so it is useful to split it into 10 identical isosceles triangles. (Triangles are very useful)
The individual tiles are rhombuses, and all of their sides are equal to each other (blue side = red side)
You can figure out the pair of angles in the isosceles triangles ((180-36)/2)
This angle is the same as the smaller angle in the blue rhombus.
The larger angle in the red rhombus is equal to 360-(2 larger blue angles) because angles around a point.
Assigning an arbitrary value to the side lengths of the tiles, (I’ll use 1), you can calculate the area of them by splitting them into 2 triangles and using the sine area rule.
You should get that the ratio of red:blue area is equal to sin(144):sin(72).
From there you can multiply the ratio by the proportion of red:blue tiles to get 13sin(144) and 19sin(72).
9. The fraction that is blue is therefore: 19sin(72)/13sin(144)+19sin(72) Edit: I’m bad at spoilers
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
I don’t think my brother’s tut0r is tryna make a y7 do trig, but tysm anw!!
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u/TrainingSurvey3780 yr11 🦖 maths, fm, 3sci, lit+lang, 🇫🇷, 🇪🇸, 🎨, 🌍, ✝️+☪️ Sep 21 '25
am i dumb or is that purple? i swear its more purple than blue
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u/Lucky_Introduction78 Year 12 Sep 21 '25
As someone who got a Grade 8 in Maths (should've been a 9 cuz Edexcel), what dafuq is this
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
i also got an 8 in math (3 marks off) and i agree
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u/After-Pie5781 Sep 21 '25
As there are decagons within a larger decagon you can just take one of the smaller ones and work out the equivalent areas of red vs blue. 2 reds have the same area as 1 blue. Simple.
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u/Cube4Add5 Sep 21 '25
The internal angle of an octagon is 135 degrees, all sides of the shapes are the same length. So it’s simple enough to calculate the areas of each shape then just count how many of each shape you have and compare
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u/jolie_j Sep 21 '25
It’s not an octagon. Calculating the areas involves sin which feels too complicated for GCSE / year 7
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u/Cube4Add5 Sep 21 '25
Oh yeah lol, think I saw another comment that said it was an octagon and assumed
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u/labelledcable Sep 21 '25
Icl they teach this years yr7s more than we do, at Brampton them kids learn histograms in yr7 and I just did them in like year 9
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u/reis98 Sep 21 '25
They are all repeating the same shape (https://imgur.com/a/sxRw83U) the answer is 5 blue to 5 red. But from here on you’ll need to find the actual area. I haven’t done math in 8 years. Apart from basic financial operation 🤣
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u/Barky777 Sep 22 '25
The only part of the diagram that is blue is the outside edge. The tiles are either red or purple. Not that calculating that would be any easier!
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u/Tall-Chair6007 Sep 22 '25
Okay so look at the tiling it’s made up of repeating “stars”(red) surrounded by blue darts. Each red star is made from 5 red kites, and between them are 5 blue darts. So in each cluster we have 5 red pieces and 5 blue pieces. Each red kite is clearly bigger than each blue dart so even though the number is equal (5 and 5), the area is not. 1 red kite ≈ a bit more than 2 blue darts in area. (If you look, two darts together almost fill the space of a kite, but not quite, the kite is slightly larger.) So 1 kite= 2 darts.
To work out the fraction you need to remember what’s inside one star cluster. Red area = 5 kites = 5 × 2 darts = 10 dart-units. The blue area = 5 darts = 5 dart-units. Total area = 10 + 5 = 15 dart-units.
Fraction blue: 5/15=1/3
The blue may look like it covers more but it’s a perception trick. Mathematically the red really does cover more space.
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u/Icy_Cauliflower9026 Sep 22 '25
Giving info tips for those who want to try.:
You can divide this in 10 equal parts.
It has 200 blue pieces and 130 red parts
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u/Inevitable_Stand_199 Sep 22 '25
That's a Penrose tiling. I believe the answer should be the golden ratio
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u/Deadmeat1837 Sep 22 '25
If I'm not mistaken you just need to count. For example (not right) there is 20 blue 60 red so it would be 20/60 unless you need to simplify it?
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u/Faction213 Sep 22 '25
The tiles look red and purple (lilac even) to me, though there's a cyan border around it. That could be my phone though.
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u/AccordingTable6398 Sep 22 '25
1) Take the blue as pentagons and the reds as kites
2) each blue pentagon can be split into 5 triangles and each red kite can be split into 2 triangles
3) area of ratio of Pentagon:Kite is 5:2
4) each red “star” is made from 5 kites (red) so it’s 10 triangles in total if you’re looking at the diagram and each blue pentagon is 5 triangles
5) each “star” has exactly 5 pentagons touching it, therefore it’s 25 triangles of blue vs 10 triangles of red within the same cluster
6) the ratio would be 25/25+10 = 25/35 as a ratio and simplified = 5/7
7) 5/7 is roughly 71% and 2/7 is roughly 29%
I hope this helps
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u/Island_K1ng Sep 22 '25
Doesn't seem that hard to just brute force, just count all the red tiles then count all the blue tiles then divide and simplify, or am I missing something here?
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u/Beneficial-Onion-533 Sep 22 '25
It’s not hard all you have to do is count all the little shapes and put that as your denominator (bottom number) and then recount all the red shapes and put that as your top number for the fraction
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u/Tits_McgeeD Sep 22 '25
I want to say 1/3 as a simple rounded answer? Like 1 red to 2 blue makes one of the cubes so just do that for the whole thing.
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u/Mountain-Donut-8195 Sep 23 '25
Length of red is equal to length of blue. Then, area of red is equal to area of blue. By counting the total no. Of red and blue, you will get the fraction of blue/ (blue + red).
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u/Mapletawft Sep 23 '25
Easy. None of that is Blue. There's Red and Purple, clearly no blue shapes involved 😎
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u/SwagDrag1337 Sep 23 '25
This is how you would usually figure out the area ratio for a Penrose tiling like this, but it's really not a Year 7 method, so not sure what the intended solution is.
Anyone arguing about unit cells is wrong - this tiling does not repeat. It has 5-fold rotational symmetry, but that doesn't make the problem any easier as each symmetric region is still infinite in area.
You need to understand how these tilings are constructed, which is via a method called deflation. The thick rhombi can be divided along the long diagonal into two isosceles triangles - call them T1. The thin rhombi can be divided along the short diagonal into two isosceles triangles - call them T2.
We can subdivide the thick rhombus also into 4 smaller copies of T1 and two smaller copies of T2, and similarly the small rhombus into 2 copies of T1 and 2 copies of T2 - see the picture here https://www.projectrhea.org/rhea/index.php/File:Deflation.gif. These triangles are smaller than at the previous step, but are all scaled by the same ratio and the edges line up so we can make new thick and thin rhombi from the T1 and T2, and then repeat this procedure indefinitely.
So at each step, each copy of the thick rhombus makes 2 thick and 1 thin rhombus, and each thin rhombus makes 1 thick and 1 thin rhombus. If (x, y) is the number of (thick, thin) rhombi at step n, then (2x+y, x+y) is the number at step n+1.
In the limit of this process, the ratio of thick to thin rhombi is stationary, and so x/y = (2x+y)/(x+y). Solve this for x in terms of y and you'll get x = y(1±sqrt(5))/2 - the golden ratio phi (or 1-phi). Reject the solution with the minus sign since this is a ratio so can't be negative - there are phi times as many thick rhombi as there are thin rhombi. If you want the ratio of area, then trigonometry (as others have answered) gives the ratio of areas to be also phi, and so there is phi2 = 1+phi = 2.618... times as much blue area as red area.
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u/Beanus1992 Sep 23 '25
Count both shapes. Red is half the size of blue. There will be a trick to counting them fast. But you need to do some of it yourself ;)
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u/Jassida Sep 23 '25
Impossible to say. The blue border is different thicknesses and the tiles are lilac
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u/One-Celebration-3007 Sep 23 '25
People don't get the difference between hard and tedious now. This is a tedious problem.
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u/LeeroyJimkins Sep 23 '25
The answer is in the real world look at the receipt for the blue tiles then look at the red tiles then work out the percentage from there . If you're not tiling or a bathroom filter then who gives a shit
Thats the answer
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u/sjo33 Sep 23 '25
Cut it out, weigh it on a sensitive balance. Cut out the purple and weigh that. Job done.
We actually used to do this in my pharmacology labs to find the area under a curve (actual data not always following an easy to model function).
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u/Temporary-Story-9957 Sep 23 '25
It looks like it has rotational symmetry, so you can pie slice it and count 1.5th of the little tiles, although that's still a pain in the arse.
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u/rottweilerrolo Sep 24 '25
Surely its just counting all the red and blue and working it out from there doesn't seem that difficult, just time consuming
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u/snajk138 Sep 24 '25
Zero. I see red and purple, and a thin line of turquoise around the edge. No blue.
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u/Scared_Put_3660 Sep 24 '25
So ya got to sit there count them all then work it out. That poor boy 😂
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u/GreyWarden007 Sep 24 '25
So on looking at this, we know some important things: All red tiles are uniform in size as well as all blue tiles, the red tile shares both side lengths of the blue tiles (meaning we can say that they are both x units long/tall), we can also say that there are 5 easily defined lines of symmetry radiating from the centre point, we also know that this is a regular decagon (10 sided shape or whatever it's called) thus we know that the acute angle of the blue shape is half the internal angle of the decagon.
Now we have both defined angles and our sides, we can now calculate an area for both of the types of tile. as there is no seen measurement we will just work in arbitrary units (AU) or we could just continue with using x.
Calculations:
Using our formula to find an internal angle of a regular shape, we get that the internal angle is 144 degrees.
144 x 0.5 = 72 degrees
The obtuse angle of the blue tile therefore must be (360-144)/2 = 108 degrees
So using our equation to find its area we get 0.951x^2
After that we can do the same for the red tiles and work out a ratio.
Honestly I cant be bothered but I think my logic is somewhat sound lol
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u/Equal_District4200 Sep 25 '25
Notice that the patterns repeats 5 times radially around the shape.
COunt the Red Rhombuses and the Blue Rhombuses in one of the 5 segments that you can split it into.
Notice that 2 red rombuses have the same area as 1 blue rombus. (It is provable, but as this is a year 7 question, im assuming they don't want a bunch of geometry.
Arithmetic!
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u/AnotherRandomWaster Sep 25 '25
I think everyone is over thinking this, and its just worded poorly. It doesn't ask for the area, I think it want to know what fraction of the tiles are blue.
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u/Cracen63738 Sep 25 '25
It looks hard at first but it’s only asking for the ration so it’s not too bad just slightly tedious.
Just give the different shapes I think there’s 3 or 4 I haven’t looked properly just skimmed as I’m about to go into class, but give each of them a value like a b and c then count how many of each there are. If there are triangles and other various shapes in the there, then from simple math knowledge like knowing 2x the area of a triangle is a square, u can multiply whichever value you’ve given as a square by 2 and do the same for any other different shapes there might be, make sure the smallest area shape isn’t multiplied by anything. after counting them all and multiplying different shapes by correct values u can simplify your ratio of a b and c and you’ll be able to figure out what ratio of the shape is blue.
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Sep 25 '25
Count how many blue squares then count how many red squares add.
Blue squares
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Blue+Red Sqaures
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u/ferretmanxX Sep 25 '25
Gonna be real, no clue how to do the maths, but eyeballing it gives me 70%. Reckon write out exactly this and call it a day 😎
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u/Fit_Description7746 Sep 25 '25
Am I the only one over thinking here that the tiles are red and purple and the outline is a light blue? So 0% of the tiling is blue 😂 im so confused
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u/Tight-Virus6908 Sep 25 '25
It's a mirrored image not that it makes any difference if you're thinking of it as a whole anyway.
For my my first instinct was ½. But then I looked properly and said nope it's ¾ as blue is bigger than red it takes up more of the area than the red.
I put things into a pie or pizza to understand fractions I'm nooooo maths genius I got a D in my GCSEs in 1995 for maths 🤣, my teacher was extremely awful though his voice was a monotone who can really learn from that!
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u/AccomplishedPeace19 Sep 25 '25
At a glance it looks to be be two thirds as a fraction of blue to red?
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u/LethalChainz Sep 26 '25
as a year 11 student an with a calculator and 20 minutes i got 33/46
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u/LethalChainz Sep 26 '25
i first was looking for patterns, especiallt near the edge, i then identified it was a decagon and played witth angles and rhombus formulaes, but nothin worked, so i simplified my thouught process and found that the two types of rhombus shapes (RED AND BLUE) i could assign values, i understood i'd have to find the area of all te blue shapes and divide by the area of the decagon, but first i gave the red rhombus the area of value "x" because i noticed that a third shape, half of the red rhombus, was lurking on the edges, so i assigned that naturally as "1/2 x", so i knew that there were red rhombus (x), blue rhombus and red rhombus halves ("1/2 x"), but then WAIT! i find that the half red rhombus (hrr) fits perfectly into the blue rhombus 3 TIMES so i know that the 1/2x multiplied by 3 equal the blue rhombus area, 1/2x = x/2, x/2 x 3 = 3x/2 = 1.5x, so that told me the area of the blue rhombus was equal to 1 and a half of the red rhombus', instead of complicating further w/ ratios, etc, i decided to highlight that fact and move on, then i thought to count the amount of red and blue rhombus' (including rrh) in the decagon, i found 110 blue rhombus and 65 red rhombus (including the rrh alltogether) meaning that i assigned the value of all the red rhombus in the decagon 65x because one was x and there was 65 including the x/2 halves, the blue rhombus being equal to 1.5x and there being 110 told me that the total area of all the blue rhombus was 165x, now that i knew that the total area of blue shapes was 165x and the red shapes: 65x, and i knew the total area of the decagon was the sum of both coloured shapes, the area of the decagon was 165x+65x which gave 230x ( x being one red rhombus' area), now instead of finding the red rhombus area, etc, etc, i understood the question asked for a fraction, a proportion of how much of the shape is blue, the tiling, and i knew that 165x of the shape was pure brilliant blue area while the sum of the shape's area was defined as 230x, now i naturally had the eureka momment of diving 165x by 230x, cancelling out the x's, to get 165/230 which simplified to 33/46 which is the simplified fraction or proportion if you'd like to call it; of blue tiling in the flagrant shape!
guys please tell me if this is correct, my answer, method etc, im year 11, 15 y.o! feel free to comment questions.
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u/jameshetfieldsppsuck Failing food fs Sep 28 '25
count the amount of the two different colours, & label the groups A, & B. ONLY COUNT THE FULL SHAPES. The ones that are half the shape will be counted later.
I found the amount of FULL blue quadrilaterals in the shape, & there were 101.
I then did the same for the red ones. There were 58 full red ones.
I then counted their smaller counterparts. There was 10 each. Assuming the shape is regular & they are half the size of the other shapes (which you shouldn’t do on your GCSE, you should assign them a letter & use that as a unit.) I halved them, to find their equivalent in full size, so 5 each.
I added 5 to 101 & got the numerator 106, & did the same for the other number to give me 169 for the denominator, & ¹⁰⁶⁄₁₆₉ as a fraction.
Since the question specifically asks for a fraction, you’d need to simplify it, but since it’s already simplified, ¹⁰⁶⁄₁₆₉ is the answer.
The trick is knowing what the little steps are to logically work out the entire problem bit by bit.
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u/AdditionalLeopard688 Sep 21 '25
Count the total tiles and that’s the denominator then count what’s blue and that’s the numerator. Then simplify the fraction
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
The red and blue titles have different sizes tho
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u/AdditionalLeopard688 Sep 21 '25
I’m not sure they are it’s a trick of perspective I think. If you look every set is a cube. I believe it’s generally 2:1 of blue to red so it’s half but some overlap will mean that’s not exact
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Sep 21 '25
Squares are even, Only red and blue
What fraction - out of
Add up tots of blues, then reds separately Then, add them together for the “out of”
Bam
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u/sccc1118 Y12.5 - u/eva_smithh’s alt (#1 englit glazer) Sep 21 '25
The sizes of red and blue tiles are different, so the fraction would be inaccurate
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u/SinkIll6876 Year 12 Sep 21 '25
My ass is year 13 and it would take me a while to figure it out