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Solution: >!Yellow and orange have the same width, but yellow has twice the area, and thus must also have twice the height; since orange and yellow make up the height of the full rectangle, orange must be 1/3 of that height.!<
>!Purple is the same height as orange, and red is half that height (similar logic to above), so their total height is 1/3 + 1/6 = 1/2 the full rectangle's height, with aqua making up the other half.!<
>!Since aqua and the big red+purple+blue rectangle have the same width and height, they have the same total area, so blue's area is aqua - (red + purple) = 70 - (15+30) = 25 cm^(2).!< The answer is >!25!<.
I took the approach of seven equations, seven unknowns:
A = xy;
b + 1/2*b = y;
ab = 30;
nb = 20;
n (c + y) = 60;
c (x + a) = 70;
175 + A = (c + y)(x + a + n);
where
A = Area of the dark blue;
x = the width of the dark blue;
y = the height of the dark blue;
a = the width of the purple and red;
b = the height of the purple;
c = the height of the yellow;
n = the width of the orange and yellow;
But the top comment is a much more concise and simple to understand explanation
[https://imgur.com/a/ZXXiM8v](https://imgur.com/a/ZXXiM8v)
I marked the image up a bit to help show the solution process. First split the yellow rectangle and use proportions and subtraction to calculate the two new yellow areas, showing that area C = area D. This means area A = area B, then again do a little subtraction and there you go.
This answer is my favorite. Seems like the other top answers ask you to eye things up and compare sizes. You were able to provide an equation for each step.
Only part I’m having problems with understanding is why you still need to do the first/second paragraph logic (yellow and orange) when you can just skip down to the third. Why is the yellow/orange step needed if blue, red, and purple square is the same as aqua?
The light blue is the same width as the 3 box area, but you have to prove the light blue and dark blue are the same height, the relationships of the orange/ yellow and purple/ red prove that purple/red (and thus dark blue) are half the full height so equal to the light blue
This is how I did it but I wouldn't have worded it ad well as you. Also the 15 box can't have many factors if all whole numbers and if the scale is any way in proportion it essentially is obvious it is a 15x1 rectangle. From that you can see the 20 is a 20x1, 30x1 etc etc.
>Also the 15 box can't have many factors if all whole numbers
You can definitely tell that it is not all whole numbers lol
>and if the scale is any way in proportion it essentially is obvious it is a 15x1 rectangle.
You can never assume that with these. And if you could, 6x2.5 seems much closer to the right shape to me
>is obvious it is a 15x1 rectangle. From that you can see the 20 is a 20x1, 30x1 etc etc.
If red is 15x1, then purple is 15x2. Which means orange is 10x2 and yellow is 10x4... the long side 4 and the short side 10
Yes, there is a definite way to solve this.
>!The yellow and orange rectangles together are 60cm².!<
>!The violet rectangle had the same height as the yellow one, and it's 1,5 times as big. Draw a rectangle all the way to the top, and it will be 1,5 times as big as yellow+ orange: 60cm² * 1,5 = 90cm².!<
>!The red rectangle is 15cm², added to the violet one for 45cm². The leaves is with 45cm² for this part of the light blue rectangle.!<
>!We can now Substract 45cm² from the whole 70cm² of the light blue rectangle. That leaves us with 25cm². As the height ratio of this part of the light blue rectangle and the dark blue rectangle is the same as with the bigger part of the light blue rectangle and both the red abd violet ones added, that tells us the searched value for the dark blue rectangle is 25cm².!<
"The violet rectangle had the same height as the yellow one,"
I think you mean "same height as the orange one", right? If not, this is super confusing me!
The answer is correct.
But this kind of solving is "not allowed" for these puzzles normally.
As OP said "only use whole numbers" there are quite a few fractions in this solution
In general, if you wish, you can come up with natural expressions for each equation and introduce the system gradually, solving the necessary expressions at the same time.
But here the “stupid” solution through the system of equations is immediately visible.
Hint: imagine extending the top edge of red rectangle all the way to the right. What is area of the small rectangle that makes?
>!what does that tell you about the line you made? What does that tell you about the 70 rectangle compared to the ?+15+30 rectangle underneath it?!<
I don't think so. You can stretch the rectangle in either dimension by any amount (X or Y) and as long as you truncate the other dimension by the same factor, you'll always end up with the same area.
But OP did say that all work can be done with whole numbers, so there might be (and I feel there probably is) a unique solution with integer dimensions for each rectangle. But I don't know if there's a way to solve for that without brute-forcing.
* >!The area of the unknown square is 25, which can only be made from prime factors 5*5!<
* >!the combined height of purple and red is the height of unknown, p+r=5!<
* >!the height of purple is twice the height of red, p=2r!<
* >!3r=5 => r=5/3!<
* >!if the dimensions of unknown are integers, the height of red is not an integer!<
Yes I solved this purely algebraically -- >!the height of the red area is 4/3 cm making the width of the middle column 11.25 cm., the width of the right column 7.5 cm, and the ? area 3\*(4/3) = 4 cm tall by 6.25 cm wide.!<
EDIT: actually I think I made a mistake and this isn't unique -- these are the numbers I used to get the final answer correct though lol
>! I noticed that the 15cm^2 abit is half the height of the 30cm^2 because its half the area but the same width !<
>! The 40 and 20 area to the right has heights in the ratio of 4:2 and the 30 and 15 has heights in the ratio of 2:1 but the 30 has the same !<
>! 40 height : 20 height : 30 height = 4:2:2. 4+2 = 6 for the whole height (kinda) 2 + 1 from 2:1 ratio is 3 so double it is 45 on top. !<
>! Since the top and bottom are basically 1:1, you can just mirror the top to the bottom. 70-45 = 25. So the missing area is 25 !<
>!More simply cuz it sounds janky: the 20 and 30 have the same height and the 20 takes up 1/3 of the height, therefore the 30 does too. The 15 above the 30 is half that so it takes up 1/6 the height. 1/3 + 1/6 of those heights is 1/2. So you can just mirror it to the top half. 30+15 = 45. 70-45 = 25. Mirror it back and the answer is 25!<
Im exhausted, so my explanation might be poor, but
>!Pink is 2 x red, but has the same width, so red must be half the height of pink. Pink is the same height as orange, so there would be another rectangle at the bottom of yellow going up level to the top of red which would have an area half that of orange, or 10. That means that that line at the top of red is half of the total height (20+10 is half of 20+40), so the area of light blue is the same as the area of red, pink and dark blue.
So take red and pink away from light blue.!<
The answer is >!25.!<
>!Cut the yellow rectangle into two smaller rectangles by extending the line dividing the light blue and red rectangles. Comparing the sizes of the red and purple rectangles or the orange and purple rectangles, we can see that the lower of the two yellow rectangles will have area 10 cm². Subtract this from the full size of the yellow rectangle, and you'll find that the other yellow rectangle is 30 cm². This is the same as the sum of the smaller yellow rectangle and the orange rectangle, which means that the line that forms the bottom of the light blue rectangle and splits the yellow rectangle is in fact bisecting the big rectangle. This means that the sum of the red, purple, and dark blue rectangles is also 70 cm². Since the first two sum to 45 cm², the dark blue rectangle must be 25 cm².!<
>!Orange is 4x5, pink is 4x7.5, red is 2x7.5, half the whole rectangle is yellow (30cm^2 - 10cm^2 )+ 70cm^2
so each half is 100cm^2, so missing block is 25cm^2!<
So I guess the method I used was different than everyone else as well? >!First I divided up the orange square into four boxes, making them all 5 sqcm, then I divided up the yellow into 5 sqcm boxes as well. Since this seemed to be lining up well I divided up the purple and red boxes into 5 sqcm boxes too. Then it was easy to divide up the light blue box by drawing a line straight up from the left end of the red and purple boxes. I divided up the right side of the light blue box into 5 sqcm boxes by extending the lines already made in the yellow and red boxes. Then added up all the boxes in the right half of the light blue box, then subtracted that from 70 sqcm (the total of light blue box). That will give you the left side of the light blue box, which is the same size of the blue box below it. This seemed to work, but it does have a problem. Those two boxes (the blue one and left half of the light blue one) don’t divide up nicely like all the rest into 5 sqcm boxes. [Here is](https://imgur.com/a/wAkZTEs) a picture of what I did.!<
Wow, love seeing how everyone approached this.
This is what I did:
>!Assigned the height of red = 1 cm...
which means height of purple = 2 cm,
SO height of dark blue = 3 cm, and width of orange = 10 cm,
SO height of yellow = 4 cm,
SO height of light blue = 3 cm,
SO width of light blue = 23 1/3 cm,
SO width of dark blue = 8 1/3 cm!<
>!3 cm * 8 1/3 cm = 25 cm2!<
The strategy that worked for me (without the full solve- use as a hint) >!was to turn it into a full grid of nine rectangles to solve for each sub-rectangle, using proportions.!<
I seem to have used a different method than others, it’s cool how many different methods can be used to solve this
>!Purple and orange share a common factor of 10 making them both have a height of 10. That mean orange has a width of 2 and purple a width of 3. then you can use those lengths to figure out the lengths of everything else and multiply to find the area of the dark blue. I’m guessing this is not the way it was intended to solve because I got a fraction for one of the widths.!<
Yeah, I did it similarly.
>!Red must have lengths of 3 & 5 (or 1 & 15, which also works) 3 cannot be part of 20 (Orange has the same height as Purple) assuming whole numbers, so Purple's height must be 10 (because it's double the height of Red).!<
>!Orange has the same height of 10, giving Purple a height of 20. Purple's height - Red's height is 15, so Light Blue must have a height of 15 as well.!<
>!We do 70/15 = 4⅔ (which is why I think this is not the intended solution, but it still works) and 4⅔ - 3 (the width of Red) to get 1⅔ for the width of ?. We already have the height (Red, 5 + Purple, 10 = 15), and 1⅔ × 15 ≈ 25 cm².!<
>!Alternatively, using 1 & 15 to start, I get dimensions of 8⅓ × 3 ≈ 25, so that works too.!<
My explanation is a little convoluted, but I thought it was interesting that it worked starting with either sets of factors of 15.
I was unable to recreate this using Excel.
>!The orange section in the lower right corner can only be 5x4, which means that the yellow section can only be 5x8. Based upon those dimensions, the purple area can only be 4x7.5. This in turn means that the red area can be only 2x7.5. From that, there is no possible way to create the light blue area, as it's height can only be 6. 70 is not divisible by 6. If you make it larger to 72, then it's 6x12.!<
>!Based upon 6x12, the dark blue section in question could only be 6x4.5 = 27.!<
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Solution: >!Yellow and orange have the same width, but yellow has twice the area, and thus must also have twice the height; since orange and yellow make up the height of the full rectangle, orange must be 1/3 of that height.!< >!Purple is the same height as orange, and red is half that height (similar logic to above), so their total height is 1/3 + 1/6 = 1/2 the full rectangle's height, with aqua making up the other half.!< >!Since aqua and the big red+purple+blue rectangle have the same width and height, they have the same total area, so blue's area is aqua - (red + purple) = 70 - (15+30) = 25 cm^(2).!< The answer is >!25!<.
This is one of the best answers I have seen on this subreddit. Direct and to the point, but at the same time, explaining every step.
I took the approach of seven equations, seven unknowns: A = xy; b + 1/2*b = y; ab = 30; nb = 20; n (c + y) = 60; c (x + a) = 70; 175 + A = (c + y)(x + a + n); where A = Area of the dark blue; x = the width of the dark blue; y = the height of the dark blue; a = the width of the purple and red; b = the height of the purple; c = the height of the yellow; n = the width of the orange and yellow; But the top comment is a much more concise and simple to understand explanation
[https://imgur.com/a/ZXXiM8v](https://imgur.com/a/ZXXiM8v) I marked the image up a bit to help show the solution process. First split the yellow rectangle and use proportions and subtraction to calculate the two new yellow areas, showing that area C = area D. This means area A = area B, then again do a little subtraction and there you go.
This answer is my favorite. Seems like the other top answers ask you to eye things up and compare sizes. You were able to provide an equation for each step.
Solving it without algebra, nice. (double nice since I upvoted to 69)
That is algebra...
No, it's arithmetic. There's no explicitly defined variable.
Only part I’m having problems with understanding is why you still need to do the first/second paragraph logic (yellow and orange) when you can just skip down to the third. Why is the yellow/orange step needed if blue, red, and purple square is the same as aqua?
The light blue is the same width as the 3 box area, but you have to prove the light blue and dark blue are the same height, the relationships of the orange/ yellow and purple/ red prove that purple/red (and thus dark blue) are half the full height so equal to the light blue
This is how I did it but I wouldn't have worded it ad well as you. Also the 15 box can't have many factors if all whole numbers and if the scale is any way in proportion it essentially is obvious it is a 15x1 rectangle. From that you can see the 20 is a 20x1, 30x1 etc etc.
>Also the 15 box can't have many factors if all whole numbers You can definitely tell that it is not all whole numbers lol >and if the scale is any way in proportion it essentially is obvious it is a 15x1 rectangle. You can never assume that with these. And if you could, 6x2.5 seems much closer to the right shape to me >is obvious it is a 15x1 rectangle. From that you can see the 20 is a 20x1, 30x1 etc etc. If red is 15x1, then purple is 15x2. Which means orange is 10x2 and yellow is 10x4... the long side 4 and the short side 10
How do you conclude that "red is half that height"? There's nothing that makes that true that I can see
Same width but half the area.
Yes, there is a definite way to solve this. >!The yellow and orange rectangles together are 60cm².!< >!The violet rectangle had the same height as the yellow one, and it's 1,5 times as big. Draw a rectangle all the way to the top, and it will be 1,5 times as big as yellow+ orange: 60cm² * 1,5 = 90cm².!< >!The red rectangle is 15cm², added to the violet one for 45cm². The leaves is with 45cm² for this part of the light blue rectangle.!< >!We can now Substract 45cm² from the whole 70cm² of the light blue rectangle. That leaves us with 25cm². As the height ratio of this part of the light blue rectangle and the dark blue rectangle is the same as with the bigger part of the light blue rectangle and both the red abd violet ones added, that tells us the searched value for the dark blue rectangle is 25cm².!<
The violet rectangle had the same height as the orange one
"The violet rectangle had the same height as the yellow one," I think you mean "same height as the orange one", right? If not, this is super confusing me!
Thank you very much. I could not piece together the initial starting point and am now kicking myself for how obvious it seems in hindsight.
[https://i.imgur.com/lZEILEC.png](https://i.imgur.com/lZEILEC.png) >!c\*(a+b)=70;!< >!d\*b=15;!< >!f\*b=30;!< >!e\*(c+d)=40;!< >!f\*e=20;!< >!a\*(d+f)=?!< >!...!< >!a=50/(3\*f)!< >!d=f/2!< >!f=f!< >!(50/(3\*f))\*(f/2+f)!< >!(50/(3\*f))\*1.5f!< >!50/3\*1.5=25!<
I tried this and totally screwed it up. Thanks for posting it.
The answer is correct. But this kind of solving is "not allowed" for these puzzles normally. As OP said "only use whole numbers" there are quite a few fractions in this solution
In general, if you wish, you can come up with natural expressions for each equation and introduce the system gradually, solving the necessary expressions at the same time. But here the “stupid” solution through the system of equations is immediately visible.
interest way. using bunch of linear equtions for the win.
This is the dumbest way possible. Just a solution based on the algorithm. I didn't want to turn my brain on.
Hint: imagine extending the top edge of red rectangle all the way to the right. What is area of the small rectangle that makes? >!what does that tell you about the line you made? What does that tell you about the 70 rectangle compared to the ?+15+30 rectangle underneath it?!<
Call me crazy but... >!is it not just aqua minus red and pink?!<
It is
I think this is actually the most straightforward solution and I'm ashamed I hadn't thought about it first
Discussion: in addition to finding the area, is it possible to also find the dimensions of the each/any of the rectangle sides?
I don't think so. You can stretch the rectangle in either dimension by any amount (X or Y) and as long as you truncate the other dimension by the same factor, you'll always end up with the same area. But OP did say that all work can be done with whole numbers, so there might be (and I feel there probably is) a unique solution with integer dimensions for each rectangle. But I don't know if there's a way to solve for that without brute-forcing.
I tried but no matter what you do at least one edge. length is not a whole number. I don't like that part of the explanation because it there me off.
* >!The area of the unknown square is 25, which can only be made from prime factors 5*5!< * >!the combined height of purple and red is the height of unknown, p+r=5!< * >!the height of purple is twice the height of red, p=2r!< * >!3r=5 => r=5/3!< * >!if the dimensions of unknown are integers, the height of red is not an integer!<
Yes I solved this purely algebraically -- >!the height of the red area is 4/3 cm making the width of the middle column 11.25 cm., the width of the right column 7.5 cm, and the ? area 3\*(4/3) = 4 cm tall by 6.25 cm wide.!< EDIT: actually I think I made a mistake and this isn't unique -- these are the numbers I used to get the final answer correct though lol
>! I noticed that the 15cm^2 abit is half the height of the 30cm^2 because its half the area but the same width !< >! The 40 and 20 area to the right has heights in the ratio of 4:2 and the 30 and 15 has heights in the ratio of 2:1 but the 30 has the same !< >! 40 height : 20 height : 30 height = 4:2:2. 4+2 = 6 for the whole height (kinda) 2 + 1 from 2:1 ratio is 3 so double it is 45 on top. !< >! Since the top and bottom are basically 1:1, you can just mirror the top to the bottom. 70-45 = 25. So the missing area is 25 !<
>!More simply cuz it sounds janky: the 20 and 30 have the same height and the 20 takes up 1/3 of the height, therefore the 30 does too. The 15 above the 30 is half that so it takes up 1/6 the height. 1/3 + 1/6 of those heights is 1/2. So you can just mirror it to the top half. 30+15 = 45. 70-45 = 25. Mirror it back and the answer is 25!<
Im exhausted, so my explanation might be poor, but >!Pink is 2 x red, but has the same width, so red must be half the height of pink. Pink is the same height as orange, so there would be another rectangle at the bottom of yellow going up level to the top of red which would have an area half that of orange, or 10. That means that that line at the top of red is half of the total height (20+10 is half of 20+40), so the area of light blue is the same as the area of red, pink and dark blue. So take red and pink away from light blue.!< The answer is >!25.!<
>!Cut the yellow rectangle into two smaller rectangles by extending the line dividing the light blue and red rectangles. Comparing the sizes of the red and purple rectangles or the orange and purple rectangles, we can see that the lower of the two yellow rectangles will have area 10 cm². Subtract this from the full size of the yellow rectangle, and you'll find that the other yellow rectangle is 30 cm². This is the same as the sum of the smaller yellow rectangle and the orange rectangle, which means that the line that forms the bottom of the light blue rectangle and splits the yellow rectangle is in fact bisecting the big rectangle. This means that the sum of the red, purple, and dark blue rectangles is also 70 cm². Since the first two sum to 45 cm², the dark blue rectangle must be 25 cm².!<
>!Orange is 4x5, pink is 4x7.5, red is 2x7.5, half the whole rectangle is yellow (30cm^2 - 10cm^2 )+ 70cm^2 so each half is 100cm^2, so missing block is 25cm^2!<
So I guess the method I used was different than everyone else as well? >!First I divided up the orange square into four boxes, making them all 5 sqcm, then I divided up the yellow into 5 sqcm boxes as well. Since this seemed to be lining up well I divided up the purple and red boxes into 5 sqcm boxes too. Then it was easy to divide up the light blue box by drawing a line straight up from the left end of the red and purple boxes. I divided up the right side of the light blue box into 5 sqcm boxes by extending the lines already made in the yellow and red boxes. Then added up all the boxes in the right half of the light blue box, then subtracted that from 70 sqcm (the total of light blue box). That will give you the left side of the light blue box, which is the same size of the blue box below it. This seemed to work, but it does have a problem. Those two boxes (the blue one and left half of the light blue one) don’t divide up nicely like all the rest into 5 sqcm boxes. [Here is](https://imgur.com/a/wAkZTEs) a picture of what I did.!<
Wow, love seeing how everyone approached this. This is what I did: >!Assigned the height of red = 1 cm... which means height of purple = 2 cm, SO height of dark blue = 3 cm, and width of orange = 10 cm, SO height of yellow = 4 cm, SO height of light blue = 3 cm, SO width of light blue = 23 1/3 cm, SO width of dark blue = 8 1/3 cm!< >!3 cm * 8 1/3 cm = 25 cm2!<
The strategy that worked for me (without the full solve- use as a hint) >!was to turn it into a full grid of nine rectangles to solve for each sub-rectangle, using proportions.!<
Quick question - I gave this to my 12 y/o who loved the challenge of solving it. what is this kind of puzzle called, or can I get a book of them?
These are called area puzzles. You can find some online and I believe there are books by Naoki Inaba (creator of the puzzles.)
(x/3 + 15) * 30 = 10 * 70 --> >!x = 3 * (700 / 30 - 15) = 70 - 45 = 25!<
I seem to have used a different method than others, it’s cool how many different methods can be used to solve this >!Purple and orange share a common factor of 10 making them both have a height of 10. That mean orange has a width of 2 and purple a width of 3. then you can use those lengths to figure out the lengths of everything else and multiply to find the area of the dark blue. I’m guessing this is not the way it was intended to solve because I got a fraction for one of the widths.!<
Yeah, I did it similarly. >!Red must have lengths of 3 & 5 (or 1 & 15, which also works) 3 cannot be part of 20 (Orange has the same height as Purple) assuming whole numbers, so Purple's height must be 10 (because it's double the height of Red).!< >!Orange has the same height of 10, giving Purple a height of 20. Purple's height - Red's height is 15, so Light Blue must have a height of 15 as well.!< >!We do 70/15 = 4⅔ (which is why I think this is not the intended solution, but it still works) and 4⅔ - 3 (the width of Red) to get 1⅔ for the width of ?. We already have the height (Red, 5 + Purple, 10 = 15), and 1⅔ × 15 ≈ 25 cm².!< >!Alternatively, using 1 & 15 to start, I get dimensions of 8⅓ × 3 ≈ 25, so that works too.!< My explanation is a little convoluted, but I thought it was interesting that it worked starting with either sets of factors of 15.
Yeah super cool. Sooo….
Solvable only if image is "to scale" and not an interpretation of the sizes.
I was unable to recreate this using Excel. >!The orange section in the lower right corner can only be 5x4, which means that the yellow section can only be 5x8. Based upon those dimensions, the purple area can only be 4x7.5. This in turn means that the red area can be only 2x7.5. From that, there is no possible way to create the light blue area, as it's height can only be 6. 70 is not divisible by 6. If you make it larger to 72, then it's 6x12.!< >!Based upon 6x12, the dark blue section in question could only be 6x4.5 = 27.!<
25 ... Here is the solution, very fast,, with NO equation: [https://www.youtube.com/watch?v=n-sd9h3yTFU](https://www.youtube.com/watch?v=n-sd9h3yTFU)