Brilliant Blue Area Question

[@bertberw8653]

This dude's videos are amazing. Let's put a box around it.

[@stellarr_x]

How exciting :)

[@STAIRDROPPER]

Iconic point of this man "how exciting" in the end

[@khalief_.]

"speaking of brilliant, lets talk about brilliant." edit: im alive edit2: im still alive edit3: im still yet alive edit4: im still alive yet again

[@Y_Chen123]

Pretty smooth 😂

[@stuchly1]

Smoothest sponsor segue ever 😂

[@TheBoeingCompany-h9z]

"speaking of our sponser lets talk about brilliant"

[@MuhammadTheOld]

oh hey

[@roma540]

That line was brilliant

[@davidwalterhall]

There was a simpler way to do this. Once you knew purple:green was 1:3 and grey:yellow was 1:2, then the portion of red below the yellow must be 14, as that column is split into rows of relative height 1:2:1. That makes the remaining corner of red, below the blue, 7. So that left column splits 7, 14, 7, and the top two combine to make 21.

[@dsw1111]

Agreed, this is the sort of puzzle you can do in your head using ratios, so I was really surprised to skip forward and find massive algebraic equations!

[@go_gazelle]

This is what I did as well. I think algebra is typically a nice touch, but once you have the ratios, it doesn't make sense to me to flip them into an equation.

[@chrishelbling3879]

I love your delivery, graphics, and clever problems. Keep going!

[@Maazin5]

I had to do a similar problem in real life recently. An assessor provided the total area of a home but the breakdown combined the kitchen+living room. I had to do this kind of problem to find the area of the living room on its own

[@migmit]

Much better problem: if blue is marked “21” and grey is marked “?” - can you find that grey area?

[@gabrielecossettini2923]

Just commenting to put this comment on top

[@raqqafeller152]

i actually tried to solve this but couldn't do it. i found 3 equations with 3 unknowns x, y and z. they were extremely difficult for me to solve, i finally put them on wolframalpha and it actually solved it in milliseconds. i would love for andy to show the solution to this problem.

[@raqqafeller152]

i wrote a comment below this one but it somehow got deleted. i also want to see him solve this.

[@ApophaticTheosis]

Yes. Disregard 35 and 105, don't need them. You can visually compartmentalize each of the sections with squares. In the 21 box, there are 3 squares. 3 squares equals 21, 4 squares equal 28, so you can determine that each square is 7. Since the grey section can fit 2 squares, 2(7) = Area of Grey section.

[@raqqafeller152]

@@ApophaticTheosis would we be able to determine all those things after disregarding the relationship between 35 and 105 though? wouldn’t we need to know the fact that the ratio between “top” section and the “bottom” section is 1/3?

[@moonverine]

I love these. They call these Area Mazes, and there's at least 2 books of hundreds of these available.

[@horizontal]

Cute problem I broke it down into easy shapes. Since 105/35 is 3 you can turn the 105 into 3 areas of 35 with the same height and width. Similarly you can turn the 28 into two 14s. The four 35 area’s height equals the height of the three 14s + the 21 and since the 35 and 14 have the same height, the 21 must also match the height of 35 area. Then you can turn the 21 into a 14 area by giving it the same width. Making the remain piece 21-14=7. And finally split the blue piece into 3 pieces matching the height and width of the 7 area, so you can multiply 3*7 for 21. Writing it is a little convoluted but it turns it into more of a logic problem with very very simple math.

[@abuhuraira6354]

i did the same .. much easier this way and quicker

[@simonharris4873]

Same here. Much simpler IMO.

[@larrydickenson8922]

Extend all of the lines from border to border and divide all numbers by 7. The answer will jump out at you.

[@adrianscarlett]

Common divisor of known areas is 7, that allows you to accurately divide the whole region into an 8x4 grid

[@EN91editz]

Nothing would force me to open youtube but your videos

[@mekaindo]

When you realize the blue region was the same as the red region, and they actually have the same area.

[@Jorgantuous]

i love questions like these that just flow so nicely.

[@mohamadalakhras9750]

Took the sat today and a question similar to one that you've solved here came up

[@JenishTheCrafter]

So glad to see a Math dude getting a sponsorship! I like to explain Math and you are an inspiration. Haha, even bprp lol.

[@h_three_five]

Wow, I solved it a completely different way. Didn't even think of yours. How exciting.

[@1959mikel]

How exciting indeed!

[@taedenite]

Since the blue box is 1:2 we can write the whole 14+28+blue area = (a+b)²since 2x and 14 have same dimensions they have same area I.e 14 there by (a+b)²= (a²+b²+2ab) (Root 28 + b)²= 28+28 +b² Which gives b = root 7

[@taedenite]

And total area is x+2x which is 7+14=21

[@7H07sAndH03s]

Where do you find these questions ??

[@augustnmonteiro]

we need an Andy Physics now hahahahha

[@caskillet]

There an easier way. Extend line from between 14 and 28 to left edge. Realize that 14 is 2/3 of 21, so the blue area to left of 14 is 7. 28 is 2/3 of 42, so blue area to left of 28 is 14. Add 14 and 7 to get blue area = 21. Put a box around it.

[@gugubughu4670]

I like this but we don't know initially that rectangle 14 is the same height as rectangle 21 so it may not be a third. Edit: actually we do: 35 is 3rd of 105. So 14 height is 3rd of the 28 and 21 heights combined. 28 is double 14s height so yes 21 is the same height as 14 and your solution works

[@tamirerez2547]

Clear, clean, and elegant solution. 👍 Side question can be: If the area of each rectangle is in cm.sqr can you find the SIDES of each rectangle?

[@KrytenKoro]

I don't think you can -- the numbers should work out no matter what the horizontal or vertical portions of each unit block are, as long as they still multiply to 7. So if each "block" is 2 horizontal and 3.5 vertical, that would produce the same areas as if they were squares with sides of 2.65

[@creepx_hd1236]

I actually got the right answer! How exciting!

[@norothegamer1006]

you deserve all the support on all these videos

[@BLCKAGLE]

More understandable in 2 minutes than my math teacher when i was in school 😂 Keep up the great work🙌

[@jascharl]

Great solve. How exciting.

[@HeirToPendragon]

I got the blue area by just looking at it logically. All the other rectangles have areas of 7 so I started with the purple width being 5 and the height of 7. This gave me 21 for the green height, a mix of the yellow's 14 height and the red's 7 height. So Gray's width is 2, red's is 3, thus blue is a 1x21 rectangle.

[@BMW520ITURBO]

Cool solution but you can also asign any values to the purple one's sides that multiplied give 35. From that point on it solves its self

[@Snorkl7879]

Easy to solve instinctively and guess/check. They’re all divisible by 7. Green is 105=21x5. Therefore vertical purple is 35/5=7; therefore horizontal grey is 14/7=2; therefore vertical yellow is 28/2=14. vertical green&purple (28) minus vertical grey&yellow (21) is vertical brown = 7, so horizontal brown is three. Therefore the horizontal side of blue is 1, and area is 1x21=21 .

[@memestrous]

I did it in a bit more complicated way. I considered 14 to be equal to xy. Then I expressed the other rectangles as xy using their coefficients. And since many sides are shared, it can be concluded that the length of the blue area is 3y and the width is 0.5x. Which wpuld make the area 1.5xy which is equal to 1.5 of 14 which is 21

[@herbthompson8937]

the first one Ive been able to do in my head so far

[@S3C5HUNAT3]

I got it right by seeing the red 21 space and blue questionable space were the same length around yellow 28 and figured if 28 is square then its sides would be even with even expansion, leading to my correct conclusion without looking at half of the numbers on the chart...K.I.S.S.

[@Tmwyl]

Finally got one!

[@tincantank5174]

I can preach for the quality of brilliant. It is amazing. Only problem is that it costs $25 a month. Yup.

[@maciejpsyk]

It can be improved. variant 1 "Rectangle G has been divided into 6 rectangles A-F. The sides of the smallest A are natural numbers. What are the sides of rectangles B-G?" variant 2 "A square has been divided into 6 rectangles A-F. What are their sides?"

[@miguelc5251]

How exciting.

[@StevenForditude]

Very fun challenge. I solved by finding the lengths of the sides. Your solution was a cool way of doing it.

[@paparmar]

I was able to minimize the algebra by reasoning (along the lines of the video) that the height of the target rectangle is 3/4 the height of the total rectangle, and that the width of the target rectangle is 1/8 the width of the total rectangle. Hence the area of the target rectangle is 3/32 the area of the total rectangle, which means the other constituent rectangles add up to 29/32 the total area. If 203 sq. units is 29/32 of the total area, the total must be 224 sq. units. That leaves 21 sq. units for the target rectangle.

[@JenishTheCrafter]

Which software do you use for this solving of equations, removing values, etc. Is it just normal white background and editing?

[@Just_Gaming20]

Is it 7 ?

[@Mrchingchingdingding]

Got it 🎉

[@arcaltoby5772]

Thanks, also, will you do a video on Exponentially Indecomposable Principal?

[@joeschmo622]

And 21 is 1/20th of 420. How exciting... I did it by ratios. 3x+x on the right side heights, then x+2x+? down the middle, so the Mystery Height of the 21 box is also x. Ratios of left/right parts is 21:35 or 3/5 since x=x for the heights. So 3/5 of 105 is 63. 14+28+blue = 63, so blue=21.

[@OrganicAlkemyst]

I found all of the lengths and heights of the boxes and then found the area of the unknown box. I know it is harder but it is another way to prove it.

[@Mozartkugel]

This is a classic example of a menseki meiro, or area maze. A Japanese type of geometric riddle based on the attributes of simple rectangles. I love these things so it took me mere seconds to find the result: 21. 🙂👍

[@Rodrigo-jd2wg]

Now what's the perimeter of the rectangle 🤔

[@HeirToPendragon]

72

[@LebrunVentre]

Could you please make a video on how we can solve these kinds of problems?

[@MerlynMusicman]

Drawing a verticle line through the 22 block,you know the 14 block is a third of the rest of that column, so the proportion of the 21 block in the column is 14. That means the bit under x is 7, and also 1/4 of that column. X being the other 3 quarters makes x 21. I've no idea if that was what you did with the maths but it didn't look like it.

[@MerlynMusicman]

... vertical line through the 21* block ... damn phone keyboard.

[@nugrafik]

What would be interesting is solving the height and width

[@YQS.]

i was gonna say there was a much simpler way of doing it but i realized nothing is to scale so you cant really use base time height for anything

[@Mrchingchingdingding]

Interesting approach, I solved by realizing that 14 shares a side each with 28 and 35. Using 28 we can't determine if the short side of 14 is 1 or 2 since a 2x7 grey square requires a 4x7 yellow square making the blue height 6 or if the dimensions are 1x14 for grey and 2x14 for yellow making the blue height 3. 35 however only shares 7 as a common factor with grey 14 meaning purple must be 7x5 and grey 2x7. But wait! That means the illustration isn't proportional because that makes the shared side of grey and purple 7 and the shared side of grey and yellow 2, meaning blue's height is now 7+14! The process continues by realizing that the height of purple is now 21, and as the sum of yellow and red we see that red's height must be 7 and it's width 3, the sum of widths from blue and yellow. This means blue width is 1, it's length is 21 and its area is or course 21 units^2

[@cpergiel]

Not the problem, but can you determine the dimensions of any of these rectangles just from the numbers given? I don't think so.

[@Antony_V]

Let's divide the blue area in A and 2A (as in the video) and call H the point in wich green, purple, gray and yellow rectangles join. We can apply the rule for a rectangle divided in 4 parts: (A+14)*105=(2A+21+28)*35 that gives A=7, 3A=21.

[@MrJJbleeker]

Are the sides of this rectangle unique? Or are there multiple possible solutions?

[@scrollogy5847]

0:11

[@MUBEENAHAMEDKABIRRIBAYEE]

This is how I solved it though... All the numbers seemed to be divisible by 7 so the areas of these rectangles are multiples of 7. In that case, the maroon block (21) is basically 3 parts of 7 sq. units, the yellow (28) is 4 and the grey (14) is 2 parts, and so on. So you can see geometrically, one part of the 7 covers the breadth of the blue area and 3 parts along the width, meaning 1x3 = 3 parts of 7 sq. units each = 21 sq. units.

[@mr.d8747]

*Here's how I solved it: I figured that the rectangle of area 28 is actually a square, from there it has side lengths √28 = 2√7. From there, the longer side of the area 14 rectangle is also 2√7, which means it's shorter sidelength will be √7. This √7 length will also be a sidelength of the area 35 rectangle, so it's other sidelength must be 5√7 to get 35 as the area. This 5√7 length will also be the sidelength of the largest rectangle, so we need to multiply this by 3√7 to get an area of 105. The shorter side of the area 21 rectangle will be 3√7 - 2√7 = √7, meaning it's longer sidelength has to be 3√7. Finally, the shorter side of the blue rectangle is 3√7 - 2√7 = √7 and the longer sidelength will be 3√7 + √7 - √7 = 3√7. This means that it's area will be 3√7 • √7 = 21, which **_is_** the correct answer as you guys could see in the video, meaning that the yellow reactangle **_is_** indeed a square. To check if our solution is correct we can see that 8√7 • 4√7 = 224 = 28 +14 +35 +105 +21 +21.*

[@That1BeegWhale]

I found a pretty neat geometry problem. Where can i contact you?

[@alexonstott4954]

My answer was so much more complicated creating a system of equations that proved the height was 4x, because i didn't realize 105 was 3*35 🤦. I guess any solution is better than no solution.

[@redfinance3403]

Hahahaha I had two variables with a and b, where the width was 35/b and the height was b for the 35 rectangle. And for the 105 rectangle it was 105/a and a. I worked with a and b all the way through and proved a=4b, which gave me the answer. Funny how I didn't see 3*35 = 105 from the beginning when I labelled stuff 😂

[@James-l5s7k]

Good one!

[@samlee5549]

Let the part of the red under the yellow and grey part be x. 14:35=28+x:105 42=28+x x=14 Thus, the part of the red under the light blue is 7. 7:14=?:(14+28) 1/2=?/42 21=?

[@karthikeyankaz3532]

The price of two suitcases and three wooden boxes is Rs. 7,200. If the cost of two wooden boxes and three chairs is Rs. 4,950. what is the cost of one chair? The answer is "data inadequate". But can you solve this...?

[@gubarsah3710]

iconic

[@VIKASHV-ml8lm]

at 0:48 he added 35 and 105 we can avoid that ,for to make simple calculations. see 1/3=x+14/2x+49 3x+42=49+2x x=7 !!!!!

[@thecheapaudioengineer]

So i thought it like .. 35:105= 1:3 And 14:28=1:2 Then it is 1:2:1 = since brown is 21 = then the total there should have been 21*4=84 Then 84=21+14+28+x x=21

[@NeatMemesDotCom]

Am I tripping or does this one could resolved without the large section? 1/3 = x+14/2x+28+21

[@UnknownGhost97]

Actually answer should had been 28 But this method is the correct answer for 21

[@mohamadalakhras9750]

Serious question (possibly dumb) why can't we add 14 and 28 then multiply it by 21 - 14

[@joeldoonan-ketteringham5174]

It's not exactly mathematics, but given that blue and maroon are the same size just a different Orientation you can quickly tell that its 21

[@SylvieWylviee]

I solved this through the following: sqrt((grey square)/2)(sqrt(yellow square)+sqrt((grey square)/2)) Or 3(sqrt((grey square)/2))^2

[@daydreamergal]

lol I got the answer but in pretty dumb way, so they didn't mention if they were all rectangles so I just took the yellow one as square and did tons of multiplication divison addition and subtraction , figured out the lengths of the sides of all the rectangles and got a final answer of 20.9 aprx which i rounded off to 21

[@deRangutang]

69 is divisible by 3. Just saying. This could have been NICE.

[@dogandogruis]

21 ?

[@ThatMevely]

i thought of 21 instantly coz i saw all the numbers being multiples of 7

[@Dinhnguyen-km6zd]

21

[@ahuman1184]

Guys I got it right!!!

[@rothgang]

what's 9 plus 10?

[@nikhilchandel2732]

Sorry for disturbing you in this video but would you like to help me to learning English. I am not native English man. And I know here are lot of native English man who can speak English. So which tip you you give to me to learn English speaking? Your suggestion will be appreciable. Thank you for bring your kind attention on my this comment and again sorry for disturbing you.

[@acool6401]

Green / purple = 105 / 35 = 3 Extend the line under 14 and 35 to the left in order to create a smaller blue box and call it “x” (smaller sub blue region) therefore… the larger sub blue region to the left of 28 is now 2x since yellow / gray = 28 / 14 = 2 Now we are ready to create an equation: 3 (x+49) = 2x + 154 Solve 3x + 147 = 2x + 154 x = 7 therefore entire blue region = 21 Alternative equation: Once you realize that green / purple or 105 / 35 must automatically render the same ratio of 3 to 1 for sections: (red + yellow + 2x) / (x + gray) Then… 3 (x + 14) = 2x + 49 Solve 3x + 42 = 2x + 49 x = 7 therefore entire blue region = 21

[@KhaidarMusic]

First one, amazing videos

[@LimeSpeedCrystal]

ur not first

[@rudilambert1065]

I'm wondering... could we not also have approached this by determining that the blue region is the same size as the red one, who's value is given? Update: yes, this one can easily be solved in your head. Without recourse to algebra, considering all the given values are multiples of 7, the whole figure can be reptesented as a grid of squares with value 7. Each colored region being made up to of several of those squares. The top row, must therefore, be the same height as the bottom one. Now its immediately obvious that, like the red region, the blue one is made of 3 squares. It must have the same value. How exciting 😅

[@ishansoni1857]

I'm not sure, but if you look close enough, the red region is slightly wider then the blue region, so we can't assume..

[@godlyBlade]

You usually cannot assume the figure is drawn to scale in these kinds of questions.

[@rudilambert1065]

​​@@ishansoni1857 @godlyBlade the values giving all being multiples of 7 the equality can be derived from the given values

[@godlyBlade]

@@ishansoni1857 They are the same width but the blue area is slightly longer giving you the impression it is thinner. Thus, we cannot assume they are the same because they are not visually the same.

[@ishansoni1857]

@@rudilambert1065 after seeing your edited comment, it seems like you just managed to find another way to solve the question 😄

[@BCuzLates]

Stpuif

[@nabil4389]

No views Bro fell of

[@STAIRDROPPER]

Why are there these kinds of comments bru💀

[@thomasfisher3985]

What are you talkin about, he still cookin 🥵

[@mrmarno7639]

I made up this equation when I was looking at the powers of 2: n²-(n-1)²+2= (n+1)²-n², in which n is any whole number. Lets try it out with n being 69. 69²-(69-1)²+2= (69+1)²-69² 4761-68²+2=70²-4761 4761-4624+2=4900-4761 139=139