How To Solve For The Area - Viral Math Problem
ฝัง
- เผยแพร่เมื่อ 23 ส.ค. 2024
- Thanks to Reio in Romania for emailing me this fun problem! What is the area? This puzzle was shared with the tagline "you should be able to solve this."
Solution to 5th grade Chinese challenge problem
• Genius student solved ...
My blog post for this video
wp.me/p6aMk-82G
Reference
Math problem from anime Steins;Gate
math.stackexch...
Subscribe: www.youtube.co...
Send me suggestions by email (address in video). I consider all ideas though can't always reply!
Why are there comments before the video is published? Get early access and support the channel on Patreon
/ mindyourdecisions
If you buy from the links below I may receive a commission for sales. This has no effect on the price for you.
Show your support! Get a mug, a t-shirt, and more at Teespring, the official site for Mind Your Decisions merchandise:
teespring.com/...
My Books
Mind Your Decisions: Five Book Compilation
amzn.to/2pbJ4wR
A collection of 5 books:
"The Joy of Game Theory" rated 4.1/5 stars on 44 reviews
amzn.to/1uQvA20
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 3.5/5 stars on 4 reviews
amzn.to/1o3FaAg
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.4/5 stars on 13 reviews
amzn.to/1LOCI4U
"The Best Mental Math Tricks" rated 4.7/5 stars on 8 reviews
amzn.to/18maAdo
"Multiply Numbers By Drawing Lines" rated 4.3/5 stars on 6 reviews
amzn.to/XRm7M4
Mind Your Puzzles: Collection Of Volumes 1 To 3
amzn.to/2mMdrJr
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 13 reviews
amzn.to/1GhUUSH
"Math Puzzles Volume 2" rated 4.5/5 stars on 6 reviews
amzn.to/1NKbyCs
"Math Puzzles Volume 3" rated 4.1/5 stars on 7 reviews
amzn.to/1NKbGlp
Connect with me
My Blog: mindyourdecisi...
Twitter: / preshtalwalkar
Newsletter (sent only for big news, like a new book release): eepurl.com/KvS0r
2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter.
Happy anniversary! Thanks for nearly 2 million views in one year! I really thought it's obvious why this equation is true (a + b) + (c + d) = (b + c) + (a + d).
Please tell us, I can't figure out why
MindYourDecisions You could've just said "by the associative and commutative laws of addition". That would've been about as fast as saying "notice that" or "it's obvious that" while still giving viewers enough information to research their missing puzzle piece.
Luke Johnson : Ah, yes, the plight of so, so many proofs in textbooks! Just spell out all the steps that are so "obvious" or "clear" -- it doesn't even take that many more words usually!
@@HeracIeid This is the first real world application of commutative and associative laws I have encountered!
Other way to see it is because the whole square is 2a + 2b + 2c + 2d. Let's know the area of each triangle, their angles and length of their legs, just for fun.
I contemplated the thumbnail of this video for 20 minutes before finally deciding to watch it.
SAME.. I stared for like 10 min.. I think the answer is 28... (16+32=48) and (20-48=28)...
Let's see what's the answer now
I just guessed and said 28 and I was right
Given triangle ABC, a = 3, b = 4, sin A = 0.6 . Find c and draw it geometrically. It's not so easy as you think!
Eytan ADAM hey i didn't know there would be trigonometry in this!
@@samdoesstuff4924 use trigonometry and algebra, and you will be surprised!
Who don't understand why the sum of opposite areas are equal, just try to count what triangles these areas consist of.
Yellow area - a, b
Red one - c, d
Green one - b, c
Blue one - a, d
So
Yellow and Red areas together consist of a, b, c, d.
Green and blue areas together cosist of a, b, c, d too.
So, it's why they are equal
Actually, it's not the "opposite areas" which are equal, because that would mean a+b = c+d. I also at first somehow read and heard "opposite areas are equal" which is just wrong. It's really "the *sum* of the opposite areas is equal". I guess more ppl just misheard/misread it and missed that critical word "sum".
TheFeldhamster you're right, I corrected
Excellent explanation. Was wondering where he pulled this magical statement from.
nice bro...
Another way to look at it would be to consider the interior point as a locus of a kind.
28? Aw man, so close! My answer was "Argentina"
Presh Talwalker: You will really need to change your name, at least the first name. I have never seen a name that I really don't like.
I got 25
Ha! I guessed what i subtract and add and and I got 28 😝
I seriously got 28....but i did not use the exact same method !
So were my grandpa’s
Another method: Designate the side of the square as 2a. Run a diagonal line between the center points of each side of the square to the adjacent sides. The square is now divided into 8 pieces. The outer 4 pieces are 45 degree right triangles with area (a^2)/2 . The other four pieces are triangles each of a different size. One triangle goes with each of the four areas the square is divided into. We denote the areas of these triangles T16, T20, T32, and T?, with the number corresponding to the identified area of the original square. Now rotate this cluster of four triangles by 45 degrees clockwise. You see that you have a square whose side is sqrt(2)*a. Now notice that based on the base and height formula for a triangle that T16+T32 = a^2. Now add the two 45 degree triangles that go with the area that's 16 and the area that's 32. You get (a^2)/2 + (a^2)/2 + T16+T32 = 16+32. Hence 2a^2=48 and a^2=24. Thus a=6sqrt(2). Now we know that T20= 20-(a^2)/2= 20-12=8. We also know from the square containing the four triangles, using the area of a triangle formula that T20+T? = a^2. Hence T?= a^2-T20 = 24-8=16. The area the problem asks for is T? plus one of the 45 degree corner triangles whose area is (a^/2)/2 which is 12. So the area the problem asks for is 16+12=28.
how does T16+T32 = a^2
Because the area of a triangle is a^2/2 and since they are two triangles with equal sides, then it's a^2/2*2, so a^2 @@account5223
"I don't know" is also a correct answer
I put that answer for my exams and I got an F! I got Fantastic!
@@-Chicken_ I think in this case F stands for "Fair"
When you can solve complex analysis problems but didn't even know where to start solving this
There is a much simpler way to arrive to the same answer. Starting with the upper left and working clockwise, let's label each area A, B, C, and D respectively. Since each line segment bisects the outer wall, then (A + C) = (B + D), giving us (20 + C) = (32 + 16), or (20 + C) = 48. Subtract 20 from both sides: C = 28.
This is way easier
Can you explain why areas A+C will be equal to B+D? Didn't understand that part..
@@girishshivshankar6339+
@@girishshivshankar6339 If you join the mid points of the sides of the square then you get a smaller inner square. Now each of A,B,C,D is composed of a triangle outside the inner square and a triangle inside the inner square. The outside triangles are all the same size. Because it's a square the sum of the heights of opposite triangles in the inner square are equal. Because their bases are also equal, the sum of their areas is equal.
@@girishshivshankar6339 Divide the square into two rectangles vertically at intersecting point. 2b is half the area of left rectangle, 2d is half the area of right rectangle. The remaining area is 2a+2c.
“The sum of opposite areas are equal” , you lost the majority of us there mate.
Yes but it is not essential to resolve the problem !
If the sum of "opposite areas" (definition, please?) are equal, why even bother to subdivide into triangles?
Aha! Don't think of this as "opposite areas are equal," think of this:
a + b + c + d = a + b + c + d, so ...
(a + b) + (c + d) = (b + c) + (a + d) merely by rearranging, and the rest follows.
@@josepherhardt164 Thanks Bro.
How does the side length matter?
If the whole square area equals to 100%, and other areas' sum is 16%+20%+32% = 68%, so the rest would be 32%, in current measures is 32 cm^2. Can this be correct?
@@MrMeGaSeNt we don't know if its 16%, just that it's 16 cm^2.
I solved it in a different way with matrices.
I first thought about a square cut into 4 even pieces (like how a window looks like) so the dot would be right in the middle
I labeled each area z
Then I thought about what would happen if u moved the intersection (the dot) directly left or right. If I moved the dot to the right, then the left 2 squares would increase the same amount the right 2 decreased , since you can't magically gain or lose more area by partitioning them differently.
So I made x the amount a square gains/loses when the dot moves left or right.
The same reasoning for the dot moving up or down. Made the change y
So upper right was z + x + y = 32
Upper left was z - x + y = 20
Lower right was z - x - y = 16
3 equations 3 unknowns. z = 24, x = 6, y = 2
The equation of the lower left square was z + x - y. Plug in and you get 28
nice
Damn, great solution.
Liked ur approach but something is a miss here.. You got ur ans coz of square or a parallelogram which have this as a property .. But according to your method(area can never be created) this can be applied to any of the figure like quadrilaterals.. N there it fails.. Try it urself.
Alex Xela
This is so much better
For the bonus: Call the point where the red area and the 72 area meet M, the point where the 10 area meets side BC N, and the point where the 8 area meets side CD P. Area(triangle ADM) + area(triangle BPM) = 1/2 area(parallelogram ABCD) = area(triangle ADN). The missing spots are the same in each side, so call the sum of their areas y and the red area x. Then x + y + 72 +8 = y +79 + 10 -> x = 9.
Thus the red area is 9.
Nice
Who told those are of same size
@@KM-om1hm It's been a couple years and I've only taken a glance at the problem but I'm gonna go with probably some combination of the definition of a parallelogram and my brain. I'll look more closely after work and try to get a more accurate answer.
@@OmniLiquid okay. Thank you
@@OmniLiquid The reason is the triangle area formula mentioned in the video. If we call the height of the parallelogram 'h', and represent the length of line segment AM as [AM], then the area of ADM is 0.5h[AM] and the area of BPM is 0.5h[MB]. Add these and you get 0.5h[AB]. Since the area of the parallelogram is defined as h[AB], the sum of the two triangles is half the total area. You can show in a similar way that the area of ADN is half the area of the parallelogram.
Hey. A very similar question was asked in Turkey's university entrance exam yesterday. I don't know if I could solve it if I didn't watch this video. I haven't got the results yet but you made me make one question more and closer to the university I want. Thanks a lot.
I hope you've made it!
Hey, I'm curious, what was the exam that you took? Was it SAT? YOS?
Hope you can make it to uni, btw.
@@abshariadamIt was YKS I think.
I think, i‘m a genius. Just continue watching and don’t even try solving, because he‘ll tell you the solution.
wow,intelligent!
Thats cheating!
By the way MY method was way faster...
Solution:
Take the square root of 76 and then scroll through the comments...
The first step I left out
Genius answer , pal , total 101% genius
What praise should I give you for such an genius answer
Einstien's brain is the mere dust of your feet , pal , genius
Benhard Wiesner, You cannot apply the square root like that......can You?
I solved it from the first try 😊
I got negative twelve.
I got 76 :c
I got 28...
Good ol' negative area.
how could yoy do that? great!
Negative 12 would work too
I really appreciate that you talk through the answer slowly so I can watch a bit and then have a go :)
I solved it by just looking at the thumbnail and randomly:
b+c=20
b+a=16
Difference is 4.
Then a+d=32-4=28
Pretty cool considered i didnt use anything lol
same way...but by suming up opposites areas...
I just simply assumed that the sum of both "diagonal sections" should be equal, that SW + NE = NW + SE, specifically NW20 + SEx = SW16 + NE32, so 20 + x = 48, so x = 28.....no need to divide every section in half or consider as triangles. Or was it mere coincidence that this worked in this case ?
It wasn't coincidence because your assumption was correct in this case, but in maths you should always prove your assumptions before using them in your calculations (as Presh did in this video).
it wouldn't be coincidence since with a point anywhere in the square coming from the midpoints of each side, you can split it up in the same way as he did with a, b, c and d without knowing the exact area. since (a + b) + (c + d) will always be equal to (b +c) + (d + a) this means that the sum of both opposite sections in any square split up that way will be equal. I hope this made sense
I Believe your approach works in every instance regardless of where that point is.
They are always the same, and it is not a coincidence. As you can see, both diagonal areas are (a+b+c+d).
You can also get to this conclusion by noting that any deviation from the center point creates a equal but opposite change to the areas of the diagonally opposite areas, so they must stay constant.
How did you come up with your assumption?
Imagine the point is coincident with one of the corners. Then it becomes more clear why your approach works.
Hell yeah ! When I finally decided to watch the video I was like, "ok might as well try to find a solution before watching" and I found 28cm2 in my head. God it's nice to get a boost in math confidence for once :')
For everyone confused why he took so long and it only took you all second, hes doing the same thing. Just proving the method you probably used is true. The sum of opposite areas are equal, so he can justify his answer, which is usally the hardest part.
I made it with a harder way through figuring out the side length of square = 4√6. After watching the video, I found it's actually a simple question.
How did you calculate the side ? ! !
He didn't
Fooling himself only
@@critisizerr245 I did the same thing, so no, he wasn't fooling himself.
中文解答:Please have a look at MY SOLUTION: th-cam.com/video/h01cab2ngM0/w-d-xo.html
i am a simple romanian: i see the world romania i get my heart warmed and like the video
A Chinese fifth grader solved the more difficult problem in less than a minute... *cries myself to sleep*
it is so easy for a Chinese like me.
And i‘m lerning integration now, although i‘m just a ninth grader :)
I think your problem perfectly encapsulates the beauty of math, in the sense that, although it's a really hard problem, it's solution is also quite simple, but not intuitive!
Beautiful problem!
For the first question, let's call the centre point "P" which links to the centre point on the four square edges to divides the square into 4 areas.
The first question then can also be solved by dividing the 4 areas in the other way: Link the middle point of each square edges to form a 45 degree square (call it the centre square). This centre square divides the 4 areas into 8 triangles. The sum of the top-right + bottom-left area, could be calculated as the sum of their four triangles. These four triangles has the same base length: the edge length of the center square! The total height of the 4 triangles is exactly the diagonal of the original square. Thus the sum 4 triangles is 1/2 x Centre square edge length x Outer square diagonal length. The same argument can be applied to the top-left + bottom-right area, and results in the same number (since the area sum will not depend on the position of point "P".)
Now we know the two diagonal area sums are the same, and the problem can be solved as 32 + 16 - 20 = 28
你好。請看我的:Please have a look at MY SOLUTION: th-cam.com/video/h01cab2ngM0/w-d-xo.html
I got Jamaica on the 2nd question, is it correct?
Hahaha emre bravo
Did you divide by two?
Huh, I got Jamaica/sqrt(Cuba) on that question.
Nah dude. U forgot to multiply by Canada and divide the remainder with Caribbean
Chaos Master26 u gotta subtract the USSR first though...
The discussion about the smaller triangles being equal was interesting and potentially useful for something else but that had nothing to do with the actual solution. Knowing that the sum of the areas of the opposing polygons being equal is all you needed to solve the problem. I didn't know that so I learned something new today.
But the smaller triangles is how he showed you that was true. He divided it into smaller triangles, with pairs of the same area, by noting they had the same base and height. Then he took the original polygons and noted what their areas were in terms of those smaller triangles. The SW polygon has an area of a+b, NW is b+c, NE is c+d, and SE is d+a. Then you can note that the combined area of the SW and NE polygons is a+b+c+d, and the combined area of the NW and SE polygons is b+c+d+a. Then you can see that those sums are the same; they're the sums of the same four equivalent areas.
The point of the video is showing that that's true, not merely telling you that that's true.
the Chinese kid who solved the end problem in under one minute answered with "Red Triangle is Supreme over all. Long live Red Triangle".
I had a similar problem on my algebra test a few years ago, but I didn't know the thing about opposite areas being equal, so I had to go the long away around.
In this case, I did the same thing by dividing everything into triangles and labeling the equal areas a, b, c, and d. Then I did some algebra. You know a+b=16, b+c=20, and c+d=32, and what you're looking for is a+d. So based on that you know that d=b+12 and a=c-4.
Then to finish, you do (a+d)=(b+12)+(c-4) and then simplify it to =b+c+8. And since we already know that b+c=20, we know that (a+d)=20+8=28
Who else checked the comments for 'clues' but realized they were so early that all the comments aren't very helpful yet... Well I did...
Gifted Guy all you can i find in this section are comments like "fresh dog walker"
Let a be the half-length of a side. Total area is T = 4a². Also, T = SW + NW + NE + SE = 68 + SE. So SE = 4a² - 68.
Let x and y be the coordinates of the point. Use the area of a triangle: A = 1/2 × base × height. For SW: 16 = 1/2 × a × x + 1/2 × a × y = a/2 × (x + y). For NE: 32 = 1/2 × a × (2a - x) + 1/2 × a × (2a - y) = a/2 × (4a - x - y). Sum these two, the x and y cancel out: 48 = 2a².
Replace in the first equation: SE = 48 × 2 - 68 = 28.
This is how I solved it as well. It turns into 3 equations with 3 unknowns
Or just 16+32=20+x
I can actually solve this initial problem way easier. I did it in under a minute. It is simple! due to the fact that all lines connect to the center of one of the lines of the square, it does not matter where you put the connection point.
in these circumstances diagonally opposing area's together are ALWAYS half of the square. So the equation to solve this is simply
16+32-20=28
Sasha Shadowhive Wow, I didn’t know that! Thanks for telling me this!
@@comershaw7610 no problem. I just don't understand why he uses such a roundabout way
Keep in mind this only works if the connection points on the lines are turn symmetrical. 2 easy examples are, like in this problem, the middle of each line or in each corner. But if the connection point is, for example, on 2cm from the left corner (if you look at it from the center of the square) and 5 cm from the right corner, this theory will still work if ALL 4 POINTS are 2 cm from their respective left corner and 5 from their right corner. If the points are no longer the same this will not work
@@sashashadowhive6128 Yeah, but try to prove that it actually is correct solution (I know it is, but I wouldn't say the proof is obvious, meanwhile, in his solution, the proof is quite obvious, it's pretty much the solution itself).
Or
d+a=??
d=32-c
a=16-b
So
32-c+16-b=??
Because b+c=20, then -b-c=-20
Therefore 48-20=28
I did something kind of similar. We know that b=16-a. c is then 4+a because they are the green triangles adding up to 20. c is also 32-d. So 4+a=32-d, or a=28-d. a+d=(28-d)+d=28.
32-c+16-b=?
32+16=b+c
48=b+c
B+C=20 from ..fig .🙄
@@kiranshetti7991 i think this is what i did, the area equality of a square divided from a central point/equally on the sides meant i just intuitively knew that each half was the same area as 16 + 32 = 48, so 20 + ? = answer, or 28
@@kiranshetti7991 you cant actually take b and c to right hand side because r.h.s is not equal to 0.
It is actually equal to ? Which is actually a+d
@@arshjeetsingh7113 Exactly he forgot ? = a+d. What he did was add b+c to a+d, which is 48.
I couldn't think of a simpler way to do it, so I set it up as 4 equations and 4 unknowns, using nothing but areas of rectangles and right triangles, and was able to grind through the algebra to solve for the total area, from which the resulting area is 28. I'm embarrassed that it took me over 1/2 hour.
I went through 6 eqns and 6 unknowns ! FML brute forcing my way through everything XD
Awesome
My first idea was 16 equations and 16 unknowns...
I predicted 28 because of the relationship of the shapes but I didn't know that it was the answer
20-16=4
32-4=28
But this thing doesn't apply to everything
I took the Relation between 16 and 32. The middle is 24 what describes the size of all squares If the Point were in the middle. And than Just 24-20=4 so the opposide has to be 24+4=28
... I think that Always works
Same
LOOL SAME! I THOUGHT IT WOULDNT WORK BUT I DID IT ANYWAYS
same I did it the same way
@@GabbaGandalf-fo7cg I feel like that is mathematically the same as the "sum of opposites are equal" method, just executed in a different way :P
3:52 You can do that also with system of equations
Yep I ended up doing that. A bit longer to do AND it doesn't really use the method the sender wanted us to use but it gets the job done !
@@exlzyor2006 Yes, I'm more comfortable with algebra than geometry. I used x = base of all of those triangles (i.e. half the length of each side of the square), y = how far below the midpoint was from centre, and z = how for to the left of the midpoint from centre. So for the "a + b" shape, my formula was x(x-y)/2 + x(x-z)/2 = 16. And so on. Fortunately, there were ways of simplifying the algebra, so it didn't take all that long to get the right answer. But still not nearly as elegant as Presh's approach.
@@aquamarine99911 elegant or not, I solved it in head very fast using different division and sum of heights. Inscribed square is a half. Areas b+d=a+c=2S because of sum of heights. S is ⅛ of the whole square. 4S = 16+32 => S=12 and square is 96.
Unknown area will be 96-16-20-32=28
Bro mine is coming 25.5cm² . Please give a response
Let equals side be x and other according to their type like y,z,k,z,l
Area of first =20cm²=x*x*y*z
Area of second=16cm²=x*x*y*k
Area of third=32cm²=x*x*z*l
Area of fourth=?=x*x*k*l
Area of 4th= Area of 2nd * Area of 3rd / . Area of 4th
(x*x*y*k)*(x*x*z*l)/x*x*y*z=x*x*k*l(which is 4th )
16*32/20=25.6 cm²
@@shadowblock5814 Look, connect middle points of the square - you'll get 4 equal isosceles right triangles (note them s) and inscribed square with side equal L√2/2, and area being a half of initial square. Inscribed square is split into 4 triangles, let's note them clockwise a,b,c,d. And correspondingly we'll note initial shapes A,B,C,D. a and c have base equal to the side of inscribed square, and the sum of their heights is equal to the side, too. The same to b and d. Thus, sum of their areas are equal a+c=b+d. Equality will stand if you add the same to both sides.
2s+a+c=2s+b+d => (s+a)+(s+c)=(s+b)+(s+d) => A+C=B+D => D=A+C-B= 16+32-20=28.
Hello there. I always loved math. I loved it so much that I became an actuary. I saw this problem a different way. I figured, if you pick any point inside of a square, and draw a line from that point to the center of every side of the square, then the sum of opposite areas has to equal half of the area of the square. I used that principle and came up with the same answer of 28.
Please have a look at MY SOLUTION: th-cam.com/video/h01cab2ngM0/w-d-xo.html
I finally solved one of these! Although my way of solving was far less elegant.
Well thank God I solved it as fast as a fifth grader in China. I can be proud of that
"these videos build confidence" not when you have no idea how to solve something
You did not need the "sum of opposite areas are equal" equation. It could be solved for 28 just by knowing that a+b=16,b+c=20,c+d=32,a+d=x. You use basic substitution across the first three equations, and you eventually get that a=28-d. As such, if a+d=x, then x must equal 28.
hi presh, I solved this problem in another way, using the same principle but without any algebra, it is a much more visual solution. To explain, Let me label the upper right region as A and the upper left as B and so forth moving counter clockwise. (so D is our unknown area)
Start by drawing in the lines between the centers of adjacent sides of the square.
We have now decomposed each region into two triangles, like in your method. But here one is in the corner of the square (and has fixed area) and the other has a base of that diagonal and a height dependant on the position of the point. let's call these the variable triangles.
Now draw a line through the point and parallel to the base of the variable triangle from region D (call this line m). Notice that this line is also parallel the base of the variable triangle from region B. What we notice as we drag the point around on line m is that the areas of regions B and D stay the same, and consequently area is just shifted back and forth between regions A and C.
If we drag the point to the middle of line m, the figure is symmetric and by that symmetry, regions A and C have the same area.
Originally they had 16+32=48 area between them, and they still must, so each has an area of 24.
Now draw the main diagonal that is also the symmetry line, and passes through our point, and is parallel to the bases of the variable triangles of regions A and C. (Call this line d)
If we drag our point around on line d we notice that regions A and C don't change area and the area shifts between regions B and D. And! we notice that we can put the point in the middle of the entire square. When we do this all areas are the same, so all the areas in this position must be 24. Shifting the point back so that region B only has an area of 20 means that 4 units of area mush have shifted from it into region D. 24 +4 = 28 done.
Actually, thinking about this further, there are loads of really cool observations here that can lead us to the same conclusion. Using the same type of argument as above we can see that moving diagonally doesn't change the total areas between diagonally opposing regions at all (A and C, and B and D). After accepting that, and the existence of the state where the point is in the middle and all regions are equal, it is obvious that D= A+C-D.
Also if we instead make cut the large square into quarters with a horizontal and diagonal line, and focus on triangles created in that picture, we can notice the following things about combined areas of adjacent sides:
If we move the point horizontally, A+B and C+D both remain constant.
If we move the point vertically, A+D and B+C both remain constant.
Which is cool, but not helpful for this problem.
Simple I had 28 in a minute!!
Where the point in the inner square is, is indifferent for the answer!!
The inner square (connect half way points) is half the size.
Now each of the two opposing triangles of the 4 triangles of the point to the edges of the inner square take half the size.
So half the size is 16+32 because half the inner square plus half the outer parts of the larger square equals half the larger square.
Now you know that
16+32=20+x and x = 28
Guessing 28 is not good enough.
TheReactor8 your explanation makes no sense sir. Your language is so confusing. Without explaining what you mean by inner square. Reread your first sentence starting with “now each” and think if anyone would be able to understand such a long sentence without label or a reference picture. Impossible.
Bă,Raio din România, mi-am spart creierul cu problema asta vreo 2 ore,apoi cand am vazut cum a fost rezolvată m-am luminat:făină de tot problema cu o rezolvare elegantă.Bravo!
04:34 problem
Let's name the points:
In between line AB is point E
In between line BC is point F
In between line CD is point G
Parallelogram ABCD area consist of two half-areas, ABF+CDF and ADF but also ADE+BEG and BCG+DEG.
Let's look at these two half-areas:
ABF+CDF and ADF
ABF area is ? + a + 72 + b
CDF area is e + 8 + f
ADF area is c + 79 + d + 10
? + a + 72 + b + e + 8 + f = c + 79 + d + 10
And then the other two half-areas:
ADE+BEG and BCG+DEG
ADE area is ? + c
BEG area is 72 + d + 8
BCG area is b + 10 + f
DEG area is a + 79 + e
? + c + 72 + d + 8 = b + 10 + f + a + 79 + e
All of those four are half-areas, so we switch and get these equations:
? + a + 72 + b + e + 8 + f = b + 10 + f + a + 79 + e
? + c + 72 + d + 8 = c + 79 + d + 10
Take all these unknown areas a b c d e f away:
? + 72 + 8 = 10 + 79
? + 72 + 8 = 79 + 10
They both give the answer ? = 9
Alternatively:
Draw the inner diamond of the square, where the 4 cornerpoints of the diamonds are the middles of the edges of the square.
All edges of this diamond are equal by the pythagorean theorem and the fact that all half-lines of the square are the same size. (1)
The angles of the inner diamond are all 90 degrees, proven by the 1/1/sqrt(2) triangles having 45 degree angles.
Focussing on the diamond. draw a line with an angle of 90 degrees from all edges of the diamond through the dot. The lines from opposite sides coincide, leaving 2 new lines.
Now the base of the triangle is the edge of the diamond, which is the same for all sides. See (1).
The height of blue + the height of green is also equal to this.
The area of the blue triangle + area green triangle = 0.5 *e dgeLength * height(blue) + 0.5 * edgeLength * height(green)
=0.5* edgeLength * (height(blue)+height(green)) = 0.5 * edgeLength * edgeLength
Repeat this for the red and yellow area's.
The area of all 1/1/sqrt(2) triangles are all the same, therefore, green area+blue area = red area + yellow area
I hope it's somewhat understandable
This is a good idea but only works if the dot is inside the centre diamond. If numbers of the problem were changed and the dot was outside the diamond this method falls apart.
Actually I had a go and it still works if you consider negative lengths/areas in this case.
That's one thing about answering geometric problems with text comments. You can't show a diagram, and describing a diagram clearly is tough.
there is another way for this example wich is figuring out the middle worth of every section(in this case its 24) wich makes the square 96 cm² bc there are 4 sections so you just subtract 68 from 96 and you get 28
But if the three known areas have a total surface of 68, how do you get to an average of 24?
@@NilfNilf1972 i went from twenty to the number wich is in the middle of 16 and 32 wich is 24
24 is not the average of 16, 20 and 32...
@@timonburkard3481 i know but in this case the average from the biggest and smallest section is the overall average
@@jolioding_2253 Thanks, i understand now what you mean. Anyway i do not understand why in general (a+b) + (c+d) = (b+c) + (a+d) ?
I am a retired Aerospace Structural Design Engineer with too much time on my hands. I solved the problem graphically by building a Catia model and measuring the 4 areas. Then I adjusted the size of the square and the position of the center point until I came up with the correct areas for 20, 32, and 16 square cm. At that point, the fourth area was 28 sq cm and the square was exactly 9.797959 cm on a side, which is the square root of 96. Brute force method exercises the brain too.
"The sum of opposite areas are equal."
Then there is no need to show that the area of a triangle is 1/2 of the base times height.
16 +32 = 20 + ?
? = 28
Unless you are trying to show that this is how it is derived.
My explanation to why the the sum of a pair of opposing areas= Sum of the other pair:
Consider a square inscribed inside the original square, with its vertices at the midpoints of the original square.
Within this inscribed square are four triangles.
Drop altitudes(heights) from the interior point to the bases of these triangles (sides of the inscribed square)
Sum of one pair of opposing heights= Side length of inscribed square= Sum of the other pair of opposing heights.
Multiplying the heights by the inscribed square side length and you will get their areas.
Hence
Sum of one pair of opposing triangles in inscribed square= Sum of the other pair
Adding the areas of the triangles generated in the space between the original square and the inscribed square, and since the four of them are equal, you get
Sum of one pair of opposing areas= Sum of the other pair
Hope that helps.
I got what you said, but for people who don't try it this way, it's hard!
I had the same approach at the first time, and it shows the true nature of the problem... but I think that Presh's solution was more elegant...
Marcos Guilherme To be frank if I were in a time-limited exam, I would have gone first for Presh's solution. It was more methodological and alelgebraic. It isn't as intuitive and requires you to write down the algebra or form them in your mind to realize the relation. Whereas I think my method aids more the geometrical understanding of the relation. It took me a while to think of my method. In an exam I definitely would recommend Presh's.
The key to solve this problem is proving that the "sum of the opposite areas of triangles are equal", which is not obvious. At @4:05, he just mentions "notice" and moves forward. That begs some explanation as it confuses lots of people.
Proof : Denote the midpoints along the sides of the square as EFGH, starting from top moving clockwise. Denote the vertices as ABCD, starting from top-left, moving clockwise. Let the intersection in the middle of the square be O. Consider the square EFGH.
Within this square -
Area of triangle EOH plus triangle FOG = (1/2*HE*h1) + (1/2*GF*h2)= HE(h1+h2)=HE*HE.
This is because in the square EFGH, h1+h2 equals one of the side, HE. And also, as it is a square HE = GF.
Similarly,
Area of triangle EOF plus triangle GOH = (1/2*EF*l1) + (1/2*GH*12)= EF(11+l2)=EF*EF.
11+l2 equals one of the side, EF. And also, as it is a square EF= GH.
Q.E.D
How do we solve otherwise ?
I agree he just glosses over it and does not explain why clearly. It is a problem with these type of videos. It's mostly a math teacher showing off and does not help the students get better.
The solution in the intersection point inside the square that we will use the x, y coordinates of the point with regard to the( 0, 0) vertex of the square and make 2 equation of the area equal =16, it will be 32=L(X+Y)by dividing the area to rectangle & 2 triangles ,L=0.5 square length, the nd equation of the area=32 it will be 64=L(4L-(X+Y)) by dividing the area to rectangle &deduction 2triangles , by substitute L=32/(X+y) in the 2nd equation X+y=(128/3)^.5 so L=32/(128/3)^.5 so the required area =4L^2-68=((4*32*32*3)/128)-68=28
Meanwhile, Autocaptions: “ Hey, this is pressed Tell Walker” 😹😹😹😂🤣
th-cam.com/video/Ik7beS1YbnM/w-d-xo.html
Here is another simple solution: move the inner point at an angle of 45degrees until you hit the vertical line that divides the square in half. The areas 20 and x will remain constant. Therefore also the sum of the others will be constant (48). Now we have a symmetrical situation on the left and on the right half of the square, so 20 + x = 48.
Why will the areas 20 and x remain same?
SM MR: because the inner point moves parallel to a line that divides the areas 20 and x in two triangles. One of them remains unchanged, the other has a constant height.
Matthias Burger There are two lines that divide those areas into triangles and they aren't parallel to each other, so how can you move the point parallel to both of them?
Mykolas Karpavičius: connect the midpoints of the sides of the square accross the areas 20 and x.
Wow! Really nice solution!!
I found this way.
If the sides are equally cut then 16+32=20+x
And x becomes 28 easy man easy
Weeb- Desu yeah I did the same say too
How it works
Weeb- Desu me too
That is literally what the video proves, but instead you just said it.
His vids are good but they are all longer than 10 minutes and i just dont watch a proof of a questions answer for 10 mins
you can inscribe a different square connecting the the middles of the sides of former square and then u see 32-y+16-y=20-y+x-y where y is the rectangular triangle
is 28 right? i think the opposite parts should always be same size. So 16+32= 48 and 20+x=48 --> 28
6man12 EXACTLY WHAT I DID
If you would answer that on your exam in my course, you wouldn't get any credit for the 28 unless you could explain WHY the sum of the opposite areas must be equal. THINKING something is true never counts.
I don't think dividing the shapes to triangles helped in solving ... only the fact that the sum of the opposites is equal needed.
that fact is derived from observing the division into triangles
I agree that it is not needed, but it can be helpful for those who don't know/remember that the sum of opposite areas are equal.
Here, you have a+b+c+d=a+b+c+d : the equality is demonstrated.
You can't proove it otherwise.
@@sanpedro5018 > thanks, I was wondering where this "sum of the opposites are equal" was coming from, but this is a lot more clear :)
Ya you are right
Can’t you just say that opposite quadrants always add to half the square? So quadrant 1 + 3 = 2 + 4
16 + 32 = 20 + n
n = 28
True, but prove it.
I just looked at it and had that thought about the sum of the diagonal areas being equal, but just couldn't recall how on earth I'd prove it, years of not trying Maths problems.
Ya thats what i thought at first
As long as the bases are separated evenly, it's true, and it's proved in the video tho
Sure. That's because the area of one of the quadrants is = ½ (distance from side midpoint to neighbour side midpoint) (length of projection of the quadrant onto the diagonal that starts in the quadrant's corners). So, the sum of the area of two opposite quadrants is ½ (distance from side midpoint to neighbour side midpoint) (length of the diagonal). Since both diagonals have the same length, quadrant 1 + quadrant 3 = quadrant 2 + quadrant 4.
I got it right without all this just in my head. My reasoning behind this is that the 16 and 20 section has a difference of 4, so I subtracted 32 by 4 and got 28. I don't know if this can apply to every other situation, but it applied here.
By the way, as a Chinese, I am sure the question at the end is for 5TH grader. Well actully that is an easy one for 5th grader. I went to Mathematical Olympiads class every weekend from 3rd grade and at 5th grade we already start some entry level calculus questions....
I don't get why the sum of opposite areas are equal. Somebody pls explain?
OhSaft The explanation is in the video, get a paper and a pen and try to replicate it
Vitor Nathan Gonçalves how do you proove it?
Once all of the a, b, c, and d areas are labelled, you can see that the sum of the areas of opposite regions is always a+b+c+d.
You could try cutting them up into triangles, and since the area of a triangle is 1/2*b*h and they share a common base, opposing triangles, when added up, would have the area 1/2*1/2(Length of cube)*length of cube. Do this for the adjacent triangle in the same quadrilateral and you’d find out that the area for them is half the area of the entire square
way too complicated but this was also my first thought to be honest :D
I`m Japanese!
I love this channel!
Assigning the point coordinates (a,b) and writing expressions for the original areas worked fine too.
//=h, ha/2+hb/2=16, h(2h-a)/2+h(2h-b)/2=32 , insert first in second and solve for h. Then you get the total area of the square 4h^2=96.
Please have a look at MY SOLUTION: th-cam.com/video/h01cab2ngM0/w-d-xo.html
I looked at the thumbnail and had a guess at 28, looked like a valid guess so I checked the video, and sure enough.
im 8th grade in turkey and i found the second one 9 under a minute: the half of the shape is [(72+8+a)+(b+x)] and also the half of the shape is (79+10+a+b) so x = (79+10)-(72+8) = 9. Is it correct?
Wow, I had been staring at it for several minutes and not getting anywhere, but you're absolutely right! For those who don't get it: the triangle with base AD and tip between B and C (containing 79, 10 and two white areas) is exactly half the area of the parallellogram. But the combination of two triangles with their bases on the two parts of AB and tips in D and on DC, together, are also half the size of the parallellogram. Those consist of the same two white areas, 72, 8 and the red area. So the red area has to be 9. Brilliant, well done!
Sovled within ten seconds, with the correct assumption that opposite segments equals half the total area. It's easy to see this is true when we have the intersection point in the exact centre (all segments equal sized squares), but if we visualise the intersection point moving from the centre to the corner, we see one segment get larger, while its opposite corner becomes smaller. Meanwhile, the other two segments change shape equally, but they do not change area relative to one another. If they never changed area relative to one another, they they must have remained the same area as they were when the intersection point was in the exact centre.
Thus, opposite segments are equal to half the total area, regardless of the location of the intersection point.
I saw the same thing from the beginning but could not believe its so simple :) so spend a lot of time by proving it math way.
I solved it slightly differently:
Let half of the square’s side length equal “L”
Now, connect the midpoints of the square’s sides to form four isosceles right triangles of side length “L”
Via the Pythagorean theorem, the length of the hypotenuses of these triangles is sqrt(2)*L
Now, consider the interior square formed by the hypotenuses of these triangles. This area is divided into 4 triangles by the given lines, each with a base along the edge of this inner square (length sqrt(2)*L)
Now, consider two opposite triangles within this area. The area of one will be 1/2*sqrt(2)*L*r1, where r1 is the distance from its base to the given point. The second will have an area of 1/2*sqrt(2)*L*r2, where r2 is the distance from *its* base to the given point.
Since r1 and r2 come out perpendicular to opposite sides of the inner square and go to the same point, they form one line of length sqrt(2)*L, expressed as r1+r2=sqrt(2)*L
Now consider the combined area of those two triangles, 1/2*sqrt(2)*L*r1+ 1/2*sqrt(2)*L*r2, or 1/2*sqrt(2)*L*(r1+r2). Substituting in “r1+r2=sqrt(2)*L”, we get 1/2*sqrt(2)*L*(sqrt(2)*L), which simplifies to L^2
Since the total area of the internal square is 2L^2, we can conclude that any two opposite pairs of triangles take up half of the area of the internal square, meaning both sets of opposite triangles have equal total area.
Now, consider the larger areas formed by the given lines. Each area is made up of an isosceles right triangle of side length L and its triangular portion of the internal square we drew. Thus, any two opposite areas are made up of two triangles of side length L and two opposite portions of the internal square. Since both sets of opposite portions of the internal square are equal, both sets of opposite larger areas are equal
From here, we say that 16+32=20+x, with x being the unknown area. Thus, x=28.
You can solve this much easier:
Add the two opposite sites together (16+32=48)
The other two sites have the same surface -> 48-20=28
And thats the easiest solution!
Please have a look at MY SOLUTION: th-cam.com/video/h01cab2ngM0/w-d-xo.html
Me: Wow, I have no idea. It probably involves math I never learned of something.
Him: So this problem can be solved using the equation A=.5bh
Me: Goddamnit
Once I saw the equal triangle, I was able to solve it. I was trying to do it algebraically.
Easier way:
Draw a line from each middle of each side to the middle of it's adjacent sides to create a (square) diamond inside the square.
Draw two perpendicular lines paralell to the sides of the diamond that pass through the point and end at the sides of the diamond.
Since the combined height of the oposite triangles inside the diamond is the same for the sw->ne and nw->se region, the sum of the area of the sw&ne regions is equal to the sum of the ne&sw region, hence 32+16=20+x
x=28
OMG THIS IS HOW I DID IT :D
If you draw lines from the interior point to the corners of the square, you end up with 8 triangles. Pairs of triangles which share the same side of the square have equal area as they have the same base and height. So you can set up the following equations (i'll use compass directions to denote the triangles of equal area). W+N = 20. N+E = 32. S+E = x. S+W = 16. (W+N)+(S+E) = 20+x. (N+E)+(S+W) = 32+16. 20+x = 48. Therefore x = 28.
So I solved it right in the most wrong way, I saw they were all multiples of 4 and similar in size. 16 was a little less than 20, so I just did a little less than 32 but a multiple of 4 and got 28 lol
I literally did the same
Wow So brilliant...its funny😂
Same
MadLad
Same here. Just visual guess.
The only person you should try to be better than, is the person you were yesterday
-Shazistic
“The sum of opposite areas are equal”
With this information in mind, just use problem data and do the math: 16+32=20+x. Nothing about blablabla triangle area was necessary
you are absolutely right. he took the long road when it could have been done in 30 seconds
4:06 the sum of opposite areas are equal. This is a beautiful and (to me) surprising general result that i find even more interesting that the solution to this particular problem.
ahh nice problem!!
Problematic(puzzle channel) ????
No. It's NOT a nice puzzle because it requires the knowledge of a geometric formula.
A puzzle that requires me to look up a math formula is not a puzzle at all - it's a MATH problem.
Disliked this video.
edmis90 I disagree with you. He doesn't simply produce a formula but explains showing different triangles.
edmis90 Your point: All question with solving geometry problem are not nice puzzle
Im from China and that was exactly how I failed my grade 5 math yet got 4.0 gpa in calculus at one of the best universities in Canada
5th grade, damm Asian
go back to china...
@@trinkCOKEorDIE bruh
The moment u started talking about triangles I was convinced I could do it and I actually did it! (with a long detour, but I got there none the less)
When I was a 9th grader, honestly, geometry was a hell for me. It irritated me whenever my math teacher drew circle, lines, and such. It was beautiful, it was an art yet we couldn't even understand what she's trying to say. Luckily, when I became a college student (math major) I've learned to love geometry and its figures.
Please have a look at MY SOLUTION: th-cam.com/video/h01cab2ngM0/w-d-xo.html
28 was my first guess after 15 seconds. Then I let myself get into the mood and my eyes deceived me into saying 20
Me too. I was like: 'hmmm.. let's go with 28' and it worked (this is btw my way of surviving math lessons)
This was literally in a math olympiad I participated. I did not finish it, but later I did🙂
A similar problem was in an olympiad I participated in as well. Are you Chilean?
Pretty smart solution. What I'd do after splitting the areas up into the triangles with areas a, b, c and d is: total area of the square T=2*a+2*b+2*c+2*d and rewrite to T=2*(a+b)+2*(c+d). We know a+b=16 and c+d=32 so you get T=2*16+2*32=96. We also know T=16+32+20+(a+d) so (a+d)= 96-68=28cm2. But stating that a+b+c+d=a+b+c+d (which is obviously true) and rewriting that as (a+b)+(c+d)=(b+c)+(a+d) to get to (a+d)=28 is pretty neat too.
I went about this in a different way: If you connect the midpoints of the sides of the square you get another square inside the square we started with, the inscribed square. Let's call the length of the sides of the big square 2a so that all the lines marked ‖ (equal-sign) have a length a. The sides of the inner square then become √2a^2. You'll see that each of the 4 areas we created has the same triangle in them, with two legs of length a and one of √2a^2. Since all 4 areas have this same triangle in them this triangle can be discarded when looking at how the areas relate to each other. This I'll use in a bit.
You'll also see we now have 4 triangles in the inner square, each with one of the sides of the inner square as their base and the other legs are the lines we drew to make the 4 areas. Let's call the top-left triangle B, the top-right one C, the bottom-left one D, the bottom-right one E and the intersection point S. Now drop a line from the intersection point S perpendicular to the base of each of the triangles B, C, D and E. These lines are the heights of those 4 triangles in the inner square and you'll see that the height of triangle B (b) plus the height of the triangle E (e) equals the length of their base. The same goes for the combined heights of triangles C (c) and D(d). So b+e=c+d=√2a^2.
As we know the area of a triangle is 1/2*base*height. These 4 triangles all have the same base so the relation between their areas depends on the relation of their heights. I've also determined that the relation between the 4 areas we created at the beginning depends on the relation between the area of these 4 triangles. Since their area depends upon their heights (as they all have the same base) I can state that the relation between the heights of these triangles and the areas we created at the beginning is b+e=c+d. So 20+e=32+16 gives e=28 which is the area we were looking for. Total area of the square then is 16+20+32+28=96 and its sides are √96=4√6 long.
I have found that there are several solutions. My reasoning is as follows: let's label the unknown area x. Then the sum x + 32 + 20 + 16 must be a square integer since we are dealing with a square form. So that the square root of x + 68 is an integer x can take several values: 13, 32, 53, 76 ... If I'm wrong, I would like to understand why.
@MindYourDecisions
a+b= 16, b+c=20, c+d=32 => a+b+b+c+c+d=16+20+32=> a+d+2*(b+c)=68 => a+d=68-2*(b+c)=68-2*20=28. No need to perplex things by means of mentioning this about the opposite areas. Cute problem thought. Thanks for sharing it!!!
Lol I took a completely different approach but got the right result. My approach wasn’t that clever, i divided all 4 areas into rectangular triangles and rectangles. I described those mathematically by using one variable for the (half) side length of the big square, one for the horizontal offset of the „point“ and one for the vertical offset. I ended up with 3 variables and equations (each for every known area) which turned out to be very easy to solve after some terms cancelled out.
This is what I did. Its not elegant, and it takes a lot more work, but it gives the correct answer, and requires no brilliance. More interestingly, it gives you the coordinates of that point in the middle. The crazy thing is just how simple it gets when things start cancelling, and you know there was a way that didn't involve manipulating all these terms around.
The point is SQRT(6) over and SQRT(6)/3 up. The entire square has a side length of 4*SQRT(6).
That's exactly what I did: I expressed 16=(stuff) and 20=(stuff) in terms of the offsets and half-sidelength, s, and was able to easily get the offsets in terms of s. Then I spent a few hours down a total rat hole failing to make any headway with 32=(similar stuff). It gave me a quartic for s which turned out not to have easy to find factors (indeed, it turns out no "easy" real roots), so I bailed out on that.
Eventually (a good night's sleep probably helped), I ignored the 32= and focused on the missing area, equating its s- and offsets-based formula with 4s² -(16+20+32) hence solving for s.
I enjoyed it, but was very grateful to come out of the other side with the right answer! I guess I'll never make a mathematician (too late to carve out another career anyway, now) :-)
Glad to see I was not alone! I used the same method too. The equations were indeed easy, and I managed to get the coordinates of the point and then 28 without using pen and paper.
+1, me too. This is the approach that simple exploration leads you inexorably down, without requiring spotting a special"trick".
1/ Given any point inside a square and connect this point to the 4 corners; the sum of the areas of the 2 opposite triangles are equal.
2/Just connect the 4 midpoints we have a smaller square of which the sum of the opposite parts have the equal area.
moreover every 4 parts has added up 4 equal triangles
so ? + 20= 32+16 ----> ?= 28 sq cm
Not sure if someone already mentioned it, but provided that sum of opposite areas are euqal, there is no need to split areas into two. All you have to do is state that A+B = C+D, therefore 16+32 = 20+D, which equals 28.
Don’t know if this is luck but I looked at the three areas and said that 16 +32 is 48 and 48 minus 20 is 28 and that’s how I got it
Mag Arts Not luck, it’s logical maths
Oh ok thanks. That actually makes me feel more confident
You don't need any triangles at all. Imagine the point starting in the middle. Since it's a square all smaller squares are equally sized. So they have a ratio of 1:1:1:1.
Now the point is being dragged into the position it has in the video. Now the relations are:
1 (for 32 cm^2) : 0.5 (from 16 cm^2) : 0.625 (from 20 cm^2) : x (from y cm^2)
The previous relations tell you, that their sum has to be 4. Mathematically speaking:
1+ 0.5 + 0.625 + x =4
x = 0.875
Now we can calculate y, the unknown area. Since we used 32 cm^2 as the reference-size we need to calculate:
y = 0.875 * 32 cm^2 = 28 cm^2
best way
32+16=20+X
I really don't exactly know what you are doing but I can assure you, that 1 + 0.5 + 0.625 + 0.875 = 3 and not 4.
So I think your proof is not entirely accurate. But maybe I am missing something. If so, could you please explain and elaborate a little further?
The 32cm^2 area is larger than one quarter of the area. We shouldn't have 1+0.5+0.625+x=4. Rather, it should sum to some unknown constant. Put another way, when we move the point from the center, that region with area "1" Is Not constant meaning later assigning it to 32 is not valid. Essentially, the first part of your argument assumes the area is 4*32, which is false.
x = 1.875
You can solve this even without the assumption "area of the opposite triangles are equal" (Which I am not even sure if the meeting point of all vertices of the triangles are not the midpoint of the square). All you need is basic algebra to solve this :
1. c+b = 20
2. a+b = 16
3. c+d = 32
4. a+d = ?
Adding 2 and 3 :
a+b+c+d=16+32=48 ==> (c+b)+(a+d) = 48
But from 1, we know c+b = 20. Substituting c+b in above eq., we get :
20+a+d=48 ==> a+d = 48-20 = 28
Therefore, (a+d) = 28 !
I looked at it a minute and assumed the opposing areas should be equal area since the segments are drawn from the midpoints of the sides and all share a point. Then I did that last line in my head and came out with 28. I was so happy to see my assumption was correct. Not sure if I was lucky or logical, but I feel smart.
When you said 28 because 32 plus 16 is 48 minus 20 is 28 and its right
Lul
The way I found out was that the answer must be times by 4.
Sameee
Me, too, but my first off-the-cuff guess was 24 :D
I think I got a simpler solution:
Since there is a square, which is divided into four areas, the product of all four areas must be able to be devided by 4.
First of all, to make it easier you look at the largest common dividor of the areas in the square. You see that all areas can be devided by four.
If you do that we have the values: 4, 5 and 8.
The unknown area must have a value which added up with the other values can be devided by four. Values like these would be 3, 7, 11,...
When you then multiply these values with 4...
3×4=12
7×4=28
11×4=44
... there is only one value that makes sense in this context, namely 7 respectively 28.
Easy to apply but difficult to explain, i hope it is understandable.
An area of a triangle is not (base x height) / 2 but (base x height x sin(a)) / 2 where a is the angle between base and height. The reason the triangle pairs have equal areas is that they have the same base and height but also the same sine value of the angle between them. This is because the angles are x and 180-x and sin(x)=sin(180-x)
The sum of opposite areas are equal, I doubt recall this, may I ask the name or where you got that part because I dont seem to be able to solve it without that and I missed where it came from
Because they're triangles leaning on each other forming a circle so to say hence their sum is always 360 no matter how many times you slice them.
Also they share same base length and same height if you mean a and a, b and b etc
If it’s a square, shouldnt it have an area thats a square number? 16+32+20 is 68. I assumed 32 because it seemed more reasonable than 13
Joseph Wood no, the side could be any real nber
Number*
Arnav Nagpal i figured that was the case after i posted the comment
Two points:
(a) He said it's a rectangle.
(b) A square need not have an area that's the square of an integer, or even an integer at all. Consider a square 1/3 on each side. Its area is 1/9.
Tehom i think he said its a square
I subtracted b+c from a+b to get a-c=(-4), then added c+d to get a+d=28. The b’s & c’s cancel out this way which confirms Presh’s math that (in a different order) 16-20+32=28
I considered an equally split square with four areas size "a" and used simultaneous equations based on which areas were gaining or losing ground horizontally and vertically. Got 28
exactly what I did