What is the area shaded in green?

No video

What is the area shaded in green?

Рет қаралды 345,323

Күн бұрын

Thanks to Jumar for the suggestion! Can you solve this interesting geometry problem?
0:00 Problem
1:10 Similar triangles
2:29 Solution
Subscribe: www.youtube.co...
Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas!
If you buy from the links below I may receive a commission for sales. (As an Amazon Associate I earn from qualifying purchases.) This has no effect on the price for you.
Book ratings are from June 2021.
My Books (worldwide links)
mindyourdecisi...
My Books (US links)
Mind Your Decisions: Five Book Compilation
amzn.to/2pbJ4wR
A collection of 5 books:
"The Joy of Game Theory" rated 4.2/5 stars on 200 reviews
amzn.to/1uQvA20
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 4/5 stars on 17 reviews
amzn.to/1o3FaAg
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.1/5 stars on 30 reviews
amzn.to/1LOCI4U
"The Best Mental Math Tricks" rated 4.2/5 stars on 57 reviews
amzn.to/18maAdo
"Multiply Numbers By Drawing Lines" rated 4.1/5 stars on 23 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 75 reviews
amzn.to/1GhUUSH
"Math Puzzles Volume 2" rated 4.3/5 stars on 21 reviews
amzn.to/1NKbyCs
"Math Puzzles Volume 3" rated 4.3/5 stars on 17 reviews
amzn.to/1NKbGlp
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.
My Blog
mindyourdecisi...
Twitter
/ preshtalwalkar
Merch
teespring.com/...
Patreon
/ mindyourdecisions
Press
mindyourdecisi...

Пікірлер: 551

@tonyhasler985 2 жыл бұрын

Another way...no better no worse By symmetry, we can see that the green area is a square. As Presh explains, there are four 30/40/50 triangles and their overlaps are 12/16/20 triangles. The portions of the 50 hypotenuse that are not part the square edge are 12 and 16, so the side of the square is 50-16-12=22. If the square has a side length of 22, its area must be 22x22=484.

@kingdomofjace1190 2 жыл бұрын

The video was uploaded 10mins ago. How is your reply 4days old ?😕😐

@artha1679 2 жыл бұрын

@@kingdomofjace1190 re-upload?

@Discobobulated 2 жыл бұрын

@@kingdomofjace1190 The video was unlisted prior, likely to his Patreon supporters, he then set it to public later.

@sahas1514 2 жыл бұрын

I did it this way

@kmsbean 2 жыл бұрын

That's how I did it in like 30 seconds

@phasm42 2 жыл бұрын

I solved it the same way. But afterwards, I noticed that using the similar triangles, you could compute the leg lengths of the purple triangles (12 and 16), which when subtracted from the length of a blue triangle's hypotenuse (50), leaves 22 as the side length of the green square.

@anjhindul 2 жыл бұрын

Exactly what I did Paul, was a bit less math lol. Similar triangles will have similar sides with the same ratios. So the 20/50 goes to 12/30 and 16/40 The purple triangles are 12-16-20. 12+16=28 50-28=22 22*22=484 faster.

@Zollaho 2 жыл бұрын

Went directly for this way. Takes 3 minutes mentally.

@professorx3060 2 жыл бұрын

Same. But I did it both ways just for fun

@Vyrlokar 2 жыл бұрын

I went for this one mentally too. Find the length of the square sides and square that for the area. You still need to figure the sides of the smaller triangles in the same way, but the end is much simpler (to my brain)

@warfyaa6143 2 жыл бұрын

That is exactly what I did.

@dayashankarsuresh57 2 жыл бұрын

It is intriguing how a perplex looking question's solutions happens to be one of the most satisfying and basic answer.

@bobbytheferret6809 2 жыл бұрын

@HASANALI-zt7xz Жыл бұрын

Maths is simple things complicated..I mean with very less concept, u can like create a 1000 questions....

@raphaelkometpiste8311 2 жыл бұрын

I solved it this way: Entire square (50*50) minus two opposing triangles (40*30) minus two opposing triangles (20*10) gives you a parallelogram with the base of 50 and an area of 1,100. Hence, the hight of the parallelogram is 22, and the area of the green square is 484.

@clieding 2 жыл бұрын

Your solution is very simple and elegant- congratulations! 🤩💐

@lellab.8179 2 жыл бұрын

I like your solution!

@juancarlosfontanafontana909 2 жыл бұрын

👍

@stevieg1227 2 жыл бұрын

Nice one!

@nidalapisme 2 жыл бұрын

A simple and elegant solution. I like it ❤️

@erikwiessmann8681 2 жыл бұрын

I got linear equations for the hypotenuses: y = -4/3x+40; y = 3/4x-7.5; y = 3/4x+20; found the intersections and then found the distance of the points at the intersections (22).

@albiewxdude 2 жыл бұрын

This is how I did it as well.

@huwpickrell1209 2 жыл бұрын

And me

@trnfncb11 2 жыл бұрын

I did that too. It's the quickest with straight lines.

@eterty8335 2 жыл бұрын

@@trnfncb11 I don't think so. I started doing the same thing and then I just realized that any side of the green square can be seen as a hypotenuse of two imaginary sides that you get by moving the two segments between one vertex of the bigger square and the points where the two closest blue triangles touch the sides of the bigger square I'm not sure I described it well but, if I'm not mistaken, one side of the green square is the hypotenuse of a rectangle triangle whose sides are (50 - 30) and (50 - 40), and you can get that just by observation

@WarmWeatherGuy 2 жыл бұрын

I set the first two equations equal to each other to find the intersection (x=22.8). From that I found the height of the purple triangle to be 9.6. The base is 20 so I could calculate the area of the purple triangles.

@SteezyMD 2 жыл бұрын

It might be a cool idea to consider adding difficulty tiers to these videos. Like “What is the area of the shaded region? (Tier II)”

@Serizon_ Жыл бұрын

@justgio599 2 жыл бұрын

Maybe simpler this way: 1) identify the small corner triangle which sides are x=50-40 and y=50-30 2) calcultate the hypotenuse which is 22 3) area by 22^22 gg

@somewhatblankpaper1423 2 жыл бұрын

clever idea that i've also thought about, but the assumption true in this case since the hypothenuse of the corner triangle doesn't form a parallel line with the square. you can prove they are not parallel with some linear equations.

@snotgarden7569 2 жыл бұрын

This one is way easier than others. The overlapping area is another 3/4/5 triangle scaled to 12/16/20. big square - ( big triangles - small triangles ) (50*50) - ( 4*(1/2*30*40) - 4*(1/2*12*16) )

@brothapipp 2 жыл бұрын

So I did a little math in the lower right corner. you have 2 end points of 2 triangles. If you draw a connecting line between the those 2 end points you get the length of one side of the shaded square in the center. Since we know that 50-30 = 20 and 50-40=10, we have 20^2+10^2= L of inner square That ended up being 10*sqrt(5) Square that to get 500 units squared. Now I am 16 units squared off, can anyone tell me where I went wrong?

@bennjanse 2 жыл бұрын

Now we are 3 that did it this way I think we correct 🍺🍺🍺

@sagnikdas6049 2 жыл бұрын

Me too🙋

@sagnikdas6049 2 жыл бұрын

I was trying to find this thread lololol

@sagnikdas6049 2 жыл бұрын

They got 22 as each side of the green square As opposed to 22.3606... or 10√5

@sagnikdas6049 2 жыл бұрын

I just got our mistake... Drat

@user-oh2kt8lf6g 2 жыл бұрын

I used vectors to measure the distance between the opposite sides of the inner square. All the coordinates are known and you only need one scalar product.

@sergeyg2926 2 жыл бұрын

How do you get the coordinates for the corners of the inner square? I do not see a way to do it without using the same similar triangles.

@user-oh2kt8lf6g 2 жыл бұрын

@@sergeyg2926 You do not need those coordinates. The distance between two parallel lines can be calculated as absolute value of scalar product of line's normal vector of unit length and a vector connecting any two points that belong to the two different lines. 1. Suppose the lower left corner of the big square to have coordinates (0, 0). 2. Imagine there are only upper left and lower right triangles placed into the square. Their hypotenuses are parallel to each other and the distance between them equals to the side length of the inner square at 0:36. 3. Upper and lower hypotenuses cross the sides of the outer square at the points with coordinates (0, 20), (40, 50) and (10, 0), (50, 30) respectively. 4. The tangent vector of each hypotenuse, either upper or lower, is (40-0, 50-20) = (40, 30). Its length is 50. 5. The normal of unit length to any of the two hypotenuses is (30/50, -40/50) = (3/5, -4/5). 6. Now, we need a vector that connects any two points at our hypotenuses. Choosing first points from clause 3, we get (10-0, 0-20) = (10, -20). 7. For distance between hypotenuses, we have d = abs((3/5)*10+(-4/5)*(-20)) = abs(6+16) = 22 8. Hence the area of the inner square 22*22 = 484.

@kjl3080 2 жыл бұрын

No! This isn’t how you’re supposed to play the game!

@sergeyg2926 2 жыл бұрын

@@user-oh2kt8lf6g Thanks for the excellent explanation - totally forgot about projecting a vector onto a unit normal.

@lumsdot 2 жыл бұрын

4 big triangles area is (30 x 40)/2 times 4. then deduct 4 times the area of the overlap triangles, these are 3 4 5 triangles with hyp of 20, so easy to calc.

@birkett83 2 жыл бұрын

I did it using cartesian coordinates and equations for two of the hypotenuse. I got the right answer but I feel dirty. Still, if you want the inelegant solution, the short version with some of the working out omitted is like this. Set up coordinate axes with 0,0 in the bottom left of the square. The hypotenuse of the bottom-left triangle passes through the points: (0, 40) and (30, 0) so it has equation y = 40 - 4/3x. The hypotenuse of the bottom right triangle has equation passes through the points (10, 0) and (50, 30) so it has equation y = 3/4(x - 10). Solving these equations gives us the coordinates (x, y) of the bottom corner of the green square (22.8, 9.6). Since the side length of the big square is 50 and using symmetry we see that the left corner of the green square is (9.6, 50 - 22.8) = (9.6, 27.2). So by pythagoras the side length is sqrt((22.8-9.6)^2 + (27.2 - 9.6)^2) which means the area is just (22.8-9.6)^2 + (27.2 - 9.6)^2 = 484. Cartesian coordinates get the job done but it requires a lot more arithmetic and the smart kids will spot the nice geometric way to do it instead.

@keithroberts1401 2 жыл бұрын

I solved it a similar way. Nothing dirty about it if it works, and sometimes you are better off doing it the long way if you know it will work instead of spending time trying to find a simpler way. I was considering using integration at first, but I soon realized it would be easy once the problem became equations..

@birkett83 2 жыл бұрын

@@keithroberts1401 Task failed successfully!

@Terminarch 2 жыл бұрын

Not sure if it's easier, but I just plotted it out. The bottom-left triangle and it's intersecting hypotenuses (hypotenusi?) are -4x/3+40 and 3x/4-7.5 and 3x/4+20 Compute intersects: (9.6, 27.2) and (22.8, 9.6) Compute length: 22, squared is 484

@huwpickrell1209 2 жыл бұрын

Yep my method too

@sleepybutawake749 2 жыл бұрын

You could just use the sin and cos of any one of the int angles. As the smaller triangles and bigger ones are similiar, the values of sin and cos of the int angles will also be equal.

@cdmcfall 2 жыл бұрын

I calculated the length of the smaller triangle's hypotenuse, giving me the lengths of the two shorter size because it's also a 3-4-5 right triangle. That gave me dimensions of 12-16-20. The length of one side of the green square is the length of the hypotenuse of the large triangle minus the lengths of the two shorter sides of the small triangle (50 - 16 - 12 = 22). I then squared that length to get the area of the square (22 * 22 = 484). Edit: Looks like I'm not the first to solve it by this method.

@falling_banana Жыл бұрын

ur method is (imho) more straightforward & ur explanation's great!

@force10guy26 2 жыл бұрын

One of these days, something I learned here might help me somewhere in life. Thanks Presh 🙏

@jaedenvanderberg3890 2 жыл бұрын

A = arctan(40/30) Ans = (50 - 20cos(A) - 20sin(A))squared Sections. 1. A 2. 50 3. 20cos(A) 4. 20sin(A), and 5. squared. My method here is to find one side length of the green cube, and from there to merely square it. 1. ‘A’ is an angle which appears often in this solution, so I removed it for ease of understanding. A is the bottom right angle of the original triangle, created when taking the arctan of 40 over 30. 2. I find a line which includes the side length of the cube, which would be the hypotenuse of the first triangle we are introduced to, whose length (through Pythagorean theorem) equals 50. 3. The triangle overlap specifically with the bottom right of the blue cube and the top left of the blue cube. In the bottom right, we see that (50-30 = 20) 20 units of the original triangle remain, which leaves 20 units unrevealed (this is a 90 degree rotation to the left of the original triangle). These 20 units form an imaginary triangle whose base is on the 50 unit hypotenuse, and it shares an angle with the original triangle, ‘A’, so trig tells us to use cos(A) = adjacent over hypotenuse, then we rearrange things to get a segment of the original hypotenuse. 4. The method here is the same as in 3, we find overlap in the top left triangle, we use the known 30 units uncovered and remove 10 from that, to know that we have a similar scenario, but the solution is to use trig again. In short, 20sin(A), since the OPPOSITE side of the original triangle (90-A) is part of this little triangle, the lower left angle would be A. 5. We are left with the side length of the green cube, thus we square our answer and get it. 484.

@mohamedsaifalahmer 2 жыл бұрын

square root (30^2+40^2)=50 40+30-50=20 as you did as similar triangle 20×40/50=16 20×30/50=12 the centre square area = (50-16-12)^2=22^2=484

@santiagoarosam430 2 жыл бұрын

Analyzing the overlap between the right triangles whose respective right angles coincide with the lower left and upper left angles of the original square, we see that the hypotenuse of the overlap measures 20 ud and therefore the similarity ratio is 20/50=2/5, so where we deduce that the legs of the flap measure 16 and 12 ud. From the above we can deduce that the side of the square measures 50-12-16=22 ud and the area sought is 22x22=484 ud²

@determinedhelicopter2948 2 жыл бұрын

I don't know where I went wrong with mine but I tried to find the hypotenuse of the small triangles in the corner that would be 20 by 10, so the hypotenuse should be sprt(400+100), and since the inner area is a square and its side is parallel to this hypotenuse I thought it would be an area of 500 where did I go wrong?

@semmathe5628 2 жыл бұрын

Square root the 500 to get the length of the hypotenuse of the 10 by 20 triangle then you have you length of the inner square sides. I did the exact same thing and I got a length of 22.36 using two decimals then I just squared it to get the area of the square

@erniecolussy1705 2 жыл бұрын

Determined Helicopter I have been puzzling about the same thing. I initially did it the same way that is shown in the video and got the same results as the video. Then I did it the way you did. I came up with the same results you did. I have been trying to figure out what invalid thinking has caused one of these answers to be wrong. If you figure this out please reply. Note when doing a pure math problem rounding should never be used. If Sem Mathe didn't round he would also get 500.

@erniecolussy1705 2 жыл бұрын

Determined Helicopter I figured out our error. First to go over the process that we used. We project the sides of inter square toward one of the corners of the outer square. We made corner triangles using the corner of the large square and the two intercept points of these two projected lines with the sides with the sides of the large square as the corners of the triangle. We calculated the length of the hypotenuse of this corner triangle which is sqrt(500). We used the length of the corner triangle hypotenuse as the length of one of the sides of the square to calculate the area of the square. This gave us sqrt(500)^2 which is 500. In finding the area of the inter square this way we assumed that the hypotenuse of the corner triangle is parallel to the side of the square. For this to be true the corner triangle hypotenuse has to be parallel to the hypotenuse of the large triangle. Which means that the corner triangle has to be similar to the large triangle. Since these two triangle look similar we assumed them to be similar. May have even justified this logic. But this two triangles are not similar. They both are right angle triangles so we can test to see if they are similar by comparing the ratio of the two shorter sides. The corner triangle has a ratio of 20/10 reducing to 2. The large triangle has a ratio of 40/30 reducing to 4/3. Since these ratios are not equal the hypotenuses are not parallel. This is where our error came from. Since these two triangles are close to similar the angles of the hypotenuses are not off by much. That is why Sem Mathe error of rounding was able to cancel our error.

@Batanem1000 2 жыл бұрын

I did it. On calculation, I got h = 20 for hidden triangle. So its l = 16 and b = 12. Those are belong to hypotenuse of big triangle. Hyp - (l +b) = a (side of small square) Means (50 - ( 12+16) )^2 = (50-28)^2 =22^2 =484

@unnameableuser Жыл бұрын

Yes, this is a very smart and quick way to solve the problem. I took a longer route and found on it not only the value of the green area, but also that it is a square with the side length equal to 22.

@ayoubkhlifi4903 Ай бұрын

me too

@geoninja8971 2 жыл бұрын

Its always great to see Presh get the same answer, my method was rather less elegant I must say..... more trigonometry to work out the square side length.

@kingdomofjace1190 2 жыл бұрын

The video was uploaded 10mins ago. How is your reply 3days old ?😕😐

@geoninja8971 2 жыл бұрын

@@kingdomofjace1190 wormholes

@safi3968 2 жыл бұрын

@@kingdomofjace1190 Probably because the video was private or unlisted with a custom link, which only supporters from places like Patreons got early.

@kojak8403 2 жыл бұрын

@@kingdomofjace1190 - he's from the future. He can't be bargained with, he can't be reasoned with. He doesn't feel pity, remorse or fear.

@mustafizrahman2822 2 жыл бұрын

@@kingdomofjace1190 He is a time traveler.

@anthony17mapoy46 2 жыл бұрын

Working backward - what a bright idea! Thank you MindYourDecisions for having such a brilliant answer!

@tpros6289 2 жыл бұрын

I figured this out in a simpler way. The triangles are all proportionally similar due to the angles of the inner square to the whole. They are 3,4,5 triangles. The long leg is 40 on a 50 square, the short leg is 30. The triangles overlap in a pattern. You can see the top triangles short leg is cut to 10 due to overlap. This creates a triangle with 20 for the hypot. and a proportion of 3,4,5. So the legs of the cutoff triangle are 12 and 16. Thats equal to the length difference of the 2 squares. 12 + 16 is 28, 50 - 28 is 22. 22 square is 484. Simple. I never calculated the area of anything except the square in question. 2500 never mattered, nor the area of the triangles, and I only subtracted 2 digit numbers.

@arghayeaye4591 2 жыл бұрын

Nice

@Khan-uv3wr 2 жыл бұрын

Bekaar

@paulparker1425 2 жыл бұрын

This is how I did it as well. Much simpler than the given solution.

@ABDxLM 2 жыл бұрын

I didn't understand what do you mean by cut off triangles

@tpros6289 2 жыл бұрын

@@ABDxLM the areas where the triangles overlap are also triangles with the same proportions to the large right triangle on the left. Those areas are cut off of the other triangles. I called those areas cutoff triangles. Theyre not actually there, in fact they are areas that are missing.

@Awesome-ct7vr Жыл бұрын

If you extend all of the internal diagonals You will get an updated diagram of 4 over lapping 3 4 5 squares. Or in another perspective... 4 smaller 3 4 5 triangles and 4 pentagons. If you name the angles alpha and beta You will find all of the 4 triangles and the 4 pentagons are congruent. Now that we know that. You can see the the side of the big square is 50 and the other side we have 40. Leaving us with the short side of the pentagon is 10 (apply that to all of them) Now we look at the most left-down pentagon see it overlaps our 40 30 50 triangle we can see the short side of the pentagon laps over the short side of the triangle Leaving us with 30-10 which is the hyppotenuse of the small triangle and equal to 20. Now we know the small triangle and the big triangle are same 3 4 5 triangle. We get ratios: 5/4 = 20/x --> 5x = 80 --> x=16 16 is the longer side of the small triangle. 5/3 = 20/x --> 5x = 60 --> x=12 12 is the short side of the small triangle (Apply to all triangles) We can see the 2 left small triangles are overlaping our bigger triangle's hyppotenuse (30 40 50) One with the short side (12) And one with the longer side (16) 50 - 28 = 22 The side of the middle square is 22 S□ = 22² = 484

@rajendraprasad936 2 жыл бұрын

this question is very easy, in small triangle the hypotenus will be 20 and according to (3,4,5) side of right angle tri the two other side will be 12 and 16 and to find side of square then 50-(12+16)=22 then square will be 22*22=484 that is easy

@tripik426 2 жыл бұрын

I solved this in a different way, much more complicated and ugly, but I didnt give up. Getting the answer and confirming it, was absolutely euphoric. THANK YOU, And most importantly, THANK ME. You know, there are those who have suggested that solving a nice math problem is like eating chocolate. I agree with that. So much.

@BohumirZamecnik Жыл бұрын

Nice. Another ways is to compute the side of the green square directly as 50 - the two sides of the purple triangle (40+30)*20/50, and then square it.

@johnchessant3012 2 жыл бұрын

Coordinates! Let the origin be the bottom-left vertex of the large square. Solve y = 40 - (4/3)x = (3/4)(x - 10) to get the bottom vertex of the small square; calculate its distance d to the center (25, 25). This is half the diagonal so the area of the small square is 2*d^2 = 484.

@spacescopex 2 жыл бұрын

Please watch MY SOLUTION: kzfaq.info/get/bejne/oZ9hja2C3au0fJc.html

@bobjordan5231 2 жыл бұрын

Knowing the similar triangle equation is going to help a lot on future posts. Thanks!

@anandarunakumar6819 2 жыл бұрын

Taking ratios give 3/5 and 4/5 as each of smaller triangles. It is easy to show the smallest triangle's length is 20, thus 12 and 16 are other two sides. Now 50- 12-16 gives 22 as length of inner square's side length. Now 22^2 = the area of inner square.

@wolftamerwolfcorp7465 2 жыл бұрын

484. You need to find the length of the triangle’s hypotenuse that isn’t overlapping/overlapped. To do that you need to find the area that is overlapped, which is its own triangle that has an easy to find hypotenuse of 20 and shares its angles with the 30-40-50 triangles. Figure out the 37 and 53 degree angles from that, if you don’t know them, and you’ll get side lengths of 12 and 16. Because each side length is represented once and only once when you subtract them from the 50 you get 22. Then because it’s asking for area multiply 22 by 22 and voila.

@gregturner2363 2 жыл бұрын

I solved it by using a real round-a-bout method. I considered the square on a grid with the lower left hand corner being (0,0). Then I came up with the equations of the lines that bound the green square. This allowed me to solve for the coordinates of the corners of the green square, which allowed me to solve for the length of the side of the green square which allowed for solving for the area. I liked your solution much better !!!!!

@unarei 2 жыл бұрын

I did the same thing, just had to solve for one intersection point and then I could rotate it to get the others

@kennethsizer6217 2 жыл бұрын

Wow! I can't decide whether you lose points or get extra credit for that! 🤣

@ReliableRandy 2 жыл бұрын

I basically did this with a system of equations. I used variables ‘x’ to denote the length of the inner square, ‘y’ to denote the shorter leg of the purple triangle and ‘z’ to denote the other non-hypotenus length of the purple triangle. With this we can easily get: x+y+z=50 z^2 + y^2 = 400 (Same method you used to determinne the purple triangle hypotenuse 20^2=400) lastly knowing that they are similar triangles it will have the same angles. Hence cos`(30/50)=cos`(y/20) -> 3/5=y/20. Three equations, three variables. Now we can solve to get the values. y=12, z=16 and x=22 making Area=484

@onrshort4583 2 жыл бұрын

Marvellous!

@vishwasmalhotra3714 2 жыл бұрын

Well I have a solution , let the left bottom coordinates be (0,0) after that you can take out equations of any two parallel lines . Then just apply the formula of distance between parallel lines . That distance will be side of square . Square the distance . Answer will be 484

@19Szabolcs91 2 жыл бұрын

I like how the challenge here only comes from the fact that the way it was originally presented purposefully obscured the lines (the triangles "overlapping). If you actually draw it for yourself without erasing the lines, it becomes super obvious.

@LarsDennert 2 жыл бұрын

I did a fast solution in my head. I just drew a hypotenuse across the bottom right corner from the intersecting points. This line being the length of the green square side. 10 squared + 20 squared of that triangle equals 500. The square root of that being the length of a side. The square being the area.

@smchoi9948 2 жыл бұрын

Coordinate geometry often guarantees a solution manageable by most of us towards problems alike. Place the outer square on the Cartesian plane where one of its corner is at the origin and the 2 sides joining it lie on +ve sides of the axes. N.B. the hypotenuse of one △ passes through (0,40) & (30,0), so the equation of the extended line (w/ slope = -4/3) is y=-(4/3)x+40. Similarly, another passes through (0,20) & (40,50), so the equation concerned (w/ slope = 3/4) is y=(3/4)x+20. The 2 lines intersect at P(48/5,136/5). By symmetry, y=-(4/3)x+40 intersects the hypotenuse of another △ at Q(50-(136/5),48/5) = Q(114/5,48/5). By 2-pt. formula, the required area is PQ² = (48/5 - 114/5)² + (136/5 - 48/5)² = 484.

@quigonkenny 5 ай бұрын

Let's look at two of the triangles along one side. We'll use the left side for convenience. The full side length is 50, one triangle side (the top one) is 30 and the other (the left triangle) is 40, so there is (40+30)-50 = 20 units of overlap. If we look at the overlap, it creates a right triangle with hypotenuse 20 (the overlap along the side of the square) and angles matching those of the original triangles, so the triangles and the overlap triangle are similar. If we look at the original triangles, since rhey have side legs of 30 and 40, they are 10:1 ratio 3-4-5 Pythagorean Triple triangles, and thus the hypotenuse is 50. Since the overlap triangle is similar, and thus also a 3-4-5 right triangle, we can determine the side leg lengths in relation to the known hypotenuse: b = 20(4/5) = 16 a = 20(3/5) = 12 The total area covered by the triangles will be the areas of the four original teiangles munus four overlap triangles, and we can determine the incovered area by subtracting this area from rhe total area of the square: A = s² - (4BH/2 - 4bh/2) A = 50² - (4(40)30/2 - 4(16)12/2) A = 2500 - 2400 + 384 = 484

@vit.budina 2 жыл бұрын

I went the long trigonometric way by doing this: (sqrt(30^2+40^2)-[sin{tan^(-1)(30/40)}*20+cos{tan^(-1)(30/40)}*20))^2

@ghseam 2 жыл бұрын

Hi, another way is to calculate the lengths of small triangles by the proportion to large ones which will be 16 and 12. So the length of green square will be 50 minus ( 16 plus 12) and the answer is 22. The surface area of square is 22 power 2 = 484

@povijarrro 2 жыл бұрын

Let E, F, G, H are point of intersect the boundary of origin square and origin 30-40-50 triangles then Area of parallelogram EFGH is 50 times the side of green square. But it is also 50^2-2*30*40/2-2*10*20/2. So if green square has side length equal to x, then 50x=2500-1200-200. So x=22 and x^2=484

@MikeRendell-zz3kq Жыл бұрын

This was a good problem (MindYourDecisions produces many good problems). I solved it (and got the right answer) an entirely different way. I measured (with math not a ruler) the distance between two of the straight lines and came up with 22. Hence an area of 484.

@anandk9220 2 жыл бұрын

This looked really weird and demoralizing to me. But watching the video solution and thinking more about this problem made me realize another easier way here. If you observe, you'll realize the fact that extending all sides of inner square on to the respective sides of outer square helps us obtain the lengths of 20, 20, 10 for each outer square side in clockwise order, starting from lower left vertex of outer square. After extending the inner square sides on the outer square, we have 2 small similar triangles with the 40-30 triangle given in figure. Applying similarity property gives upper and lower part length (of 40-30 triangle hypotenuse) to be 16 and 12 respectively. Since hypotenuse = 50, Remaining length = Side of inner square = 50 - (16 + 12) = 22 So, Area = 484 square units

@cherylclarke4316 2 жыл бұрын

How does the remaining length equal to 10

@tychhimou5791 2 жыл бұрын

The blue triangles are congruent

@jimmykitty 2 жыл бұрын

*2:38** After illustrating this simplified diagram, I was able to the problem without watching the entire video* 😹 *Love from Bangladesh* 🇧🇩🇧🇩❤❤🌿

@advaykumar9726 2 жыл бұрын

Hello

@farhan5103 2 жыл бұрын

Same😊 I am also Bangladeshi

@mustafizrahman2822 2 жыл бұрын

Wow! Einstein.

@jimmykitty 2 жыл бұрын

@@advaykumar9726 Hii.. Legends are everywhere 🤩

@jimmykitty 2 жыл бұрын

@@mustafizrahman2822 Wow! Isaac Newton 🤩

@playgroundgames3667 2 жыл бұрын

417 rounded SA: A^2 + B^2 = C^ | vC = c, c divided by three is the side length of each side of the green square. ⛽ 34.75 is the length of each side of the green square.

@angrytedtalks 2 жыл бұрын

Alternatively, the side length of the inner square can be found using pythagoras on diminishing triangles: 50-16-12=22 Therefore inner square is 22²=484

@AArrakis 2 жыл бұрын

Did the same

@twinnklle 2 жыл бұрын

I got answer to be 500! I connected the two points: the 40 mark of one triangle and 30 mark of the opposite triangle. Here, the sides will be 10 and 20 respectively. As hypotenuse is parallel to the side of the square and cut by opposite sides of the square, the hypotenuse squared is the area of the square, i.e. 10^2+20^2 =500! Please correct me if i am wrong…

@spiderjump 2 жыл бұрын

I solved for the lengths of the smaller triangles using similar triangles which are 12 ,16 and 20 50 - 12-16= 22 22 x22 = 484

@araptuga 2 жыл бұрын

I did it by thinking of two of the original triangles on opposite corners (upper right and lower left) being removed. The original square has area 2500. The two triangles would form a 30x40 rectangle of area 1200. So the leftover strip (from upper left to lower right) has area 1300. The length of each parallel strip is 50 (hypotenuse original triangle). If we subtract from that the length of the little green square (call it L), then the remaining part is (50-L). Now, consider the two remaining quadrilaterals that, along with the green square, make up the strip. They are identical dimensions. Slide one down until it overlaps with the other one, forming a rectangle with width L and length (50-L), or area (50L - L^2). Next, consider that those two quadrilaterals overlap. Each will overlap by an area equal to the small right triangle in lower right, of length 20 and height 10. Together they form a rectangle of area 200. OK, now we're able to put all that together: Area strip = Area of ("green square" + "blue rectangle" - "overlap rectangle"). Or: 1300 = L^2 + 50L - L^2 - 200 1100 = 50L 22 = L Thus area green square = L^2 = 22^2 = 484

@tutoygonzales293 2 жыл бұрын

Got the answer in a different approach. Getting the area of 2 triangles in opposite sides, top and bottom, will give you an areas equal 1200, 600each. Then rmoving those triangles will give you hexagon, with that, it will give you 2 triangles with 20 and 10 as sides. You can get areas equal to 200, with that, you can just subtract it with the remaining area and you can get yhe side of 22.

@9machine4you 2 жыл бұрын

1:10 - 2:29 please do more of this type of stuff :)

@N8570E 2 жыл бұрын

There is another way, determined after the main calculations: 1. After determining the hypotenuse of the small triangle. 2. Determine the length of each side of the square. And of course, it being a square, determining the length of one side is adequate. 3. The purple triangle is a 3 by 4 by 5. But it needs to be multiplied by 4, to give 12 by 16 by 20. 4. Subtract each 'small' side from 50. 50 - 12 - 16 = 50 - 28 = 22. 5. 22 squared is 484. MindYourDecisions (a.k.a., Presh Talwalkar), thank you for your efforts. May you and yours stay well and prosper.

@Narennmallya Жыл бұрын

Another way is this right.. bigger left most corner triangle is similar to the purple one towards the left top side So, X/40 = 20/50 this x = 16 And the other side of the smaller triangle becomes sqrt((20)^2 - (16)^2)= 12 So the side length becomes 50-12-16=22 So area of the shaded portion becomes square or 22 = 484

@bheriraju722 2 жыл бұрын

Hypotenuse of smaller triangle is 20 Other 2 sides of same triangle are 16 and 12 because it's scaled 345 right triangle So side of inner square is 50-16-12=22 units Area = 22 x 22 = 484 square units

@shadrana1 2 жыл бұрын

At 0:46,label the green square ABCD in clockwise direction from eastern apex, Set up grid (0,0) at the SE apex of the large square. Consider extended line AB, y=mx+c m=30/40=3/4 At x=0,y=20 20=3x/4+c,20=0+c,c=20 Therefore,y=3x/4+20................................(1) Consider the extended lineAD, y=mx+c, m=40/30=-4x/3 since AD is perpendicular to AD, At x=0,y=40, 40=0+c,c=40, y= -4x/3+40...............................................(2) the x and y are common at the point where AB and AD meet, Equate (1) and (2), 3x/4+20= -4x/3+40 Multiply by 12, 9x+240= -16x+480 25x=240,x=240/25=48/5 Substitute into (2), y= -4/3*48/5+40= -64/5+200/5=136/5 Point A is A(48/5,136/5)........................................(A) Consider the extended line BC, y=mx+c, m= -4x/3, BC is perpendicular to AB, At x=50,y=10, 10= -4*50/3+c= -200/3 +c c=230/3 Therefore, y= -4x/3+230/3......................(3) Point B is where AB and BC intersect, Equate (1) and (3), 3x/4+20= -4x/3+230/3 3x/4+4x?3=230/3-60/3 Multiply by 12, 9x+16x=680 x=680/25=136/5 Substitute x into (1), y=3*136/4*5 +20=408/20+400/20=808/20=202/5 Point B is B(136/5,202/5).........................................(B) Let S=side length of green square and S^2 =area of green square, By Pythagoras, S^2= {(136-48)/5}^2+{(202-136)/5}^2=(88/5+66/5)^2=(7744+4356)/25=12100/25 =484 square units and that is our answer. This is a good place to stop.

@donaldasayers 2 жыл бұрын

Small triangles similar to big triangles, scale them and that gives the side of little square as 22.

@crane8035 2 жыл бұрын

You can directly obtain the sides length by doing a lil work on the hypotenuses’ overlap

@PeterPan-dz7mu 2 жыл бұрын

You can also find the length of the square's sides by subtracting the height of the triangle twice from the diagonal of the big triangle.

@saranshbharti3875 2 жыл бұрын

Since there is similarity in triangles, you can easily figure out that the green region is indeed a square. The purple region is a similar triangle with hypotenuse 20, which gives the other sides to be 16 & 12. The side of the green square comes out to be 50-16-12 = 22, which gives the area to be 484 in lesser time.

@eyeofthasky 2 жыл бұрын

0:49 after looking at it for 3 seconds i thought: wellwe need the length of a side of this inner square... and as the triangle on the bottom right is in a manner overlapped, that the remaining width is the length of that square, i just need to determine that shape. just following the outer border around the corner, i can calculate that width since thats another triangle whose hypothenuse is parallel to the sidelength of the green square, with 90° angle so pythagoras applies too. 50-30, the remaining 20 is one leg, 50-40=10 the other, so with A=s²=(10²+20²)=500

@uditchoudhary6740 2 жыл бұрын

Very simple question bring some tough questions

@doko239 2 жыл бұрын

I got the answer in a slightly different way; once I got the ratio of the small and large triangles, I multiplied the side lengths of the legs of the large triangle by the ratio to get the lengths of the small triangle. Because of the way the diagram is laid out, the length of the side of the central square is equal to the hypotenuse of the large triangle minus the legs of the small triangle, which is 22. 22 squared is 484.

@len3160 2 жыл бұрын

A simpler approach : The height of the blue triangles = 24 and aligned with the diagonal, being 50*sqrt2. Green square side length is diagonal - 2*triangle height, = 22.7, and area = 484.

@georgezhang865 2 жыл бұрын

How do you know the purple triangle is 4 times smaller? I mean it’s kinda obvious but is there a mathematical way to proof?

@BriBear 2 жыл бұрын

3:46 how did you find the hypotenuse equals 20?

@Noobish_Monk 2 жыл бұрын

Oh heck always find out that I just don't know the theory for solving these problems. Then you always name things like "similar triangles" in this video and I then solve the problem. Thank you!

@ayoubkhlifi4903 Ай бұрын

it's enough to solve it just if you know that there is a sin = opposite/hypothenuse as i did and you will find the rule of the similar triangles even if you don't know it exists as a rule,

@OrenLikes 2 жыл бұрын

Very nice! I placed the big square on a coordinate system with the bottom left = (0, 0), Found the function of two parallel diagonals and one of the other two, Found the x's and then the y's of the two intersects - ends of one side of the small square, Found the distance between them = side length, Squared that distance = area of the small square. 3-4-5 right-triangle came in handy. Scaled down by 10 to make it easier - got 25ths..., Multiplied by 25 to make it easier, Got side length 55, then multiplied by 10 and divided by 25 back to the initial non-scaled: 22. Squared that, and got the correct answer of 484. 3-4-5 right-triangle is a special case of the Pythagorean Theorem - so knowing the Pythagorean Theorem is enough - for anyone who doesn't know similar triangles and x^2 areas (which could be deduced as you showed).

@ayoubkhlifi4903 Ай бұрын

if you didn't know about the similar triangles as me,it will be enough to work with the tan of this two angles B and A ,and you will arrive to the theorem,for exemple let's take a and b sides of the small triangle and c and d of the big one,the tan(A) =tan(A) imply from the first and the second triangle: a/b=c/d

@ayoubkhlifi4903 Ай бұрын

and of course there is the third side the hypothenus h but i didn't work with it to find the rule of similar triangles

@yvessioui2716 2 жыл бұрын

I love listening to your videos and trying most of them. I did try this one and got it wrong. It forces me to investigate my own assumption(s) in building my case. That is one side of listening to your demo. I found out why I was wrong and happy to have understood why. As info on the 1st procedure I tried : I slide the green square at the bottom right corner of the big square. Found from 50 -40 the lenght of one leg of the small triangle in that corner and found out the other leg of that triangle, 20 (50-30). The hypotenuse of that triangle being x, the x we need to square for the area of the green triangle. So x=10 squared + 20 squared=500. I was wrong because I assumed too much, specially that green square ending flush with the big square on the corner.

@agoogleaccount5207 Жыл бұрын

My process for the answer was to find the outer lengths of the bottom right shape, find the hypotenuse of those outer lengths, then square it, since the green area is a square and the hypotenuse would be a length of it. One length of the triangle would be 20 since The entire shape is also a square so 50 - 30 = 20, for the other length, I guessed that because of how all the blue shapes interact with the green square, the lengths on the left would be equal to the right, if so then the second length would be 10. Then using Pythagorean, c^2 is 500 meaning the green square has an area of 500.

@balakrishnakarri6264 2 жыл бұрын

I have simplest way... Hypotenuse of triangle is 50 Short leg of small triangle is 12 (small triangle sides 20,16,12 ) Long leg of small triangle is 16 Hypotenuse of triangle (30,40,50) consists long and short legs of small triangle and side of green square so 50 = 12+16+ side of green square Side of green square = 22 Area of green square = 22×22= 484

@shubhambairagi7866 2 жыл бұрын

How u got A2=600×(20/50)^2 ....any formula bro ....I did not understand this part in this video plzz tell me

@sk.43821 2 жыл бұрын

Compare with 1:51 following. Especially listen to sentence at 2:19 "This justifies the principle that if the triangles have *sides* in a *ratio* equal to *x* their *areas* will be in a proportion that's equal to *x squared* ." The *sides* of the small triangles (because they are similar to the big triangles) are all in a *ratio of 20/50* to the sides of the big triangles. [small triangle hypotenuse = 20, big triangle hypotenuse = 50 ; small triangle legs are 20/50 x 40 = 16 and 20/50 x 30 = 12] That's why the *areas* of the small triangles are in a proportion of *(20/50)^2* to the areas of the big triangles.

@user-ch3wv1ly1w 2 жыл бұрын

Whenever I solve a problem from this channel correctly my self esteem just goes up

@md.abdulmomenadil8437 2 жыл бұрын

This is the first math on your video that I could do by myself ☺️☺️☺️ it was a pretty easy one .. though i didn't know your method before. I did simple trigonometry and got the measurements of the triangle by sin@ and cos @

@MyHydralisk 2 жыл бұрын

Didn't knew/forgot the formulae for similar triangles area....and took a very long way to solve this. So I've built a rhombus out of those four purple triangles with the side of 20. And I know the area of rhombus is a^2*sin(A). Angle A is double of purple angle (doesn't matter witch), so sin(A)=2*sin(Purpleangle)*cos(Purpleangle). That sin and cos can be equated from big triangles as 30/50 and 40/50 (since they are similar). So sin(A)=24/25. Then rhombus area=20*20*24/25=384. Then I pretty much did the same, but instead of +4 purpleareas I did +1rhombus area (which is same things). The length one goes without one simple formulae :D

@riccardofroz 2 жыл бұрын

My Approach was sligthly easier I think. The hypotenuse of the triangles is equal to 50. The small overlapping triangles are similar to the triangles since they have exactly the same angles and their hypotenuse are equal to 20 (as per the video). Here is what I would do next: The other two sides are 12=30*20/50 and 16=40*20/50. Then to find the side of the internal square you simply need to subtract 12 and 16 from 50, that is 22 which squared is 484.

@bobajaj4224 2 жыл бұрын

484 = 22^2.. and another method is to use a coordinate system and find the equation of the 4 lines which is pretty easy, then find the intersection points :)

@spacescopex 2 жыл бұрын

Please watch MY SOLUTION: kzfaq.info/get/bejne/oZ9hja2C3au0fJc.html

@greysun27 2 жыл бұрын

I had to leave a comment because this is the one of the first questions I wholeheartedly solved and got right on this channel. I completely re-drew the figure on a coordinate plane so that the bottom left corner was at the origin. I then found linear equations for all 4 diagonal lines. I noticed the smaller 4 triangles and found the intersection of what I'm going to call line 3 and line 4. Those lines and the y axis made up the far left triangle that sat on the y axis. Finding the insection of this triangle was essential finding the height on the triangle which was 9.6. The base of the triangle was 20 as it was the distance of the 2 y-intercepts (0,40) and (0,20). Making the area of the smaller triangle (20*9.6)/2= 96. I then found the area of the larger triangles and subtracted the overlap. I did this: 1200+1200-(4*96)= 2016 Finally I took the total area of the square and subtracted my calculated area. 2500- 2016= 484

@tychhimou5791 2 жыл бұрын

The first glance at the cover picture, I was trying to figure out if the blue triangles are congruent...

@axbs4863 2 жыл бұрын

If the square is centered or something couldn’t you just subtract the distances to get to get the diagonal is equal to 30 and then solve for the sides to get the area? Or am I missing something

@funtimewithhuskyman5601 Жыл бұрын

I have solved it using a bit defferent way, but how can you say that the hypotenuse of that purple triangle equals to 20?

@markmark2080 2 жыл бұрын

I was only off by 6.5, by measuring it with my tape measure in 1/32nds of an inch...I should have used calipers.

@user-pu5dd8tf7c 4 ай бұрын

Here is another simple method without calculating any triangle area: Sides of the right triangle: a = 30 b = 40 c² = a² + b² = (3⋅10)² + (4⋅10)² = 9⋅10² + 16⋅10² = 25⋅10² c = √(25⋅10²) = 5⋅10 = 50 (hypotenuse) Angles of the triangle: sin(∠A) = a/c = 30/50 = 3/5 sin(∠B) = b/c = 40/50 = 4/5 The overlapping triangle is a right triangle with legs x, y, and the hypotenuse z, and it has the same angles as the larger triangle, i.e. both triangles are similar. The hypotenuse of the overlapping triangle is equal to: z = a - (50 - b) = 30 - (50 - 40) = 20 sin(∠A) = x/z 3/5 = x/20 x = 20⋅3/5 = 4⋅3 x = 12 sin(∠B) = y/z 4/5 = y/20 y = 20⋅4/5 = 4⋅4 y = 16 The green square has sides equal to: s = c - x - y = 50 - 12 - 16 = 22 The green square has therefore an area equal to: s² = 22² = 484

@donaldhobson8873 2 жыл бұрын

Or just use that the purple triangle are 12 16 20 so the green square is 22 x 22.

@hnahler 2 жыл бұрын

And the green square has sides of length 22 = sqrt(484). Using the similarity of triangles, the small triangle has length 20, 16, 12. Looking along one of the tangents to the green square, the tangent has a length of 50, the two sections other than the middle section touching the green square have lengths 16 and 12 from the small triangles (adjacent and opposite). Therefore, the side of the green square is 50 - 16 - 12 = 22 and it's area is 22^2 = 484. - As you said, yours is one way to solve the problem. This one is using the same starting point (similarity of triangles).

@pavelgatnar7164 2 жыл бұрын

The small triangle has legs 12 and 16, subtract it from 50 and you'll get the side of the green square. Then the area is 22^2=484.

@smthB4 2 жыл бұрын

Exactly how I did it.

@shahriar1111 2 жыл бұрын

Me too

@PuzzleAdda 2 жыл бұрын

Riddle - I have 3 brothers and each brother has three brothers. How many are we? Answer - kzfaq.info/get/bejne/ad1pes2Z192dp6c.html

@billhill897 2 жыл бұрын

the ratio of the hypotenuses of the small triangle and the large triangle is 2/5 therefore the legs of the small triangle are 12 and 16. Therefore the side of the of the green square is 50-(12+16) = 22. 22 squared is 484

@andyiswonderful 2 жыл бұрын

I solved it algebraically using the equations of the straight lines, and solving for the coordinates of the corners of the inner square. Also got 484.

@honeygala6187 2 жыл бұрын

Area uncovered by triangles = area under overlap. Hence square = 4* 96 = 384

@calholli 2 жыл бұрын

I did the end differently........ I added two full 600 triangles.......... Then I took a 600 triangle and subtracted the two overlapping purples... 600 -96 -96 = 408... So 50x50= 2500... 2500 - 600 -600 -408 -408 = 484

@ThAlEdison 2 жыл бұрын

I found the length of the legs of the purple triangle. The side of the green square ends up being the length of the hypotenuse minus the longer leg of the purple triangle and also minus the shorter leg of the purple triangle. 50-16-12=22, 22^2=484.

@ramosapien 2 жыл бұрын

Presh, thank you for your videos. Could you tell me what program do you use to make these animations? Thanks once more!

@milantokar704 2 жыл бұрын

So simple solutions I can see there :( I gone through trigonometry => angles in triangle; then lengths of small triangle sides; and these subtracted from 50. Final number the same, but that wayyy...

@akmil807 2 жыл бұрын

So my thinking was, each of the outer edge parts are 10×50=500 That leaves four overlapped corners being 10×10=100 so the entire edge would be 4×500=2000 Minus 4×100=400 meaning that the edge parts were 1600. The rest can be divided by 2 because of the way it works with the edges, concluding that it would be 900/2=450. Why doesn't this work? What is the thing I missed?

@242math 2 жыл бұрын

understand the process, very well explained, thanks for sharing

@dhannukumar2504 2 жыл бұрын

Coordinate geometry is the Another method for solve this problem let the point like (30 , 0)... show on and find the eq. Of the and then also there intersection like ( _ , 96 ) and find the value of the small area 😀 and continue...

@fongalex6639 2 жыл бұрын

Very good video to apply fundamental concepts to solve a question with medium difficulty

@GDPlainA 2 жыл бұрын

I drew some lines and I see 4 smaller triangles that are similar to the big triangles. By observation, we have the sides 12, 16 and 20 and the area is 96. Then we find the blue area which is 4(big triangles)-4*96=2016 Then 50^2-2016= 484

@GDPlainA 2 жыл бұрын

I actually found another way which is to find the side of the green square which is to subtract the hypotenuse of the big triangle by the 2 sides of the small triangle which would be 22. then square it and it becomes 484

@xz1891 2 жыл бұрын

My way, ez Show all hidden lines, note all rt triangles are 3-4-5 rt ones. Square side's dividing pts divide the length as 20, 20, 10, thus small rt triangle has two sides of 12 and 16, thus the side length of the inner square (must be, due to symm) =50-12-16=22, thus its area =22*22=484

@ricardoescareno8135 2 жыл бұрын

Alternatively, but also using the similarity of triangles, we could calculate the sides of the green square. It's the hipotenuse of the big triangle (50), minus the legs of the small triangles (12 & 16). Thus, the square's sides are equal to 22, and its area is 484

@wizrom3046 2 жыл бұрын

I also calced thenside of the square, using a typical engineering solution sin to get the distance from outer corner to the side of the square, then double that and subtract from the big square diagonal distance.