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.
@kingdomofjace11903 жыл бұрын
The video was uploaded 10mins ago. How is your reply 4days old ?😕😐
@artha16793 жыл бұрын
@@kingdomofjace1190 re-upload?
@Discobobulated3 жыл бұрын
@@kingdomofjace1190 The video was unlisted prior, likely to his Patreon supporters, he then set it to public later.
@sahas15143 жыл бұрын
I did it this way
@kmsbean3 жыл бұрын
That's how I did it in like 30 seconds
@dayashankarsuresh573 жыл бұрын
It is intriguing how a perplex looking question's solutions happens to be one of the most satisfying and basic answer.
@bobbytheferret68092 жыл бұрын
Ok
@HASANALI-zt7xz Жыл бұрын
Maths is simple things complicated..I mean with very less concept, u can like create a 1000 questions....
@phasm423 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.
@anjhindul3 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.
@Zollaho3 жыл бұрын
Went directly for this way. Takes 3 minutes mentally.
@professorx30603 жыл бұрын
Same. But I did it both ways just for fun
@Vyrlokar3 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)
@warfyaa61433 жыл бұрын
That is exactly what I did.
@raphaelkometpiste83113 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.
@clieding3 жыл бұрын
Your solution is very simple and elegant- congratulations! 🤩💐
@lellab.81793 жыл бұрын
I like your solution!
@juancarlosfontanafontana9093 жыл бұрын
👍
@stevieg12273 жыл бұрын
Nice one!
@nidalapisme3 жыл бұрын
A simple and elegant solution. I like it ❤️
@erikwiessmann86813 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).
@albiewxdude3 жыл бұрын
This is how I did it as well.
@huwpickrell12093 жыл бұрын
And me
@trnfncb113 жыл бұрын
I did that too. It's the quickest with straight lines.
@eterty83353 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
@WarmWeatherGuy3 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.
@SteezyMD3 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
no
@snotgarden75693 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) )
@tripik4263 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.
@anthony17mapoy463 жыл бұрын
Working backward - what a bright idea! Thank you MindYourDecisions for having such a brilliant answer!
@justgio5992 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
@somewhatblankpaper14232 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.
@force10guy263 жыл бұрын
One of these days, something I learned here might help me somewhere in life. Thanks Presh 🙏
@19Szabolcs913 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.
@ФилиппЛыков-д8е3 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.
@sergeyg29263 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.
@ФилиппЛыков-д8е3 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.
@kjl30803 жыл бұрын
No! This isn’t how you’re supposed to play the game!
@sergeyg29263 жыл бұрын
@@ФилиппЛыков-д8е Thanks for the excellent explanation - totally forgot about projecting a vector onto a unit normal.
@jaedenvanderberg38902 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.
@shadrana13 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.
@lumsdot3 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.
@bobjordan52313 жыл бұрын
Knowing the similar triangle equation is going to help a lot on future posts. Thanks!
@9machine4you3 жыл бұрын
1:10 - 2:29 please do more of this type of stuff :)
@CristinaR-1112 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.
@ayoubkhlifi49035 ай бұрын
me too
@N8570E3 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.
@sleepybutawake7493 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.
@birkett833 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.
@keithroberts14013 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..
@birkett833 жыл бұрын
@@keithroberts1401 Task failed successfully!
@smchoi99482 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.
@cdmcfall2 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_banana2 жыл бұрын
ur method is (imho) more straightforward & ur explanation's great!
@santiagoarosam4302 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
@geoninja89713 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.
@kingdomofjace11903 жыл бұрын
The video was uploaded 10mins ago. How is your reply 3days old ?😕😐
@geoninja89713 жыл бұрын
@@kingdomofjace1190 wormholes
@safi39683 жыл бұрын
@@kingdomofjace1190 Probably because the video was private or unlisted with a custom link, which only supporters from places like Patreons got early.
@kojak84033 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.
@mustafizrahman28223 жыл бұрын
@@kingdomofjace1190 He is a time traveler.
@anandk92203 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
@gregturner23633 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 !!!!!
@unarei3 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
@kennethsizer62173 жыл бұрын
Wow! I can't decide whether you lose points or get extra credit for that! 🤣
@mohamedsaifalahmer2 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
@Terminarch3 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
@huwpickrell12093 жыл бұрын
Yep my method too
@tpros62893 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.
@arghayeaye45913 жыл бұрын
Nice
@Khan-uv3wr3 жыл бұрын
Bekaar
@paulparker14253 жыл бұрын
This is how I did it as well. Much simpler than the given solution.
@ABDxLM3 жыл бұрын
I didn't understand what do you mean by cut off triangles
@tpros62893 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.
@johnchessant30123 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.
@spacescopex3 жыл бұрын
Please watch MY SOLUTION: kzbin.info/www/bejne/oWqUi32HraqDgsU
@vishwasmalhotra37142 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
@eyeofthasky2 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
@LarsDennert2 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.
@anandarunakumar68193 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.
@rajendraprasad9363 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
@Batanem10003 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
@ReliableRandy3 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
@onrshort45832 жыл бұрын
Marvellous!
@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.
@Awesome-1001 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
@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.
@wolftamerwolfcorp74653 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.
@jimmykitty3 жыл бұрын
*2:38** After illustrating this simplified diagram, I was able to the problem without watching the entire video* 😹 *Love from Bangladesh* 🇧🇩🇧🇩❤❤🌿
@advaykumar97263 жыл бұрын
Hello
@csegrowth3 жыл бұрын
Same😊 I am also Bangladeshi
@mustafizrahman28223 жыл бұрын
Wow! Einstein.
@jimmykitty3 жыл бұрын
@@advaykumar9726 Hii.. Legends are everywhere 🤩
@jimmykitty3 жыл бұрын
@@mustafizrahman2822 Wow! Isaac Newton 🤩
@balakrishnakarri62643 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
@СофьяСедова-р8ч3 жыл бұрын
Whenever I solve a problem from this channel correctly my self esteem just goes up
@vit.budina2 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
@shadrana13 жыл бұрын
Take the 30,40,50 triangle at 4:55, the two purple triangles are congruent 20,16,12 triangles Looking at the hypotenuse of the 30,40,50 triangle you have x=50-12-16=50-28=22, Area of green square =x^2=22^2 =484 square units and Bob's your uncle.
@bheriraju7222 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
@playgroundgames36673 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.
@ghseam3 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
@magibalank93662 жыл бұрын
That was a very satisfying answer
@tutoygonzales2933 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.
@len31602 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.
@OrenLikes3 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).
@ayoubkhlifi49035 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
@ayoubkhlifi49035 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
@quigonkenny9 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
@doko2393 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.
@araptuga3 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
@greysun272 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
@povijarrro3 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
@yvessioui27163 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.
@budybradley82572 жыл бұрын
Wow. That’s a nice solution.
@saranshbharti38752 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.
@RadekBuczkowski-h2y7 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
@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.
@hnahler3 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).
@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
@kashyapsakshamkumar3 жыл бұрын
Did exactly the same. Glad to be able to solve this!
@Aiden-xn6wo3 жыл бұрын
This is a really nice problem, and a really nice solution.
@kienwenchang71083 жыл бұрын
Specific-size triangles placed inside specific-size square would yield specific-size non-overlaps. If No need to determine total area of overlaps, no need. 1) we know the area of the large square 2) we know how many triangles and their total area 3) we know these triangles are placed in Certain formation to produce a non-overlaps square.
@kienwenchang71083 жыл бұрын
Always Assume Simplest Case! (When Very Little Apparent Information or Description is Given)
@Noobish_Monk3 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!
@ayoubkhlifi49035 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,
@brothapipp3 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?
@bennjanse3 жыл бұрын
Now we are 3 that did it this way I think we correct 🍺🍺🍺
@sagnikdas60493 жыл бұрын
Me too🙋
@sagnikdas60493 жыл бұрын
I was trying to find this thread lololol
@sagnikdas60493 жыл бұрын
They got 22 as each side of the green square As opposed to 22.3606... or 10√5
@sagnikdas60493 жыл бұрын
I just got our mistake... Drat
@sadtear7963 жыл бұрын
I had solved it in other way and got the same answer ,that was amazing,thank u sir
@spiderjump3 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
@billhill8973 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
@riccardofroz3 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.
@angrytedtalks3 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
@AArrakis3 жыл бұрын
Did the same
@twinnklle3 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…
@calholli3 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
@hippophile3 жыл бұрын
After spotting the small triangles are similar 12-16-20 triangles, then small square has a side 50 - 12 - 16 = 22. So the area is 22x22=484. Easy peasy.
@dhannukumar25043 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...
@jack-xf6il3 жыл бұрын
The original question was "what portion of the square is not covered by the triangles"? First of all I got 19.36%, but then on re-listening I think the right answer is the white bit in the middle that then turns green. When he says "in other words" he changes the meaning of the question and then answers the new question.
@НиколайНБ3 жыл бұрын
Выделил маленький треугольничек, чья площадь повторяется в пересечении больших прямоугольных треугольников, нашел его гипотенузу, элементарно, равную 20. А далее через подобие большого и малого составил пропорцию площадей как отношение сторон, выраженных через гипотенузы и косинус и синус. Но так как угол общий то они сократились и осталось отношение квадратов гипотенуз. Маленькую мы знаем - 20. А большую найдем и будет 50. Далее отношение будет 4/25. Площадь большого 600( 1/2*30*40). Находим маленького - 96. И тогда площадь квадрата будет 2500-2400+5*96=484
@MyHydralisk3 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
@242math3 жыл бұрын
understand the process, very well explained, thanks for sharing
@bobajaj42243 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 :)
@spacescopex3 жыл бұрын
Please watch MY SOLUTION: kzbin.info/www/bejne/oWqUi32HraqDgsU
@clieding3 жыл бұрын
The difficulty of a problem is relative to the level of one’s mathematical development. A problem that earlier was difficult later becomes trivial as one acquires more skill and understanding. The problems presented on this channel are intended for a wide audience whose members are of various ages with various levels of skill, experience, and knowledge therefore categorizing them as „easy“ or „hard“ is meaningless. Most of us come here to learn, share, be challenged and entertained. Those of higher skill should refrain from discouraging or mocking those who have not yet reached the same level of understanding. [The references, often posted here and elsewhere, to nationality and race suggesting some presumed superiority are ignorant and offensive and reveal the insecurities, envies, arrogance and ugly irrational tribalism of those who engage in such behavior.]
@MyHydralisk3 жыл бұрын
What about darvinism? Survival of the fittest? Say, if in one not so big part of the world (i.e. country or region), historicaly, for a long period of time, mathematicians (or runners, or heavylifters, or cunning lads) were in favor, had more means to procreate and establish their genetic heritage - wouldn't this influence current generation? Globalism and social dynamism (which can lead to less variations in genome of humankind per territory) are with us only for couple decades, not even one generation is fully impacted. There is some merit and rationalism to said references - that doesn't mean it should support superiority notions at all, we are all beautiful and each of us makes humankind brigther, but labling any such reference as revealing 'insecurities, envies, arrogance and ugly irrational tribalism' is a bit irrational in itself. That is, if we follow 'survival of the fittest' theory...
@clieding3 жыл бұрын
@@MyHydralisk When I said : „references“ I meant in the limited context of the intentions of such posters and their unqualified statements that members of various nationalities and races are intrinsically „smarter“. I stand by my original evaluation that such arrogant, mean spirited, and irrationally tribalistic pronouncements are deeply motivated by insecurity and envy. They are an attempt to raise one‘s sense of self worth by mystically claiming to possess the qualities, skills, attributes and achievements of individuals because one happens to belong to some arbitrary grouping such as race, religion, gender, nationality etc. Each human being is unique. Each has a unique set of genes as well as a unique life experience, personality, thoughts, abilities etc. Quantifying the qualities of collections of human beings is mere statistical analysis and therefore reveals little about specific individuals.
@Johnny.Fedora3 жыл бұрын
@@MyHydralisk, 1. You don't understand what generalization (and thus prejudice is. 2. You don't understand what natural selection is, and when it is and isn't appropriate to apply it. Keep in mind the old saw, "When your only tool is a hammer...."
@MyHydralisk3 жыл бұрын
@@Johnny.Fedora Sorry, I didn't quite understood your point.
@cstrutherskgs3 жыл бұрын
Tribalism is a natural human evolutionary tendency. Discouraging it is not always easy or constructive if done improperly. It can easy just lead people to form other stronger ‘tribes.’
@sagnikdas60493 жыл бұрын
Edit: just noticed my mistake. See comment for explanation I got 10√5 (22.360...) as each side and 500 as the area as opposed to 484 which is 22 squared Seeing that it's the same right angle rotated each side, I calculated the side of the square by finding the hypotenuse of a smaller triangle from the corners of the big square (the sides are 10 (i.e. 50-40) and 20 (i.e 50-30) )
@sagnikdas60493 жыл бұрын
10√5 is the hypotenuse of the apex triangles or the line connecting the 2 end points of 2 parallel 30/40/50 triangles.... But the catch is that I was assuming that the sides of the green square is equal to that hypotenuse... But that's not correct. If the small overlapping triangles have sides 12/16/20... They aren't isosceles. So the hypotenuse 10√5 is not parallel to the side of the green square. With 10√5 as the hypotenuse, consider another right angle to solve the side of the green square (which would be the perpendicular/height); The base here is (16-12)= 4 So the side is √((10√5)^2 - 4^2) = √(500-16) =√484 =22 Since area of green square is side squared, answer is 484
@sagnikdas60493 жыл бұрын
I realised my mistake while I found someone else doing the same thing that I did... Guess it is easier to spot somebody else's mistake than your own 👀
@md.abdulmomenadil84372 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 @
@fongalex66393 жыл бұрын
Very good video to apply fundamental concepts to solve a question with medium difficulty
@ThAlEdison3 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.
@BriBear2 жыл бұрын
3:46 how did you find the hypotenuse equals 20?
@andyiswonderful3 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.
@crane80353 жыл бұрын
You can directly obtain the sides length by doing a lil work on the hypotenuses’ overlap
@PeterPan-dz7mu3 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.
@donaldasayers3 жыл бұрын
Small triangles similar to big triangles, scale them and that gives the side of little square as 22.
@milantokar7043 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...
@rxt7403 жыл бұрын
Very ez , subtract 12 +16 from 50 then square it to get 22^2 = 484
@xz18913 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
@pavelgatnar71643 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.
@smthB43 жыл бұрын
Exactly how I did it.
@shahriar11113 жыл бұрын
Me too
@PuzzleAdda3 жыл бұрын
Riddle - I have 3 brothers and each brother has three brothers. How many are we? Answer - kzbin.info/www/bejne/aaiceJ2ep9xsrdU
@AldoRPX3 жыл бұрын
Si se prolonga la hipotenusa del triángulo superior, se forma un pequeño triangulo: Hipotenusa = 20 Lado menor = 12 Lado mayor = 16 Lado del cuadrado = 50 - 12 - 16 = 22 Área = 484
@ricardoescareno81353 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
@wizrom30462 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.