Triangles In A Row
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- เผยแพร่เมื่อ 9 ม.ค. 2021
- Can you solve for the shaded area?
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Presh your decisions, I'm Mind Talwakar
Good one😂😂😂
Hello holo the wise wolf
I am a fidget spinner
lmao
Presh I'm decisions, your Talwakar mind
i was close. the solution i got was, "Thursday".
Weird. I got Heron.
I just got Abraham Lincoln
yea idk
th-cam.com/video/vUcNDYlBtoc/w-d-xo.html
hey whats all the soundtrack he used in this vid?
Very close 😂😂😂😂😂
This is what I did:
Step 1) drop a perpendicular from the vertex of 4rth triangle to its base
Step 2) find the angle formed
Step 3) give up and watch the video
@insert username would you like to share?
th-cam.com/video/vUcNDYlBtoc/w-d-xo.html
Same
find the angle formed? the perpendicular?
Did amyown else see what nonsense itnis that he does with s and part sof 3 and 4 it makes no sense..
For those who don't understand 1:23 step: look at the big triangle with side S and 3S in that minute. The ratio of its sides is 1/3. So the second unknown side can be determined by solving equation x / 2S = 1/3, so x = 2/3S. Just look at this as another big triangle, but without a part of it (without first yellow triangle and the white part next to it). Similar with the third side. x / S = 1/3, so x =1/3S.
Yes but how do you proof the ratio
The ratio of two of the same sides is always equal for similar triangles. That’s part of the definition of similar triangles
@@takyc7883 Think I get it now: because the larger triangle has top side length 3S and end side S, the similar triangle with top side 2S has end side 2/3 S.
I calculated equilateral triangle side by
s=2/3*3^3/4*A=2/3*3^3/4*√6=3.722
then calculated each triangle area by law of cosines and sines and it was
small triangle = 0.51
med triangle = 4.49
large triangle = 2
sum = 7
Here is a simple non trig finish from 1:37: color all the other 3 triangles adjacent by a vertex below the long diagonal line shown in the figure green. (basically the 3 triangles congruent to the 3 yellow triangles) Then our answer is just 1/2 (sum of areas of yellow and green triangles). Consider each of the 3 rhombi (made of 2 equilateral triangles each and will have area 2 * 6 = 12) that contain 1 yellow and 1 green triangle. In the leftmost rhombus, the yellow triangle has a base 3 times that of the green triangle so its altitude will be 3 times that of the green triangle making 3/4 the total altitude of the rhombus. This means by the triangle area formula, the ratio of the area of the leftmost yellow triangle to the leftmost rhombus will 1/2 * 1 (base) * 3/4 (height) = 3/8 and we multiply by the area of the rhombus (12) to get 4.5. The ratio of the area of the leftmost green triangle (which has the same area as the rightmost yellow triangle ) will be 1/2 * 1/3 (base) * 1/4 (height) = 1/24 giving an area of 0.5. Now, for the yellow triangle in the middle rhombus, it has a base 2/3 that of the rhombus base and a height 1/2 that of the rhombus by similar triangles meaning the ratio of the area of the yellow triangle to that of the middle rhombus will be 2/3 (base) * 1/2 (height) * 1/2 = 1/6 * 12 = 2. Thus the total area of all yellow triangles is 4.5 + 0.5 + 2 = 7.
And here I am, coordinate bashing.
U hv used rhombus property,You r genius
Daaaamnnnn
th-cam.com/video/vUcNDYlBtoc/w-d-xo.html
Does anyone know how he arrived at the values, ¾S, ⅔S and so on. if I know that, the rest is just simple ratio calculations.
2:54 You don't need to go into floating point so early. If you'd kept this as 9/2, that 9 would've cancelled nicely with 9s which will soon appear in denominators.
Please see my solution: th-cam.com/video/LuGuZt4M--g/w-d-xo.html
I solved it! Didn’t even have to construct the other triangle. Feels great!
Are you kidding how?
How ,,,, did you used integration ,,,, please reply?? By the way my name is also Atharva 😀😀
If you notice unshaded triangle has congruent yellow shaded triangles within it. Start from left bottom(smallest) and then move to the right. After that I used similar triangle properties like presh and solved the rest of the sum .
@@atharva__soni nice to meet you fellow Atharva.
I appreciate it when you ask your viewers to pause the video to try and solve it. I almost got this one but I missed the part where you need to draw another equilateral triangle to get the interior side lenghts.
so glad that this channel is still alive and does posts vids!! i really enjoy watching and learning from it
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
please solve IIT questions ❤️
Please do so
Most underrated comment
Correct
we need Jee mains and Advance problem
An intuitive solution without trigonometry:
The leftmost yellow part has area 1/2 • (3/4 • s) • h. This accounts for 3/4 of the area of the equilateral triangle (1/2 • s • h). Thus, it has an area of 3/4 • 6 = 4.5 or 9/2
The side length and height of the middle one is scaled down by 2/3 and those of the rightmost one by 1/3. Together they have area 9/2 • (1 + (2/3)^2 + (1/3)^2) = 9/2 • 14/9 = 14/2 = 7
exactly
yeah, trigonometry is crap. Geometry is the way
There are several ways to solve this problem without trig.
Like? Can you suggest some
@@tanaykumar9933 it is hard for me to write geometric proofs here.
I saw some others who wrote 2 ways that i thought of.
@@tanaykumar9933 When he makes the first constructions and says "lets calculate the area using ..." you don't need to do that. If you consider the largest yellow triangle shown, he has shown that one of the lengths is 3/4ths S. If you consider *that* side as a the "base", you can see that the yellow triangle and the remainder triangle form two triangles that sum to an area of 6. That means that the yellow triangle is 3/4ths 6 and the remainder is 1/4th -- because both triangles have the same height and share the base of a triangle who's area we do know. The remainder of Presh's argument just continues to use similarity.
Agree. No trig is needed.
@@udic01 modern Fermat
There's a much easier way to solve it, no radicals or trig ratios required.
1. Let the area of the small yellow triangle be y. As you say in the video, the ratio of side lengths of the large, medium and small similar yellow triangles are 3:2:1, so the ratios of their areas is 3^2:2^2:1^2=9:4:1, hence their areas are 9y, 4y and y and their total area is 14y.
2. The two small white triangles in the bottom left together make a triangle similar to the largest white triangle that is exactly half of the 6 equilateral triangles. The large triangle has area 6(6)/2=18, and the similar small triangle has side lengths scaled down by 1/3, so it's area is 1/(3^2)=1/9 of the large triangle, which is 18(1/9)=2.
3. The small similar white triangle (whose area is 2) and the big yellow triangle (area 9y) together have an area that is equal to the area of an equilateral triangle (area 6) and the small yellow triangle (area y) together, so 2 + 9y = 6 + y. Solve to get y = 1/2.
4. Sub y=1/2 into the total area of 14y to get 7.
Also what I did. This method does work even if the starting triangle is not equilateral.
2M subscribers
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It came
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This is the first time I ever saw a video of yours released 4 hours ago.
It happened because I subscribed you.
1:23 I don't understand why this holds true
Look at the triangles differently, i.e., the shaded region + the part to its right.
similar triangles (AA)
Isn't it simply using Thales' Theorem?
@@thoughtfuljanitor6627 I think this is more of parallel lines cut by a transversal line. Encountered this in high school.
@@jamirimaj6880 it's Thales theorem
Congrats on 2 million subs! Love your videos! ❤
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
Hello, I am an elementary school student living in Korea. I subscribed to this channel because I like math, and this video problem is really fun.
I solved this problem, too.
I respect you.
Please upload fun and informative videos from now on.
Thank you.
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
This was nice 👏 . An excellent channel to pass time 😊
This is the only guy I have ever seen who makes youtube video but never talks about subscribing and to like video
Great let's have him a hit of 2m suscribers
This comment actually made me subscribe the channel
😬
Done it in memory without sine or something like that. There's whole bunch of similar triangles. Middle is 4S, large is 9S, yellow is S(9+4+1)= 14S. 6 - 9S + S = (1/3)^2 *([6 * 7/2] - 6/2). We get 8S = 4 , S=0.5. Yellow is 14*0.5=7.
Nice
Really good puzzle. Thanks!
Congrats on 2M subs dude!
The same procedure works for n general triangles in a row, with the total area of the triangular cutoffs being (2n-1)A/6, where A is the area of a single triangle.
Very cool..... I didn't solve this, but enjoyed the solution....
what????
4 days ago??? WHAT
@@TheoSin maybe he's from another time zone/planet where there are longer days
@Rafly Yanuar see his name.. he's a ninja
On youtube, we can make a video unlisted and that video can only be accessed by people who get the link. Unlisted videos can then be set to published at a later time.
Thanks for the video!
Good stuff as always
simple way start from 1:37, once you get there, focus on the left most triangle that area is 6. Notice that the yellow part and the non-yellow part have the same height, then their area ratio is the same as their side length ratio, which is (3s/4)/(s/4)=3/1, let the yellow area be x, we have x+x/3=6, and that x = 9/2. then continue with the ratio method in the video to get other triangle area ,we end up having 9/2+9/2*4/9+9/2*1/9 = 7
Cool problem!
The general solution would be (2n+1)*A/6. Where "A" is the area of the triangle and "n" is the number of yellow triangles, which in this case is three.
My question is:
Is there a shortcut to find this expression directly?
The three yellow triangles are similar and their sides are in ratio 3, 2, 1, thus their areas are in ratio 9,4,1.
The left yellow triangle is 3/4 as large as the equilateral triangle (one side is 3/4 as long as the side of the equilateral triangle, while the corresponding hight coincides with the hight of the equilateral triangle).
And we are done.
A1=¾·6=9/2
A2=4(A1)/9=2
A3=(A1)/9=1/2
A1+A2+A3=7
Demostrate why the side lenghts are in ratio 3-2-1
@@giacomozancarini4176 because the right sides are 3/4 2/4 and 1/4 respectively of the side of the equilateral triangle, it can be proved with Thales since the sides are parallel.
I hope one day, that disturbing heartbeat will be gone and I can watch those videos again with sound
Solved this using trig and coordinate geometry but your method is quite elegant!
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
Joke of the day:
What did the farmer say when he lost his tractor?
*"Where's my tractor?"*
Why did I laugh
This wasn't at all funny
@@karangupta1825 you just don’t get it, this joke has soooo many layers to it
@@reversev9778 dude then please me explain me those layers
@The Ghost Samurai what no why
for n triangles, the area of the shaded area above the line is S*(2n-1)/6, where S is the area of each triangle.
If you think in terms of triangle height rather than side length, you don't need the sin or apex angle. Using the side marked S as the base of the triangle, the larger yellow triangle has perpendicular height equal to 3/4 of a large white triangle's height, so its area is 3/4 of 6, without depending on the 60 degree angle.
Got this without trig. To work out the area of the big triangle just subtract the small triangle underneath from 6.
To get THAT small triangle consider that, adding one new equilateral triangle to the end makes the big sloping trianle (right acrosshte diagram) one half of 8 triangles = 1/2*8*6 = 24. The small triangle is 1/4 the side so 1/16 then area - i.e.1/16*24= 3/2.
The rest of my workings also using similar tringles was the same.
My wife and I found a solution that doesn’t involve trig:
**Edit: video solution linked in thread**
we were able to find the area of the smallest triangle by realising it is the difference between two triangles that were similar to the largest triangle (the 4 lengths of the bottom of the triangles, to the top of the right triangle) and the second largest (the 3 lengths of the bottom of the triangles, to the top of the 4th, and back). Considering the latter triangle has an area of 18, it can be found that the smallest of the three yellow triangles has an area of 1/2 and the other two continue from there.
(It would be a lot easier to show our solution if not in a TH-cam comment...)
Well make a video and give a link, that would be a good watch
@@yashvashistha6004 agreed I'm very curious about your method
I found the same solution than yours, and I also think it would be easier!
Very nice! Your solution requires nothing more than A=½bh and simple principles of scaling triangles.
How fun that you and your wife work on these together!
How do we know that the triangle you mentioned are right triangles? I don’t think that is given, how do we figure it out?
I tought it was one of those simple questions that you do in memory by adding and substracting surfaces etc
Please see my solution: th-cam.com/video/LuGuZt4M--g/w-d-xo.html
We can use Intercept theorem (Thales’s) for computation the heights of yellow triangles here. This way is much simpler and the task goes to the “solved in mind” category.
Please see my solution: th-cam.com/video/LuGuZt4M--g/w-d-xo.html
That is exactly what I did in mind by minding my decisions.
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
A more complicated answer than you'd think at first glance.
I did it using co-ordinate geometry, setting (0,0) at bottom left and working out the intersection of the diagonal line with each triangle but only the first 3 intersection points are needed due to symmetry.
Pages and pages of calculations later I ended up with the wrong answer 31/4 but I did get the area correct for the largest and smallest yellow triangles right but I made a mistake with the middle triangle (I got 11/4 as its area, not 2). I later went back and found my error so I corrected it and got 2.
Your solution was simpler and less prone to error. I would have run out of time if this was on a test because my solution took almost an hour to solve. I wouldn't have had time to double check my work.
Your solution are so fascinating
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
Thanks for recommendations I will try it
To find the area of the first, leftmost triangle, observe that it is 6 minus the area of a triangle similar to the one made by the diagonal and two sides of the overall rhomboid. The overall rhomboid has an area of 48, the big triangle 24 and sides of the two triangles are in a ratio 1:4, thus the smaller triangle has an area of 1,5. 6-1.5=4.5.
I found a way without sine. If you just turn and stretch the coordinate system a bit (and forget about the useless forth triangle, that is empty), you end up with three rectangles with sidelength 3 (height) and 4 (width). Thus, if you also correctly shifted the cutting lines the yellow triangles have the same width and height of 1, 2, and 3 respectively.
thanks! these videos really help me out at math
What you thought about san cruz "where gravity lack
Hoooo!! solved it the exact way Presh did!!
Please excuse me.. if anyone already suggested this simpler alternative .. there are 682 comments by now when I read this.Here is a much simpler alt one that needs no pen no sin theta stuff...... ;).. lets mark A,B,C,D,EF,F on the base and W, X,Y,Z at the top. Area of triangle AZE =24. Let’s mark points line AZ cuts each triangle on the right sides from left to third triangle as g,h,i. now we know triangle AgB is one fourth in dimension to triangle AZE. So it’s area is 24/16 =1.5. So area of AWg I.e.first yellow part is 6 - 1.5 = 4.5. Now we also know Bg:Ch:Di:EZ= 1:2:3:4 so conversely, ratios of Wg:Xh:Yi=3:2:1. So ratio of triangles marked in yellow I.e. 9:4:1. So total area of all yellow triangles = 14/9 of first (left most yellow highlighted triangle) so that is 14/9 * 4.5 = 7.y
How did you get the fractions for the smaller triangles sides?
Visually they look right, but I am not aware of how you are going from visual to firm fractions...
Just use the Thales's Theorem.
So take for example the triangle that has for sides the top line, the far left line(before he draws the final extra triangle) and the line we drew through the whole shape. The bases that he later finds to be 2/3s and 1/3s make new triangles inside that big triangle with all of them ending at the point where the middle and top lines intersect. Those are similar to each other because s, 2/3s and 1/3s are parallel. We can find the ratio of these triangles because we already know the three top segments are equal to s. The triangle with base s has a side on the top of 3s, the triangle with base 2/3s has a side of 2s and the triangle with base 1/3s has a side of s. The other sides are solved similarly but he needed to make that extension triangle.
Congrats for 2M subs.
Your mother must be proud of you.❤️❤️🙏🙏
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
@@madhukushwaha4578 I have visited and subsribed your channel 😎😎
Stop botting plz
I used that the slop of line is s*sqrt3/7, then array yellow/array white in first triangle =3:1 (we can also use homeothetic transformation and sin 60), so yellow in firs triangle is4,5. From homothetic transformation - second yellow is4/9 * 4,5=2, third yellow is 1/9 *4,5. we have finally 4,5+2+0,5=7
This is one solution I can't do in my head! Also features some rules I wasn't familiar with/forgot.
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
Ikr I tried finding side length of triangle by saying that 2s * s√3 = 12, then 2s^2√3 = 12, but then it got confusing when I was solving for x in that so...
Brilliant solution!!
And one more little question:
Who are the 22 viewers that didn't like the video??? Why on earth???
I didn’t downvote, but other people probably did because Presh made it overly complicated by using trigonometry and Thales theorem. A simple solution actually requires nothing more than A=½bh and simple principles of scaling triangles. (Not the first time this has happened on MYD.....)
@@j10001
They dislike because of the difficulty of the problem.
Ok, maybe, I see.
Thank you for your answer.
A small thing... As long as it's a given that the area of one of the initial triangles is 6, it doesn't really matter that they are equilateral. The side lengths and angles cancel out of the calculations. Let's say that all the horizontal sides have length a, all the upper-right - lower-left sides are length b and all the upper-left - lower-right sides have length c. The area of any one of those triangles is (b * c * sin A) / 2 = 6. The area of the left-most yellow triangle will be (b * 3c/4 * sin A) / 2 = 6 * 3/4 = 4.5.
Watched the full video without understanding a single thing🙂🙂
Amezing?🤣🤣🤣
Now I challenge you to do the same exercise but for "n equilateral triangles" in which the area of 1 equil. triang. is "A"
Using the same methodology as in the video, let S be the side length of the equilateral triangles, and as such A = sin(60)/2 * S^2.
The lengths of the left edges of the yellow triangles then follow this sequence (from largest triangle to smallest):
(n-1)/(n-1) * S, (n-2)/(n-1) * S, (n/3)/(n-1) * S, ... 2/(n-1) * S, 1/(n-1) * S.
The right edge lengths follow a similar sequence:
(n-1)/n * S, (n/2)/n * S, (n-3)/n * S, ... 2/n * S, 1/n * S.
The area of each triangle is 1/2 * L * R * sin(60), where L and R are the triangle's left and right edge lengths. Plugging in said values and simplifying gets the equation m^2/(n(n-1)) * sin(60)/2 * S^2 = m^2/(n(n-1)) * A, where m is the scale factor of the individual triangle (an integer from 1 to n-1).
Thus, the sum of all the areas is A/(n(n-1)) * Σ [m=1 -> n-1] (m^2)
According to a quick google search, Σ [x=1 -> y] (x^2) = y(y+1)(2y+1)/6. Plugging in y=n-1 and simplifying gives the equation Σ [m=1 -> n-1] (m^2) = n(n-1)(2n-1)/6.
Plugging this in simplifies the formula of the triangles' area to A(2n-1)/6
This formula holds when n > 1.
I have (n-1/2)*A/3 but that looks pretty wrong when n=1
With n regular triangles we will have n-1 yellow triangles to sum. The yellow triangles are all omothetic with centre the upper right vertex so the areas grow with the square of the distance from the centre. Let's call the areas T_k, than the yellow area would be (1+4+···+(n-1)^2)T_1.
For Thales' theorem on parallel lines cut by a transverse the sides of the regular triangles starting from the left will be cut at 1/n-th, 2/n-th and so on, so the biggest yellow triangle will have area T_(n-1) = (n-1)/n · A.
We also know that T_(n-1) = (n-1)^2 T_1.
We conclude that the yellow area is
(1+4+···+(n-1)^2) / n(n-1) · A
for any given n.
An easy solution is the following: the three yellow triangles are omothetic with the centre in the upper right vertex. Since the biggest one is three times more distant from the centre than the smallest one it has the sides 3 times larger and the area 9 times larger than the smallest one. By analogy the middle one has sides 2 times larger and area 4 times larger than the little one. Let's call them A, B and C in increasing order, so A+B+C = A+4A+9A = 14A.
Now, for Thales theorem on parallel lines cut by a transverse we got that the diagonal line cut the side of the first triangle from the left at 1/4th, the second at a half and the third at 3/4th. Than we can conclude that the area of C, the big yellow triangle, is 3/4 of the regular one of area 6, because it has same height and base 3/4th. So C = 3/4·6 = 9/2 and A = 1/2. Finally, the yellow area which is 14A is equal to 7.
I liked it so much, good puzzle, good solution. Thanks..
Anladıysan anlatır mısın
@@googleben5361 benzerlikten gitmiş
As already noted, we can just use similarity to solve the problem. Probably the easiest would be to look at the small white triangle below the big yellow one completing it to the equilateral triangle. Notice that this small white triangle is similar to the big white triangle with the diagonal as one side, the side length ratio is 1/4 and the area ratio is thus 1/16, given the area of the big white triangle is 24 that can be easily seen (6*6/2+6=24), the area of the small white triangle is 3/2 (24/16=3/2). Therefore the area of the big yellow triangle is 9/2 (6-3/2=9/2), and using similarity of yellow triangles (area ratios of 9 and 4 for the big and middle triangles) we can find the total yellow area being 7 (9/2+4*1/2+1/2=7).
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
The added line segment divides the first equilateral triangle into 2 triangles. We can get the area of the upper yellow area, without using trigonometry, by computing the area of the lower triangle and deducting from 6. The lower triangle has 1/4 of the area of the whole equilateral triangle, or 1.5, because the heights are in the ratio 1:4 and bases are the same. Thus, the yellow triangle has area 6 - 1.5 = 4.5.
The proof of the height ratio is as follows. Construct vertical line segments representing the heights of the whole equilateral triangle and the lower triangle. Two of the four resulting right triangles formed, the two with the heights as their left sides, are similar by angle-angle-angle. Presh shows later that the hypotenuse of the smaller triangle is 1/4 of that of the larger triangle. Because the triangles are similar, the heights (corresponding sides of similar triangles) are in the same ratio as the hypotenuses, 1:4.
Amazing solution. Almost get the correct answer with similar steps, but made an error when calculating the second yellow triangle
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
I also used a method without trigonometry. Interstingly it shows also, that the 4 triangles don't have to be equilateral, we get the same result for isosceles triangles. The principle I used is finding similar triangles and calculating ratios of their areas by using the fact that in similar triangles the ratio of the area equals the square of the ratio of the side length.
Please see my solution: th-cam.com/video/LuGuZt4M--g/w-d-xo.html
2:06 From here there's a simpler way to calculate the first yellow triangle's area:
Think of the right side (3/4 s) as the base. The height is the same as of the given equilateral triangle (GET) and the base is 3/4 of the GET's base. So the area is simple 3/4 of the GET's area, so 3/4 of 6.
So Presh, mind *your* decisions. 😁
Please see my solution: th-cam.com/video/LuGuZt4M--g/w-d-xo.html
Hope you get 2 million subscribers in some days
If you take the base of the yellow triangle as 3S/4, then it has the same height as the equilateral triangle but only 2/3rds the base. 3/4 x6=4.5.
The area of the largest yellow triangle can be more simply calculated by treating the upper right side as the shared base of both the yellow and equilateral triangle, and noting that the areas will be proportional to the lengths of the bases.
Please see my solution: th-cam.com/video/LuGuZt4M--g/w-d-xo.html
Good proof. Drew the similar triangles and found the area of the largest of the similar triangles to be 3/4 of 6, so 4.5. Then used the area factors of 9:4:1 to find the total area to be 14/9 of 9/2 = 7.
Please see my solution: th-cam.com/video/LuGuZt4M--g/w-d-xo.html
Could you write the problem in the description. I like to do these at work
I way over complicated this
really should take a revision on geometry, because I didn't get what you meant by" in parts of 3" at 1:23
I tried to solve it using almost similar method, I didn't use trigonometry knowledge:
First of all, go to the image shown at 1:40
1. Take the top left half of the whole parallelogram. The whole triangle's area is 8*6/2=24. Now the triangle is divided by 3 equidistant parallel lines, and the corresponding areas of those similar triangles are A=24, B=24*{(3/4)^2}, C=24*{(2/4)^2} and D=24*{(1/4)^2}.
2. Now ignore the leftmost and rightmost equilateral triangles and consider the relatively smaller parallelogram. Area of the parallelogram is now 6*6=36
Now again take the top left half of the parallelogram. Notice that it has also been sectioned by equidistant 2 lines. So the areas corresponding to thoes similar triangles are X=36/2=18, Y=18*{(2/3)^2} and Z=18*{(1/3)^2}
3. Now consider the area of {(X-B)+(Y-C)+(Z-D)}
It would be the answer, sorry I am skipping the final calculation as I am a little lazy😛, please inform me if I have made any mistake.
If you did not understand the reason for the fractions of s, you need to look at the parallelograms formed by 2 triangols: since they are cut by a straight line whoch goes from the lowest point to the highest one, the points of intersection of the end of the parallelograms must be a certain fraction of the initial side (now that I explained it I understand why he didn't, it is harder than I thought to explain it, if you did not understand what i wrote it is very understandeble)
Many of the problems Presh does are from an Olympiad Math book really popular in India, btw I had Coord bashed this problem lol
Such a beautiful solution
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
Does this imply that for any scale of this problem, area x will equal 1.25 the area of the base triangle?
A useful thing to know, I don’t know where you might need it, but hey, looks good if you pull the answer out of thin air!
Yes
The lengths of the hypotenuses are 1, 3/4, 2/3, 1/2, 1/3, 1/4 from the left, so the area ratio is multiplied by the side ratio and becomes 3/4, 1/3 1/12. ∴6(3/4+1/3+1/12)=9/2+2+1/2=7
Build a big triangle with all four small ones as base. It consists of 16 small triangles. The area under the cutting line is half of a parallelogram made of 8 triangles. So the area above the cutting line is 3/4 of the whole triangle...
Sir ur geometry videos are awesome..can u pls provide some algenra stuff??
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
Do Olympiad questions next.
Hii! You may have to visit this channel for more harder questions #onlymathlovers .
Area of a Triangle
B × H = A = 6
Area of Trapezium
H(A + B)/2 = 42
Area of 1 triangle = 6
Area of 2 triangle = 12
Area of 3 triangle = 18
Area of 4 triangle = 24
Area of 5 triangle = 30
Area of 6 triangle = 36
Area of 7 triangle = 42
My fav thing to do is just guess by looking at it, this time I was spot on with 7 XD
I thought of giving it a try, but then I decided to MindMyDecision
Hii! You may have to visit this channel for more harder questions #onlymathlovers ...
I can't do this, I go back watching youtube movies with playing kittens
7
( (3/3 * 3/4) + (2/3 * 2/4) + (1/3 * 1/4) ) * 6 = (9/12 + 4/12 + 1/12) * 6 = (14/12) * 6 = 7
Draw lines parallel for the dissecting line from every corner of the triangles. Easy to see each left side is divided into 3 parts and each right side is divided into 4 parts.
I used trigonometry to solve before realizing the simpler solution using similar triangles! What a nice problem :)
Please see my solution: th-cam.com/video/LuGuZt4M--g/w-d-xo.html
It is kind of fun to solve with 'many linear equations', and (not remembering) working out the area of a generalized irregular quadrilateral. (in this derivation, I only assumed each little triangle is an isosceles; moreover, for convenience I assigned '4' as the base, and 3 as height, to give an area of 6. However, any values equally 6 could have been used!)
If a quadrilateral has lengths (𝒂, 𝒃, 𝒄 ) along its base, and heights (𝒎, 𝒏) 'heights' (in the height sense of triangle nomenclature), then the area of the generalized quadrilateral works out to be ½( 𝒎(𝒂 + 𝒃) + 𝒏(𝒃 + 𝒄) ).
From the rising-slope line (𝒚 = ³⁄₁₄ 𝒙 ⊕ 0 ) which defines the intercepting line passing thru all smaller triangles, and the respective up-facing or down-sloping lines of each of them:
𝒚₁ = -³⁄₂𝒙 + 6
𝒚₂ = +³⁄₂𝒙 - 6
𝒚₃ = -³⁄₂𝒙 + 12
𝒚₄ = +³⁄₂𝒙 - 12
𝒚₅ = -³⁄₂𝒙 + 18
We can find all the intercepts fairly easily…
𝒙₁ = ⁷⁄₂; 𝒚₁ = ¾
𝒙₂ = ¹⁴⁄₃; 𝒚₂ = 1
𝒙₃ = 7; 𝒚₃ = ³⁄₂
𝒙₄ = ²⁸⁄₃; 𝒚₄ = 2
𝒙₅ = ²¹⁄₂; 𝒚₅ = ⁹⁄₄
From which one can directly figure the area of each 'little yellow bit':
area 1 = 6 - ½4( 𝒚₁ = ¾ )
area 1 = 4.5
Figuring the (𝒂, 𝒃, 𝒄, 𝒎, 𝒏) for each next quadrilateral gives
𝒂 = 𝒙₂ - 4
𝒃 = 𝒙₃ - 𝒙₂
𝒄 = 8 - 𝒙₃
𝒎 = 𝒚₂
𝒏 = 𝒚₃
Then the area of the middle yellow triangle is
area 2 = 6 - ½( 𝒎(𝒂 + 𝒃) + 𝒏(𝒃 + 𝒄) )
area 2 = 2.0;
And the same treatment for the littlest yellow triangle
𝒂 = 𝒙₄ - 8
𝒃 = 𝒙₅ - 𝒙₄
𝒄 = 12 - 𝒙₅
𝒎 = 𝒚₄
𝒏 = 𝒚₅
area 3 = 6 - ½( 𝒎(𝒂 + 𝒃) + 𝒏(𝒃 + 𝒄) )
area 3 = 0.5
Thus the sum of them all is
yellow = 4.5 ⊕ 2.0 ⊕ 0.5
yellow = 7
And that is a great spot to stop!
Michael Penn
@@advaykumar9726 I like his style!
@@robertlynch7520 Me too!
2:01 why complicate the work, the yellow area is just (1)*(3/4)+(2/3)*(2/4)+(1/3)*(1/4) to one equilateral, dividing triangle is more intuitive than pulling out sine just to write the area and substitute it
How do we get the side length values like 2/3S, 1/4S and so on?
Simple:
- total area of figure without yellow parts is 24 as it consists, if you rearrange the white pieces, of 4 big triangles.
- The leftmost little white piece is homogeneous to the total white piece with vertices that are scaled 1/4, so area is 1/16. 24/16 = 3/2. Therefore the leftmost yellow piece is 6-3/2=9/2.
- Same logic: the middle yellow piece has same shape as leftmost yellow one with vertices scaled to 2/3 so area is scaled by 4/9, therefore area is 4/9 * 9/2 = 2. Rightmost yellow piece has vertices 1/3 so area is 1/9 * 9/2 = 1/2.
- Add all: 9/2 + 2 + 1 / 2 = 7
I am new to studying math in a different language but i am trying to understand this is actually fun and challenging and i am learning some new ways to solve math problems r this the kind of math problems u guys have in school?
@MindYourDecisions Can you design such Function which has pair set as domain (Pair set must contains Additive inverse of each other)?????????????????
If someone else can do it too???????????
這題真不錯，出在升學入學考都是好題目，解法也自由
I did it just using similar triangles. The triangle on the bottom is 24. The triangle on top is 18. Using all the similar triangle in each section I did 18 - 13.5 + 8 - 6 + 2 - 1.5 = 7
Diese wurde mir bei meiner Mathematiksprüfung gefragt, ich wünschte dieses Video wäre früher hochgeladen worden :).
As stated by others here, this problem can be generalized to n △s (n>=2), each of area A, and they only need to be isosceles for the sides not on the straight line segment that contains one side of each △.
A verification by co-geom.:
Place the whole picture depicted above on the Cartesian plane, so that the aforesaid straight line segment lies on +ve x-axis, with endpoints at (0,0) & (2nb,0), where 2b is the base length of each △ (the base is taken to be the "zero-slope" side on the aforesaid segment). The height of each △ is 2(A/2b) = A/b.
The eqn. of the "crossing" line L that joins (0,0) (the left base corner of the 1st △) & ((2n-1)b,A/b) (the top corner of the n-th △) is given by
y = mx where
m = (A/b)/[(2n-1)b], i.e. y = A/[(2n-1)b²] x ...(i).
The eqn. of the line Uₖ that contains the +ve -sloped side of the k-th △ (1
Remarks:
~ M.I. can prove, for any +ve integer N, 1²+2²+...+N² = (1/6)N(N+1)(2N+1), as used in the last part.
~ Algebra for simplifying the expression from the "Shoelace Formula" to "[A(n-k)²]/[n(n-1)]" is straightforward: terms cancel out rapidly.
generalize this result for n adjacent equilateral triangles and find the limit when n goes to infinity.
@MindYourDecisions Please consider it.
Can you design such Function which has pair set as domain (Pair set must contains Additive inverse of each other)?????????????????
If someone else can do it too???????????
My solution: (some theory proved by the video so I don't repeat)
using Ratio we can know the largest to small triangle's length is 3:2:1
By they are similar Triangles so that means their Area ratio will be: 9:4:1.
The Area of Largest :6*3/4= 4.5 (very easy to get)
second : 4.5*4/9 =2 ,small : 4.5/9 =0.5 (using ratio)
total: 7
We have 6 triangles, 6 area each one, so the total is 36; the yellow area is the half of the half, it means, a quarter; so the yellow area is 36/4 = 9
Where did you learn this from?