What is the area of the red stripes?
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- čas přidán 15. 06. 2024
- This is a brilliantly fun geometry problem!
0:00 problem
1:00 similar triangles
3:00 solution
References
Similar triangles area (Math StackExchange)
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Euclid similar triangles (book of proofs)
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Euclid similar triangles (Clark University)
mathcs.clarku.edu/~djoyce/jav...
Triangle problem on Scholr
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Triangle problem on Quora
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Triangle problem on Purple Comet! Math Meet April 2018 (problem 10)
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Triangle problem on Brilliant (google webcache)
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Make it a parallelogram. Then you have 5 blue stripes and 4 red stripes, all of the same size. So red area = 4/5 blue area. 5 seconds and no need for a pen.
U think your so smart with ur fancy explanation for something so simple and creative. Its people like you that increases the average iq of the global population. I hope your'e (i types that right!) HAPPY WITH YOURE SELF!!
Oh wow i planned a parallelogram in my head but totally forgot they had to be same size
Great solution!
Works great with odd number of stripes. Requires a bit more work with even number of stripes.
Man ifb now, it took me 10s to figure it out
Make a parallelogram by doubling triangle.
The parallelogram is divided by 9 sectors, 4 red and 5 blue.
The 5 blue areas are 290, so 4 red areas are 232.
The red area of triangle is 116.
Exactly. Why they are making things more complicated than they are
Very good
I just did 4/5 of 145 is 116 but pretty much same concept. Since A=bh and the h is equal for all and they are the triangles are similar you can cut all the other stuff out
Yep, this is how I did it.👌🏻👌🏻
I did the same. It is so many times the case that this deadwalker suggests crazy complicated solutions to problems, while trivial ones like this exist.
I did it by the "count the number of triangles each layer" method.
No idea how to prove that it works, but i saw that too.
Yes. Basically start by picturing 3:35. That leads immediately to 5:37, and you're done.
Same. Saw the picture and "counting" seemed to be faster than thinking for a better solution :P
@@truberthefighter9256 I was at first surprised by the limited information, such as no angles, but then quickly realised this means that the solution is general, so mentally adjusted the triangle to make it equilateral. From there it's easy.
I used the trapezoid formula of (a+b)*h/2. First blue is (0+1x)*h/2, second is (2x+3x)*h/2, and so on. Red is (1x+2x)*h/2, etc. We can discard the *h/2 remainder as it appears in all of them. Once we sum up both colors, we get 145 = 45x, and ? = 36x, and thus ? = 4/5 *145
Same I also used trapezoid formula
That is also how I did it.
how about double it to a cube? 5 blue layers are 145*2, then 4 reds go to 116
Me too, literally the same same way... 😊
same
An esier approach, is by shifting figure, move topmost and join with bottom part of same color, the result would be a pralellogram, and areas will be in the ration 2.5:2 = 145:x
X= 116
Exactly how I did it! Was way easier 😆
me too
I agree this seems like Presh is using a sledgehammer to crack a nut ! However the general principle is worth studying
I did it with the parallelogram also, but simply considered that there are 4 red stripes to 5 blue stripes, so the red stripe area is 20% less than the blue stripe area. I did in my head in less time than it took him to get to the main problem. He made it way too complicated
@@Kevin-S perhaps, but his method was much easier to prove
You can also create a parallelogram by adding in a copy of the same triangle rotated upside down. Now the coloured strips have become nine congruent parallelograms, 5 blue and 4 red. This means that the red area (within the overall parallelogram) is 4/5 as large as the blue area, i.e. 232. So the red area within the triangle is half of that, 116.
Glad to see my approach was easier than his for once.
mine too ! .. next time i will be glad if its *correct* too ^v^
What was your approach
@@VishalSingh-ov9oo assumed then middle stripe hast to be the difference..
@@gerdkah6064 it has to be difference of what.
I was thinking but don't get the logic and you can see in 3 continues strip area is in ratio 1:3:5 so can be please explain more
But I understand one more solution give another comments
I thought of a similar approach at first, but then I realized you can turn the triangle into a right triangle by moving each "infinitesimal horizontal layer" of the triangle, and then you can turn that right triangle into a rectangle. This rectangle has 9 equal stripes, 4 red and 5 blue. Using these ratios, you can use the blue area to find the total area and the red area.
exactly what I did, I'm glad I'm not the only one :) One of the few things I remember from school is that you can basically move the top point of a triangle along a horizontal line and it's are won't change, so you can make this into a right triangle and just calculate the area (i used percentages)
I took a different approach for this problem: I started by having the top blue section be 'Z' and then notice how the same area keeps being added with each section beneath it (so the second row is Z+X, then Z+2X, and so on). The blue sections add up to 5Z+20X = 145, or Z+4X = 29. Adding the red sections yields 4(Z+4X) or 4(29) = 116
It’s satisfying to hear you say ‘and that’s the answer’ so positively
For how many triangle questions I have solved, I can say if stuck similarity is the way out.
And sometimes congruency too
I solved it by common sense came to 120 pretty close
I compared both the sides and guessed the answer.
This was one of the easiest question lol 🤣. Just use similarity
Thanks for showing where the "x^2" come from (the Euclid proof)... and also showing the rest of the stripes in the triangles are simply multiples of the area of the original triangle on top. :)
The little details will usually give that special someone, that "Ah HA!" moment... when they get it.
Shouldn't really be surprising of course. Any two dimensional shape will scale in area by the linear scale squared. This is because it's scaled in two separate dimensions. Likewise scaling a three dimensional shape in all directions will yield cubically increasing volume
@@PattyManatty Indeed. I don't know if I ever learned an official A₂ = A₁·x² formula or anything, it just seems intuitive from what areas and volumes (indeed, any n-dimensional volume) _are._
That was such a beautiful solution, I loved it.
This is best channel for problems of maths.
Thank you Presh, for this amazing question .. Your videos help me a lot in pushing myself and gaining my interest in mathematics.
Let’s see how many subs I can gain from this comment currently at 23
@@galileogaming.5606 gajab...
@@galileogaming.5606 idk about subs but you will get Russian vodka for sure!
Disclaimer this channel is highly appreciative💐💐
Let’s see how many subs I can gain from this comment currently at 23
Hi. I figured this out real quickly by extending to a triangle by adding a rotated triangle. Then its trivial to calculate the area of one stripe. Also a general solution with 2n+1 stripes is really easy, and the proportion of red and blue stripes.
INCREDIBLE!!! Thanx you very much, mate! 🌹💐🌺
Wonderful problem and ecstatic that I could solve it on my own. 🙂👍🏼 Brought back great memories of similar triangles from school
What an amazing question ♥️
I have done no. Of blue strips given 5 and their total value is 145 I considered the value of each of each strip = x = 145/5 = 29 and so the number of red strips = 4 and so their total value 29×4 = 116 thats like I done it
Ur method is not applicable at all times. This is just lucky solution
i too did the same lol
@@balakrishnakarri6264 I know dont speak so rudely the question is given in aritmetical pattern thats why I used it I know you guys who are always for dominating and making a satisfaction to yourself that you know all thing
@@strikerstone obviously as it is an arithmetical pattern it can be applicable here
@@swattikdas7777 i mean if there is a trapezium then it is not correct.
we can also solve this by rearranging the 5 blue portions into a triangle since let the height of a single blue portion be 'x' then the height of the rearranged triangle will be 5x the area of this triangle will be 5xB(where B is the base of the rearranged triangle) now the area of the original triangle will be 9xb if we subtract both the areas we get 4xB as the area of red also 5xB = 145, xB = 29 therefore 4xb=29x4= 116
got the same result with a simpler method 145/5=29, 29x4=116. Your videos are enjoyable. Thank you!
This time you made it very complicated. As usually I have troubles finding the solution, I solved the problem figuring out the Nèstor Abad's method by myself in 2 minutes.
I just did 145 divided by the number of blues which result to 29 then multiply it by the number of red to get 116. It's in a ratio that's why ratio and proportion can do the trick
Yep. Specifically, the similar triangles result in the sum of the red ratio numbers and blue ratio numbers both forming an arithmetic series with the same average term but different numbers of terms.
Yup, this is how I did it and it took like 4 seconds.
@@chibearsfan313 you did it that way but in no way did you prove by doing it that it was the correct answer. You got lucky.
Me too I was surprised when the answer was right
@@stevenwilson5556 and here I thought I just found an easier way of solving this problem
thanks for nice pretty problems and your beautiful solutions
My solution:
Copy the triangle and do a 180 rotation with one of them to form a parallelogram, then transform it into a rectangle, so you end up with a rectangle composed of 5 blue and 4 red stripes with equal area. The red area is then 4/5 of the blue area. Divide by 2 to delete the copy of the triangle and you get the final area: (2/2 *) 4/5 * 145 = 116.
Being very skilled and clever gives a wide range of tools to solve a problem. My personal toolbox is very basic, but as it appeared to be enough for this one, the way I solved it is then much simpler ^^
Thanks for the video!
I solved it with many rectangles and triangles. Cool how there are many ways in mathematics that arrive to the same solution. Thanks for the puzzle.
Cleaver use of similar triangles principle to form a tricky question. Thanks for sharing.
Finally a problem i solved within a minute
Hi Presh! First of all, thanks for these videos. They bring me joy and spark my curiousity on the regular!
About the solution of this problem: that's certainly a way to solve it, though just counting the fitting triangles (like shown at 3:30) was easier for me. But in fact I solved it in an even simpler way:
Diuble the triangle, creating a parallelogram. Cut off and append a triangle from the parallelogram to make a rectangle. 2•145 now equals 5/9 of the whole area, so the red area is 2•145/5•4/2=145/5•4=29•4=116.
Of course you knew that and you just wanted to show a new solution which uses a rarer used method, right?
This is actually the best geometry problem I have ever seen.
This is one of the easiest geometry problems
I find it nice that these are always a challenge to solve, and that I'm at least able to solve one on occasion. I knew this one because I noticed (while playing triangular peg solitaire, of all things) that when you break a triangle into smaller similar triangles, that which make up the entire original triangle, the number of smaller triangles is a square number, because of the pattern in how each row/diagaonal is constructed. 1 right side up plus 0 upside-down, 2+1, 3+2, etc., being a consecutive series of odd numbers; a characteristic of squares is that they are also the total sum of a consecutive series of odd numbers that starts with 1.
My solution was similar to the second one, but I used only parallel lines and ASA similarity.
I drew parallel lines to the other two sides to split the entire triangle into the triangle in the top stripe. The second stripe would contain 3 times the area of the first stripe, the third one 5 times the area of the first stripe, all the way till 17 times for the bottom stripe.
I used the Gauss summation trick to figure out that the blue stripes had 9×5 triangles, and the red stripes had 9×4 triangles. I canceled out the 9’s and got to the answer 145×(4/5)=116.
A good thing about this method is that it can work with an even number of stripes, unlike the parallelogram trick. If there were 8 stripes, the Gauss summation trick would say there are 7×4 blue stripes and 9×4 red stripes. Instead of there being the same average and different number of terms, there are the same number of terms and a different average. So you can cancel out something in either case.
Dr, Talwalker there is a really easer and simpler way to solve the area problem . Cut the triangle in half flip one half upside down . Join it to the other half so that it makes a parallelogram. Both the red and blue strips are now equal in area . There are 4 red strips and 5 blue so the area of the red is just 4/5 the blue : (4/5) *(145)= 116. Real simple did it in my head. PS Sir I love your Channel Thank you
Wow, great explanation & good problem. Very well done.
How can I study your total books?
you are one hell of a teacher ,,my third video of yours, many more to come
I really like this explanation, both for the squared proportion relationship and the (obvious once shown) dividing triangle. I solved it by visual inspection slightly differently: A rectangle of triangle height times triangle base, with extended stripes, would have blue:red areas in the ratio of 5:4. Since the two parts of the rectangle outside of the original triangle can be assembled to make the starting triangle, the ratio of the stripes area remains the same for the triangle. Hence 4*145/5 =116 gives the area of the red stripes.
"And that's the answer!".
Thanks!
I love this channel
x^2 proof from Euclids elements leads to a pretty Pythagoras proof. Start with a right triangle sides a, b & c (the hypoteneuse). Draw a perpendicular from the hypoteneuse to the right angle. you now have 3 triangles - the original and two smaller ones. It's easy to show they are all similar, so their areas are proportional to the length of any side squared. Their hypoteneuses are the original triangle sides a, b & c. As the areas of the two smaller triangles make up the big triangle, you can show a^2 +b^2 = c^2
Within 20 seconds... for the first time in my life I am a math queen😄
Nice problem as well as animation 👌👌
more simple if We consider a rectangle with same base and height. area of tringle will be half of the rectangle's area, same goes for red and blue strips. so we can say calculation work for rectangle will also work for tringle. so total 9 parts parts for rectangle 5(blue) and 4(red) so area of RED is = (145/5)X4 = 232 (and 116 for tringle) . 116 half of 232.
The trapezoidal stripes form an arithmetic progression with each red sector being the average of the 2 surrounding blue ones. So you can simply say that the red area is the blue area times 4/5. You do not need to work out what the individual areas are.
Yea, I solved it with the telescope progression 1/2(b*h)*9 - 1/2(b*h)*8 ... + 1/2(b*h) = 145 = 1/2(b*h)*25 - 1/2(b*h)*20 = 5/2(b*h) = bh = 58.
so full triangle = 9/2(58) = 261, so red = 261 -145 = 116.
Hi Presh! This is a nice problem!:)
We can also solved the problem by easy trick that we imagine the triangle was 90°/45°/45° so that if we doublecate it, it will be square and we can use thae ratio between 5/145 =4/x so x= 116 and if we cut the square by half it is the same ratio.
More easy to draw an equal flipped triangle on the biggest side of this triangle, so we get parallelogram. In the parallelogram we will have 5 blue strips and 4 red stripe, each of which is equal to any another. So in the parallelogram blue shape consists of 5 strips and red shape consists of 4 strips, so square of red shape is 4/5 of square of blue shape in parallelogram. Since blue square in the parallelogram is doubled blue square in the triangle, and the same is valid for red square, then square of red shape in triangle is 4/5 of square of blue shape in triangle, i.e. 4/5 * 145 = 116
I did it by an Euler-like trick: Much like the sum of the first 5 counting numbers can be quickly added together by realizing the first and last numbers (1 and 5) average out to the middle number (3), the blue stripes average out to the middle blue stripe (i.e., the area of the top blue and the bottom blue combine to 2* the middle blue; same thing for the 2nd and 4th blue stripes). This means that the total blue area = 5* middle blue = 145, so the middle blue stripe = 29.
The middle (2nd and 3rd) red stripes ALSO average out to the middle blue stripe, as do the top and bottom red stripes. This means that the total blue area = 4 * middle blue = 4 * 29, so the total red area= 116.
I finally solved my first problem on this channel without seeing solution
I love the fact that I could solve this in my head using a different method. 😃 (clue: unfold the triangle along it’s diagonal)
Same, it tool me under a minute to do it in my head. I thought it was wrong since i'm not that good at math and usually fail these.
i did it over proportions:
the triangle was divided into 9 equal parts.
5---145
4---x
x=145*4/5
x=116
Great solve! Here is my proposal to solve this:
First we will fit the triangle into a rectangle that will have the area ( Length - b x width - 9xa where "a" is the equal space divider o the triangle ).
So this means the following:
b x 9a = 2 ( 145 + red area).
From this we can deduce that the blue part of the new constructed rectangle is:
5 ( b x a) = 2 x 145 = 280 =>
b x a = 280/5 = 58 =>
a = 58/b
With this in mind we go back to the area of the triangle which is:
(b x 9a)/2 = 145 + red area
We replace "a" with the above relation and we will get the following:
[b x (9x58)/b]/2 = 145 + red area =>
We cross b from the equation =>
(9x58)/2 = 145 + red area =>
red area = 261 - 145 = 116
My approach was to divide the large triangle into 2 right triangles of equal height. The smallest blue triangle at the top as divided into 2 triangles, both with height h, and the one on the left having a base of b1 and the one on the right having a base of b2. So the top blue layer = 1/2 b1 * h + 1/2 b2 * h.
Then, I essentially made a grid with widths of b1 on the left triangle and widths of b2 on the right triangle. Then just count the total numbers of rectangles and triangles in each layer and add up for both colours. The second layer (red) was made up of 2 triangles with bases b1 & b2, and also one rectangle with a base of b1 and another rectangle with a base of b2.
Working down, I get a total of 4x b1 triangles + 4x b2 triangles for red, plus 16 b1 rectangles and 16 b2 rectangles.
So R = [4 * (1/2) * b1 * h] + [16 * (b1) * h ]+ [4 * (1/2)(b2) * h] + [16 * b2 * h]
That can be simplified to:
R = 16 * b1 * h + 16 * b2 * h
R = 16 * h * (b1 + b2)
blue gets 5 sets of triangles and 20 sets of rectangles, which works out to:
B = 22.5 * h * (b1 + b2) = 145
or
h * (b1 + b2) = 145/22.5
substitute in that value for [h * (b1 + b2)] into the first equation, and you get:
R = 16 * 145 / 22.5 = 116
I like the solution for elegance in the small triangles but as others have posted; 1/2 bh is basically half of a rectangle so double the blue area and take the 5/9 ratio, solve for the band height (58) and then multiply by 4 (232) and divide by two (116).
Solved it using different method than demonstrated. Good little brain teaser. Well done.
Hi Presh, I did a similar trick as Nèstor: As we have 9 segments, the surface of a triangle = b*h/2, the total area of the equivalent rectangle = 9*n*y (where n=height of the segment, y the width on the bottom). So the surface of blue portion is 2*145 = 5*n*y = 290. n*y = 58.
We now have the red portion which is 4*n*y = 4 * 58 = 232. As we are looking at the triangle, we divide it by 2 and have 116.
This works only if the segments are odd. If even, then one has to use your trickery. BR
Draw diagonals which are parallel to the sides of the triangle from each intersection point between a stripe a side of the triangle. This divides the triangle into a lattice of small triangles that are all similar. The first blue row has 1, the first red row has 3, the second blue row has 5, and so on down to the last blue row which has 17. So the total blue area = 145 = (1+5+9+13+17) triangles. The total red area has (3+7+11+15) triangles. Thus the red area is 145/(1+5+9+13+17)*(3+7+11+15) = 116.
Thanks for the proof, Presh but, correct me if I'm wrong, a simpler method of calculation would be :-
There are 9 stripes. 5 blue and 4 red. The areas are in a proportion of 5 to 4. 5 blue will contain 5/9 of the total area of the triangle. The red, 4/9 of the total area of the triangle.
Therefore, by dividing the area of blue by 5 and multiplying by 4, we get the answer. (145/5)*4=116
The only drawback is that this method doesn't work for an even number of stripes.
Something about parallelograms, congruency and other stuff I haven't had to deal with in 20 years. But I got it in under a minute.
That was a good one
Hi Presh...I have been enjoying your channel for many years but here's my first comment. I found using area formula for trapezium to be fastest method...so here's how I went about it
It just needs 2 formulae
A of triangle = 1/2h x base
A of trapezium = 1/2 h x (sum of parallel sides)
1. Name all the parallel sides starting with bottom one as as a, b c,...all the way up to i
2. Now if we see the total of blue portions will turn out to be
B = 1/2h ( a+b+c+d+f+g+h+i)
Similarly area of red portion is equal to
R = 1/2h ( b+c+d+e+f+g+h+i)
So really
B = 1/2ha + R
So replacing B for 145
R = 145 - 1/2ha....(1)
Now considering the whole triangle with height equal to 9h we get total area of traingle i.e. B+R = 1/2(9h) x a
So replacing B for 145
R = 9/2ha - 145...(2)
Equating both values of R, we get
9/2ha - 145 = 145 - 1/2ha
And solving it we get
1/2 ha = 29
And so
R = 145 - 1/2 ha = 116
first note that any triangle formed by having the bottom of a stripe as its base and the top of the big triangle as its highest point is similar to any other triangle constructed by using any other stripe as the base.
for similar triangles, if one length on one triangle is X times as long as the equivalent length on the other similar triangle, then all lengths on the first triangle is X times as long as the equivalent length on the other triangle.
therefore, since the area of a triangle is height*width/2, if one triangles sides is X times longer than the other, then the first triangles area is X*height*X*width/2.
the ratio between such areas is (X*height*X*width/2)/(height*width/2) = X^2 since the height*width/2 part cancels out.
the length of one triangles sides divided by the side of a triangle formed by taking the previous stripe as its base is (N+1)/N, since the stripes are divided by equidistant lines.
thus, let the first stripe (which also happens to be the first triangle) have area A.
the next stripe has area equal to the triangle formed by its stripe as a base minus the previous triangle, 2^2*A-A = 3A
next stripe, by similar logic is 3^2*A-2^2*A = 5A
next stripe, 4^2*A-3^2*A = 7A
in fact, this sequence will simply be equal to the difference between the squares, and the difference between the squares also happen to always equal the odd numbers, therefore this sequence will continue simply being the odd number multiples of A.
thus, the blue area is equal to (1+5+9+13+17)*A
and the red area is equal to (3+7+11+15)*A
we know that the blue area is 145
(1+5+9+13+17)*A = 145
45A = 145
A = 145/45
simplify
A = 29/9
insert into equation for red area
(3+7+11+15)*(29/9)
(36*29)/9
4*29
116
QED
Solved it in 1 minute, I imagined a triangle congruent to this triangle to form a parallelogram. So the given area of the blue lines was 145 , by doubling it, its 290 . So, area of one blue line is 58. As we know, in Parallelogram both red and blue lines are equal, so I multiplied 58 with 4 ( number of red lines. Which gives us 232)
Now I converted the parallelogram into a triangle, so area of red lines would be 232/2=116
My fairly unique solution
For convenience, reshape the triangle into a rectangular one. We will call the top blue stripe a half, which is our x. The first blue stripe is made of one such half, the second of 5, the third of 9, the fourth of 13 and the fifth of 17 (just keep adding 4). Now we can make a simple equation:
x + 5x + 9x + 13x + 17x = 145
45x = 145
x = 3.(2)
Knowing the value of the half, we can calculate the area of the red stripes with this simple equation:
y = (3 + 7 + 11 + 15) * 3.(2)
y = 36 * 3.(2)
y = 116
There's probably much easier solutions but that's what I thought of first
To my surprise you made a different approach to mine, with the same result.
I simply added an upside-down mirror of the main triangle, resulting in a rhombus with 290 "blue area".
The red area is now 4/5 of the blue, or 232.
Divide by 2 and the result is also 116.
I went from the bottom up, noting the base/height of each triangle incrememted by 1/9, ended up with an equation that looked like 145 = 1/2.9b/9.9h/9 - 1/2.8b/9.8h/9 + 1/2.7b/9.7h/9 .... +1/2.b/9.h/9.
After grouping like terms, solved for the area of the total triangle bh/2 = 261, so red =116..... cool problem.
My approach was to use a mini calculus approach, use the Sum of odd Numbers is N^2 + subtract or add the 1/2 of the triangle at the end of the calculus rectangle of width 1x1, 1x3, 1x5, 1x7 & 1x9 (=5 odd numbers for the Blue) - and then equate this area to 145 and find the similar area of the Red. Sum of Blue 1 to 9 odd numbers - 1/2 unit of triangle = 5^2 - 5*(0.5) = 22.5. Sum of 1 to 8 odd numbers + 1/2 unit of triangle = 4^2 + 4*(0.5) = 16. Therefore, if 22.5 = 145, then 16 = X, solve for X, X= 116. X = 145 x (n-1)(n-0.5)/((n*(n-0.5)) - reduced = 145 x (n-1)/n = 145 x 4/5. This is the same result as presented by Nestor Abad
It's amazing question
Amazing amazing amazing we need to explain the idea quickly in order to understand it faster
If you draw a line down from the top vertex to the base, and then a line halfway up that extending to each edge, you can see an elegant solution. By folding down the top two triangles, you get a rectangle with 2.5 blue strips and 2 red strips. Therefore area of red = 145*0.8 = 116
I made vertical cuts half way up each side of the triangle, then rotated the outer triangles 180 degress, so the bottom corners of the original triangle now met the top corner. Now it was a rectangle, so easy to compute red area = (4/5) * blue area.
My approach was
1. draw height into the triangle (this creates two new smaller triangles, one on the left side and one on the right side)
2. draw the Perpendicular bisector of the height (this creates four new smaler triangles, two on top, two on bottom)
3. rotate the two top triangles by 180 degrees and move them to the sides of the two bottom triangles (this creates a rectangle)
4. 5/9 of the rectangle are blue (this means 9/5*145=216 is the area of the rectangel / the original triangel)
5. 4/9 of the rectangle are red (this means 4/9*216=116 is the red area of the rectangel / original triangel)
My approach:
We can transform the triangle into a right triangle by performing a shear, which will preserve the area. Specifically, if the top vertex of the triangle is (x1,y1), then the linear transformation defined by (1,0) --> (1,0) and (x1, y1) --> (0, y1) is area preserving because the determinant is 1. Hence, we can assume WLOG that the triangle is a right triangle.
Drop a coordinate plane such that the vertices of the triangle are at (0,0), (x1, 0), and (y1, 0). Now consider the rectangle defined by these three points. This rectangle has a second triangle in the upper right section that is congruent to our original triangle. We can continue the red and blue stripes to cover the whole rectangle, giving us a total blue area of 290. Now, shuffling the stripes around allows us to deduce that the area of the red stripes is 4/5ths the area of the blue stripes, so the red stripes produce an area of 232. Cutting this in half gives us the area of the red stripes in the triangle, which will come out to 116.
Wow this is so nice
Put a reverse triangle to form a parallelogram. The blue part is 290 and it includes 5 small parallelograms. The red part includes 4. So the answer is 145 x 4/5.
Good method.
Looks good problem combining geometry and finite series...
Here is how I did it: The total blue area is 145 spread over 5 stripes. That gives us an average area of 29 and since the height of each stripe stays the same, this can legally be used as an average. (To be precise: the width would be 29/height of each stripe but since we will multiply by that height later on, we can as well leave it like this)
Now there are 4 red stripes with average area 29 (and if we decided to divide by the height, here is where we would have to multiply by it) and four of those give us an area of the red stripes with 4*29=116
Since the stripes grow linearly in area, the stripe in the middle has the average area of both the blue and the red stripes. That's why the red area is 4/5* blue area.
I solved it differently, but this approach is 10x better than mine. This is brilliant!
@@michaellautermilch9185 Thanks! Glad to help :)
Did that one in my head. Differently and got it
There are easier methods as mentioned in the comments below. But, if you insist on using the similar triangles approach, you can use the Thales theorem instead. FYI, Thales theorem is the basis for proving similar triangles
A much quicker method:
By inspection (easy to prove) the middle band of the 5 blue is the average band size. And the blue bands 1 above and 1 below average to the size of the middle band, as do the blue bands 3 above and 3 below.
So the average size of the blue bands = the average size of the red bands. There are 4 red and 5 blue, so the answer is 4/5 * 145 = 116.
Exactly, nice solution
I made the triangle right angle (only width and height matter anyway) and then cut the lower right part to make a rectangle. The area ratio is like the ratio of the lines, 4/5. 145*4/5=116.
Interesting solution.
I found the solution using a simpler method. There are 9 stripes 5/9 of those stripes has an area of 145, which means 145 is 5/9 of 261. The red area is 4/9 of 261 which is 116.
With a triangle the smaller stripes at the top will average out with the larger stripes at the bottom (There has to be some mathematical theory for this, or something that disproves it). Think of it this way, the area of an object is equal no matter the shape (ex - square/rectangle or triangle, etc). If you were to morph a triangle into a rectangle, you would need to move some of the area from the larger stripes to the smaller stripes, which is how they average out.
Amazing!!
MYD: "This is a formula we all learned in geomety"
Memory: we've been tricked, we've been backstabbed and weve been quite possible bamboozled
Love ur videos prash from india
Let’s see how many subs I can gain from this comment currently at 23
I solved it by visualizing it in my head. The area of a triangle only depends on its base and height, not on the angle of it's corners, so if you change this to a right angled triangle the area's are unaffected. If you then make a copy and rotate it 180 degrees over the center point of the hypotenuse you get a rectangle with 5 blue stripes and 4 red stripes, hence the total area of the red stripes is 4/5 times that of the blue stripes; 116.
(If changing the triangle to a right angled triangle in your head is a step too far you can also split the original triangle into 2 right angled triangles and apply the copy+rotation trick on each of them separately to get to the same rectangle)
Since area of any triangle is 1/2 b*h you can substitute a right triangle, double it to make a square, take 4/5 the area of your known rectangles then divide by half again.
We notice that there is those relations between two consecutive levels (from top to bottom):
- There are two more triangles
- On of the two triangles points upward, the other downward (So, a parity trick is possible)
We notice that every two levels, we add 2 red triangles and 2 upward triangles, so there are equal number of red and upward triangles at each red level.
Number of upward triangles is given by the sum1+2+3+...=n*(n+1)/2
At the 8th level, we thus have 36 upward triangles or 36 red triangles.
We notice that every two levels, we add 2 downward triangles and two blue triangles, so there are equal number of blue and downward triangles at each red level.
At the 10th level (mind that we are offset by 1 as downward triangles start at level 2), we have thus 45 downward triangles or 45 blue triangles.
In total, we have: 81 triangles.
145/45*36=116
Many people seem to duplicate the triangle and flip it upside down. Instead I cut it in half vertically and rotated half of it to make a rectangle which is easier to calculate.
Hello, I used a method using the trapezoid area equation.
First I drew the height of the big triangle and named it 9*h (as h is the height of each sector of the graph). Then I named every horizontal line from the top to bottom respectively, a b c d e f g i and j.
The area of the big triangle is A=j*9*h/2
Let's focus now on the Blue area. It contains a little triangle at the top and 4 trapezoids. The area is B= h/2(a+b+c+d+e+f+g+i+j) and it's equal to 145.
The red area contains 4 trapezoids and it is written as R=h/2(a+b+c+d+e+f+g+i).
We can see that it is so similar to B but it doesn't have j so R=B-h*j/2.
Thus we can say that R=B-A/9 (the red area is equal to the blue area - the total area over 9)
We also know that A=R+B and we can solve the system of 2 equations and we get that R=116 knowing that B is 145
Draw a vertical from the apex to base line. Rotate one half through 180 to form a rectangle.
145/5=29, 4*29=116
This was super easy for me :P
Triangle area is 1/2 b * h. Top row of triangle plus bottom row = 1 row of rectangle of b*h. Second top plus second bottom is 1 row of rectangle of b*h etc. That gives 2 red rows and 2 blue rows of a square excepting middle row of triangle. The middle row of the triangle if you divide horizontally in two can use same principle giving half a blue row. So the ratio is 2.5 blue to 2 red. (145/2.5)*2=116
i love how so many comments are....." i solved it in a much easier way"
I solved it too.......but made it way more complicated....im using b1, b2, b3, etc for triangle and trapezoid bases.......
Naming the height of each of the stripes as h and b as the base of the large triangle/bottommost trapezoid, and calculating B = (A[small triangle] + A[blue trapezoids]) and R = A[red trapezoids], you can get a system of equations that resolves to B = 10bh and R = 8bh => R = 0.8B = 116.
I mentally constructed a horizontal line at mid-height of the triangle; then, visualized the upper part rotated and set along side the lower part. Then noticed that, being a parallelogram was constructed, all that needed to be done was find the ratio of red to blue -- getting 2 to 2.5, or (multiplying by 2) 4 to 5. So, 4/5 times 145 is 116.