0.12034 13:35 I did it differently. We know the three squares' total length, 6X, but we don't know the right and left triangle lengths which are halves of equilateral triangles. We know the height of the altitude of the equilateral triangle, sqrt 3/4 or 0.866 (from sqrt of 1^2- 1/2^2, Pythagorean Theorem) and its base, 1. Therefore, the ratio is 0.866/1 or the base is 1.1547 ( 1/0.866) its altitude. Since the right triangle's altitude is 3X, its base is 3.464X (3X times 1.1547). And hence its length is 1.732X (3.464X/2). Since the left triangle's altitude is X, its base is 1.1547X (X times 1.1547). And hence its length is 0.57735 X (1.1547X/2). So TOTAL length of the two triangles + 3 squares : 0.57735X + 1.732X +6X=1 8.309X =1 X= 1/8.309 X=0.120346 Answer 13:35
Так мило с твоей стороны, дорогой! Вы классные 👍 Я рада, что вам понравилось! Пожалуйста, продолжайте делиться предматчевым каналом со своей семьей и друзьями. Береги себя, дорогая, и оставайся счастливыми
Tão legal da sua parte, querida! Você é incrível 👍 Fico feliz que tenha gostado! Continue compartilhando o canal premath com sua família e amigos. Cuide-se querido e continue abençoado😃 Amor e orações dos EUA!
The similarity ratio between the two triangles located on both sides of the squares is: X/3X=1/3. Since the sum of their bases is 1-6X, the base of the triangle on the left is 3(1-6X)/4 and the base of the triangle on the right (1-6X)/4. Since it is half of an equilateral triangle and its height is X, we can state that: X=(√3)(1-6X)/4 → X(4+6√3)=√3 → X=√ 3/(4+6√3)=0.120345
As ∆ABC is an equilateral triangle, the angle at all three vertices is 60°, and all sides are the same length 1. As the line segments ED and FG are the sides of squares whose adjacent sides are the baseline of the triangle, ∠EDB and ∠AFG are right triangles. This means that ∠BED and ∠GAF are 30° angles, and ∆EDB and ∆AFG are 30-60-90 special right triangles and thus similar.As a 30-60-90 right triangle, the hypotenuse is 2x the length of the short leg, and the long leg is √3 times the length of the short side. As 30-60-90 right triangles, the hypotenuse is 2x the length of the short leg, and the long leg is √3 times the length of the short side. By observation, AB, which we know to be of length 1, is also of length 6x + AF + DB. These line segments are the short legs of their respective triangles. Let DB = y and AF = z. Therefore: 6x + y + z = 1 ---- (1) Triangle ∆EDB: DE = √3y x = √3y y = x/√3 Triangle ∆AFG: FG = √3z 3x = √3z z = 3x/√3 = √3x 6x + (x/√3) + (√3x) = 1 18x/3 + √3x/3 + 3√3x/3 = 1 x(18+4√3)/3 = 1 x = 3/(18+4√3) x = 3(18-4√3)/(18+4√3)(18-4√3) x = (54-12√3)/(324-48) = 6(9-2√3)/276 x = (9-2√3)/46 = 9/46 - √3/23 ≈ 0.12
So nice of you dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
That's a fun and easy problem sir My solution: We can easily see that the angle GAF is equal to 60 degree, therefore AF = GF / tan 60 = 3x / sqrt(3) Similarly, DB = x/sqrt(3) We also have AB = AF + DB + 6x = x(6 + 4/sqrt(3)) = 1 Therefore x = 1/ (6 + 4/sqrt(3)) Edit: You can also solve this if you noticed that a 90-60-30 triangle is actually just half of a equalateral triangle, effectively making the problem solvables just by using 6th grade math.
I feel like a dumbass sometimes, I look at the thumbnail and go “Holy shit, let’s see how he solves this” but then I read the information and see the 60 30 angle triangle 🤦♂️
Thank you guru. I think AF= half of GF. ie AF = 3x/2. For same reason DB = x/2. So the sum of base line = 3x/2 + 3x + 2x + x + x/2 = 1. So 16x = 2. Then x = 1/8 = 0.125
So nice you Paul. You are so generous. Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃 Love and prayers from the USA! 😃
For those like me who don’t know their triangle ratios, here’s a hint: use cos(30) and sin(30) on your beloved calculator! That’s what I did (I know, shame on me hhhh) thank you sir for giving me something fun to do in these depressing times. Kudos
Good grief, he makes a meal of this one, and gets the wrong answer. On either side there is a 30 degree triangle. Cos 30 degrees is a half, so on the right the distance from the square to the apex of the triangle is x/2 and on the left 3x/2. So 8x = 1 and x = 1/8 or 0.125 exactly. 《30s IME.
At first glance it seems difficult but a closer look and analysis, it can be easily solved using trigonometry. Below is a simple formula to solve it: 3X cos 60 degrees + 3X +2X + X + X cos 60 degrees = 1. One equation, one unknown, therefore, solvable. 😊😊
@@DhirajSingh-mn6zu Yeah, there's a little mistake in the equation I made after carefully looking at the diagram. I tried to solve it by myself without watching the solution in the video. Now I realize that there should be more steps in getting the 2 unknown horizontal segments near the squares 3x and x. First, get the hypotenuse of the small triangles beside 3x (say "y"), by using sin 60 degrees= 3x/hypotenuse1. Then get the horizontal segment "y" beside 3x by using the equation y=hypotenuse 1 cosine 60 degrees. Do the same procedure, for the horizontal segment (say"z") beside x , by using sin 60 degrees= x/hypotenuse 2. Small horizontal segment z = hypotenuse 2 cosine 60 degrees. After getting the 2 unknown horizontal segments beside the squares 3x and x, (which are y and z),we now have the equation y+ 3x+ 2x+ x + z= 1. One equation, one unknown (since we already know the values of y and z) , therefore solvable. Thanks.
Thank you sir
You are very welcome Gowri!
Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊
@@PreMath wow
0.12034 13:35 I did it differently. We know the three squares' total length, 6X, but we don't know the right and left triangle lengths which are halves of equilateral triangles.
We know the height of the altitude of the equilateral triangle, sqrt 3/4 or 0.866 (from sqrt of 1^2- 1/2^2, Pythagorean Theorem) and its base, 1. Therefore, the ratio is 0.866/1 or the base is 1.1547 ( 1/0.866) its altitude. Since the right triangle's altitude is 3X, its base is 3.464X (3X times 1.1547). And hence its length is 1.732X (3.464X/2).
Since the left triangle's altitude is X, its base is 1.1547X (X times 1.1547). And hence its length is 0.57735 X (1.1547X/2).
So TOTAL length of the two triangles + 3 squares :
0.57735X + 1.732X +6X=1
8.309X =1
X= 1/8.309
X=0.120346 Answer 13:35
Thank for PreMath
Thanks Taha dear for the input. You are awesome 👍 Take care dear and stay blessed😃
Классная задача! С удовольствием смотрю Ваши видео!
Так мило с твоей стороны, дорогой! Вы классные 👍 Я рада, что вам понравилось! Пожалуйста, продолжайте делиться предматчевым каналом со своей семьей и друзьями. Береги себя, дорогая, и оставайся счастливыми
@@PreMath U are very great sir.
喜歡這個視頻，講解清楚， 雖然，我不太懂英文，但是，我都可以理解；畫面清晰，不會隨便擋住畫面。
very helpful ... Thanks Mr PreMath for making me think. Solving is exhaustingly fun.
Muito bom! Saudações de Brasil!
Tão legal da sua parte, querida! Você é incrível 👍 Fico feliz que tenha gostado! Continue compartilhando o canal premath com sua família e amigos. Cuide-se querido e continue abençoado😃
Amor e orações dos EUA!
Amazing video 👍
Thank you so much for sharing😀
Thanks dear for the input. You are awesome 👍 Take care dear and stay blessed😃
That was a really nice task! Thank you for sharing different methods of solving it.
Explained well to math learners. Trigonometry is also explained.
Happy learning.
The similarity ratio between the two triangles located on both sides of the squares is: X/3X=1/3. Since the sum of their bases is 1-6X, the base of the triangle on the left is 3(1-6X)/4 and the base of the triangle on the right (1-6X)/4. Since it is half of an equilateral triangle and its height is X, we can state that: X=(√3)(1-6X)/4 → X(4+6√3)=√3 → X=√ 3/(4+6√3)=0.120345
As ∆ABC is an equilateral triangle, the angle at all three vertices is 60°, and all sides are the same length 1. As the line segments ED and FG are the sides of squares whose adjacent sides are the baseline of the triangle, ∠EDB and ∠AFG are right triangles. This means that ∠BED and ∠GAF are 30° angles, and ∆EDB and ∆AFG are 30-60-90 special right triangles and thus similar.As a 30-60-90 right triangle, the hypotenuse is 2x the length of the short leg, and the long leg is √3 times the length of the short side. As 30-60-90 right triangles, the hypotenuse is 2x the length of the short leg, and the long leg is √3 times the length of the short side.
By observation, AB, which we know to be of length 1, is also of length 6x + AF + DB. These line segments are the short legs of their respective triangles. Let DB = y and AF = z. Therefore:
6x + y + z = 1 ---- (1)
Triangle ∆EDB:
DE = √3y
x = √3y
y = x/√3
Triangle ∆AFG:
FG = √3z
3x = √3z
z = 3x/√3 = √3x
6x + (x/√3) + (√3x) = 1
18x/3 + √3x/3 + 3√3x/3 = 1
x(18+4√3)/3 = 1
x = 3/(18+4√3)
x = 3(18-4√3)/(18+4√3)(18-4√3)
x = (54-12√3)/(324-48) = 6(9-2√3)/276
x = (9-2√3)/46 = 9/46 - √3/23 ≈ 0.12
Tried it without using trig but could not work it. Had to use the trig and your step-by-step solution. Appreciate this question and your solution.
Nice
So nice of you dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Thank you so much 💕
Another star example, I identified the non trig method but got a bit mixed up with manipulating the root 3. Hey ho more practice needed.. thanks again
I did it the same way. Thank you sir.
nice question👍👍
So nice of you Arvind dear! You are awesome 👍 I'm glad you liked it! Take care dear and stay blessed😃
That's a fun and easy problem sir
My solution:
We can easily see that the angle GAF is equal to 60 degree, therefore AF = GF / tan 60 = 3x / sqrt(3)
Similarly, DB = x/sqrt(3)
We also have AB = AF + DB + 6x = x(6 + 4/sqrt(3)) = 1
Therefore x = 1/ (6 + 4/sqrt(3))
Edit: You can also solve this if you noticed that a 90-60-30 triangle is actually just half of a equalateral triangle, effectively making the problem solvables just by using 6th grade math.
Some Old Horses Carry A Heavy Ton Of Apples
Que buen canal
I feel like a dumbass sometimes, I look at the thumbnail and go “Holy shit, let’s see how he solves this” but then I read the information and see the 60 30 angle triangle 🤦♂️
Thank you :)
Where is non trig solution ?
Thank you guru.
I think AF= half of GF. ie AF = 3x/2. For same reason DB = x/2. So the sum of base line = 3x/2 + 3x + 2x + x + x/2 = 1. So 16x = 2. Then x = 1/8 = 0.125
So nice you Paul. You are so generous.
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Love and prayers from the USA! 😃
For those like me who don’t know their triangle ratios, here’s a hint: use cos(30) and sin(30) on your beloved calculator! That’s what I did (I know, shame on me hhhh) thank you sir for giving me something fun to do in these depressing times. Kudos
i watched and liked the video
Great
....now You are mi Yogirai!
thanks
Pleaso do skip the obviuos and get to the point!
AF + 3X + 2X + X + DB = 1; CONCIDER RIGHT TIANGLE AFG & RIGHT TRIANGLE BDE and apply trig function tan , we will get the value of x
The middle square is not necessary; it is an extra information
Sorry, the middle square is also necessary
👍👍👍
(x+3*x)/√3 + 6*x = 1
or x = 1/ (4+6√3)
x=（9-2√3)/46
👍🏻
0,12.
(9-2√3)/46 units is the exact answer.
ngl it really is much a nicer diagram
Good grief, he makes a meal of this one, and gets the wrong answer.
On either side there is a 30 degree triangle. Cos 30 degrees is a half, so on the right the distance from the square to the apex of the triangle is x/2 and on the left 3x/2. So 8x = 1 and x = 1/8 or 0.125 exactly.
《30s IME.
Pretty Mathematics problem.
That was too easy! I solved it in one minute.That video is too long
Easy problem I can do this problem in minute
Very smart Raj👍Take care dear and stay blessed😃
At first glance it seems difficult but a closer look and analysis, it can be easily solved using trigonometry. Below is a simple formula to solve it:
3X cos 60 degrees + 3X +2X + X + X cos 60 degrees = 1. One equation, one unknown, therefore, solvable. 😊😊
Please learn and then comment
@@DhirajSingh-mn6zu Yeah, there's a little mistake in the equation I made after carefully looking at the diagram. I tried to solve it by myself without watching the solution in the video. Now I realize that there should be more steps in getting the 2 unknown horizontal segments near the squares 3x and x. First, get the hypotenuse of the small triangles beside 3x (say "y"), by using sin 60 degrees= 3x/hypotenuse1. Then get the horizontal segment "y" beside 3x by using the equation y=hypotenuse 1 cosine 60 degrees. Do the same procedure, for the horizontal segment (say"z") beside x , by using sin 60 degrees= x/hypotenuse 2. Small horizontal segment z = hypotenuse 2 cosine 60 degrees. After getting the 2 unknown horizontal segments beside the squares 3x and x, (which are y and z),we now have the equation y+ 3x+ 2x+ x + z= 1. One equation, one unknown (since we already know the values of y and z) , therefore solvable. Thanks.
Thank you Sir