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Relaxing Math
Добавлен 14 окт 2023
Maths is not as difficult as you think. So, you can be successful in math! Thank you for watching
Area of Square
Thank you for watching❤️
Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉
🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're like: "I'm not smart enough, I'm never going to get this well". I'm going to tell you right now : "You can be successful in math, but it takes time and it takes effort and most importantly it takes great, clear and understandable math instruction.
I have a ton of videos as well, but anyways I'm trying to reach as many people as...
Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉
🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're like: "I'm not smart enough, I'm never going to get this well". I'm going to tell you right now : "You can be successful in math, but it takes time and it takes effort and most importantly it takes great, clear and understandable math instruction.
I have a ton of videos as well, but anyways I'm trying to reach as many people as...
Просмотров: 41
Видео
Area of a triangle
Просмотров 2,1 тыс.Месяц назад
Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're like: "I'm not smart enough...
A great question for a "True" Genius
Просмотров 2 тыс.Месяц назад
Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're like: "I'm not smart enough...
Area of a blue region
Просмотров 287Месяц назад
Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're like: "I'm not smart enough...
Area of Square
Просмотров 2,3 тыс.Месяц назад
Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're like: "I'm not smart enough...
Area of a blue region
Просмотров 877Месяц назад
Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're like: "I'm not smart enough...
Tricky question job interview
Просмотров 68Месяц назад
A great interview question. Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're...
Area of a triangle | Tricky interview question
Просмотров 82Месяц назад
Area of a triangle? Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're like: "...
2 Medium Pizzas or 1 Large (Interview Question)
Просмотров 39Месяц назад
Which is the better deal? Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're l...
A Green Circle
Просмотров 2,2 тыс.Месяц назад
Area of a green square? Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're lik...
A Square and a Triangle
Просмотров 1,3 тыс.Месяц назад
Area of a red square? Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're like:...
Area of a Red Square
Просмотров 1,6 тыс.Месяц назад
Hope you're doing well. Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're lik...
3 Circles
Просмотров 672Месяц назад
Area of the red circle? Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're lik...
Find the Area of Shaded Region
Просмотров 52Месяц назад
Hope you're doing well. Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're lik...
Area of a Rectangle
Просмотров 370Месяц назад
Hope you're doing well. Thank you for watching❤️ Happy to see you here. My name is John and I am a math teacher. My goal is to try to make math clear and understandable and interesting, but really my passion comes from helping people. 😉 🌍Now, what I try to do on this channel. I've seen people struggle in math, because they think they can't do math. They're really hard on themselves, they're lik...
Changing up the problem as you go is just utter BS. The tips of the triangle are not required to be in the center. You provide no proof that they are, and you provided no notation at the start of the problem that those are the center. You are just a fucking liar!
This is silly. If you rotate the triangle in another position, the problem will be incomplete
Would have been nice to say, I don't know..."these points at the centres of the squares". You know, actually state the problem.
That would be a complete statement of the problem, the way is presented in incomplete. If you rotate the triangle even outside the squares, the problem is unsolvable because is incompletely defined, silly.
I think it would be helpful to somehow indicate that the tips are at the center of the squares. But I love what you are doing. thank you!
30/45 = 2/3. Area of the circle is proportional to the squares of the diameter. 2 * 2/3 * 2/3 = 8/9 < 1. The area of 2 smaller pizzas is less than the area of the bigger one.
I calculated this without drawing. I just let the common side of the squares to cross the triangle and break it into 2 triangles with common side (it's size is 3 - 3/7), and sum of the areas of 2 yellow triangles with the same base and different heights will be (1/2) * base * (sum of heights) = 1/2 * 18/7 * 7 = 9
Write your solutions here 👇👇👇 Thank you for watching❤
You should have told us that the vertex of the triangle were at the centres of the square. With that piece of information it becomes easy. Without it, it's impossible.
This would be easier with vectors. if we take the center of the smaller square as origin, the other coordinates of the triangle are (3,3) and (7,1). Half the crossproduct is 9.
You did not state from the outset that the triangle is constructed by joining the centres of the squares, and without this crucial piece of information it is not clearly not possible to find any solution and with it the solution is trivial.
@@KenFullman Exactly. He looked at the drawing and ESTIMATED that the vertices are at the centers of the squares, basically solving it graphically lol. What if the centers are at 2.95 and 4.01? Basically the problem lacks some information to have a definitive solution.
Question was easy yet your answer in wrong. I dont know your email or i would have send you the answer. Your answer: 10.5... Correct ans: 13.73 Thank you!!
The answer given in the video rounded to 6 decimal digits is 10.9577, which is correct. You can check using the following Mathematica code: Integrate[Boole[x^2 + y^2 - 16 < 0 && (x - 4)^2 + y^2 - 16 > 0 && (x + 4)^2 + y^2 - 16 > 0], {x, -5, 5}, {y, -5, 5}]
Here's a simpler computation. Label the top point of triangle 1 as A. Label the bottom point of triangle 4 as B. Label the right point of triangles 2 and 3 as C. Triangle ABC is equilateral with side length twice the height of the small equilateral triangles : a = r*sqrt(3). If we remove triangle ABC from the central circle we get 3 zones, each exactly matching half of the area enclosed in the intersection between 2 circles. So we need 4 of them to match the total area to remove from the circle. A(blue region) = A(circle) - 4/3 * ( A(circle) - A(triangle ABC) ) = 4/3 * A(triangle ABC) - 1/3 * A(circle) A(triangle ABC) = 1/2 * sqrt(3)/2 * a^2 = 3*sqrt(3)/4 * r^2 = 12*sqrt(3) A(circle) = pi*r^2 = 16*pi A(blue region) = 4/3 * 12*sqrt(3) - 1/3 * 16*pi = 16*sqrt(3) - 16*pi/3
Just by looking at the thumbnail, you can see that the drawing is completely wrong. There are two possibilities: (a) the yellow "right angle" is not a right angle, or (b) the "square" is not a square, but only a rectangle. The reason being: there is only one right-angled triangle whose hypotenuse is 5, and that would be the yellow triangle, but also the white triangle in the top-right corner. Which means that the square side is 4, and the area of the square is 16. The rest is BS. The drawing with the measurements as shown in the thumbnail is impossible to construct. Very cheap "maths channel".
Nice problem. Since you pointed out the rhombus pattern, we can also use it to calculate the area of the small blue sectors in this manner Area(green sector) x 2 - Area(rhombus), which will give us the area of 2 of the smaller blue sectors.Then you can add Area(two smaller blue sectors) + Area(two green sectors) to get the whole area of one of the overlapping parts of middle the circle. Then it's just (pi r^2 ) - 2 x Area(overlap) for the area of blue region. I only suggested this method too because I am lazy and rhombus area is easier to calculate (4 x 4) :)
Find area of 6 equilateral triangles with each side 4. (48) Find area of circle with radius 4, (16X3.14159=50.26544 ) Subtract one from the other. (2.26544) Divide that by 6 (.37757333)and multiply by 4 (1.51029)and subtract that number from the area (16)of the 2 triangles. answer : (14.489) This answer did not take 12 minutes and 42 seconds.
Interesting problem. Seems like everything after 5:36 is needlessly complicated. A solid blue section is just the dashed-green section minus two dashed-blue regions. So 8pi/3 minus 2 x (8pi/3 - 4sqr3), or 8sqr3-8pi/3. Then double it because there are two solid blue sections.
You solution is great. Much love and respect, Sir!
What is your first language?
Thank you for watching❤ Have a great day! What do you think about this question?
This was a fun problem to solve. However, there was one problem I encountered when I tried a different approach. I tried to take the area of the bigger segments, which are the occupied space in the middle circle but if you cut them in half lengthwise, to get the area of the occupied space, but for some reason, the answers are not right. I would always get that the area of the 4 segments combined are 4(16pi/3 - 4√3), or roughly 39.308, which doesnt match up with the correct answer. Do you have any idea why? or am I doing something wrong in this solution
Also weirdly enough, I also came up with a solution much different but coincidentally, it gave me the answer but off by several decimals. I think this was pure coincidence and I was surprised when I found this
Not intended as a personal attack but as an advice: you talk fast with a particular intonation and accent. Clearly your command of English vocab and grammar is very good, since you are able to talk really fast. But I would work on the correct pronunciation and a more natural rhythm. As it is, it sounds a little bizarre. Of course, if your persona is the "nutty professor" then keep doing it :) (no comments on the math, it's good content for the purpose of "relaxing math")
It a very bad drawing. Your answer is wrong the yellow triangle is a 3.4.5 . Looking at picture your top right triangle is also 3.4.5 so this makes your square 4x4 which is 16.
A right triangle with hypotenuse of 5 can have other side values. The 3.4.5 triange is just a famous one.
@@hrayz Also, because the hypotenuse of the lower-left triangle is already 4, the other two sides of that triangle must each be less than 4, meaning the side of the square is less than 4 (since it's one of those shorter legs of the lower-left triangle).
I have a lot of videos as well, but anyways I'm trying to reach as many people as possible and I need your help. So, please consider hitting that subscribe button. It helps me a lot. Thank you for your support. Have a great day!
the “right” angle not being right is driving me insane
How is it "relaxing" math when the words are coming out at the speed of light? Chill, my friend! Let the information sink in...
great video dude, this was really interesting
Thank You Mister! Have a great day!
Wow, this was a very simple approach compared to mine. Here was what I did: I labeled the bottom-right angle of the yellow triangle "b" and the angle below that "a". Now, if you draw a line straight down from the topmost vertex of the triangle to the bottom edge of the square, the length of that line would have to be equal to the length of the bottom edge of the square. This equality can be expressed as: 5sin(a+b)=4cos(a). b can be easily found and plugged into the equation. Now, all that's left is to solve for a, and then plug a back into either 4cos(a) or 5sin(a+b), square that value, and you have your answer. At least, I think this is right.
clever
Thank You for watching! What do you think about this approach? If this video is helpful for you, don't forget to leave a like and subscribe , as that definitely help me out now for this particular video. Thank you for your support!❤❤❤
u cant just say its an equilateral triangle because you dont immediately know that touchpoints and centers of the circles lie on the same line
it'd be impossible to solve if they didn't and it's pretty visually obvious they do
If two circles are tangent, the point at which they touch is always on the line between their centers. If you solve the system of equations for the two tangent circles and the line going through their centers, the tangent point is the solution (I can write out the math later if you'd like).
@@chrissabal7937 i know, but its not obvious is it? i was just saying that he should have mentioned it in the video (no need for the proof just the mention)
Hello Mr Relaxing Math. At 2:26 you said that the area of each three parts is 1/6 then became pi/6 Why is it 1/6 and then became pi/6? I forgot the math stuff related to it. Thank you for your help
The equation for the area of a circle is πr², r=1 and 1² = 1 So the area is equal to π Since the angle of the section of the circle is 60° it's 1/6 of the whole thing. 1/6 times π equals π/6
360 degrees is full circle, and each corner of an equilateral triangle is 60 degrees. 60/360 = 1/6
Pretty simple! So the area of a circle is r^2xpi -> our r = 1 so we can pretty much take out the r^2 part, left is that the area of a circle with radius 1 which is just pi. Now to get the area of a partcial circle area we just do alpha/360° and then multiplied by pi (if radius stays 1). In our example a = 60° and 60°/360° is just 1/6. so we have (1/6) x pi which we can simplefy it to (pi x 1)/6 and then to pi/6. Then we just take that 3 times and subtract it all from the Root of 3 and bam all done :]
Different approach: Diagonal from bottom left to top right. The diagonal, from the point where the circle touches the square, is 2√2. This is equal to 1 + √2 times the radius of the circle.
Great approach, Sir!
(√2 + 1) ⋅ r = 2√2 r = 2√2 / (√2 + 1) r = 2√2 ⋅ (√2 − 1) r = 4 − 2√2 A = π ⋅ r² = (24 − 16√2) ⋅ π ≈ 4,31.
3.14 squared, whatever, surely , radius half the diameter, so is 1, 1 squared is err still 1 , times pi, 1 x 3.14 ..is 3.14 units ..took me longer to type it ....?
Yet another simple and geometric solution. Let's call the edges of the big square OABC (counterclockwise). Let's name the centre of the big square D and the center of the green circumference G. Now let's draw the 2 circumferences with radius OD=2√2 and OA=OC=4. The first one intersecates the big square at the points E and F. EA=FC=DG=r (in other words, G MUST be a point on the circumference with radius 4 because it's the only point equidistant from the big square and its centre). Hence r=4-2√2
The distance from the square center to its edge is: r + r/√2 = 2, hence r = 4 - 2√2. The area π*r² = π*(4 - 2√2)².
Great approach ,Sir!
If you draw a diagonal through the square it will.be tangent to the circle at the centre point. The side of the square is tangent and the same length. The radius is then the side minus this length. A neat little problrm.
It is simpler to make construction of 1 radius at 45° & another at 0° summing up to 2. So R(1 + 1/√2) = 2. R = 2/(1 + 1/√2). Rationalizing this gives r = 4 - 2√2. Squaring this gives R² = 24 - 16√2 Area = (24 - 16√2)π.
Great approach Sir!
Physicist's approach: assume green circle as big as a 2 by 2 box. area is therefore 2 * 2 = 4 units.
I think there's a quicker but less rigorous way. If you know the hypotenuse of the triangle on the left is five then the pythagorean triples say the other sides must be 3 and 4. Whichever of those sides is shorter must be 3, so radius 5 - 3 = 2 and 2 squared is 4.
Great solution, Sir! Have a great day!
I reckon that you could've better multiplied out (4-2sqrt2)^2? Maybe? I'm not sure it would have let us be able to better understand it.
I haven't seen that equation for figuring out the radius before, perhaps you could derive it?
Simple. Draw triangle with inscribed circle. Construct lines from triangle vertices to the circle's center. This cuts the triangle up into 3 triangles. The bases of these 3 triangles ate the sides of the original triangle. Construct lines from the tangent points on the circle to the circle's center. These are radii, & the heights of the 3 little triangles. Each triangle's area is (1/2) base * height. All the 3 heights are the same, so total area of all 3 = total area of original triangle = (1/2)(side a + side b + side c))(radius) = (1/2)(perimeter)(radius).
Do you happen to be colour blind (1:53) or shape blind (description) perhaps? Or was it a bad copy paste? :p
Ah, that's what you look like! I recognised your voice :D
Which is the better deal? Write your thoughts Thank you for watching❤
Ridiculous to pull an equation like that out of your rse as the solve
We can generalize this: A rectangular triangle with sides a,b and hypothenusa c, has area ab/2 The square inscribed has side x and area x² The smaller triangles are congruent with the big one. Their sides are proportional by x/a, x/b and x/c respectively, so their areas are proportional by the squares (x/a)², (x/b)² and (x/c)² This leads to the following equation ab/2 = x² + ab/2 * ((x/a)²+(x/b)²+(x/c)²) factoring out ab/2 while resolving the sum of fractions between the brackets: 1 = 2x²/ab + x²(a²b²+b²c²+c²a²)/a²b²c² multiply by the denominator of the last term a²b²c² = x² (2abc² + a²b²+b²c²+c²a²) bringing 3 terms together and resolving for x² x² = a²b²c² / ((a+b)²c² + a²b²) (you can make this canonic by further replacing c² with a²+b²)
Yeah but.. how do you derive the formula ?
This is all know formula from math book. What are you talking about?
It is also possible to solve only usign the pythagorean theorem, observing that half the diagonal of the 4×4 square is equal to the radius (r) of the circle plus the diagonal of a r×r square. You can visualize this by drawing the radius of the circle for each relevant point on the border of the circle
1:33 - you jumped a step - you are making an assumption that dropping a perpendicular from the point of intersection of the circle to the square leads to an equal division of one side of the square into x/2. You need to prove it.
You can (mostly) do it in your mind if you work with areas directly. Just imagine zooming in untill the small triangle on top is as big as the original one. Then you know the area of the enlarged square will be 25, because is side is now 5. So the ratio of the area of the top triangle to the area of the square is 6/25. You can do the same for the other small triangles, and you get the ratios 6/16 for the left one and 6/9 for the right one. The area of the large triangle is the sum of all the small triangles and the square, so it will be: 1+6/25+6/16+6/9=6×1369/3600 times larger than the square. And so the area of the square is 3600/1369=2.63 Doing all the calculation in your head is a bit tricky, but if you approximate a bit you can get an answer of 24/9=2.66 (which is not bad)
Great approach , Sir!
No need to solve simultaneous eqns. A. Left triangle has hypotenuse of 5a/4 B. Top triangle has smallest side of 4a/5. A+ B = smallest side of large triangle = 3. Thus (3/5+5/3)a = 3 a= 60/37 Taking the side length 4: (4/5+5/3)a = 4 a = 69/37, no problem. Simple.
You approach is great, Sir! Have a great day!
I'm hopeless at maths...I spotted the red square, but after that...😆(great explanation!)
Thank you, Sir! Have a great day!
Hello everyone. Thank You for watching! If this video is helpful in any way, don't forget to like and subscribe as that definitely help me out now for this particular video. Thank You for your support, I really appreciate it. Have a great day!❤❤❤
thanks for the great content and dynamic explanation, very good video👍
Thank you for your kind comment. I really appreciate it. Have a great day, Sir!
What is the correct answer? What do you think? Also, Thank you for your support. Have a great day!🥰