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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.

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

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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!!

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.

What is your first language?

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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.

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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

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

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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

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

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.

(√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.

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.

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.

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?

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

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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 ?

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)

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.

I'm hopeless at maths...I spotted the red square, but after that...😆(great explanation!)

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thanks for the great content and dynamic explanation, very good video👍

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