IS IT POSSIBLE to calculate the BLUE AREA?!
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- Опубліковано 10 лют 2025
- In this video, I demonstrate how the blue area remains unchanged regardless of the value of the radius of the innermost circle. It might seem strange at first, but it turns out to be relatively simple to calculate that the area always has the same value. Can you figure it out too?
best math teacher! I have learned so much from you Jonas
Great video! Keep em coming!
Thanks!! I will try 😁
First one of these where I got to the answer without watching all the way through... 25*pi straightaway. I got it right!
Very nice! :D
Great!🎉🎉 More, please.
Thanks! I’m on it!
If you reduce r to 0. Then you just have A=(R/2)^2 * pi . Or am I missing something?
This is correct, and beacuse R=5 (if r=0), then the area is 25pi.
But this is no proof that this is true for other values of r.
Yea but then u have no proof that the area will always be the same
To justify using this value for other values of r, I think you can imagine spinning the chord around the inner circle, and argue the d-area swept out by it during a d-theta is the same as for that swept out by the diameter of the r=0 case during the same d-theta (it is just translated away from the origin). IOW establishing an area-preserving map between r=0 and r=something cases
good video!
The shorter way: R^2-r^2=5^2
πR^2-πr^2=25
nice!
♥️♥️♥️
it looks right from the animation, but is there actually an explanation as to why ehen we change the inner circle radius, the shaded area doesnt change?
Yes, because when we cancel rˆ2 and -rˆ2 at the end, all instances of r are eliminated from the expression. This means that the value of r does not affect the value of the blue area.
I guess it isn't evident from the animation but both the outer and the inner radius is changing simultaneously.
As explained in the video,
R² - r² = 25
Or we can say R = √(25+r²)
(Assuming only positive values)
So, for any value of r we get a corresponding value of R for which this situation holds true (that the length of the tangent chord is 10).
And hence for any value of r there is a value of R such that the blue area is 25π.
cool
So for any line of length L the area would be (L/2)²π
It's like a pseudo diameter or something....
Yes, very cool I think!
Ok paused at 0:46 if the area is same regardless of the value of r, then we can give r any value we want. Like zero.
In that case the blue area becomes the entirety of the bigger circle(since smaller circle has an area of pi*0²=0), and the big circle has radius=5, giving us 25pi.
Except I feel like I would need to prove that the area remains same for all values of r. And while I dont know how to do that, I can give it another value, 12. At which point inner area is 144pi, the outer circle area is 169pi(due to 12-5-13 triangle), giving us a blue area of 25 pi again.
While this is not a proof, it is interesting. So, lets proove it.
I assume if I gave the 2 circle's radiuses a relation of x²+25=(x+y)², tossed the variables around, Id get 25=(x+y)²-x²
x+y is the big circle radius, and the blue area is (big radius² - small radius²)*pi, meaning, yeah, the blue area is always 25.pi
Additionally we can proove that the blue area is always equal to the square of half the length of the line. Times pi.
Very nice!