Can also be solved using 'Special Case' argument. Consider the radius of the inner circle infinitely small, in this case the length of "50 cm" automatically becomes the DIAMETER of the outer circle. And since in this case the radius is 25cm, hence the area becomes 625 pi cm^squared.
@@kumardeals2903 If you consider the inner circle to be extremely small, as if it doesn't even exist, it will be as if the tire was made only of the outer circle, no hole for the wheel. Because this solution works only for the case a = 0, it is called a special case.
@@henricobarbosa7634 Just to be clear, only if there were no inner circle. However small it is it won't become diameter unless there is no inner circle.
This is not a "seemingly impossible" problem. It's actually pretty simple, although I take no position on whether it's appropriate for grade school kids. The question may look deceptively difficult at first because it's one of those odd problems in which, instead of solving for two or more variables separately, you solve for the *relationship* between variables and plug the whole result right into a formula.
A = ¼πc² = ¼π50²= 625π cm² Area of circular ring A=¼πc². Same formula of circle area, but putting the chord "c" instead the diameter "d" Border conditions: c=d ---> Circle area c=0 ---> Círcular ring disappears
I got the answer so easily, 25^2×pie= 625×pie.... Actually it's all about how you deal with the questions... That's why problem solving skills and concept building is so necessary
Wow! Mind Blowing Answer! 👍 I was using the hard way/formula: a = 50, A = area 2s = a + a + a = 3a ➡️ s = 3a/2 r = A/s = (1/2)•a•a•(1/2)√3 ÷ (3a/2) = a/2√3 (inner circle)* R = a / 2sin60° = a/√3 (outter circle)** Black Area = πR² - πr² = π•(R²-r²) = π(a²/3 - a²/12) = πa²/4 = π•50²/4 = 625π ✅ or = (a/2)²•π Complicated Me 😅
So the real trick here is that it seems impossible because there are many circle configurations that fit the described measurements, but ALL of them actually yield the same result for the area. It would be impossible to identify a single correct radius as there are an infinite number of working values, but we don't need a single correct radius in the first place.
@ Why so? I never even learned about tangent at all yet. Then there was a guy that said that he could remove the hole in the tire infinitely so the radius of the circle would be 25 and then i would be able to get 625pi. This one is something i understood but never learned that could be done
@gamgamesplayer4506 tangents were introduced to me in 10th class. I'm not sure if you were taught same stuff as me(like in your country, the course may be different) . Btw, I tagged you mistakenly.
Nice trick. At first sight everybody thinks that is impossible. We can get that 50 cm value at any tyre size. BUT The surprusing thing is: The area of the tyre will be always the same with that value. That is the fact what we dont know and our faces look like 😮
buddy come to india and take addmission in cbse we have a lot of such question in 10th grade at first you think how to solve but this question comes for 2-3 marks only making it not that much hard solve sin^6A+cos^6A = 1-sin^2Acos^2 (it is dam easy than you think)
convert sin^6+cos^6 to (sin^2)^3 + (cos^2)3 then use the a^3 + b^3 formula you will get sin^4+cos^4+sin^2cos^2 because sin^2+cos^2=1 then (sin^2+cos^2)^2 - 2sin^2cos^2 + sin^2cos^2 simplifying that should give you the rhs
@randomguy-dky yep you are perfect but when you do by other methond that is (a+b )(a^2+b^2-ab) it will make bit mess and by doing further you have to solve a^4+b^4 we can only solve trigo only if we have a lot pratice but still using some short tricks in cbse you won't get mark
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Can also be solved using 'Special Case' argument.
Consider the radius of the inner circle infinitely small, in this case the length of "50 cm" automatically becomes the DIAMETER of the outer circle. And since in this case the radius is 25cm, hence the area becomes 625 pi cm^squared.
This is good!
How come 50cm be diameter of the outer circle if it doesn’t pass through the center?
@@kumardeals2903 If you consider the inner circle to be extremely small, as if it doesn't even exist, it will be as if the tire was made only of the outer circle, no hole for the wheel.
Because this solution works only for the case a = 0, it is called a special case.
If there is almost no inner circle, the rod will become the diameter.
@@henricobarbosa7634 Just to be clear, only if there were no inner circle. However small it is it won't become diameter unless there is no inner circle.
Outer radius = R, inner radius = r.
The pythagorean theorem gives = R² - r² = 25².
Area = pi(R² - r²) = pi.25²
This is not a "seemingly impossible" problem. It's actually pretty simple, although I take no position on whether it's appropriate for grade school kids. The question may look deceptively difficult at first because it's one of those odd problems in which, instead of solving for two or more variables separately, you solve for the *relationship* between variables and plug the whole result right into a formula.
As soon as you made right triangle, I was like oohhhh
A = ¼πc² = ¼π50²= 625π cm²
Area of circular ring A=¼πc².
Same formula of circle area, but putting the chord "c" instead the diameter "d"
Border conditions:
c=d ---> Circle area
c=0 ---> Círcular ring disappears
I simply assumed the inner circle was zero radius so the cord became the diameter of a solid tyre and used pi x r x r.
Same
What?? Why not set the inner circle to radius 0, and then calculate the area of a circle with radius 25
R^2 - r^2 = 25^2
Tire area = (25^2)π
In such questions we are given with the information which is just enough to find what is asked
I got the answer so easily, 25^2×pie= 625×pie.... Actually it's all about how you deal with the questions... That's why problem solving skills and concept building is so necessary
Question from 10th grade, ch 10 (NCERT books)
Wow! Mind Blowing Answer! 👍
I was using the hard way/formula: a = 50, A = area
2s = a + a + a = 3a ➡️ s = 3a/2
r = A/s = (1/2)•a•a•(1/2)√3 ÷ (3a/2) = a/2√3 (inner circle)*
R = a / 2sin60° = a/√3 (outter circle)**
Black Area = πR² - πr² = π•(R²-r²) = π(a²/3 - a²/12) = πa²/4 = π•50²/4 = 625π ✅
or = (a/2)²•π
Complicated Me 😅
I too solved it in the same way but I used the chord bisector theorem
So the real trick here is that it seems impossible because there are many circle configurations that fit the described measurements, but ALL of them actually yield the same result for the area.
It would be impossible to identify a single correct radius as there are an infinite number of working values, but we don't need a single correct radius in the first place.
But tires have the bumps on the edge, thus we need more info to actually solve it
Let's it's really worn out and smooth like you dad's head.
2:41 pi(b^2-a^2) = pi(25^2+a^2-a^2) = 625pi
Sorry, but this is actually a pre-school question.
Preschool learns abcs bro.
I just finished 10th grade and yet wasn't able to solve it
@gamgamesplayer4506 sorry for you.
Preschoolers may not be able to solve this problem but certainly a highschooler can.
@ Why so? I never even learned about tangent at all yet. Then there was a guy that said that he could remove the hole in the tire infinitely so the radius of the circle would be 25 and then i would be able to get 625pi. This one is something i understood but never learned that could be done
@gamgamesplayer4506 tangents were introduced to me in 10th class. I'm not sure if you were taught same stuff as me(like in your country, the course may be different) . Btw, I tagged you mistakenly.
Pythagoras and basics get you far
That is really clever.
I thought you will use chord chord power theorem to solve it
not needed
Yeah
Before I watch, I'm assuming Pi*625, which is about 1963.5.
Did this in my head.
This is from class 10 maths
Nice but only got the sidewall area not whole tire.😂
Hint: Pythagoras' theorem (勾股定理)
Nice trick. At first sight everybody thinks that is impossible. We can get that 50 cm value at any tyre size.
BUT
The surprusing thing is: The area of the tyre will be always the same with that value. That is the fact what we dont know and our faces look like 😮
special case?
No way I got that right on the first try lol
buddy come to india and take addmission in cbse we have a lot of such question in 10th grade at first you think how to solve but this question comes for 2-3 marks only making it not that much hard solve sin^6A+cos^6A = 1-sin^2Acos^2 (it is dam easy than you think)
convert sin^6+cos^6 to (sin^2)^3 + (cos^2)3
then use the a^3 + b^3 formula
you will get sin^4+cos^4+sin^2cos^2 because sin^2+cos^2=1
then (sin^2+cos^2)^2 - 2sin^2cos^2 + sin^2cos^2
simplifying that should give you the rhs
@randomguy-dky yep you are perfect but when you do by other methond that is (a+b )(a^2+b^2-ab) it will make bit mess and by doing further you have to solve a^4+b^4 we can only solve trigo only if we have a lot pratice but still using some short tricks in cbse you won't get mark
Piece of cake 5 seconds answer that too I had to write on paper waste
It was easy
863.9378 it is veri esi math wanli 3 sicond
Veri smole math
ok