This is an interesting problem because the radius of each arc segment can vary and as long as the chord segment is 16 units long, the blue area will remain the same.
In the beginning, you state that the chord is tangent to small semi-circle. However, you don't define the chord as a parallel line to the "line" that is the diameter of large semi-circle. This is a remarkably good geometric problem. But, the chord geometry needs to be fully given. The tangency to small semi-circle can take on infinitely many orientations. In the "middle" of the talk, you ASSUME the chord is parallel to diameter line. Love your channel and solution methods. Thanks !
He doesn't assume that. And that assumption is therefore not necessary. Edit: it just looks like that because he draws the radii vertically, but they don't have to be.
Because the chord is tangent to the smaller circle, it is perpendicular to its radius. The chord is also perpendicular to the radius of the larger circle that bisects it because that's a property of chords. So, the question is, how do we know that the centers of the two circles are colinear on the same diameter? By symmetry, they have to be. But yeah, you might want to say something to justify saying that the distance from the center of the larger circle to the chord is the same as the distance from the center of the smaller circle to the chord. It can be proven, but I'd agree that it's not immediate.
@@rickdesper "Because the chord is tangent to the smaller circle, it is perpendicular to its radius" .....Yes. This is very obvious from line tangent definition to to curvatures. "The chord is also perpendicular to the radius of the larger circle that bisects it" ......This is not intuitive and I will have to look at "chord" geometry proof(s) and I don't believe this to be true (for now). Thanks Rick.
The chord is 16 units. Let the white semicircle shrink down to nothing and the 16-unit chord becomes the diameter of the blue semicircle. The radius is 8 and the area is ((8^2)*pi)/2 = 64 pi /2 =32 pi. This is true for any value of the small semicircle radius r. There is no maximum value for r; but r = sqrt(R^2 -- 64). Setting r = 0 gives the same answer as above. Cheers. 🤠
That's how I did it as well, no need for complex calculations, just the observation that given what we were told, the blue area must be constant regardless of the sizes of the circles.
@@alternativeglasto We don't know apriori that enough information is given for the problem to be solvable. This approach assumes that, but the video shows that this assumption is not necessary.
I think you've made a mistake. How could the chord length that is not drawn at the diameter possibly be equal to the diameter? An accidental proportionality has accrued that is not consistent in general. If I am wrong, well great. Common Archimedes show me up.
@@jonathanfanning9558 We don't know the radius of the white circle and it can't be calculated so the solution can't depend on that value if the problem is solvable. Therefore we can assume that radius of the white circle can be zero. Avbryt Svara
the first two lines you draw are not necessarily the same length, for them to be the same length we would need to know that the chord is parallel to the base of the larger semi-circle but we are not told this and it is not possible to prove from the given information
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Nowhere in the initial problem is it stated that the chord is parallel to the base line. Hasn't the lack of this condition prevented this solution, since this condition is required for the solution.?
This type of puzzle with very few data is often simple. In this case, let's assume there is no missing data. So, the absolute radius of the two circle don't matter. Suppose the little one be zero, then 16 is the diameter of the great and the blue shaded region is just a semi-circle of radius 8.
If there is one solution it must be independant of the size of the smaller semicirkle. So, by putting the smaller cirkle =0 the blue region equals rhe area of a semicirkle with radius=8 i.e. 32xPI.
Dear Noor Alam, that's a big dilemma! I haven't seen any quality book yet. If I come across with any good resource, sure I'd love to share with you. Don't spend too much money on any book. Meanwhile keep watching PreMath channel. We work very hard to put quality videos for our valued viewers. Thanks for asking. You are awesome. Keep it up 👍 Love and prayers from the USA! 😀
Considering that small semicorcle can have any radius, imagine that small circle extremely small. Then shaded area will equal to area of large semicircle with radius = 8. =} 32π
Very neat problem. If you try to solve for r or R, you'll get nowhere. Neither r nor R can be pinned to a specific value, even though (R^2 - r^2) is itself determined.
Puisque le diamètre des 2 demi-cercles ne dépendent pas de la longueur du segment 16, alors on peut faire D=16 et r=0 et on trouve immédiatement la surface bleue, non? Je préfère utiliser S=Pi.D*2 / 4
It's assumed that the cord is parallel to the diameter of the larger semicircle. What's the basis for that assumption, other than that's how it appears to be drawn? Unless I'm missing something, if it's not, your solution falls apart.
No need for Pythagore, just trigonometry ... R cos a = 8 and R sin a = r PI/2 (R2 - r2) = PI/2 R2 (1 - sin2 a) = PI/2 R2 cos2 a = PI/2 64 = 32 PI (took me 5 seconds)
Shouldn't it be stated in the known facts about the problem that the chord is actually perpendicular to the radius? 16 units isn't enough information, and the chord could still be a tangent of the smaller circle while not being parallel to that bottom part of the larger semicircle.
Which radius? It's perpendicular to the radius of the smaller circle because the chord is tangent to the circle there. And it's perpendicular to the radius of the larger circle that bisects it: that's an easy theorem of geometry.
Another in-the-head quickie. Spoiler alert. No variables needed. Simply imagine the smaller semicircle shrinking to nothing, so that 16 becomes the diameter of the larger semicircle. Then its radius is 8 and its area is 8²π /2 = 64π /2 = 32π square units.
It's wrong We cannot be sure that the side of the triangle is 8 because we do not know that the chord is parallel to the diameter of the great circle It's could be bigger or less than 8
Let R be the radius of the larger semicircle and r be the radius of the smaller semicircle. r•r + 8•8 = R•R R^2 -r^2= 64 Area of shaded part = pi/2(64) = 32pi
I love this beautiful task.
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This is an interesting problem because the radius of each arc segment can vary and as long as the chord segment is 16 units long, the blue area will remain the same.
Excellent!
Thanks for your feedback! Cheers!
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That is interesting 👍🏻
I hate these kinds of counterintuitive problems. They make my brain hurt.
In the beginning, you state that the chord is tangent to small semi-circle. However, you don't define the chord as a parallel line to the "line" that is the
diameter of large semi-circle. This is a remarkably good geometric problem. But, the chord geometry needs to be fully given. The tangency to small
semi-circle can take on infinitely many orientations. In the "middle" of the talk, you ASSUME the chord is parallel to diameter line. Love your channel and
solution methods. Thanks !
He doesn't assume that. And that assumption is therefore not necessary. Edit: it just looks like that because he draws the radii vertically, but they don't have to be.
Because the chord is tangent to the smaller circle, it is perpendicular to its radius. The chord is also perpendicular to the radius of the larger circle that bisects it because that's a property of chords. So, the question is, how do we know that the centers of the two circles are colinear on the same diameter? By symmetry, they have to be. But yeah, you might want to say something to justify saying that the distance from the center of the larger circle to the chord is the same as the distance from the center of the smaller circle to the chord. It can be proven, but I'd agree that it's not immediate.
@@rickdesper "Because the chord is tangent to the smaller circle, it is perpendicular to its radius" .....Yes. This is very obvious from line tangent definition to
to curvatures. "The chord is also perpendicular to the radius of the larger circle that bisects it" ......This is not intuitive and I will have to look at "chord" geometry proof(s) and I don't believe this to be true (for now). Thanks Rick.
The chord is 16 units. Let the white semicircle shrink down to nothing and the 16-unit chord becomes the diameter of the blue semicircle.
The radius is 8 and the area is ((8^2)*pi)/2 = 64 pi /2 =32 pi. This is true for any value of the small semicircle radius r.
There is no maximum value for r; but r = sqrt(R^2 -- 64). Setting r = 0 gives the same answer as above.
Cheers. 🤠
Wow, that is very interesting observation 👍🏻
That's how I did it as well, no need for complex calculations, just the observation that given what we were told, the blue area must be constant regardless of the sizes of the circles.
@@alternativeglasto We don't know apriori that enough information is given for the problem to be solvable. This approach assumes that, but the video shows that this assumption is not necessary.
I think you've made a mistake. How could the chord length that is not drawn at the diameter possibly be equal to the diameter? An accidental proportionality has accrued that is not consistent in general. If I am wrong, well great. Common Archimedes show me up.
@@jonathanfanning9558 We don't know the radius of the white circle and it can't be calculated so the solution can't depend on that value if the problem is solvable. Therefore we can assume that radius of the white circle can be zero.
Avbryt
Svara
Your explanation is simple ,easy to understand. I am watching your mathematics videos in comparison to others.
the first two lines you draw are not necessarily the same length, for them to be the same length we would need to know that the chord is parallel to the base of the larger semi-circle but we are not told this and it is not possible to prove from the given information
Very well explained👍
Thanks for sharing😊😊
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why is it assumed that the chord is parallel to the diameters?
It's given in the problem.
Very easy to understand. I wish my math teacher 20 years ago was teaching like this 🤣. It's never too late to learn math.
Amazing, and beautifully explained. Much appreciated.
Thanks for video.Good luck sir!!!!!!!!!!!!
Thank you too
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Thanks to Sir Master 🇧🇷
good insight!
💪
Thankyou so much sir for your hardwork 🙏🙏
You are very welcome!
So nice of you.
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Marvellous. Once you provided the value of Rsq-rsq my face lit up as it all fell into place.
thank you 💐💐👌🏻👍
Is it possible to also find the area of the white semicircle given only the chord length of 16?
I wish I had this guy for Maths in my matric🤗
Will you please suggest some books name in which this types of such complicated problem are given?
Nice question sir 🙏🙏
Glad you think so!
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Very interesting geometry question
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Nowhere in the initial problem is it stated that the chord is parallel to the base line. Hasn't the lack of this condition prevented this solution, since this condition is required for the solution.?
This type of puzzle with very few data is often simple. In this case, let's assume there is no missing data. So, the absolute radius of the two circle don't matter. Suppose the little one be zero, then 16 is the diameter of the great and the blue shaded region is just a semi-circle of radius 8.
Yes but that assumption is presumptive and not in in keeping with a generalised solution.
Blue shaded Area :
A = ⅛ π C²
A = ⅛ π 16²
A = 32π cm² ( Solved √ )
Moc pěkný příklad.
Nice explanation 😍
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This problem is so amazing
et tout ça sans avoir besoin de calculer les rayons: bravo :))
You should mention what number you are using for Pi , you mention 3.14 but you used the Pi key on your calculator to create the answer 100.53 sq. un.
Just let r = 0. Then 16 is the diameter of the bigger circle. Hence, the area is
Pi times 8^2/2
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Genius .
Exactly. I immediately had the same idea.
oh that's fiendishly clever
OK, but that does not prove that you get the same answer for all values of r.
If there is one solution it must be independant of the size of the smaller semicirkle. So, by putting the smaller cirkle =0 the blue region equals rhe area of a semicirkle with radius=8 i.e. 32xPI.
One believes at first it‘s impossible to solve, but no matter what the two radii are, the area is always 32pi!
Exercício parece ser difícil, mas na realidade é fácil
sir can you suggest any math book for practise
Dear Noor Alam, that's a big dilemma! I haven't seen any quality book yet. If I come across with any good resource, sure I'd love to share with you. Don't spend too much money on any book. Meanwhile keep watching PreMath channel. We work very hard to put quality videos for our valued viewers. Thanks for asking.
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Parece difícil pero el uso de las fórmulas lo hace accesible.
Considering that small semicorcle can have any radius, imagine that small circle extremely small. Then shaded area will equal to area of large semicircle with radius = 8. =} 32π
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Intersecting chord theorem: (8)(8) = (R-r)(R+r) = R^2-r^2. Multiplying both sides by π and dividing both sides by 2 gives 32π.
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Love it
Very neat problem. If you try to solve for r or R, you'll get nowhere. Neither r nor R can be pinned to a specific value, even though (R^2 - r^2) is itself determined.
Puisque le diamètre des 2 demi-cercles ne dépendent pas de la longueur du segment 16, alors on peut faire D=16 et r=0 et on trouve immédiatement la surface bleue, non?
Je préfère utiliser S=Pi.D*2 / 4
Non! What if r is not equal to 0? Does it still work? That needs to be proved not just assumed.
So whats the value of R and r ?
It's assumed that the cord is parallel to the diameter of the larger semicircle. What's the basis for that assumption, other than that's how it appears to be drawn? Unless I'm missing something, if it's not, your solution falls apart.
Я решил так же - сразу догадался, что искать нужно не сами радиусы, а разность их квадратов.
Hi PRE MATH, I have a question for you . Solve it √(129-90√2)=?
El problema se resuelve igual que el de una corona circular 》Azul =(1/2)Pi((16/2)^2)=32Pi.
Gracias y saludos.
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No need for Pythagore, just trigonometry ...
R cos a = 8 and R sin a = r
PI/2 (R2 - r2) = PI/2 R2 (1 - sin2 a) = PI/2 R2 cos2 a = PI/2 64 = 32 PI
(took me 5 seconds)
Amazing infinitely combinations lead the same result.
Area = 0.5*pi*(Rsq - rsq). Rsq = rsq + 8sq. Rsq - rsq = 64. Area = 100.53
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👍
Shouldn't it be stated in the known facts about the problem that the chord is actually perpendicular to the radius? 16 units isn't enough information, and the chord could still be a tangent of the smaller circle while not being parallel to that bottom part of the larger semicircle.
Which radius? It's perpendicular to the radius of the smaller circle because the chord is tangent to the circle there. And it's perpendicular to the radius of the larger circle that bisects it: that's an easy theorem of geometry.
Another in-the-head quickie.
Spoiler alert.
No variables needed. Simply imagine the smaller semicircle shrinking to nothing, so that 16 becomes the diameter of the larger semicircle. Then its radius is 8 and its area is 8²π /2 = 64π /2 = 32π square units.
32 × π = 100.53
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@@PreMath Thank you sir may you also progress
i just moved the small circle to center.
then much easier.
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It's wrong
We cannot be sure that the side of the triangle is 8 because we do not know that the chord is parallel to the diameter of the great circle
It's could be bigger or less than 8
I noticed this also. We are not given that parallel as a fact. It is just an assumption. The diagram is not fully constrained.
I never would’ve figured that out. 😢
If it does Not depend of the radiuses so make the white one zero and you will have the solution asap.
This time I didn't find the solution !!!! Something must be wrong with me.
too complicated. push down the 16 segment, until the tiny circle becomes 0 radius. and just calculate the are a of the semicircle of diameter 16.
Let R be the radius of the larger semicircle and r be the radius of the smaller semicircle. r•r + 8•8 = R•R
R^2 -r^2= 64
Area of shaded part = pi/2(64)
= 32pi
Thanks for sharing! Cheers!
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Am I the only one who doesn't get it? 'Fess up guys . . . haha
Love it
If it does Not depend of the radiuses so make the white one zero and you will have the solution asap.