Again, you took a relatively simple problem, and made it much more complicated than it needed to be. A good teacher makes things simple and easy to understand, and doesn't introduce added complexity unless it is necessary. Just connect the bottom center and the top right to the intersection point and each other. This creates 2 congruent right triangles with legs 1/2 and 1. Either angle can be determined with an arctangent, then the other angle is 90 - that angle. Then add together the areas of the wedges formed and subtract the total area of the triangles.
@@SyberMath I enjoyed the solution you presented here. I have encountered this exact problem before but didn't think to solve it this way. To those who struggle to solve this problem, someone who introduces added complexities may not be a good teacher, but to those who can solve it, someone who introduces a different solution is a good teacher. You did exactly that for me. Keep up the good work.
I started this way, got the limits of integration, looked hard at the integrand, and said (because I'm lazy...) "There HAS to be an easier way!" [And there was. :-) ]
I didn't see anyone in the comment section use a formula for the area of a segment of a circle to solve the problem... the area of a segment of a circle *A = ½·r²(x - sin x)* where r is the radius and x is the measure of the angle in radians.
For segment A, r = 1 and x = α and sin α = ⅘, so area A = ½·1²(arcsin ⅘ - ⅘) = ½·arcsin ⅘ - ⅖.
For segment B, r = ½ and x = π - α, so sin x = sin α = ⅘, and area B = ½·½²(π - arcsin ⅘ - ⅘) = ⅛(π - arcsin ⅘ - ⅘).
The shaded area A + B = (½ arcsin ⅘ - ⅖) + (⅛π - ⅛·arcsin ⅘ - ¹/₁₀) = ⅛π - ½ + ⅜·arcsin ⅘ ≈ 0.2404347884... quick and relatively painless.
Another way ->since you know the common chord between the two circles with different radii is to use the formula for chord area from chord length C=2/sqrt(5) -> A=r^2(2*arcsin(C/2)-(C/2)). Know the common chord is C=2/(sqrt(5)) and the two radii are 1 and .5. For areas A and B respectively. A lot of ways to do this problem. None terribly easy.
@@SyberMath Oops! Aseg=(.5 * R^2 * ( alpha - sin(alpha)) where alpha = 2 * arcsin(C/(2*R)) C is Chord Length and R is radius. To summarize given fixed chord length common to two circles with different radii the Aseg is a function of radius and chord length. So for your A and B A=.5*(.927 - sin(.927))= .0636 with alpha = .927 B=(.5)(.25)(2.214-sin(2.214))= sqrt(2)/8 =.17675 with alpha = 2.214 A+B = .24035 Not an elegant approach but I believe still valid.
@8:30, You didn't need to go thru tan(alpha) = tan(beta) to get alpha = beta. If you looked at the two angles (whose measures you called alpha & beta), you see that the two sides of one angle are perpendicular to the two sides of the other angle, which means that the two angles are congruent. (result straight from geometry). The only other thing I did differently was, in using the area of sector formula, I kept all angles in radians. (hey, what mathematician likes degrees, anyway?! LOL!) Fun problem. Thx!
Hello everyone! This puzzle is easy to draw but not that easy to solve! I know many people will go for the Calculus option because it's so trivial if you know Calculus, of course! I haven't seen this puzzle on RUclips. If you do, please let me know. I did find this webpage which is full of interesting details. So make sure to check it out: math.stackexchange.com/questions/2638152/how-to-find-the-shaded-area It's the same thing with different numbers. Any thoughts? Plz share...
I'm bad at geometry but (relatively) good at calculus, so I almost always turn things into calculus problems if I can! In this case it's not hard to work out the intersections points (1/5,2/5) and (1,0) and the equations of the arcs. Then it comes down to (in Mathematica notation) Integrate[ Sqrt[1/4 - (x - 1/2)^2] - 1 + Sqrt[1 - (x - 1)^2], {x, 1/5, 1}]. The integral of Sqrt[a^2 - (x-a)^2] is done by x - a = a sin(t) etc.
I was trying to solve this using analytics and integrals. I succeeded, but I wasn't able to calculate the integrals I created. What a great deal. : ) But maybe at least Wolphram Alpha would tell me the result based on this. : )
Split the integral into three parts; the two square root parts can be done using complete the square followed by trig sub to get √(1-sin²Θ) and then using trig identities to bring it home.
If you were to use calculus the approach below is fairly simple: Starting from the left hand corner and going clockwise label each vertex as: (-1,-1), (-1,0), (0,0) and (0,-1). The equations of the circles are: (x + 1/2)^2 + (y + 1)^2 = 1/4 x^2 + y^2 = 1 Solving we obtain x = 0, -4/5. The upper curve is just y2(x) = -1 + sqrt(1/4 - (x + 1/2))^2 and the lower curve is y1(x) = -sqrt(1 - x^2) implies The desired area A = integral (-1 + sqrt(1/4 - (x + 1/2))^2 + sqrt(1 - x^2)) dx on the interval [-4/5, 0]. For both square roots we use a trig substitution. For A_(1) = Integral y1(x) dx on [-4/5, 0] we let x = sin(theta) implies dx = cos(theta) implies A_(1) = 1/2 * (arcsin(4/5) + 12/25). For A_(2) = integral sqrt(1/4 - (x + 1/2))^2) dx on [-4/5, 0] we let x + 1/2 = 1/2 * sin(theta) implies dx = 1/2 * cos(theta) implies A_(2) = 1/8 * (pi/2 + arcsin(3/5) + 12/25). Putting this all together and letting w = arcsin(4/5) implies A = -1/2 + pi/16 + 1/2 * w + 1/8 * (pi/2 - w) = pi/8 + 3/8 * arcsin(4/5) - 1/2.
Too hard to follow because cursor is invisible. Is it possible for further videos to see your cursor to be able to clearly see what exactly you are pointing at?
I have better mathod for solving this question but I am not able to explain by only writing the comment. But I will try, So firstly take coordinate (0,0) at mid point of lower side which will be center of semicircle and le r( = 1/2) be radius. Then for another circle center will be ( r, 2r) and radius will be 2r ( = 1 ) write equation of circle and we will find that they will meet at (1/2 , 0) and ( 0, 1/2) Then take area of semicircle and quarter circle and add them also add area of remaining part and subtract area of square , then we will got area of shaded region because it is added twice but subtracted one time.
I don't think ( 0, 1/2) is an intersection point. Can you check? You can post a picture on twitter: twitter.com/intent/tweet?text=@SyberMath or share a link like imgur.com/...
As the results from the equations you have the points of the chord, so you have the length of the chord (using Pythagoras) , say L, then you have to calculate the area of 2 circular sectors with the same chord length L, but not the same subtended angle. Very easy from then on.
Great video as always!! Can you do some permutation and combinatorics problems? I thought they were boring for a long time, but now, I find it to be very interesting... try them once, :D
This approach does not use calculus so it's accessible to people who hasn't studied calculus yet. When I don't write the degree symbol, people make a big deal about it! 😁
I enjoyed so much even if i think the first step you did it we can avoid it Beacaus you can proof that we have a kite and from it we can get the sin(180-alpha) and sin(alpha) .......!!!!!!
I don't like geometry bcoz of the extra construction that we need to do to find the ans,I cant get idea for construction easily,and the video was good with many tricks😁
Yeah, geometry puzzles are not that popular and hard to make. (This one is an exception. It was very easy to draw but hard to solve) That's probably why I try to stay away from them. Plus there are some really good puzzlers on twitter and they come up with amazing puzzles.
@@SyberMath yes i split the "area" with a straight line chord. One side has an area of 0.17... i thought i could double it to get the final answer. That was a mistake.
at the start when you picked your origin you just said "this". i barely noticed you had drawn a little cyan dot. none of the other math youtubers would have made that mistake. you need to be clear at all times. also you referred to the circles ambiguously. as if it was always obvious which you were referring to. again, it was obvious to be. wouldn't be to many others people will probably attack me and defend you because I'm not bending over backwards to be polite here. but consider this very good advice from someone who knows a thing or two about teaching stem.
@@SyberMath Similar right triangles of the "1-2-√5" type are all over in this puzzle. All we need for an elementary solution is the two other angles of this triangle... 🤔
Again, you took a relatively simple problem, and made it much more complicated than it needed to be. A good teacher makes things simple and easy to understand, and doesn't introduce added complexity unless it is necessary.
Just connect the bottom center and the top right to the intersection point and each other. This creates 2 congruent right triangles with legs 1/2 and 1. Either angle can be determined with an arctangent, then the other angle is 90 - that angle. Then add together the areas of the wedges formed and subtract the total area of the triangles.
I find it very interesting to solve the same problem using different techniques. So it's easy to check and compare...🤔
I'm not a good teacher because I'm not even a teacher. As @maths247366 put it:
"I like math and I am here to share few exciting things with you."
😉🙃😁😎
@@SyberMath
I enjoyed the solution you presented here. I have encountered this exact problem before but didn't think to solve it this way. To those who struggle to solve this problem, someone who introduces added complexities may not be a good teacher, but to those who can solve it, someone who introduces a different solution is a good teacher. You did exactly that for me. Keep up the good work.
@@SyberMath No you are a good and nice teacher thanks for your videos and keep on
Your solution is definitely more elegant but, there's no need to offense.
In your coordinate system, you can of course also integrate from 1/5 to 1 the difference of the two curves.
exactly what i was thinking 👍
I started this way, got the limits of integration, looked hard at the integrand, and said (because I'm lazy...) "There HAS to be an easier way!" [And there was. :-) ]
I didn't see anyone in the comment section use a formula for the area of a segment of a circle to solve the problem... the area of a segment of a circle *A = ½·r²(x - sin x)* where r is the radius and x is the measure of the angle in radians.
For segment A, r = 1 and x = α and sin α = ⅘, so
area A = ½·1²(arcsin ⅘ - ⅘) = ½·arcsin ⅘ - ⅖.
For segment B, r = ½ and x = π - α, so sin x = sin α = ⅘, and
area B = ½·½²(π - arcsin ⅘ - ⅘) = ⅛(π - arcsin ⅘ - ⅘).
The shaded area A + B = (½ arcsin ⅘ - ⅖) + (⅛π - ⅛·arcsin ⅘ - ¹/₁₀)
= ⅛π - ½ + ⅜·arcsin ⅘
≈ 0.2404347884... quick and relatively painless.
Nice!
Another way ->since you know the common chord between the two circles with different radii is to use the formula for chord area from chord length C=2/sqrt(5) -> A=r^2(2*arcsin(C/2)-(C/2)).
Know the common chord is C=2/(sqrt(5)) and the two radii are 1 and .5. For areas A and B respectively. A lot of ways to do this problem. None terribly easy.
Thanks for sharing! Can you give me more details on that formula?
@@SyberMath Oops! Aseg=(.5 * R^2 * ( alpha - sin(alpha)) where alpha = 2 * arcsin(C/(2*R)) C is Chord Length and R is radius.
To summarize given fixed chord length common to two circles with different radii the Aseg is a function of radius and chord length.
So for your A and B
A=.5*(.927 - sin(.927))= .0636 with alpha = .927
B=(.5)(.25)(2.214-sin(2.214))= sqrt(2)/8 =.17675 with alpha = 2.214
A+B = .24035
Not an elegant approach but I believe still valid.
@8:30, You didn't need to go thru tan(alpha) = tan(beta) to get alpha = beta. If you looked at the two angles (whose measures you called alpha & beta), you see that the two sides of one angle are perpendicular to the two sides of the other angle, which means that the two angles are congruent. (result straight from geometry).
The only other thing I did differently was, in using the area of sector formula, I kept all angles in radians. (hey, what mathematician likes degrees, anyway?! LOL!)
Fun problem. Thx!
Hello everyone!
This puzzle is easy to draw but not that easy to solve! I know many people will go for the Calculus option because it's so trivial if you know Calculus, of course!
I haven't seen this puzzle on RUclips. If you do, please let me know. I did find this webpage which is full of interesting details. So make sure to check it out:
math.stackexchange.com/questions/2638152/how-to-find-the-shaded-area
It's the same thing with different numbers.
Any thoughts? Plz share...
I'm bad at geometry but (relatively) good at calculus, so I almost always turn things into calculus problems if I can! In this case it's not hard to work out the intersections points (1/5,2/5) and (1,0) and the equations of the arcs. Then it comes down to (in Mathematica notation)
Integrate[ Sqrt[1/4 - (x - 1/2)^2] - 1 + Sqrt[1 - (x - 1)^2], {x, 1/5, 1}]. The integral of Sqrt[a^2 - (x-a)^2] is done by x - a = a sin(t) etc.
That's very common! I'm not very strong in geometry, either! 😁
I got excited but then I remembered I know calculus and could go full autopilot and get the solution :(
No shortcutting! 😁
How do you solve it using calculus?
Integrals
I was trying to solve this using analytics and integrals. I succeeded, but I wasn't able to calculate the integrals I created. What a great deal. : ) But maybe at least Wolphram Alpha would tell me the result based on this. : )
It will! 😁
Split the integral into three parts; the two square root parts can be done using complete the square followed by trig sub to get √(1-sin²Θ) and then using trig identities to bring it home.
If you were to use calculus the approach below is fairly simple:
Starting from the left hand corner and going clockwise label each vertex as: (-1,-1), (-1,0), (0,0) and (0,-1).
The equations of the circles are:
(x + 1/2)^2 + (y + 1)^2 = 1/4
x^2 + y^2 = 1
Solving we obtain x = 0, -4/5.
The upper curve is just y2(x) = -1 + sqrt(1/4 - (x + 1/2))^2 and the lower curve is y1(x) = -sqrt(1 - x^2) implies
The desired area A = integral (-1 + sqrt(1/4 - (x + 1/2))^2 + sqrt(1 - x^2)) dx on the interval [-4/5, 0].
For both square roots we use a trig substitution.
For A_(1) = Integral y1(x) dx on [-4/5, 0] we let x = sin(theta) implies dx = cos(theta) implies A_(1) = 1/2 * (arcsin(4/5) + 12/25).
For A_(2) = integral sqrt(1/4 - (x + 1/2))^2) dx on [-4/5, 0] we let
x + 1/2 = 1/2 * sin(theta) implies dx = 1/2 * cos(theta) implies
A_(2) = 1/8 * (pi/2 + arcsin(3/5) + 12/25).
Putting this all together and letting w = arcsin(4/5) implies
A = -1/2 + pi/16 + 1/2 * w + 1/8 * (pi/2 - w) =
pi/8 + 3/8 * arcsin(4/5) - 1/2.
Too hard to follow because cursor is invisible. Is it possible for further videos to see your cursor to be able to clearly see what exactly you are pointing at?
There's no cursor
But you're right! I should make it more clear what I'm pointing at
I have better mathod for solving this question but I am not able to explain by only writing the comment. But I will try,
So firstly take coordinate (0,0) at mid point of lower side which will be center of semicircle and le r( = 1/2) be radius. Then for another circle center will be ( r, 2r) and radius will be 2r ( = 1 ) write equation of circle and we will find that they will meet at (1/2 , 0) and ( 0, 1/2)
Then take area of semicircle and quarter circle and add them also add area of remaining part and subtract area of square , then we will got area of shaded region because it is added twice but subtracted one time.
I don't think ( 0, 1/2) is an intersection point. Can you check? You can post a picture on twitter:
twitter.com/intent/tweet?text=@SyberMath
or share a link like imgur.com/...
@@SyberMath I don't use twitter
@@SyberMath intersection points will be (0,1/2) and (1/2,0)
I got it right afterall, I just forgot that what I had counted is the non shaded area in that smaller circle, and that I had to subtract it
Sounds good!
As the results from the equations you have the points of the chord, so you have the length of the chord (using Pythagoras) , say L, then you have to calculate the area of 2 circular sectors with the same chord length L, but not the same subtended angle. Very easy from then on.
Oh, I see! That makes sense!
the range of Inverse cosine function is [0, pi], I think where you divided by 360 degree should be replaced by 2*pi
Couldn't the range be [0,180]?
@@SyberMath it can, but normally people use range [0, pi]
Before watching I tried this problem myself, but got stuck to solve the integrals. Great video ❤❤
That's great!
Nice video, I need to get more exposed to these kind of topics
Thank you!
You can find the integeral using trig sub
@@skwbusaidi nice I make calculus videos on my channel btw
Can we do area under da curve method🤔🤔🤔
Absolutely! You can do it to check if you get the same answer
Exercise : Solve the shaded area (easy mode)
Exam: Solve the shaded area (THIS video)
Great video as always!! Can you do some permutation and combinatorics problems? I thought they were boring for a long time, but now, I find it to be very interesting... try them once, :D
Thank you! They are not my type but I will take a look
@@SyberMath a different approach to this problem is now available on my channel
There's a different way to approach this and it will come on channel @I Challenge You wait for it 😉😀
We'll see 😁
Put it on an axis and its very easy with 1 integral
Fourth is an Ordinal
Quarter is not
American English is exceptionally opaque and ambiguous
Yes it is!
I would have just integrated over the area. Don't use degrees it just complicates things.
This approach does not use calculus so it's accessible to people who hasn't studied calculus yet. When I don't write the degree symbol, people make a big deal about it! 😁
Really enjoy your channel. Keep up the good work!
I enjoyed so much even if i think the first step you did it we can avoid it
Beacaus you can proof that we have a kite and from it we can get the sin(180-alpha) and sin(alpha) .......!!!!!!
That's right! I could've used some symmetry. I actually thought about it
Although good job !!
Area = integral of y1-y2
Y1=1/4-(x²-1/4)
Y2=(√1-(x²-1))+1
Integral range is from 1/5 to 1
Where Y1 is the equation of semicircle in terms of y
And Y2 is the equation of circle interns of y
Isn't there a square root in Y1?
@@SyberMath ya yes sir printing error.the general eq for semicircle is (x-1/2)²+y²=1/4
It can be written as y=√(1/4-(x-1/2)²)
@@SyberMath area is integral of
(√(1/4-(x-1/2)²))-(√(1-(x²-1))+1)
From 1/5 to 1.
Wow u accept my recommendation for geometry
Was it your recommendation? 🤔
@@SyberMath oh , sorry I recommend this problem on other channel , btw thanks
You should do the geometrical way when do these kind of problem, integral is a kind of brute force approach and uninteresting
I agree
Ahh i thought there might be a trick but there isn't it seems, we need those inverses of cosine.
Sybermath the unit length solver😜
😁
شكرا عله توضيح رياضيات
You're welcome!
I love geometryyyyy
Great! Geometry is fun1
Me: Finally a calculus problem
SyberMath: Hold my Baldor.
Love this videos, really nice job!
Thank you! 😊
*SYBERMATH* 👍
😊
I don't like geometry bcoz of the extra construction that we need to do to find the ans,I cant get idea for construction easily,and the video was good with many tricks😁
Yeah, geometry puzzles are not that popular and hard to make. (This one is an exception. It was very easy to draw but hard to solve)
That's probably why I try to stay away from them. Plus there are some really good puzzlers on twitter and they come up with amazing puzzles.
Is it 0.41?
I did it again and got 0.35...smh.
Now i know what i did wrong...
Awesome!
@@SyberMath yes i split the "area" with a straight line chord. One side has an area of 0.17... i thought i could double it to get the final answer. That was a mistake.
Area = (1/4)sin-1(2/sqrt(5))+sin-1(1/sqrt(5))-(1/2) = 0.240435
at the start when you picked your origin you just said "this". i barely noticed you had drawn a little cyan dot. none of the other math youtubers would have made that mistake.
you need to be clear at all times. also you referred to the circles ambiguously. as if it was always obvious which you were referring to. again, it was obvious to be. wouldn't be to many others
people will probably attack me and defend you because I'm not bending over backwards to be polite here. but consider this very good advice from someone who knows a thing or two about teaching stem.
Thank you for the advice! You're right about that.
this question can also be solved using integral calculus ....initial steps are similar as you did......i think
That's right! Is there a synthetic method that does not use trig?
Thank for sybermath
You're welcome!
都已經算出兩個交叉點也寫出兩個圓的方程式
我還以為要用積分來算面積了呵呵呵
Another great explanation! Longer than usual.
Thank you! Yeah, it was longer than usual. Sorry
Too, too long and not enough proper diagram designations. Too, too talky!!!
I expected that you would solve this elementary geometry instead of analytical geometry coz I am just a 10th grader.
There must be an elementary solution which I haven't looked into yet
@@SyberMath Similar right triangles of the "1-2-√5" type are all over in this puzzle.
All we need for an elementary solution is the two other angles of this triangle... 🤔
You can't use completely simple geometry, you need a little bit of trigonometry to get the angles, but it's not nearly as complicated as he made it.
@@drbeevs8808 yaa. Trigonometry can be understood. But I don't understand a bit of calculus.
Easy, but hard
I would have used integration
I know. Most people would do
👍
This is really easy
Hmm
Easy questions
You must be good at geometry puzzles!
Dude, why don't you simply use Calculus?
That's too standard! 😁
Sad calculas noises
😁
that seemed to be at least a thousand times more complicated than it needed to be, jesus christ man