I've been watching your content for some while now and finally for the first time in a long time, I manage to solve one! Thanks for making my interest in Math come back again, it was one of my favorite subjects back in my days in school
I think I can see a very simple way of doing this. Spoiler alert. The green area will be the area of an equilateral triangle of side-length 4 minus the area of a semicircle of diameter 4. Area of triangle = 4√3. Area of semicircle = 2π. Green area = 4√3 − 2π = 2(2√3 − π) square units. Numerical equivalent ≈ 0.645 square units.
@@User-jr7vf That doesn't really explain why solving the geometric problems is more satisfying that solving arithmetic or algebraic problems. It's a personal preference for geometry I'm talking about. I think it's related to the what you see is what you get nature of geometry. There's something immediate about its transparency.
My first thought was drawing a square around the circles for calculating. Then I saw I realised it wasn't possible to find the solution this way. Suddenly I saw it. I've drawn a triangle like in the video and find the solution as shown in the video. And of course I found the solution myself and then I checked if I've done it the right way.
... Good day, Coincidentally I derived the general formula of the area in between 3 identical circles with radius R (lol) ... so Area = [ SQRT(3) - pi/2 ] * R^2 ... finally applying to your problem, where R = 2 ... A(Green shaded region) = 4*SQRT(3) - 2*pi ... thanks for your presentation .... best regards, Jan-W
1)calculer la hauteur du triangle avec Pythagore. 2) calculer la surface d'un cercle. 3) calculer la surface du morceau d'un cercle inclus dans le triangle. Le triangle étant équilatéral, les angles sont de 60° ou 1/6 de l'aire du cercle. On multiplie l'aire des trois quartier de cercle par 3 et on soustrait de l'aire du triangle.
Without a calculator or putting pen to paper I calculated the area of the equilateral and the 1/6th circles and ended up with 4root3 - 2pi. The calculator gives that as 0.645(3dp) sq units. I'm fairly ancient and trying these puzzles helps keep my mind alert and active.
Angles Are the Same. No Need Phytagoras there, Pal. 😓 Area = (1/2)•4•4•sin60° =4√3 Sectors = 3×(60°/360°)×π×2" = 2π Green = 4√3 - 2π = 0.645 units" Anyway, I give U #161st thumb's Up 🙂 👍
Many approaches are possible to find the solution to this problem! Glad to see the Trigonometry! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀
Haven't seen it yet, my guess is you construct an equilateral triangle ABC, then subtract the area of the three segments? The angle is 60°, convert into radians and get the three areas, subtract from ABC's area
No peeking: Consider triangle ABC. Invoking Pythagoras, altitude = 2 sqrt(3); base = 4; and area = (1/2)(2)(sqrt(3))(4) = 4 sqrt(3). The three 60-degree circle segments inside the triangle add up to 180 degrees, which is half the area of one of the circles, or (1/2)(4)(pi) = 2 pi. The green shaded area is the difference, or 4 sqrt(3) -- 2 pi. Staying sharp, one problem at a time.... 🤠
Is there a reason we end up approximating to the numerc value? When I did math(s) at school we were taught that unless the question specifically asked for a decimal representation we had to leave the problem with its actual exact value. So our answer would be 4(sq.root3)-2pi. If we converted that to a decimal we would be marked down. It wouldn't be wrong as such, just not properly right. It is however over 40 years since I sat at a school desk. Note - any of our tech/science courses would expect a decimal, but math(s) was a 'pure' subject, not subject to the vagaries of spherical cows in a vacuum.
He does derive the exact solution first. I was taught to estimate the answer first, so that I could evaluate the reasonableness of my answer. The green area looked to me to be about a tenth the area of one of the circles. The area of each circle is about 12. So I guess the answer should be about 1. His answer was within a factor of 2 of my guess, so my comfort level would be reasonably high. There is no way I could see this by looking at the exact solution.
I didn't know the sector area formula and to be honest I don't really understand why it works. But since we deal with an equilateral triangle each angle is 60 degrees and therefore each sector area is one sixth of a circle. So 3 times the area of a circle divided by 6 =3/6 the area of a circle =(r^2*π)/2 green region =area of trianle minus (area of circle divided by 2) =2√3-(r^2*π)/2 (r=2) =2√3-4π/2 =2√3-2π =2(2√3-π)
The approximation is what's killing me here! Because I like exact! So I'm thinking because Pi and 3^⅓ are so Fractal so too speak... I don't know, just throw the resulting 0.645 square units out the curved space window. I'm just trying too be funny...I absolutely love your channel.
It doesn't make sense to me write the outcome as a decimal fraction. pi is just pi and you can't add it with any other number to get a nice result. I would write it like this; green area= 2(2√3-π)
Before watching: equilateral triangle with sidelength 4 (2r) has the area 4/2*2*sqrt(3) (1/2*base*height) or 4*sqrt(3). From that we deduct 3 circle segments of 60 degrees each or the area of half a circle with the radius 2 or 1/2*4*pi or 2*pi. Thus, the solution is 4*sqrt(3)-2pi or 2(2Sqrt(3)-pi).
Solution: The middlepoints of the 3 circles form a triangele with 3 equal sides of s = 2+2 = 4. The area of this triangle is: s*s/2*√3/2 = s²*√3/4 = 4²*√3/4 = 4*√3. The area of the 3 circle sections with the angle of 60° is together the area of a semicircle. And the area of the green shaded region is: 4*√3-π*2²/2 = 4*√3-π*2 ≈ 0.6450
Well, I screwed it up. I tried to calculate the outer triangle with Pythagoras, but didn't pay attention to what was the hypotenuse. Haven't been doing this a lot since my school days... never too late to (re-)learn things!
For lack of time: 1 equilateral triangle CBA, length of each side = 2*2. then 3 sectors, 60° each --> Area corresponds to π*r²/2 .... etcetera. Please forgive the hurry.
Los centros de los círculos de radio r=2, tangentes entre sí, corresponden a los vértices de un triángulo equilátero de lado 2r=4 → Área zona verde = (Superficie del triángulo equilátero) - (Superficie de tres sectores circulares de r=2 y ángulo de 60º) = [(4x4√3/2)/2] - [π2²(3x60/360)] =(4√3) - (2π) =0.6450 Gracias y un saludo cordial.
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Thanks for bringing my memory of the Sector Area back
I've been watching your content for some while now and finally for the first time in a long time, I manage to solve one! Thanks for making my interest in Math come back again, it was one of my favorite subjects back in my days in school
I think I can see a very simple way of doing this.
Spoiler alert.
The green area will be the area of an equilateral triangle of side-length 4 minus the area of a semicircle of diameter 4.
Area of triangle = 4√3.
Area of semicircle = 2π.
Green area = 4√3 − 2π = 2(2√3 − π) square units.
Numerical equivalent ≈ 0.645 square units.
ht of triangle ABC = sqrt(16-4). Area ABC = 0.5*4*3.46 = 6.93. Green Area = = Area of triangle ABC - 3*60 deg part of circle. = 6.93 - 2pi = 0.645
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Enjoyed solving this one. Solving these geometric problems is especially satisfying for whatever reason.
Excellent!
Glad to hear that!
Thanks for your feedback! Cheers!
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
yep,it makes you feel smart
@@User-jr7vf That doesn't really explain why solving the geometric problems is more satisfying that solving arithmetic or algebraic problems. It's a personal preference for geometry I'm talking about.
I think it's related to the what you see is what you get nature of geometry. There's something immediate about its transparency.
@@muttleycrew I got it
My first thought was drawing a square around the circles for calculating. Then I saw I realised it wasn't possible to find the solution this way.
Suddenly I saw it. I've drawn a triangle like in the video and find the solution as shown in the video.
And of course I found the solution myself and then I checked if I've done it the right way.
Well done !! Clearly presented !
... Good day, Coincidentally I derived the general formula of the area in between 3 identical circles with radius R (lol) ... so Area = [ SQRT(3) - pi/2 ] * R^2 ... finally applying to your problem, where R = 2 ... A(Green shaded region) = 4*SQRT(3) - 2*pi ... thanks for your presentation .... best regards, Jan-W
That was so beautifully explained. Splendid job Pre-Math.
Excellent thanks
1)calculer la hauteur du triangle avec Pythagore. 2) calculer la surface d'un cercle. 3) calculer la surface du morceau d'un cercle inclus dans le triangle. Le triangle étant équilatéral, les angles sont de 60° ou 1/6 de l'aire du cercle. On multiplie l'aire des trois quartier de cercle par 3 et on soustrait de l'aire du triangle.
Thanks. I got 0.648 square units. For getting area I used Heroin formula= rad.48
That one is easy to see how it is done. I figured it out in about 30 seconds, including the sectors of the three circles making half a circle.
Good Morning MASTER
Thanks Sir
A Hug from Rio de Janeiro
So nice of you, Alex
Thanks for your continued love and support!
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
Thanks for video.Good luck sir!!!!!!!!!!!!!
Thanks for your continued love and support!
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
Awesome👍
Thank you so much for sharing.😊
Thanks for your continued love and support!
You are awesome. Keep smiling👍
Love and prayers from the USA! 😀
Without a calculator or putting pen to paper I calculated the area of the equilateral and the 1/6th circles and ended up with 4root3 - 2pi. The calculator gives that as 0.645(3dp) sq units. I'm fairly ancient and trying these puzzles helps keep my mind alert and active.
Yay! I solved it.
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Nice... I like these, thanks again 👍🏻
S=4√3-2π≈0,64
Angles Are the Same.
No Need Phytagoras there, Pal. 😓
Area = (1/2)•4•4•sin60° =4√3
Sectors = 3×(60°/360°)×π×2" = 2π
Green = 4√3 - 2π = 0.645 units"
Anyway, I give U #161st thumb's Up 🙂 👍
Many approaches are possible to find the solution to this problem!
Glad to see the Trigonometry!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Haven't seen it yet, my guess is you construct an equilateral triangle ABC, then subtract the area of the three segments? The angle is 60°, convert into radians and get the three areas, subtract from ABC's area
not seen it yet but the square root of 3 minus pi over 2? Isn't that how to calculate the area between 3 mutually tangent unit circles?
No peeking:
Consider triangle ABC. Invoking Pythagoras, altitude = 2 sqrt(3); base = 4; and area = (1/2)(2)(sqrt(3))(4) = 4 sqrt(3).
The three 60-degree circle segments inside the triangle add up to 180 degrees, which is half the area of one of the circles, or (1/2)(4)(pi) = 2 pi.
The green shaded area is the difference, or 4 sqrt(3) -- 2 pi.
Staying sharp, one problem at a time.... 🤠
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
@@PreMath Thanks. As I may have said before, the USA needs all the love and prayers it can get.
Is there a reason we end up approximating to the numerc value? When I did math(s) at school we were taught that unless the question specifically asked for a decimal representation we had to leave the problem with its actual exact value. So our answer would be 4(sq.root3)-2pi. If we converted that to a decimal we would be marked down. It wouldn't be wrong as such, just not properly right. It is however over 40 years since I sat at a school desk. Note - any of our tech/science courses would expect a decimal, but math(s) was a 'pure' subject, not subject to the vagaries of spherical cows in a vacuum.
I also find it odd that he is always approximating his otherwise exact answers.
He does derive the exact solution first. I was taught to estimate the answer first, so that I could evaluate the reasonableness of my answer. The green area looked to me to be about a tenth the area of one of the circles. The area of each circle is about 12. So I guess the answer should be about 1. His answer was within a factor of 2 of my guess, so my comfort level would be reasonably high. There is no way I could see this by looking at the exact solution.
I find that reducing it to single number can be helpful for a "reality check" of the result. So probably, I'd give both, the exact and the numeric.
I didn't know the sector area formula and to be honest I don't really understand why it works. But since we deal with an equilateral triangle each angle is 60 degrees and therefore each sector area is one sixth of a circle. So 3 times the area of a circle divided by 6
=3/6 the area of a circle
=(r^2*π)/2
green region
=area of trianle minus (area of circle divided by 2)
=2√3-(r^2*π)/2 (r=2)
=2√3-4π/2
=2√3-2π
=2(2√3-π)
Belo exercício muito bom
The approximation is what's killing me here! Because I like exact! So I'm thinking because Pi and 3^⅓ are so Fractal so too speak... I don't know, just throw the resulting 0.645 square units out the curved space window. I'm just trying too be funny...I absolutely love your channel.
Thanks for your feedback! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
It doesn't make sense to me write the outcome as a decimal fraction. pi is just pi and you can't add it with any other number to get a nice result. I would write it like this; green area= 2(2√3-π)
Bravo. Great problem and solution!!
2*sqrt(3)*radius - pi/2*radius^2 ~ 0.645
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Atriangolo-3*settcirc=4*4sin60/2-3(4pi)/6=4sqrt3-2pi
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Before watching: equilateral triangle with sidelength 4 (2r) has the area 4/2*2*sqrt(3) (1/2*base*height) or 4*sqrt(3). From that we deduct 3 circle segments of 60 degrees each or the area of half a circle with the radius 2 or 1/2*4*pi or 2*pi. Thus, the solution is 4*sqrt(3)-2pi or 2(2Sqrt(3)-pi).
La solución es 4√3-2π mediante cálculo mental.
area of an equilateral triangle is sqrt 3/4* a^2.
Solution:
The middlepoints of the 3 circles form a triangele with 3 equal sides of
s = 2+2 = 4.
The area of this triangle is: s*s/2*√3/2 = s²*√3/4 = 4²*√3/4 = 4*√3.
The area of the 3 circle sections with the angle of 60° is together the area of a semicircle. And the area of the green shaded region is:
4*√3-π*2²/2 = 4*√3-π*2 ≈ 0.6450
Solved it the same way (without expounding on the height of the triangle, obviously 😅).
Thanks for your feedback! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Well, I screwed it up. I tried to calculate the outer triangle with Pythagoras, but didn't pay attention to what was the hypotenuse.
Haven't been doing this a lot since my school days... never too late to (re-)learn things!
For lack of time: 1 equilateral triangle CBA, length of each side = 2*2. then 3 sectors, 60° each --> Area corresponds to π*r²/2 .... etcetera.
Please forgive the hurry.
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Los centros de los círculos de radio r=2, tangentes entre sí, corresponden a los vértices de un triángulo equilátero de lado 2r=4 → Área zona verde = (Superficie del triángulo equilátero) - (Superficie de tres sectores circulares de r=2 y ángulo de 60º) = [(4x4√3/2)/2] - [π2²(3x60/360)] =(4√3) - (2π) =0.6450
Gracias y un saludo cordial.
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
As a formula: the green area is r² (√3 - π/2) where r is the radius of the circles.
Thanks Professor. Another great problem.❤🥂
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Love and prayers from the USA! 😀
The answer is the difference of area of the triangle and the semicircle=(4x4 root 3/2)/2-pix2^2=0.645 approximately.
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A=((√3)-π/2)r^2
Hi sir
Hello dear
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これは簡単すぎた。
良い
ありがとう
Ok, now solve it using the area of a hexagon, rather than a triangle.
*0.645* [no units] but Im not ashamed to admit it - my covid riddled brain took 3 failed attempts before I got it
These videos are slow even when playback speed is set to 200%.