I have to draw all the polygons from 3 sides to 8 sides for my project. Your videos really helped me. I couldn't have asked for a better tutorial. Thank You sir.
So interesting, I can see how this works now. From the line AB you have determined the point Q. The line AQ is the golden ratio (Phi) of AB. The opposing corners of a pentagon e.g. line AD is the result of multiplying the side length of the pentagon by Phi (1.618....)
I am building a 5 sided cupola to go on a pentagon shaped gazebo. It is an inverted radius conical cupola. I needed this one formula to lay it out. Thank you.
Ah seriously, thank you so much. I've already watched two of your videos and it's been going great! Taking Tech Draft class was probably a bad one on my side but I can't help but try to widen my knowledge at some point.
Thank you very much for your wonderful video! it is so brilliant! Since not all methods are exact, I was curious to find out whether the angle QBC is indeed 72 degrees. So, I tried to proof it as follows: First, use the right-angle triangle MBP to compute the length MP. For clarity, denote the base AB = b. This yields MP=0.5*b*sqrt(5). We now compute the length AC which equals AQ. Since AQ = MP + 0.5*b, it follows that AC = 0.5*b*(sqrt(5) + 1)). The triangle ABC has a base length of AC and two equal sides of length b. Using the cosine rule: AC^2 = 2*b^2 - 2*b^2*cos(gamma), where gamma is the angle ABC. This yield cos(gamma) = 0.25*(1 - sqrt(5)).= -0.3090. gamma = acos(0.3090) = 108 degrees. The angle QBC is then 180 - 108 = 72 degrees. It remains to be proved that point D is indeed the top of the pentagon. This is proved by the fact that the triangles ABC and BCD are congruent and thus BC = CD. Arthur, in contrast to other methods, your method is exact! Thank you again for your crystal-clear video which inspired me to do this exercise with tools we learned 50 years ago. Let me know if I missed out on something.
Awesome! Thank you for taking the time to create this video and post it. This is the perfect information I was looking for that will allow me to construct my device.
thnx......i actually know a method tht can draw a pentagon too easier though...but i cannot remember.....it all starts w/ cross and circle.....but urs are a bit complex but still easy to understand....
Thank you for your Reply. With respect, if you look on from about 3:30 you will see that points C and E are each located of by the intersection of 2 arcs, one of which has radius (AB), and the other an arc of greater but unspecified radius passing through point D. DC and DE are not directly referred to distance AB.
@glyn hodges Sorry, I have been 'off-air' for a while and have just discovered your detailed and comprehensive Reply. Thank you. I look forward to following through your reasoning, and will respond in due course.
Is it regular, though? I like it, but you have to check the angles to make sure it's not slightly squashed vertically or horizontally. Many people, showing various methods, are failing to check the angle measurements to make sure they are all 108°. One method gave me two 106°(at the bottom), two 109°, and 110° at the top. Even with the sides all equal, it still can have symmetrical variability in it's angle measurements.
I saw the other version where you inscribed it in a circle. That would guarantee the angles to all be equal at 108°. So, you'd have to find the center of this Pentagon and draw a circle around it to see if all the corners fall on the line.
There is an easier method of drawing an accurate pentagon using the characteristics of the pentagon cross. The horizontal arm bi-sects the vertical at the phi ratio of 38-62. and is 5 pc longer than the vertical so if the latter is say 20cm, the former will be 21cm. Once you have drawn the cross, join the arms to the head and determine the side length. centre this on the bottom of the vertical and join the feet to the arms. Simple.
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
I got scared looking the intro , i though it was the actual construction😂
I'm glad it wasn't just me lol.
lol
Lol good one
Heptagon
Shaikh Shoeb HAHAHAHA i agree
I have to draw all the polygons from 3 sides to 8 sides for my project. Your videos really helped me. I couldn't have asked for a better tutorial. Thank You sir.
So interesting, I can see how this works now. From the line AB you have determined the point Q. The line AQ is the golden ratio (Phi) of AB. The opposing corners of a pentagon e.g. line AD is the result of multiplying the side length of the pentagon by Phi (1.618....)
Bruh🙄
Great.....to see how RUclips can be utilized to do useful stuff....
I am building a 5 sided cupola to go on a pentagon shaped gazebo. It is an inverted radius conical cupola. I needed this one formula to lay it out. Thank you.
Wowwwwww. It actually works. It's like magic.😮. May God bless you😇 and Thank you sir.😋
Thank you sir apne mere submission me help ki 🤗
This is BETTER than watching a Hollywood blockbuster movie.!!
Thank you Aurther Geomitry!!!You helped me with my module in my math!it was so hard until I saw this!!😄😄😄
Thanks!
Thank u so much have a test tomorrow and this helped me loads fam thanks
Oh wow, it turned out great. Thank you so much !!
Ah seriously, thank you so much. I've already watched two of your videos and it's been going great! Taking Tech Draft class was probably a bad one on my side but I can't help but try to widen my knowledge at some point.
OMG! this type of videos are amazing for do arts homeworks.
Thank you very much for your wonderful video! it is so brilliant! Since not all methods are exact, I was curious to find out whether the angle QBC is indeed 72 degrees. So, I tried to proof it as follows: First, use the right-angle triangle MBP to compute the length MP. For clarity, denote the base AB = b. This yields MP=0.5*b*sqrt(5). We now compute the length AC which equals AQ.
Since AQ = MP + 0.5*b, it follows that AC = 0.5*b*(sqrt(5) + 1)). The triangle ABC has a base length of AC and two equal sides of length b. Using the cosine rule: AC^2 = 2*b^2 - 2*b^2*cos(gamma), where gamma is the angle ABC. This yield cos(gamma) = 0.25*(1 - sqrt(5)).= -0.3090. gamma = acos(0.3090) = 108 degrees. The angle QBC is then 180 - 108 = 72 degrees.
It remains to be proved that point D is indeed the top of the pentagon. This is proved by the fact that the triangles ABC and BCD are congruent and thus BC = CD. Arthur, in contrast to other methods, your method is exact! Thank you again for your crystal-clear video which inspired me to do this exercise with tools we learned 50 years ago. Let me know if I missed out on something.
That's Awesome, I always wanted to draw a pentagon without using a protractor.
Im doing technical drawings and I missed this class so thank you
4 years later and you're helping me pass my last year of school
A god amongst men,thank you
Thank you so much sir....26/10/2021....Really help me with my school work
awesome video helped me so much god bless ya bro
This help me a lot in my submissions
This was a tough one kept getting confused over the changing radius and where to put the compass point, had to rewatch a fair few times
Thanks man, your better than my teacher
I'm experiencing a problem. My teacher gave me a length of 20cm and my compass can't go that far, what should I do?
I've been saved
Thank youuuuu
thanks you, it's so helpful
Sir, Thank you so much. Because I was badly in need of a Pentagon. This video is easier than others.
Right
You say right
thank you, keep up the good work in teaching and helping students.
Thanks for your tutorial.
Omg so simple and easy. I got it on the first try. Thanks.🤗🤗
Please give us a video on "Drawing Pentagon Using Perpendicular bisector method"
Copy that. Thanks!
@@ArthurGeometry Welcome 🥰
I have a hmwrk assessment to do and this helped a lot
Phyc Ward SAME.
Let me guess you're here because you have an EGD exam coming soon I know...I know same reason
Ommmggg ur a live saver !!
Finally found one that works, much appreciated thank you so much!
Awesome! Thank you for taking the time to create this video and post it. This is the perfect information I was looking for that will allow me to construct my device.
THANK YOU SO MUCH u explained it so well
you are the best one
thnx......i actually know a method tht can draw a pentagon too easier though...but i cannot remember.....it all starts w/ cross and circle.....but urs are a bit complex but still easy to understand....
Thank you so much!😊✍
🙏💕 thank you you help all students👦📖🎒👦📖🎒👦📖🎒
Love from Nepal
Was stuck on an assignment and this helped me out. Gracious 😇
Thank you this will help me on my drawing s
Very good and easy steps
Nice man
Wonderful... this is so helpful, i have been struggle to draw a pentagon now i got it exactly. Thanks more your video helped me out.....
enjoy this pentagon is very beautiful, thanks very much.
cute pentagon!
A very interesting construction. Thank you for it. Have you a geometrical proof that ED and DC are equal in length to AB?
They were constructed with that length on the compass!
Thank you for your Reply. With respect, if you look on from about 3:30 you will see that points C and E are each located of by the intersection of 2 arcs, one of which has radius (AB), and the other an arc of greater but unspecified radius passing through point D. DC and DE are not directly referred to distance AB.
@glyn hodges Sorry, I have been 'off-air' for a while and have just discovered your detailed and comprehensive Reply. Thank you. I look forward to following through your reasoning, and will respond in due course.
ruclips.net/video/h5nU6v2p5Qw/видео.html
Please watch like comment
very useful, good job
Thank you
Thank you so much sir... 👍🏻
Thank youuu❤❤
Intro🔥🔥🔥🔥🔥
Thank you very much ❤
Thankz alot....u really explain very well and it's very easy to understand
Thank you it helped ma a lot
Pentagon process is lit
Am I mistaken, or is this an approximate construction based on the very crude approximation of √5 by √8?
Is it regular, though? I like it, but you have to check the angles to make sure it's not slightly squashed vertically or horizontally. Many people, showing various methods, are failing to check the angle measurements to make sure they are all 108°. One method gave me two 106°(at the bottom), two 109°, and 110° at the top. Even with the sides all equal, it still can have symmetrical variability in it's angle measurements.
I saw the other version where you inscribed it in a circle. That would guarantee the angles to all be equal at 108°. So, you'd have to find the center of this Pentagon and draw a circle around it to see if all the corners fall on the line.
Super it helped me a lot
Thank you so much man i love you
Fun fact: i am from italy and i understood YOUR explanation and not my teacher one LOL
LEGEND!!!
Sir...this video is good
Thank you so much ✨
thanks for this video
Thanks it was awesome
Helped a lot thanks
Clear information
Great work!
Thank you.
Rlly appreciate it
waoooo... thank uuuu sooooo much buddy... great help.... great help....
thanks! worked for me
Very helpful
Thank you sooo much
Thank you very very much sir
Thanks alot you helped me in My Project☺️🙏🙏
Is there any other alternative method?
+Brinda Parthasarathy Hi please find attached a link to an alternative method called "Inscribed Pentagon"
ruclips.net/video/9NmO1Bq-oWg/видео.html
Very helpful thank you
Thanks a lot, Sir! Really helpful and easy to learn.
Nice sir 👌
Oh you GODSEND
Thanks a lot for this my Pentagon turned out perfect!
first class instruction,thank you.
Sick bro
Thanks dude , this vid help alot
Thanks for video
really helpful
thanks you
Thank you so much
thanks to this i passed my test yesterday
understandable video, nice
Thank you very much sir
the best construction vid ever thnx a looooooooooooooooooooooooot!!!!!! LOL
technically it is not a construction in the euclidian sense, more a drawing.
thank you very much for your easy tutorial!
yayy!! i did it thanks to youuu
Thanks
There is an easier method of drawing an accurate pentagon using the characteristics of the pentagon cross. The horizontal arm bi-sects the vertical at the phi ratio of 38-62. and is 5 pc longer than the vertical so if the latter is say 20cm, the former will be 21cm. Once you have drawn the cross, join the arms to the head and determine the side length. centre this on the bottom of the vertical and join the feet to the arms. Simple.
Thank you sir