The Reaction - check out chapter 5 of journey through genius by William Dunham. I think you can find it online for free. It’s a pretty interesting proof!
When you kept playing with the factorization rules at around 6:00, I already figured out how to prove Heron's formula. I tried to verify the formula years ago using the sine rule (A = ½bc sin t), but the equation got very complicated until I didn't know how to simplify it. This video shows the importance of mastering algebra, especially when it comes to solving simple problems like this.
If you look closer, you're actually indirectly proving the trig stuff (especially the cosine rule) along the way, you're just not explicitly stating the identity.
As a non native english speaker, (chinese for that matter, very different) he is thinking of cancelled as the base verb, and adding ed to make it past, but he is making a past tense go into like a double past tense, so he says cancelleded
I've come up with a formula for the area of triangles using hard algebric geometry. It takes the sides squared as inputs, so it works best on a carthesian plane. A,B,C are sides squared A=1/4 * sqrt(- A2 - B2 - C2 + 2(AB+BC+CA)) it uses pretty big numbers so it's better to use a calculator or use it in a program... But I'm sure it can be transformed into heron's and viceversa.
Thats great video i always thought this theorem was long and needed so much effort so i never been curious about it and rarely used it but you changed my mind keep up the good work
You know the one who is credited with the invention of zero is Aryabhatta, but this dude (Brahamagupt) was the one who first gave rules to actually use zero for calculations. His formula shown here is one of the first applications of setting the other side equal to zero to solve a problem. He also has contributions in fields like linear algebra, trigonometry and astronomy. Here's a link to his wiki page if you're interested in knowing more: en.wikipedia.org/wiki/Brahmagupta
I've squared the second quantity under the root and struggled with the algebra but finally I looked at what I had which is a fourth degree polynomial in terms of "a" and solved for "a" squared and took the square root and rearranged the solutions to get the product of the final 4 quantities Really amazing problem that I can actually solve.
@@noahtaul It does, you can't cancel an undefined part in your expression just by multiplying it with zero. Instead, the whole expression becomes undefined. It's similar to e.g. 0*1/0, it doesn't equal 0 or 1, it's undefined.
In any Δ ABC, the Cosine Rule gives cos(C) = (a²+b²-c²)/(2ab). So, sin(C)= √[-cos²(C)] =√[(2ab)²-(a²+b²-c²)²]/(2ab). ∴ area(ABC) =(ab/2).sin(C) =√[(2ab/4)²-{(a²+b²-c²)/4}²] which can be facto -rized to give Heron's formula. But who need's Heron's formula! For the 5,6,7 triangle; the area = √[{2(5)(6)/4}²-{(5²+6²-7²)/4}²] = √[(60/4)² - (12/4)²] = √[15² - 3²] = √(225 - 9) = √216 = 6√6. .
8:26 There is a mistake here, right? The square of (a - b) cannot give you (a^2 + 2ab - b^2). It gives you (a^2 - 2ab + b^2), if you look at the identities.
By me best herons formula prof is to prof volume of equilateral triangle , and then any other treangle as resized equilateral in two directions, so this can be used as prof for n-dimensional triangles volume
I actually discovered this formula in religion class by accident when I was playing around with 1/2ab * sin(C) and cosine rule (to find the angle used in the area formula and then use inverse trig identity). Thankyou for sharing this.
This formula is actually taught in 9th grade to us, here in India, at a age when we don't know trigonometry. So, this is a really helpful way to understand Heron's formula
There's a simpler version. Through law of cosines, we have cos(A)=(a^2+b^2-c^2)/2ab. Then, we have sin(A)=sqrt(1-cos^2(A))=(1-cos(A))(1+cos(A)) and you can easily finish the proof using 1/2 bc*sin(A). It's the same algebra as above except you skip a lot of steps.
u8y7541 the nice thing about the method in the video is that it uses only basic algebra and no trig functions. depending on where you live, you learn this kind of algebra before you learn about trig functions (at least I did), so for that reason I would consider this method more elementary
Just use Sin and Cos formula and set the sum of squares equal to 1. Area = bc/2 sinA and cosA = (b^2 + c^2 - a^2) / 2bc 1 = sin^2 A + cos^2 A = (2 Area / bc)^2 + ( (b^2 + c^2 - a^2)/2bc )^2 (2 Area / bc )^2 = sin^2 A = 1 - cos^2 A = (1 + cosA) (1 - cosA) = (2bc + b^2 + c^2 -a^2) (2bc - b^2 -c^2 + a^2) / (2bc)^2 So, (4 Area)^2 = [ (b+c)^2 - a^2 ] [ a^2 - (b-c)^2) ] = [ (a+b+c) (b+c-a) ] [ (a-b+c) (a+b-c) ] = [ (2s) (2s - 2a) ] [ (2s - 2b) (2s - 2c) ] since 2s = a + b + c Therefore, Area ^ 2 = [ s (s-a) (s-b) (s-c) ]
1/2*a*b*sin(C) is much simpler imo. (C is the angle between the sides a and b) you can easily derive it geometrically if you draw h on the side a: sin(C)= opposite/hypothenuse = h/b so h=b*sin(C)
The first part of the proof is so simple and straightforward yet I have never been able to do it on my own (maybe I did the first part of the proof, but I know for sure that I was never able to prove this formula which bugged me since I always feel uneasy using formulas that I can neither prove rigorously or have some good intuitive understanding why they should be true without knowing the rigorous proof. Just implementing/using a formula that I have read in a textbook always felt like cheating)
@@castilloguevaragiancarlomi6952 I know formulas for half angles, I knew how to derive all trigonometric formulas I have been working with. But I couldn't derive Heron formula. That's what bugged me using it felt like cheating.
Using Brahamgupta formula i.e. sqrt(s-a)(s-b)(s-c)(s-d) = area Since all triangles can be circumcirled by a circle and d = 0 in a triangle Therefore area = sqrt(s)(s-a)(s-b)(s-) Hence proved ~anish gupta....in a hope of reply from you
After doing some stuff on WolframAlpha, I got this: Area = (1/2) a b sqrt(1 - (a^2 + b^2 - c^2)^2/(4 a^2 b^2)) I used a lot more trig though: C=cos^-1((a^2+b^2-c^2)/(2ab)) h = a sin C Area = b h/2
I think these might not be as clear to me as other people, because the red and black pens look the same to me because I am red green colour blind. I can't believe I have been actually watch you since you started and I only just worked out you are even using two different colour pens. I mean it's in the name! ❤️🖤💀
Great, this video proved 3 formulas at the same time, one formula attributed to a Chinese mathematician from the 13th century, then a formula found by Kahan anb finally the Heron's formula.
When I was in 9th there is where I learned heron formula & as note I found brahmagupta's formula & I'm amazed that just putting d=0 you can get heron's equation.. Man Indian Mathematician were too good at that time I always love to learn more & more about them
Whenever I proved this formula, the class, and I, always felt that it was best to stop when we got to the stage A= sqrt((a+b+c)(a+b-c)(a+c-b)(b+c-a) because this is much easier to apply to any problem. Try it!
Hey! Great to see your proof of Heron's formula. The way I know is based on formula A = bc*sin(a) where a-(alpha) is angle between sides b and c in a given triangle. Would love to see more geometry on your channel.
I never tried to prove it, because I thought it was very more complicated. Maybe it is, but with you, all becomes simpler. Hope to see you soon at the black board back !
The same way Heron's formula works for triangles and Brahmagupta's works for quadrilaterals, I wonder if there's a general pattern for any polygon with n sides. I assume that the proof for the quadrilateral formula comes from cutting the quadrilateral into 2 triangles and applying Heron's twice, so theoretically it's possible to derive a formula for a pentagon and so on.
Έρικ Κωνσταντόπουλος because a regular polygon isn’t a general polygon, it’s only one special case. Just like this video doesn’t prove the area for an equilateral, but some random triangle.
At 10:38 shouldn't theta be the average of the two opposite angles, rather than the sum? That's what wolfram says, anyway. It's also the only way to get the right area for a square
The proof is simple and easy.A lengthy and hard proof using geometrical construction is given in the book "Journey through Genius" written by William Dunham.
I've always used the law of cosines to prove it, but this is pretty slick! Thx bprp
Did they even have the law of cosines when Heron proved this?
The Reaction - No, but I believe his argument was purely geometric rather than algebraic.
@@BrainGainzOfficial I completely overlooked that they might have not even had algebra either. It would be nice to see how he did it.
The Reaction - check out chapter 5 of journey through genius by William Dunham. I think you can find it online for free. It’s a pretty interesting proof!
@@thereaction18 How do you mean they didn't have algebra? They obviously had it, at least geometric algebra
When you kept playing with the factorization rules at around 6:00, I already figured out how to prove Heron's formula. I tried to verify the formula years ago using the sine rule (A = ½bc sin t), but the equation got very complicated until I didn't know how to simplify it. This video shows the importance of mastering algebra, especially when it comes to solving simple problems like this.
damn bro chill out
Thats a nice proof without any trigo involved making it clean and simple😄
If you look closer, you're actually indirectly proving the trig stuff (especially the cosine rule) along the way, you're just not explicitly stating the identity.
Your tone is really great - that is half the battle of being a good teacher. Great video I enjoyed it.
"HE RUNS" formula
RUclips's captions in a nutshell
i agreeeeeeeeeeeeeeeeeeeeeeeeeeee
I love how he pronounces “cancelled” as “canceldid” so much
Cancelled it😊
splendid canceldid!
@Yosif Abbas and I can't believe I actually posted that
As a non native english speaker, (chinese for that matter, very different) he is thinking of cancelled as the base verb, and adding ed to make it past, but he is making a past tense go into like a double past tense, so he says cancelleded
2:42
I got asked to do this as an interview question
It took some, to say the least...
Did you pass
I’ve been wanting to see a proof of this formula for a while now. Thanks for showing this great proof!
I've come up with a formula for the area of triangles using hard algebric geometry. It takes the sides squared as inputs, so it works best on a carthesian plane.
A,B,C are sides squared
A=1/4 * sqrt(- A2 - B2 - C2 + 2(AB+BC+CA))
it uses pretty big numbers so it's better to use a calculator or use it in a program... But I'm sure it can be transformed into heron's and viceversa.
I'm a backbencher sir,but your every explanation is just so easy to understand ♥️
so you are dum
The formula for area of quadrilateral was shocking.
Wow! Good information.
You are doing good.
Thats great video i always thought this theorem was long and needed so much effort so i never been curious about it and rarely used it but you changed my mind
keep up the good work
10:38 Never heard of that, but COOL!
You know the one who is credited with the invention of zero is Aryabhatta, but this dude (Brahamagupt) was the one who first gave rules to actually use zero for calculations. His formula shown here is one of the first applications of setting the other side equal to zero to solve a problem. He also has contributions in fields like linear algebra, trigonometry and astronomy.
Here's a link to his wiki page if you're interested in knowing more: en.wikipedia.org/wiki/Brahmagupta
I proved Heron's formula a few years ago with SOHCAHTOA. This proof is much nicer and more concise. Great video, BlackPenRedPen!
I've squared the second quantity under the root and struggled with the algebra but finally I looked at what I had which is a fourth degree polynomial in terms of "a" and solved for "a" squared and took the square root and rearranged the solutions to get the product of the final 4 quantities Really amazing problem that I can actually solve.
Presh Talwalkar's fans will be complaining of you not using Gougu's theorem :-)
Anyway, you are the king of RUclips math-teachers!!
Wow, I'd never heard of Bretschneider's formula at 10:38, that's weird! How do you prove it? It reduces to Heron when d=0.
bretschneider?
@@sx86 generalized Brahmagupta's = Bretschneider's
TBF, it doesn't exactly reduce to Heron's formula because of the way θ is defined (it would be undefined).
Έρικ Κωνσταντόπουλος Well it doesn’t matter what theta is because d=0 kills the cos^2(theta) part.
@@noahtaul It does, you can't cancel an undefined part in your expression just by multiplying it with zero. Instead, the whole expression becomes undefined. It's similar to e.g. 0*1/0, it doesn't equal 0 or 1, it's undefined.
I was just curious as to how this was derived and this derivation is neat!
I'm so happy I found this, stay safe
The formula is introduced in Heron's book Περί Διόπτρας, where he proves it by using the inscribed circle, an elegant geometrical proof
That's the proof that I was hoping he'd do.
Blackcursorwhitecursor
I never knew about this formula, and the proof is really easy but I found this video extremely entertaniing
In any Δ ABC, the Cosine Rule gives cos(C) = (a²+b²-c²)/(2ab).
So, sin(C)= √[-cos²(C)] =√[(2ab)²-(a²+b²-c²)²]/(2ab). ∴ area(ABC)
=(ab/2).sin(C) =√[(2ab/4)²-{(a²+b²-c²)/4}²] which can be facto
-rized to give Heron's formula. But who need's Heron's formula!
For the 5,6,7 triangle; the area = √[{2(5)(6)/4}²-{(5²+6²-7²)/4}²]
= √[(60/4)² - (12/4)²] = √[15² - 3²] = √(225 - 9) = √216 = 6√6.
.
احسنتم وبارك الله فيكم وعليكم والله يحفظكم يحفظكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .
Please do a video explaining the Bretschneider's formula at 10:38
Brahmagupta's***
@@randomdude9135 I think he was High enough..
Random Dude no, Brahmagupta’s formula is only for cyclic quadrilaterals, and doesn’t have the last cosine term.
No blackpenredpen spelt it wrong lol
Thank you! I’ve always wondered about the proof!
Cuuute! it's something even young students can do to really stretch their algebra skills hehe it's easy but with some algebra tricks 😊 nice
u r just great, thanks for making our studies easier, soon my exams, and so blessed to have found ur channel)))
8:26 There is a mistake here, right? The square of (a - b) cannot give you (a^2 + 2ab - b^2). It gives you (a^2 - 2ab + b^2), if you look at the identities.
There's a minus right outside that distributes through.
Hi, on Wikipedia it says that Heron originally proved this using cyclic quadrilaterals; please could you make a video on that? Thanks so much.
It can be derived from a particular case of the generalized half angle formula. Se here: ruclips.net/video/WbkQHnNthg8/видео.html
By me best herons formula prof is to prof volume of equilateral triangle , and then any other treangle as resized equilateral in two directions, so this can be used as prof for n-dimensional triangles volume
This video is absolutely perefect form my math project!!!!! TYSM!!
Old School Style blackpenredpen!
I actually discovered this formula in religion class by accident when I was playing around with 1/2ab * sin(C) and cosine rule (to find the angle used in the area formula and then use inverse trig identity). Thankyou for sharing this.
3:47 is the cosine rule!
Very Easy to understand...Thank you
We can use cosine formula a2= b2+c2-2abcos(c).Work out since and use area formula A= 0.5absinc
This formula is actually taught in 9th grade to us, here in India, at a age when we don't know trigonometry.
So, this is a really helpful way to understand Heron's formula
But iirc we didn't have the proof, just like 2 exercises from NCERT which made the formula imprint in our heads
@@0VexRoblox Yes exactly, no proof was given to us then
I love this proof. Pls make more videos like this
Perfectly explained. Loved the video. Thank you so much😊
There's a simpler version. Through law of cosines, we have cos(A)=(a^2+b^2-c^2)/2ab. Then, we have sin(A)=sqrt(1-cos^2(A))=(1-cos(A))(1+cos(A)) and you can easily finish the proof using 1/2 bc*sin(A). It's the same algebra as above except you skip a lot of steps.
u8y7541 the nice thing about the method in the video is that it uses only basic algebra and no trig functions. depending on where you live, you learn this kind of algebra before you learn about trig functions (at least I did), so for that reason I would consider this method more elementary
Just use Sin and Cos formula and set the sum of squares equal to 1. Area = bc/2 sinA and cosA = (b^2 + c^2 - a^2) / 2bc
1 = sin^2 A + cos^2 A = (2 Area / bc)^2 + ( (b^2 + c^2 - a^2)/2bc )^2
(2 Area / bc )^2 = sin^2 A = 1 - cos^2 A = (1 + cosA) (1 - cosA) = (2bc + b^2 + c^2 -a^2) (2bc - b^2 -c^2 + a^2) / (2bc)^2
So, (4 Area)^2 = [ (b+c)^2 - a^2 ] [ a^2 - (b-c)^2) ] = [ (a+b+c) (b+c-a) ] [ (a-b+c) (a+b-c) ] = [ (2s) (2s - 2a) ] [ (2s - 2b) (2s - 2c) ] since 2s = a + b + c
Therefore, Area ^ 2 = [ s (s-a) (s-b) (s-c) ]
In Brahmagupta's formula I think θ is the sum of two opposite angles divided by two. I really like this video.
Great video! Please do more proofs!
1/2*a*b*sin(C) is much simpler imo. (C is the angle between the sides a and b)
you can easily derive it geometrically if you draw h on the side a: sin(C)= opposite/hypothenuse = h/b so h=b*sin(C)
But Heron's formula doesn't need any trig at all.
Thank you
I've been dreaming about learning proof of this formula some 5 years now
or Area=1/4 sqr((P(P-2a)(P-2b)(P-2c)) P is The perimetr of ABC
It can be used that but looks like most people uses Heron since it has simpler formula
okay?
My favorite ways to write it are
(4S)^2 = (a+b+c)(a+b-c)(a+c-b)(b+c-a)
and
S^2 = xyz(x+y+z), where
p = (a+b+c)/2
x = p-a
y = p-b
z = p-c
My most favourite proof.
The same way i also derived this formula .....It's suprising to me that I can think like the blackpenredpen....
The first part of the proof is so simple and straightforward yet I have never been able to do it on my own (maybe I did the first part of the proof, but I know for sure that I was never able to prove this formula which bugged me since I always feel uneasy using formulas that I can neither prove rigorously or have some good intuitive understanding why they should be true without knowing the rigorous proof. Just implementing/using a formula that I have read in a textbook always felt like cheating)
You can try this formula faster knowing a little trigonometry (half angle)
@@castilloguevaragiancarlomi6952 I know formulas for half angles, I knew how to derive all trigonometric formulas I have been working with. But I couldn't derive Heron formula. That's what bugged me using it felt like cheating.
@@smrtfasizmu6161 Sorry I think I did not read your comment well my native language is Spanish
Using Brahamgupta formula
i.e. sqrt(s-a)(s-b)(s-c)(s-d) = area
Since all triangles can be circumcirled by a circle and d = 0 in a triangle
Therefore area = sqrt(s)(s-a)(s-b)(s-)
Hence proved
~anish gupta....in a hope of reply from you
I love youuu, so helpful, u just expplained it so simply and clearly
After doing some stuff on WolframAlpha, I got this:
Area = (1/2) a b sqrt(1 - (a^2 + b^2 - c^2)^2/(4 a^2 b^2))
I used a lot more trig though:
C=cos^-1((a^2+b^2-c^2)/(2ab))
h = a sin C
Area = b h/2
That's the same thing
@@jadegrace1312 but a lot less simplified
Nice work. In the generalized Brahmagupta’s formula angle aplha correctly is half of sum of opposite angles.
I think these might not be as clear to me as other people, because the red and black pens look the same to me because I am red green colour blind. I can't believe I have been actually watch you since you started and I only just worked out you are even using two different colour pens. I mean it's in the name! ❤️🖤💀
Great, this video proved 3 formulas at the same time, one formula attributed to a Chinese mathematician from the 13th century, then a formula found by Kahan anb finally the Heron's formula.
Why someone has to downvote educational videos like these?
Monk Klein true, its ruin our efforts
What 3B1B is a patron? Damn!
Yes.
When I was in 9th there is where I learned heron formula & as note I found brahmagupta's formula & I'm amazed that just putting d=0 you can get heron's equation.. Man Indian Mathematician were too good at that time I always love to learn more & more about them
Whenever I proved this formula, the class, and I, always felt that it was best to stop when we got to the stage
A= sqrt((a+b+c)(a+b-c)(a+c-b)(b+c-a) because this is much easier to apply to any problem. Try it!
Sorry, I forgot to write the divide by 4!
My man please prove Stewart's theorem.
Sir you are very talented
Trigonometry proof is the best
I'm going to use this for right triangles from now on and nobody can stop me
Teacher at school:- You're challenging me?
Awesome explanation!
Nice prove
Hey! Great to see your proof of Heron's formula. The way I know is based on formula A = bc*sin(a) where a-(alpha) is angle between sides b and c in a given triangle. Would love to see more geometry on your channel.
Beautiful. A really excellent explanation.
ThanksMaster
Finally a proof I understand: :’)
I enjoyed very much. Thanks for making such nice videos!
Nice upload
Embezzlement can be very well demonstrated using geometry
3:47 is literally the cosine rule, look closer.😂
Your proof is much better and self-contained than Taylor Swift's!
I never tried to prove it, because I thought it was very more complicated.
Maybe it is, but with you, all becomes simpler.
Hope to see you soon at the black board back !
I have a few pre recorded videos in my usual setting. Hopefully the current situation gets better soon for everyone.
3b1b is one of your patrons? That's awesome!
I've proved this, its really easy, define the hypoteneuse as c=sqrt(a^2-h^2)+sqrt(b^2-h^2), A=(hsqrt(a^2-h^2)+hsqrt(b^2-h^2))/2
You dont know a value for h
YEEESSSS. I LOVE IT.
Now how the hell did Heron ever figure that out?
The same way Heron's formula works for triangles and Brahmagupta's works for quadrilaterals, I wonder if there's a general pattern for any polygon with n sides. I assume that the proof for the quadrilateral formula comes from cutting the quadrilateral into 2 triangles and applying Heron's twice, so theoretically it's possible to derive a formula for a pentagon and so on.
Best maths teacher
Now do the formula for the area of a pentagon.
Dr. πm has made a video where he proves the formula for the area of a general regular polygon. ruclips.net/video/B07EgGGL6q0/видео.html
Έρικ Κωνσταντόπουλος It works, but it does not generalize.
@@angelmendez-rivera351 How does it not generalize?
Έρικ Κωνσταντόπουλος because a regular polygon isn’t a general polygon, it’s only one special case. Just like this video doesn’t prove the area for an equilateral, but some random triangle.
@@SB-wy2wx But a pentagon is a regular 5-gon.
At 10:38 shouldn't theta be the average of the two opposite angles, rather than the sum? That's what wolfram says, anyway. It's also the only way to get the right area for a square
Ah! Yes, you are correct!
Thanks for pointing this out. I just pinned your comment so others can see it. Thank you.
@@blackpenredpen Apparently it got unpinned
Bruno Moreira because op edited the comment
@@blackpenredpen Needs to be repinned lol
Thanks!
thank you so much!
Thank you for this great ful video
I am in 10th grade and this is the first video of BPRP that I understood well
The proof is simple and easy.A lengthy and hard proof using geometrical construction is given in the book "Journey through Genius" written by William Dunham.
Thank you
I love the phrase "invite into the square root house", I never thought of thinking of it that way.
thanks for the amazing content
Such a gorgeous proof ✔
It's really good sir.i want more mathematical proof sir
Good bro, your videos are amazing. Please try in upcoming videos to solve
Derivative of x!
splendor, bro carry on
rip the small whiteboard :(
ns video btw
It is digital board anyway (you write it on a computer)
this is insane
I always wanted to know this.
Wow amazing formula 😍
Whats brahmagupta's formula
Yeah ik it gives the quad area but pls elaborate it
NICE EXPLANATION SIR. YOU ARE GREAT. #themathsgurudev