Heron's Formula Proof (the area of a triangle when you know all three sides)
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- เผยแพร่เมื่อ 10 ก.พ. 2025
- We can find the area of any triangle with Heron's formula when we know the sides of the triangle. Here we will see how to prove the heron's formula, which is a classic trigonometric result. And because you like Hero's formula, you probably will also like the proof of the following...
Law of sine and cosine: 👉 • Classic math proofs of...
Pythagorean Theorem 👉 • Pythagorean Theorem & ...
Pythagorean triple generator👉 • finding ALL pythagorea...
🛍 Shop my math t-shirt & hoodies: amzn.to/3qBeuw6
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
Your tone is really great - that is half the battle of being a good teacher. Great video I enjoyed it.
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.
"HE RUNS" formula
TH-cam'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’ve been wanting to see a proof of this formula for a while now. Thanks for showing this great proof!
The formula for area of quadrilateral was shocking.
Wow! Good information.
You are doing good.
I'm a backbencher sir,but your every explanation is just so easy to understand ♥️
so you are dum
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
I got asked to do this as an interview question
It took some, to say the least...
Did you pass
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.
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.
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'm so happy I found this, stay safe
Presh Talwalkar's fans will be complaining of you not using Gougu's theorem :-)
Anyway, you are the king of TH-cam math-teachers!!
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)))
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
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
I never knew about this formula, and the proof is really easy but I found this video extremely entertaniing
Thank you! I’ve always wondered about the proof!
In Brahmagupta's formula I think θ is the sum of two opposite angles divided by two. I really like this video.
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!
احسنتم وبارك الله فيكم وعليكم والله يحفظكم يحفظكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .
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
Perfectly explained. Loved the video. Thank you so much😊
I was just curious as to how this was derived and this derivation is neat!
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.
Let's use a combination of elementary trigonometry and Al-Kashi's theorem:
1) Elementary trigonometry : S = (ab/2)sinθ (where θ is the angle between a and b)
2) Al-Kashi's theorem : c² = a² + b² -2abcosθ
Additionally, use the identity: 1 - cos²θ = sin²θ
Thus, we derive an expression for cosθ in terms of S, and another expression independent of S. By equating the two, a bit of algebra yields:
16S² = -(a^4 + b^4 + c^4) + 2a²b² + 2a²c² + 2 b²c²
Next, apply Euler's formula (which is straightforward to verify):
-(a^4 + b^4 + c^4) + 2a²b² + 2a²c² + 2 b²c² = (a+b+c)(a+b-c)(a-b+c)(-a+b+c)
This leads directly to Heron's formula.
Please refer to my five math books for middle and high school, which illustrate the fundamental simplicity of mathematics, presented on my TH-cam channel.
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.
We can use cosine formula a2= b2+c2-2abcos(c).Work out since and use area formula A= 0.5absinc
I love youuu, so helpful, u just expplained it so simply and clearly
This video is absolutely perefect form my math project!!!!! TYSM!!
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.
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.
.
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
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: th-cam.com/video/WbkQHnNthg8/w-d-xo.html
I love this proof. Pls make more videos like this
Very Easy to understand...Thank you
Thank you
I've been dreaming about learning proof of this formula some 5 years now
Nice work. In the generalized Brahmagupta’s formula angle aplha correctly is half of sum of opposite angles.
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.
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
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
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
I got the formula at 4:47 - which turns out to be the same - but thought I had made a mistake. Anyway, it really bugs me that there is not an explanation of “why” the simplified formula holds. It is just so incredibly simple and elegant. We need someone to resurrect Euclid! Or does the inscribed circle mentioned below provide the "ah ha"!?
And now to start thinking about the area of an arbitrary quadrilateral! I wouldn’t be surprised if the algebraic topologists are completely on top of this.
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
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) ]
Great video! Please do more proofs!
Old School Style blackpenredpen!
I love the phrase "invite into the square root house", I never thought of thinking of it that way.
3:47 is literally the cosine rule, look closer.😂
Blackcursorwhitecursor
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! ❤️🖤💀
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.
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
The same way i also derived this formula .....It's suprising to me that I can think like the blackpenredpen....
I enjoyed very much. Thanks for making such nice videos!
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.
on 8:00 you turn a^2 + 2ab - b^b into (a - b)^2.
isnt (a - b)^2 supposed to equal a^2 - 2ab + b^2?
someone explain what happens there
He turned -a^2+2ab-b^2 into -(a-b)^2
It is -a^2 there so it is turned correctly
Beautiful. A really excellent explanation.
10:33 what is that box?
AgaBe_
it's a sign of the end of the proof.
@@みく-t8h I thought it was Q.E.D.?
AgaBe_
Q. E.D. is not used very much recently for
the following reasons.
1.it's troublesome to write
2.the language that it became the cause
is Latin,so it is incomprehensible
on the other hand,the box comes to be used
well because it's easy to write.
sorry in my poor English.I'mJapanese
@@みく-t8h ok thanks (ありがとう) :)
10:39 this isn't the Brahmagupta Formula, this is the Brettsneider formula, in the Brahmagupta, there isn't the (-abcd*cos theta) however it work just on cyclic quadrilateral.
Except this, the proof is good.
thats why it says generalized
it generalizedthe other one
not that it is generalized and frkm that guy
Now at 10:40 how about generalize for any n gon or if that's too hard the next level - area of a pentagon in terms of semiperimeter.
Good bro, your videos are amazing. Please try in upcoming videos to solve
Derivative of x!
Hi, nice video buddy! Can you answer me this question? How many planes are defined by one line and 3 collinear points that do not lie on that line
Sir you are very talented
Damn I always forget about Heron's formula and it's so useful! I totally could have used this on my Calc 2 homework a few weeks ago
Whats brahmagupta's formula
Yeah ik it gives the quad area but pls elaborate it
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'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 EXPLANATION SIR. YOU ARE GREAT. #themathsgurudev
Thank you for this great ful video
I love your videos! Can you tell my what programs do you use to record the screen and what app/program do you use to write? Do you use mouse for writing?
Guess it is Ms Paint-like since it has a brush like and it is possible to use Bandicam but He is using Mac so...
Probably he wrote it using pen since if using mouse, it wouldn't be so good
At 3:18, you can do a bonus proof of the law of cosines easily by seeing from the picture that b₁ = a cosγ and rearranging. I don't know how I missed this one back in April!
thanks for the amazing content
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!
What 3B1B is a patron? Damn!
Yes.
My man please prove Stewart's theorem.
My most favourite proof.
Can you show the proof for 10:38 ?
@bluepenredpen I have a doubt well, are all numbers equidistant from infinity?
It's really good sir.i want more mathematical proof sir
thank you so much!
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 was at grade 8 when my teacher asked to prove heron's formula for a triangle's area. Idk but I somehow managed to prove it, now, three years later, I can't even do simple banking problems (SI and CI
YEEESSSS. I LOVE IT.
Now how the hell did Heron ever figure that out?
It's been decades since I used algebra so I've forgotten a few things. At 6;00, how did he get (2ab)2?
it equals 4a^2b^2. Multiply that by the 1/4b^2 and it equals a^2 which is what he initially had
Finally a proof I understand: :’)
I am in 10th grade and this is the first video of BPRP that I understood well
pls do the trigonometrical proof also using sine and cosine formulae
the first thing up until 4:00 could have worked with Pythagora's Theorem in the triangle formed by a, b1 and h, too, i think
edit: wait, no, hold on, we couldn't have found out what b1 is equal to, sorry, ignore this comment
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.
I always wanted to know this.
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. th-cam.com/video/B07EgGGL6q0/w-d-xo.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.
3b1b is one of your patrons? That's awesome!