AWESOME Formula - AREA of a TRIANGLE (Herons Formula)
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- Опубліковано 21 сер 2024
- TabletClass Math:
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Math help with finding the area of a triangle using Herons formula. For more math help to include math lessons, practice problems and math tutorials check out my full math help program at tcmathacademy....
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The 'quick' introduction lasted 3 minutes and 48 seconds, that's 26.54% of the whole video. I don't understand these long, meandering introductions on your videos. I love watching math techniques, I just find these marathon intros to be pointless.
Same here.
It's even worse his intro is actually 5 minute exactly. What a jerk.
@@lapulapu7422
I'm beginning to think that he is pushing these videos to a certain minimum time for the sake of how many commercial breaks UA-cam employs. If so, there should be some sort of limitation for these people who purposefully create bloated content just to make money.
too much "plug" about himself!!😢
I could not agree more!
Name the vertices of the triangle as follows. Botom left is A, top is B and bottom right is C
The area of the triangle is
(1/2)*c*b*sin(A) =(1/2)*4*6*sin(A).
By the cosine rule
5² = 4² + 6² - 2*4*6*cos(A)
⇒ cos(A) = (4² + 6² -5²)/(2"4*6) = 27/48
⇒ A = 55.7711° (4 decimal placed)
⇒ sin(A) = sin(55.7711°) = 0.8268 (4 decimal placed)
So, Area of ΔABC
= (1/2)*4*6*sin(A)
= (1/2)*4*6*0.8268
= 9.9216.
Interestingly, this approach can be used to prove Heron's formula. If instead of computing the area with numbers we use the side length of a, b, c, then we could have found sin(A) from cos(A) using the fact that sin²(A) + cos²(A) = 1,
so sin(A) = √(1 - cos²(A)).
We let s = (a + b + c)/2 and then we apply some algebra on s, a, b & c in the equation, noting how to factorise the difference between squares (twice) to find sin(A).
From there it is an easy step to find the area of the triangle using area = (1/2)*c*b*sin(A) and we end up with Heron's formula.
My wife and have 4 advanced degrees between us including PhD’s. I have a professional license to teach 7-12. My mentor was teacher of the year in America. I’ve taught mathematics mastics at virtually every level up to entry level graduate studies. This man represents everything that’s wrong with math education. He droned on for more than half the video without saying a single relevant word. He then distracts the student with useless references to law of sines/cosines which at this level would be utterly opaque to the students. Then “magically “ a formula appears. Absolutely no intuition motivating it, and absolutely no mathematical reasoning as to how this formula is derived. He then instructs students to blindly apply this formula, and does a poor job of that as well. Moreover, he speaks for 15 minutes of which around 3 are actually relevant to the problem. To believe that this drivel is within the attention span of a teen student is delusional. This isn’t remotely on the spectrum of modern mathematics education. This could be used as an example of what is wrong in math education in this country. Somebody stop him before he disenfranchises another child from learning to reason mathematically.
I agree completely!!! Most is drivel as you said.
I agree! Tabletclass Math should, at the minimum, compute the height of the triangle, which is a common side to 2 right triangles with hypotenuses a and b, and sides, let's call them p and q, where p + q = c. He can readily eliminate p and q to generate a formula for the area in terms of a, b, and c. With a fair amount of algebra, that formula will simplify to Heron's formula. He could show the algebra without going through the steps.
This guy should get a job at Walmart or something. He should not be teaching math anywhere. The formula is clever, although it could be derived using algebra and geometry. I wish I had heard of it before. But to drone on and on about nothing. Yikes I'd have fallen asleep in his class.
*john donne* I agree. I played the video at x2 speed and it *still* took forever for him to get to the point.
Agree. He is repelling rather than attracting interest. But he is excellent in self-advertising of his own "achievements".
Not knowing Heron’s Formula, I went the long way around the barn. I drew in the altitude as you did. Now I have two triangles with a common height and bases of X and 6-X. I used the Pythagorean Theorem to write the equations for h^2 of each triangle, set them equal to each other and solved for X. Using that information, I calculated the areas of the two triangles, added them together and got the same 9.92 answer, it just took a little longer (maybe). FYI, h= 3.31 and X=3.75
Nice.
Very clever
That’s exactly what I did and it produces the result
The base which is 6 cm is divided between the two triangles to 2.25 and 3.75
Height is sq. Root of 10.9375 = 3.307189
Area is 1/2 of 6x 3.307189 = 9.9215
Dan Myers and mraoufmd
What? "TableClass Math" talked too much, but you didn't talk enough, for me, at any rate. How about adding a few steps? I took Algebra in High School--successfully-- about 65 years ago, but I don't see how you "solved for x."
I owe you an apology, sorry!
Heron's formula actually gives the simplified solution 15/4 * sqrt 7 = 9.92...
All the short cuts mentioned in the comment section start from the illusion that the upper angle is 90 degrees. The drawing is deceptive and the angle is close, about 83 degrees.
So if you double the triangle it won't give a new triangle because the left leg won't continue in a straight line (180 degrees).
There is no need for that angle to be a right angle or any other specific angle. Any triangle can be translated to a parallelogram by taking an exact copy any placing the two hypotenuses together, the area is then the length x the height divided by two.
@@footloose6382 You are absolutely right but that was not what my comment was about. Anyway.
There is no.illusion. You just made a bad, and easily avoided, assumption.
Unless the problem is explicitly stated or depicted as involving a right triangle, the correct assumption is that it's instead the general case.
In this particular example, we can also quickly verify that it's NOT a right triangle because the squares of the sides don't sum.
@@starfishsystems You are preaching to the choir.
Yeah it's not a right angle triangle.
a^2 + b^2 not equal to c^2.
The mirror triangle to get a rectagle seem to work too. I've been taught that way but all forgotten now. LOL
Find one of the angles by using the cosine rule. Then get the area using the sine rule: Area = 0.5 a x b x sinc
I wouldn't call that 'using the sine rule' , since you are just using the definition of sin C. In my mind there is a distinction between rules and definitions, but they are often blurred.
The sine rule, analogous to the cosine rule, is also known as the rule of sines or law of sines is a way to method to 'solve' a triangle and can be expressed by the equation
a / sin A = b / sin B = c / sin B .
Fun problem, convert sin C to an algebraic expression. Hint, sin (arccos (x)) = √ 1 - x^2. I think it comes out to be equivalent to heron's area formula, after some algebra.
Also to nitpick further that should be , Area = 0.5 a x b x sin C , where angle C is opposite side c , notice the caps on C.
Just use cosine rule to get the an angle, then use ½abSin(C)
The simplest way would Be to take another triangle of the same dimensions and rotate it and join it along the current base of 6.
This results in a rectangle with a L= 5, W=4. L x w = 20 and half of that = Area of 10 for the original triangle.
Like Dan Myers, I started out with the Pythagorean Theorem and tried to use what I call substitution of equations to find the height. I was too lazy to work it all out but my gut feeling is that by using that method, you might wind up deriving Heron's Formula. I'm not sure of that. It's just intuition. But at the end of the day, I thank Heron and I thank you for telling me about that formula.
In any Math exam I ever took, writing down a formula for the desired result and then calculating it would get you one point for knowing how to apply it to the specific case, and maybe a second point for doing the arithmetic correctly, out of a possible max of probably 4 or 5 points -- *IF* the formula is taught in the curriculum.
If it _isn't,_ (and I never learned this one in 12 years of school and three of university) then writing down this formula and calculating the answer would probably get you no marks at all because the whole point of Math tests is not to come up with an answer like a History exam but to _work it out._
That was my reaction. Look! I rote learnt a formula mum. This is not maths. Worse, this guy is teaching children.
i mean... if you don't want to use herons formula you can just use pythagoras theorem and figure out the height of the triangle *shrug*
@@erikmarkus7467I mean... if you miss the point of a comment you can just post something completely off topic 🤷♂
@@christopherbedford9897 I have no idea what your problem is for writing down this formula and using it to calculate the area of a triangle of known sides - that is what the formula is for. I learned it when I was a freshman at highschool. There's not much to "work out" when the problem is stated as 'calculate the area of a triangle with these side lengths'.
Just like given a problem of calculating the area with side a and it's altitude. You just write down a*h/2 and punch in the numbers. You don't actually prove the formula every time by completing a rectangle and taking it's half. Why would you need to prove or work out Heron's formula? You just use it because it's elementary...
IF however you don't like using it, then just use Pythagoras theorem to calculate the h first out of the given three sides. I imagine you still wouldn't include a proof of the theorem, just go ahead and use it.
@@erikmarkus7467 The "problem" is very clearly stated in my original post: Mathematics is not a rote learning subject, it's logic and working it out subject. If you want to learn formulas and rattle them off, good for you, but you'd have been better off studying a subject like History or Law. But even then, the idea behind learning those subjects is to understand the world around you abd apply your knowledge of one situation to analyse a different one.
If you think the object of Maths is to find a number that represents the area of a geometric shape your education failed you sadly.
I used to teach Heron's Formula, along with pythagos' Teorem and trigonometry. We'd use markers of three types as mentioned and students assignment was to calculate an area , on the football oval , to use all these formulas. Yhat was year 11, here that"s pre last year of high school. Good fun. Lots of number crunching if the problem is a fraction or decimal. So we set up problems that s added to an equal number, dicided by 2 to get s. Easier to teach if it's not as cumbersome as this one.
Area of shapes involving rectangle, square and triangles
My maths teacher 12 to 18 was fantastic. Home work every night was a published maths test of 90 minutes.99% of my cohort passed O and A levels (UK).
Go right to the point
I flipped the top down and made a rectangle but also had to flip sideways so that it was a perfect rectangle 4 by 5. Makes 20 square and had to half it making the correct answer 10 square.
Easy peasy!!!
That's what I thought at first. Works for a perfect right angle... ex 3,4,5 ... 3sq +4sq=5sq. But this is not quite à right angle, I guess.
4 5 6 means 4 and 5 are at 90degrees
Draw a line opposite 4 and 5 at the same measurements.
Now you have a 4x5 rectangle, 4X5=20
half of 20 is 10 (cos you want to know the area of the triangle which is half the rectangle)
The answer is 10
You mistaken it with 3-4-5 triangle - which is right triangle.
4-5-6 is NOT right triangle: quick check
4^2+5^2 = 41 not equal to 6^2=36. so Pythagoras theorem is not working here...
I fell asleep with the intro alone,.
I wish I could like your comment 1k times.
@@IbiRere Hello Mr Olakunie,thank you much,You are from Africa,Kenya i assume.
@@junpinedajr.8699 I'm from Nigeria.
@@IbiRere Thank you Mr Olakunle.😀😃😄😊
You can use also cosine law by finding unknown height, then apply the formula of finding the ares of triangle..tnx
Which would be easier without a calculator?
But where is the demonstration of the heron's formula
Showing students how to use a formula is not teaching mathematics. Problem solving and formula derivation is always the best approach. It will develop a true understanding.
A bullet-point presentation always works best for my super-charged neurons.
But waste of time
I got the answer to 10 in a lot simpler way = 4x5 = 20 which is the area of a square of 4 by 5 and halve it to get the area of the triangle = 10 - simple!!!! To find the area as described in this video is why so many students get so bored with maths YES MATHS!!! The abbreviation of mathematicS is MATHS!!! ;-)
iS NOT A RIGHT ANGLES TRIANGLE
Nice formula but you didn't show how this formula was derived like what is "s" referring to. You just used the formula and plugged the numbers and I don't see mathematical justifications as to where the formula came from. But thank you and I will appreciate it if you can prove the validity of the formula.
agree100%
the area of a 5x 4 rectangle would be 20 so the triangle is half of that therefore 10????
Good and clear presentation !
It's very easy with the Formel of Heron. The area of a triangle iş:
A=[u*(u-a)*(u-b)*(u-c)]^½ -->
u= (a+b+c)/2
Easy, make two right Triangle and calculate the area as 1/2 x base and the height. Then multiply by 2
First you find the half of 4+5+6=15 that is 7.5. Then you find the square root of the product 7.5(7.5-6)(7.5-5)(7.5-4)=7.5*1.5*2.5*3.5=98.4375, that is 9.9215.... square units. Thank you!
I got 9.9215674. Knotwilg got 9.92... Any comments?
Thanks, and it didn't take 14 minutes.
Got it!
@@4DMovie Your answer is not really correct. Knotwig is correct. If you had said, "I got 9.9215674..." or if you had said, "I got 9.9215674 correct to seven decimal places" you would've also been correct.
I got 9.92156741649221 correct to 14 decimal places
Olivier Habineza
a+b+c/2=height? Since when? Where did you get that formula? If you use it on a 3X4X5 triangle you get 12/2=6, which is Not anything of any value.
Since it is a right triangle, simply multiply te length times the width and divide by 2.
I just did that in my head got to 9 in less than a minute.
After over 5 minutes of rambling, he finally gets to the problem.
Hey John , thank you for all the tutoring, I really enjoy you videos and learn from them . I have a question , how to figure the cross section of a rectangle, to check for square
Measure the 2 diagonals and compare.
I saw it as half of a rectangle of length 5 and height 4; so the rectangle's area is (5 x 4) = 20, so the triangle is half of that, or 10.
I thought the exact same thing, but apparently, that only works if the angle at the top is 90 degrees, and that cannot be true if the sides are exactly the lengths shown.
@@jerrysherman2743 Yes you can be sure that the angle is 90. In a r angled triangle, c (in this case 6) = the sq. root of the sum of the squares on the other 2 sides. So, (4x4 + 5x5) = 42, so the sq. is 6.48 which is 6 in building terms or in any country where profit is more important than accuracy. As such, the area is 10, half of the rectangle that is 4x5.
@@fedup3449 In math, accuracy is important, regardless of the country. In this case, the square of the triange in the formula is 9.9215674. Plugging that value into the formula to find the height, the height comes to 3.307189.
You need to finish this by also asking what is the height, as this will require you to solve equation with 2 unknowns.
I took your formula also first, than I checked it by this : 25-x^2=16-(6-x)^2 than h=(25-3.75^2)^0.5
Total the three sides divide by 2 , call the result s.
Subtract the respectively sides from S.
Múltiples all 3 value, then find sq rt of product.
Múltiples s by sq rt of earlier product.
Incoherent.
Thanks a lot John! Very useful formula!!! From Sardinia, Italy
Looks like a right angled triangle so (4x5)/2 = 10. Can't be given Pythogoras (as hypotenuse would be 6.4 not 6) Your exact measure of 9.9 ain't too far away from 10.
very good. Thanks
Nice video how to make math complicated.
Idk about you but this helped me
since there is a 90 degree corner i fig it was half a square. therefoe 4 X 5 = 20 /2 = 10
If this is a right angled triangle, then imagine putting two of the same triangle together, hypotenuse to hypotenuse, now you have a square 4 x 5, giving an area of 20. So just the one Triangle would be half of this area …so the area of the Triangle is 20/2 = 10.
Tri’ Area = (Adj’ x Opp’)/2
"Focus is the key!" - Return to sender!
Thanks
The answer is 15/4•sqrt(7). Converting it to a decimal is only an approximation. That should be made clear.
You need to solve this long hand. I don't know how you got the answer!
There is a formula, square root of s minus a, multiplied by sminus b, multiplied by s minus c, where s is the sum of a plus b plus c.
i need my popcorn listen your long explanation by solving this equation.
Thanks for sharing!
Outstanding video
I am new subscriber to your channel
Thank you for the free Videos
How about applying cosine law in trigonometry?
amazing
how could he have come up with the formula
You need to find the height first. Once you've found the height, use this formula
area = (base) (height)/2
1 mint. is sufficient for this question to solve but 15 mints. amazed***!
Finding the area of a triangle is easy.
Double it. Put 2 triangles side X side
makes a rectangle. Calc the area of
the rectangle & divide by 2.
DONE.
!
Are you sure about that?
That only works for right triangles.
Thank you so much
Thank you so much. I might not be in your class but for some reason this helped me a lot so thank you
A simple problem described in a complicated way
great and refreshing math lesson
Took me back to school .Very clear,thank you Sir.Vetri South Africa 🙏🇿🇦🙏
The heron's formula is a must for students of math and geometry.
its even better when you derive it ...
Boy, oh boy, that Heron was one smart bird!
Thanks!
I guess it's cheating but, I opened AutoCAD, drew a line 6" long. Draw a 4" radius circle with center at the left end of 6" line. Draw a 5" circle with center at the right end of 6" line. Draw lines from each end of 6" line to the intersection of the circles. Turn the lines into polylines and join them. List the triangle and read 9.92 sq inches from the list.
I am special woman with special needs. Graduate from high school through special program designed to help people like me. I am still struggling with basic math like additions , subtraction....and so on. Do you have any program that can help me master these basic math
Good job. What software are you using to display the solution?
Thank you.
paint
I was taught the proof of the Heron's formula in Hong Kong at Grade 9 in average Math class.
axa + bxb = cxc works for 90 degree triangle.
4x4 + 5x5 = 6x6
16 + 25 = 36, true
So, use legs 4 and 5 for base and height.
4x5/2 = 20/2 = 10 square units
16 + 25 = 41, NOT 36.
Thanks.
Went too quickly.
Thinking of a 3-4-5 and didn't do the math.
I'll return to prob
false. 16 + 25 = 41
furthermore, if you want to be able to tell if a triangle is acute, obtuse or right angled with no trigonometry. It is possible to do that.
Yes, is important to take notes. But sometimes the problem start with the difficulty understanding notes.
Yes
But we can also use the heros fomular
I can see why pupils in this teachers class would be bored out of their brains within a very short time. It was sooo drawn out.
And I wouldn’t pay much attention to a math teacher who doesn’t know how to pronounce Pythagoras.
Sorry, I don't have the patience to wait for the solution and I was hoping to see the development of the formula.
Not knowing Heron's formula, I did this the hard way. I calculated h of the triangle (3.3) and performed the regular A=(1/2)bh.
JHIll
How did you calculate the altitude?
@@quabledistocficklepo3597 Shorthand: ()2 means the contents of the parans is squared.
(x)2 + (h)2 = (4)2 = 16
(h)2 = 16 - (x)2
(6-x)2 + (h)2 = (5)2 = 25
(h)2 = 25 - (6-x)2
(h)2 = 25 - 36 + 12x - (x)2
(h)2 = 12x - 11 - (x)2
16 - (x)2 = 12x - 11 - (x)2
16 = 12x - 11
27 = 12x
x = 27/12
(h)2 = 16 - (27/12)2
= 16 - 729/144
= 2304/144 - 729/144
= 1575/144
h = 39.68/12 = 3.31
There is faster method to solve this problem: a=4, b=6, c=5
AxC/B=h , h= 4x5/6= 10/3
Area = 1/2x base x h= 0.5 x 6 x10/3= 10
Ans: 10
It's a right triangular...a picture worth a thousand words 🤔
I swear i had an interest in knowing how TCM was going to work it out but I did not know when I fell asleep, completely, and in broad daylight.
Did I get the area right?
Please I will like you to coach my daughter that is on grade 7.
What a little beauty!
Half the base multiply by altitude.
You talk too much.
"KISS" statement.
"Keep It Simple Stupid"
Very effective resource for teaching
Students loose interest by different means.
9.9215
how about 4 times 5 =20 divide by 2 =10sq ft
why not just ame a 4 x 5 rectangle and take half of that: 4x5=20/2=10.
I saw the same answer 3 seconds after seeing the diagram..Why the 7734 make it so complicated?
Your information is great just TOO MUCH BABBLE!!!!
Correction
Should be square root of s(s-a) (s-b) (s-c)
I could have said the same thing in 1 minute not 14! How did he come up with that formula. That's what I wanted to find out,
Ans Area = s(s-a)(s-b)(s-c) where s= (a+b+c)/2.
need sq root of that
Why call it as Heron's formula? This is an adoption of the most elegant formula by Maharshi Brahma Gupta for calculating the area of a cyclic quadrilateral.
Interesting ❤
Pythagore and simple algebra give the same answer, Herons or not!
Maybe SAVE all of the personal info for last, for those interested & all.
4+5+6=15 ÷4=3.75×3.75=14.0625
The equivalent formula for a cyclic quadrilateral (i.e. a four sided shape whose vertices lie on an inscribed circle) is
area of ABCD
= √[(s - a)(s - b)(s - c)(s - d)], where
s = (a + b + c + d)/2.
This is Brahaguta's formula.
*Note:* if the length of side d is shrunk down to zero, then the four sided figure becomes a three sided figure, I.e. a triangle and the above formula reduces to area =√[(s - a)(s - b)(s - c)(s - 0)], where
s = (a + b + c + 0)/2,
which is area = √[s(s - a)(s - b)(s - c)],
where s = (a + b + c)/2. I.e. Heron's formula.
Although this reduction to Heron's formula appears to be only valid for a circumscribed triangle (I.e. a triangle whose vertices on lie on a circle), this formula is true for all triangles, as every triangle can be circumscribed by a circle.
As another poster (Dan Meyers) said, it can also be figured out the 'long way'. In the abstract sense this would be by solving a triangle having sides 𝒂 = 4, 𝒃 = 5 and 𝒄 = 6. Drawing a vertical line from the apex of 𝒂 and 𝒃 down to 𝒄, with one side (the 𝒃 side, say) being called 𝒛 for that bit, and the other side being (𝒄 - 𝒛) we can show a few things
№ 1.0 𝒉² = 𝒃² - 𝒛²
№ 1.1.1 𝒉² = 𝒂² - (𝒄 - 𝒛)²
№ 1.1.2 𝒉² = 𝒂² - 𝒄² + 2𝒄𝒛 - 𝒛²
Setting № 1.0 and 1.1.2 together allows us to cancle 𝒛² leaving
№ 1.2 𝒂² - 𝒄² + 2𝒄𝒛 = 𝒃² … rearranging
№ 1.3 𝒛 = (𝒄² + 𝒃² - 𝒂² ) ÷ 2𝒄
№ 1.4 𝒛 = 3.7500;
That's helpful! Now the height 𝒉 can fairly easily be found in turn:
№ 1.0 𝒉² = 𝒃² - 𝒛²
№ 2.0 𝒉 = √( ( 4𝒃²𝒄² + (𝒄² + 𝒃² - 𝒂²)² ) ÷ 4𝒄² );
№ 2.1 𝒉 = 3.3072;
Then the area ½ base • height becomes
№ 3.0 Area = ½𝒄𝒉
№ 3.1 Area = ¼√(4𝒃²𝒄² - (𝒄² + 𝒃² - 𝒂²)² );
Plugging in the (𝒂, 𝒃, 𝒄) = (4, 5, 6) then we get
№ 3.2 Area = 9.922;
And there it is!
Heron's Formula is definitely much easier to remember, for sure.
But long-hand works too.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
A= 1/2 bh
bxh ÷2 = A
Can't i just divide the 6 to 2 and then make the triangle 2 right triangles. Then use Phythagorean Theorem.
Then use the formula 1/2 • b • h?
Will the results be the same?
Erreur, la surface A du triangle abc est égal à la moitié de la surface S du rectangle abca' obtenu en menant une parallèle à ab passant par c et une parallèle à bc passant pa a ces 2 parallèles se coupent en a' et forme le rectangle abca' dont la surface S est abxbc= 20
Ac étant la diagonale de ce rectangle la suface A du triangle abc est égale à la moitié de la surface S du rectangle abca. Donc A= S/2 = 20/2 = 10
When I look closely, I see a rectangular triangle. Why? 4^2+5^2=6^2 (Phytagoras). The area of this recangular triangle is therefore (4*5)/2=10.
Wrong. It's not a right triangle.
@@kevinbranshaw7294 Er, RIGHT! a 4,5,6 triangle IS a right angle triangle.
@@kevinbranshaw7294 You are right. I made a miscalculation. 16+25=41 not 36. Blame on me 🙂
@@jonathanfrayne9501 3, 4, 5 is a right angles triangle, not yours.