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.

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

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.

Just use cosine rule to get the an angle, then use ½abSin(C)

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.

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).

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

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

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).

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._

Area of shapes involving rectangle, square and triangles

You can use also cosine law by finding unknown height, then apply the formula of finding the ares of triangle..tnx

Thanks for sharing!

great and refreshing math lesson

Thank you so much

Thanks a lot John! Very useful formula!!! From Sardinia, Italy

Thank you so much. I might not be in your class but for some reason this helped me a lot so thank you

very good. Thanks

A bullet-point presentation always works best for my super-charged neurons.

Good and clear presentation !

Go right to the point

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.

Thanks!

I just did that in my head got to 9 in less than a minute.

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.

How about applying cosine law in trigonometry?

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

I fell asleep with the intro alone,.

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.

Good job. What software are you using to display the solution? Thank you.

Thank you for the free Videos

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

Thanks

Great!

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.

Since it is a right triangle, simply multiply te length times the width and divide by 2.

Took me back to school .Very clear,thank you Sir.Vetri South Africa 🙏🇿🇦🙏

You need to finish this by also asking what is the height, as this will require you to solve equation with 2 unknowns.

You need to solve this long hand. I don't know how you got the answer!

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!

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.

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!!! ;-)

i need my popcorn listen your long explanation by solving this equation.

Outstanding video I am new subscriber to your channel

A simple problem described in a complicated way

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.

But where is the demonstration of the heron's formula

"Focus is the key!" - Return to sender!

The heron's formula is a must for students of math and geometry.

Boy, oh boy, that Heron was one smart bird!

Interesting ❤

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.

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***!

Nice video how to make math complicated.

The answer is 15/4•sqrt(7). Converting it to a decimal is only an approximation. That should be made clear.

After over 5 minutes of rambling, he finally gets to the problem.

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.

Yes, is important to take notes. But sometimes the problem start with the difficulty understanding notes.

the area of a 5x 4 rectangle would be 20 so the triangle is half of that therefore 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

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 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.

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.

Awesome teaching,👍

amazing how could he have come up with the formula

What a little beauty!

I was taught the proof of the Heron's formula in Hong Kong at Grade 9 in average Math class.

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

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.

Great lesson

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?

since there is a 90 degree corner i fig it was half a square. therefoe 4 X 5 = 20 /2 = 10

Thanks you're and awsome teacher I loved it

Quarter time of your lecture goes in introduction!!!

Good lesson. Thanks.

how about 4 times 5 =20 divide by 2 =10sq ft

Correction Should be square root of s(s-a) (s-b) (s-c)

It's a right triangular...a picture worth a thousand words 🤔

Half the base multiply by altitude.

Back up from area to calculate height... for fun...

Sorry, I don't have the patience to wait for the solution and I was hoping to see the development of the formula.

Did I get the area right?

But we can also use the heros fomular

Ans Area = s(s-a)(s-b)(s-c) where s= (a+b+c)/2.

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

U.K. That was really a fun way to solve the problem. Some comments were highly ridiculous but please overlook them. Those comments are from those who will usually complain about their Maths Tutors as boring. They are those who think they are smarter than Archimedes. Maths should be fun and you made it look so. I have watched a few of your clips and I am yet to give a negative comment. You are a 5 🌟Tutor. I am considering acquiring your Algebra and Geometry Notes to help my boys at home. Looking forward to your Calculus Notes. Cheers

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,

Pythagore and simple algebra give the same answer, Herons or not!

Maybe SAVE all of the personal info for last, for those interested & all.

I have a question

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.

4+5+6=15 ÷4=3.75×3.75=14.0625

A= 1/2 bh bxh ÷2 = A

Yes he is so talkative on self centeredness not needef

ooops. thought it was right angle at first..

Doubling the triangle into a rectangle = 20 cm /2 = 10

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

Hahaha somebody fall asleep but not me i enjoy it. Tks