Finding the Area of a Non-Right Triangle

[anestismoutafidis]

Thank you for showing the Heron's method. At first, I calculated the area of the triangle by calculating the height of it (a•sinβ) and then by the modificated formula for calculating the area of the dragon figure ( A= 1/2 •e•f)÷2 <=> 9, 8m^2. It took a relativ long time, of course. I am grateful to know now about the Heron's formula.

[AG-idiv]

In India, we're taught this concept usually in Year 9. Though its been a good 3 years since I studied it, it was good to find a refresher after all this time.
Thanks Professor!

[emmanuelmwesigye]

You are a great teacher. I have learnt from you.

[mariosantos]

The important thing is the logic behind the formula/procedure. I just don’t understand why math teachers kept on demonstrating the solution to a problem without explaining the logic behind. I suspect they don’t understand too.

[anthonylawrence]

That is absolutely brilliant, I don’t know why it never occurred to me to perform the operation in this manner.

[xdragonk]

I just noticed that for a three-sided polygon to be called a triangle, the sum of it shortest sides need to be more than the longest side. It can never be smaller than. And if they are equal, it's just a straight line.

[mikaeloverfjord]

Deriving the formula yourself you will see how elegant this formula is.
(You will have to know the Area Formula A=ab*sin(C)/2 and the trigonometric identity sin(x)^2 + cos(x)^2 = 1 and the Law of Cosines cos(C)=(c^2-a^2-b^2)/(-2ab) )

[cmk]

Literally used this two days ago in an exercise in numerical mathematics! The task was to program the triangulation of a circle to compute its Area

[aarav]

When I was in 5th grade I thought that every triangle formula was bh/2. What I was in 9th grade I learnt all the formulas of a triangle

[mittelwelle__khz]

As an aside, there is an interesting aspect of any 'area-like' property of a geometric shape you calculate based on any "distance-like" properties. (Three in case of a triangle, unless you constrain the triangle by one or two angles, when in the latter case the only remaining distance-like property must occur squared.)

By "area-like' I not only mean the triangle's area, but also things like the area of the largest inscribed circle or the smallest circle touching the three corners, and "distance-like" not only are the three sides but may be also two sides and the distance of a corner to the triangle's COG.

Due to the rule that areas in affine projections scale quadratic to linear scaling of distances you can apply "dimensional analysis" (as best known from physics[*]) to give a formula you derived a quick first sanity check.

As I didn't know about Heron's formula I was two or three seconds a bit worried seeing the area calculated as square root taken from the product of THREE distance properties. BUT then it became clear that the "scalar-like" looking multiplier (8) also was derived from a distance-like property, so everything looks fine.

*: it should be understood that "dimensionally analysis", if it comes to a WRONG result, can only show a formula is DEFINITELY wrong, but if it comes out right the formula can still be wrong.

Btw: another quick "sanity check" of that formula would be if you replace the largest side according to the Pythagoras formula and see if the result comes out as ½(a•b) then.

[dilaudid]

Wikipedia explains it's derivation. You can figure it out using the Pythagorean theorem making this triangle into two right triangles, with the shared side = x, and the 7 meter sides = y and 7-y, etc etc. Its an elegant formula from thousands of years ago.

[gregnixon]

Hearing this is like finding a forgotten photo in a drawer.

[karcha]

Thanks Brahmagupta. He was the one who discovered "Heron's Formula"

[loridave]

Wow. I need to think this thru to see if it should be intuitive. Way cool.

[hassankazimaziz]

Hi sir, what kind of projector that you are using? Thanks for these incredible videos

[mohammadchaudhry]

Itis the shortest way of solution and really very smart way of solution. Of course my mathematical knowledge is very very old since 1969, but if we calculate the attitude of this triangle we can calculate the area of the triangle.
Altitude multiplied by base of the triangle ( here it is 7 m) divided by 2 .
But the gentleman man has suggested is the shortest method of solution.

[farhanfouadacca]

Never heard of Heron’s formula before. Thanks

[kiburamaha]

Thank you sir! 🙏 One question? Is there a derivation or a proof for the heron's formula?

[amousawie]

The square root for instance.
Most people think of it as if it a house with a rooftop, and the idea of seperating the square root to a surds form is nearly impossible. Because, how would you break the roof top of your house.
This is why you need two to three seminars to every math class and the class room must not feel like a court room, in that a mistake would result in prison term.

[DamplyDoo]

I actually was the one who told this to Heron. Bro took credit