AWESOME Area Formula For TRIANGLES!
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- เผยแพร่เมื่อ 9 เม.ย. 2024
- How to find the area of a triangle with Herons Formula. Learn more math at TCMathAcademy.com/.
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9 minutes of this video should’ve been enough not just for talking about the formula, but for deriving it as well.
You can get there using the cosine rule to find one angle then use area=1/2absinC.
I am probably not you regular viewer. I run puzzle evenings at a local bar and your videos give me inspiration to create mathematical puzzles.
I have what I call a beautiful solution to this problem. I divide the triangle into two right angle triangles. The bases are X and (6 - X). Calculate the height of both triangles in terms of X. These are obviously equal. Hence you can get X. ‘25- X^2 = 36 - (6 - X)^2. The X^2 drop out. Then it becomes simple substitutions.
Thanks for the inspiration!
If you try it with a Pythagorean triple, the numbers from Heron's formula and the traditional A=.5bh are nice neat numbers.
58yo and math honours in HS, but I don't think I was ever introduced to Heron's formula. Cool! Love your channel.
great lesson
The area of an equilateral triangle is x^2 divided by 4 and multiplied by v3 where x is equal to a side of the triangle. An interesting question is whether the ancient Greeks who proved Heron's formula ever used it to determine that formula for the area of an equilateral triangle.
s = (a + b + c) / 2 which is half the perimeter.
In the case of an equilateral triangle, all sides are equal, so a = b = c = x.
Substituting these values into Heron's formula, we get:
A = v(s(s - x)(s - x)(s - x))
To find the semi-perimeter for an equilateral triangle, we have:
s = (x + x + x) / 2
s = 3x / 2
Substituting this value into the formula, we have:
A = v((3x/2)((3x/2) - x)((3x/2) - x)((3x/2) - x))
A = v((3x/2)(x/2)(x/2)(x/2))
A = v((3x^4) / 16)
A = (x^2v3) / 4
My guess is they did know that simplified formula because of the proof I just provided or maybe they initially stumbling into the simplified formula after noticing that each time they used Heron's formula to find the area of an equilateral triangle, the answer always equaled a side squared, divided by 4 and multiplied by the square root of 3.
interesting way to do it, I don't remember if I was ever taught this formula, I certainly don't remember it after getting on for 60 years since I would have been doing geometry at school. Had I tried to do it on my own I would have worked out the angles, then I would have used the sine rule to work out the vertical and multiplied that by the length of the base and divided by two. The angles are easy to work out as ratios of 180degrees since you know the length of all three sides, but your method is probably quicker than mine would have been.
Take the two shortest sides 4x5/2 = 10 which is close enough to 9.9 for the ladies I go out with. The other longer sides would be the hypotenuse of the triangles.
10 is not close enough.
@@lant7123 Before calculators (B.C.), in the mid 1970s, we used to slide rules. The accuracy was at best to 3 digits. There were many incredible things designed using those tolerances. I designed then taught electronic circuits for 40 years and one of the best and worst tools used was a calculator. The calculator had you believe that the correct answer would be 9.9.. with umpteen digits in order to be correct. Even in today's world, every electronic component has a tolerance (1%, 5% or even higher). All the tolerances of the individual components add up. When measuring a circuit, if the measurement was 10.1V or 9.7V, that was close enough and the circuit still worked as it should. So 10 is close enough. Just a real life example.
Repetition makes perfection.
If
p = a^2 b^2 + a^2 c^2 + b^2 c^2 and
q = a^4 + b^4 + c^4, then
area A = 1/4 root(2p - q)
Теоремата на Херон.
AWESOME Area Formula For TRIANGLES:
Lengths of three sides of the triangle are 6, 5, 4; Area of this triangle = ?
Starting from bottom left to right of ∆ABC:
AB = a = 6, BC = b = 5, CA = c = 4,
Draw a line CD from the top of ∆ABC vertically down to line AB at point D.
Forming two straight triangles (90 degree), ∆ACD (left) and ∆BCD (right):
Let: Bottom of ∆ACD, AD = x; Common altitude of both triangles, CD = h
h² = c² - x² = b² - (a - x)², 4² - x² = 5² - (36 - 12x + x²); x = 27/12 = 9/4
h² = c² - x² = (c - x)(c + x) = (4 - 9/4)(4 + 9/4) = [7(25)]/16; h = (5√7)/4
Area of ∆ABC = (1/2)(6h) = 3h = 3[(5√7)/4] = (15√7)/4 = 9.922 square
While estimating the amount of sealant to buy, I used Heron's Law several years ago to calculate the area of a driveway using the dimensions of the stress petitions that were all parallelograms.
Make it a parallelogram then divide by two.
Yes, we all know, the area of the triangle is ... Base x Height / 2. But the height is not a given value
9.9 is an approximation. The true area is 3¾√7.
Final answer d) 9.9
I choose answer b) 15V7 / 4
Triangle is half of a square , so area of square height times width divided by 2
...half of a rectangle... and some very special rectangles are squares but most of them are just rectangles.
No, this is only true if the triangle is a right triangle where the side lengths make up a Pythagorean Triplet where a, b < c, and a^2 + b^2 = c^2 which is false for 4, 5, 6. This triangle is "half" of a paralelogram, but that's trivially true of every single triangle.
It was not a right triangle, so it was not half of a square or even half rectangle.
We're going to get 9.9… you should say "approximately" which is neither stated verbally nor written out, so your answer is wrong. That is only a decent approximation to the actual answer.
He should have said approximately 9.9
Correct, in this case the ~ sign would be appropriate because the answer is an irrational number.
x = (a^2 - b^2 + c^2)/2c
Area = (c*sqr(a^2 - x^2))/2
Replace a, b, c by their value
Area = 9.9216 exact !
Almost exact. Using fractions, you get the quite exact 15 * sqrt(7) / 4, about 9.921567
If you need to find a Right Angle out in the field, the numbers 4, 5, and 6 will give you that right Angle, A= 1/2(bh) will work.
No, you do not get a right angle with 4, 5, 6. These are not a Pythagorean Triplet. 4^2 + 5^2 = 16 + 25 = 41 which is not 6^2 = 36.
thanks for formula, but this vid could have been 1 min long
Thanks for making a 1-2 minute explination into only 9 mins instead of 10-20 hours like you usually do
That’s rude.
extend the triangle into a rectangle with sides 4 and 5. multiply 4x5 to get area of rectangle = 20. half of rectangle, the triangle, is 10..
But the area isn't 10. Good estimate though.
And extending this triangle won't give you a rectangle.
Why 4 x 5 and not 4x6?
You can't eliminate the base of 6. You could have a parallelogram with upper and lower bases of 6 and sides of either 4 or 5, but you still wouldn't know the height.
here:
b (base) = 6
h (height) = angle between 4-5 down perpendicular to line 6
using pythagoreans theoem
4^2= h^2 + b.1^2 eq.1
5^2= h^2 + (6-b.1)^2 eq.2
16=h^2+b.1^2 eq.1
25=h^2+36+b.1^2-12b.1 eq.2
subtract eq.1 from eq.2
25-16=36-12b.1
25-16-36=-12b.1
(25-52)/(-12)=b.1
(-27)/(-12)=b.1
b.1= 2.25
b.2=6-2.25
b.2=3.75
h^2= 4^2-2.25^2 eq.3
h^2= 5^2-(6-2.25)^2 eq.4
??
eq.3
h^2=16-5.0626
=10.9375
h =3.3072
eq.4
h^2=25-(6-2.25)^2
=25-14.0625
=10.9375
h =3.3072✔️
Area: = (1/2)bh
= (1/2)(6)(3.3072)
= 9.9216 units-sqd
there's a finding here that's begging to be found.
first, a crude triangle drawing:
A
b c
C m a B
where
ABC are angles.
sides abc are opposite angles ABC.
The "m" marks where a perpendicular ("p") to side a from corner A intersects side "a."
with these is mind let's find the formulas for "p."
b^2= p^2 +(Cm)^2 eq.1
c^2= p'2 +(Bm)^2 eq.2
note:
Cm + Bm = a
(Cm+Bm)^2 = a^2
Cm^2 + Bm^2 + 2CmBm = a^2
correct, but the problem is, no one said 4 and 5 are perpendicular though
@@foreigneralatroush8453 didn't assume 4 &.5 were perpendicular. What is perpendicular is a NEW line down from TOP corner between 4 & 5 that intersects bottom line at 90°.
Here are the "pretty sophisticated mathematics"...
Two little right angled triangles inside the given 4-6-5 triangle with base is 6 = x + (6-x) and height h:
x² + h² = 4² and ( 6 - x )² + h² = 5² (simple Pythagorian Rule)
-> ( 6² - 12x + x² ) + h² = 25 or 36 - 12x + ( x² + h² ) = 25 then substitute x² + h² = 16
-> 36 - 12x + ( 16 ) = 25 -> - 12x = 25 - 36 - 16 = -27 -> x = 27/12 = 9/4 units
x² + h² = 16 -> h² = 16 - x² with x = 9/4 -> h² = 16 - 81/16 = ( 256 - 81 ) / 16 = 175/16
-> h = V175/V16 = 5V7 / 4 units ( ~ 3.3 )
Now the area is A = ½ . b . h = ½ . 6 . h = 3 . h = 3 . 5V7 / 4 = 15V7 / 4 units² ( ~ 9.9 )
If this is sophisticated, what about second order differential equations???
@@panlomito CLEARLY... we ALL enjoy these math challenges
@@tomtke7351 It's a daily treat !
Thanks for the video but at the end You didn’t explain good enough
Strange thing is many of these formulas including calculus date back even before Christ & have long since been lost never to be recovered.
How would know that a formula was created, then lost and *never* recovered?
I’m pretty sure 2a x 2b x 2c = Area and I got 960.
Never mind
You are permitted to delete responses if you change your mind 🙃
@@WideCuriosity Yeah I know, I was meaning for John, the “Tablet Math Class” Creator to see so he wouldn't have unnecessarily been driven to maybe teach on what I messed up. That's all I did that for.
But thanks for directing me on what I might have not known.
3:32 DEAR GOD IN HEAVEN! Quit trying to confirm how intelligent you are & get to the FKN EXPLANATION!
We understand where you are going. This should have taken 3 minutes.
Too much repetitive talking.
Don't talk to much just do the actions 🙄🙏
Get to the point!
How boring….too much waffle !!!
You can never have too much waffle🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇🧇