Why is πr² the formula for a circle's area? 

Fun fact, the letter π was chosen to describe the ratio of circumference/diameter because of the word 'περιφέρεια' (~periphery) that is the 'περίμετρος' (perimeter) of a circle.

Just for fun, I pulled out my old Calculus book from college and integrated using the equation for a circle (after solving for y to get y= sqrt(r^2-x^2). I simply integrated from 0-1 (quadrant 1 of a circle with radius r with its center at (0,0)) and multiplied by 4. Sure enough...Area still equals (pi)r^2 !

I value much more this type of content and historical background investigation than solving nearly-impossible questions/puzzles. Amazing video!

I love how excited you get, Presh! Keep it up! 👏👏👏

Thank you!! I've just done it!! Upon learning of William Jones 1706 work you referenced, I went to Wikipedia and found out that this work (with the first use of the Greek letter pi for representing the ratio between a circle's circumference and its diameter) is titled "Synopsis Palmariorum Matheseos." Then within mere minutes, I found myself buying a copy of it on Amazon!! Thank you so much, Presh, for enlightening me!!!!

I like this format as an addition to the usual problems.

Keep on posting these videos 🙏

Math history is amazing. This was excellent! Thank you.

Here is a simple method: Picture the concentric rings from the third approach. The innermost ring has a circumference of zero (it's basically a point). The outer ring has a circumference of 2πr. Now add up all the circumferences by integrating 2πr from 0 to r. It's a very simple integral that results in πr^2

Regarding the last example, I was thinking that you can integrate the different circumferences for the radius going from 0 to r, which is \int_{0}^r 2xpi dx= r^2pi.

Wow, thank you so much for enlightening on very fundamentals of formula for a Circle's area.

This is just amazing. Getting to know this now after all these years

Loved this, thanks!

If you want the area of a non-circular ellipse, then you can use the formula πab, where a and b are the major and minor axis. For a circle, a and b would be the same.

Great summary video!

Nice proofs. Very fascinating.

I wondered about this very thing years ago. With the relatively modern benefit of calculus you can prove the relation easily enough, but fascinating to see the ancient arguments which somehow I missed out on. Thanks!!

The animations in this video were very good, Presh!

archimede's method was so good. felt like a plot twist. thats why i love maths

Great video, Presh. Love your videos, but this one is one of my favorites ... really well explained, thank you!

The last one resembles the shell method of integration (concentric circles). The formula for the area of a circle is the integral of the formula of the circle's circumference.

I really like methods 2 and 3. This was a great video!

Great video. Thanks.

The answer to this question is quite easy, but a little in involved. You are give a straight line on a flat plane of a length, doesn’t matter how long but the whatever the length is you subdivided the line into 2 equal parts, each part represents a unit scale (for simplicities sake). We establish the center as the origin and one reference point at one end of the segment. The length of the line is 2, but we should imagine an esoteric line traveling back to the reference point so in actuality the line is length 4. What is the enclosed area. To have an area you need some sort of enclosure in s second dimension, but the two lines superposition so there is zero. Segments =2 Cum. Len = 4 Area = 0 So at the origin we bisect segment and stretch it so that now we have a square joining points equal distance on two orthogonal axis. The 4 points form 4 chords. We pretend we don’t know the length or any thing about sines and cosines. So if the line going from the reference point and back is a flat loop and we stretch it out the center we are adding length into to each half a segment. 2/2 = 1. The length the line is traveling towards is 1 unit from the circle and since its point was at the origin the distance traversed of the new point is 1. Thus new point it how far from old point, let’s just do one. The reference point is 1 unit in original dimension say a, in the new dimension created the new point is b So length is SQRT((a(new) - a(ref))^2 + (b(new) - b(ref))^2) = SQRT ((0-1)^2 + (1-0)^2) = SQRT(1+1) = SQRT (2). So the tilted square is composed of 4 chords, each chords on an imaginary ⭕️. What is the area under the chords? To arrive at the area we need to project a new radii that bisects each chord. At the intersect of each chord we have a bisector that junctions with 2 equal half chords in this case it’s the SQRT(2)/2=SQRT(1/2). This bisector length is SQRT(1- SQRT(1/2)^2) = SQRT(1/2). Thus treating the chord as the base the area = halfchord * bisector which is SQRT(1/2)^2 = 1/2, the combined area under the chords is 4*1/2 = 2 # c p A/c A 2 2.0 4 0 0 4 1.4 5.7 0.5 2 The answer is right here we just cannot see it yet. I will point it out where I hid it. When I said we will use the chord as the base this means we used the bisector as the height, so I did a change of basis. But the I calculated based on the halfchord and bisector, which is correct, but as we will see that in calculating the area we’re removing piecemeal half the perimeter. I will do one more. So we are going to create 8 new chords. Again we start at our reference point (1,0). So we have defined the length of the halfchord or bisector as SQRT(1/2). The new chord is SQRT((bisector-1)^2 + (halfchord-0)^2)^2. The segment outside of the bisector is (1- SQRT(1/2))^2 = 1.5 - SQRT(2) the square of the halfchord is SQRT(1/2)^2 Chord “⭕️/8” = SQRT((1.5 - SQRT(2))+(1/2))=0.7653, its bisector = 0.923 and the area/ea = 0.3535 and total area is 2.828 If we repeat this process about 23 more times eventually we get to a point where the bisector is indistinguishable from 1 on most devices and does not change, each new chord is equal to the previous halfchord. But unchanging is the per chord calculation of area, area is 1/2 base x height. So the question is why this is the case. The ⭕️ is an abstraction created by humans, it’s actually points on a line connected by chords. A rectangle is another abstraction created by humans. We defined this as 4 rectilinear line segments and using this system we define simple area. But the area swept by each line segment,chord on circle, is not rectilinear. When we bisect the chord we generate 2 right triangles, when the diagonal sides are abutted to each other they are rectilinear and we have a calculable area, but in doing that, in 2-D space the displacement of the chord in simple area with one dimension now superpose half upon itself. So there is one detail here we need to make this work. In the beginning of the problem I set the parameter of the argument. Whatever the length is I am assigning a new length of 2 new units, I then mystically stretched the line by 2 units so that I could fill in an empty area. These are artificial processes, they are not real, they begin in the imagination. In doing this I forced the radius to be 1, in the third step I forced out another dimension. So let’s argue the original line length is 200, I set its length to 2, I stretched its length to 400 (4) so it would return to its reference point, then I processvely stretched it in another dimension to 628.32. In essence I forced most of the line into 2 dimensional space by creating a curvature. The force is 100 outward in every direction. And so to make this work in need to multiply the area I made by the area scale factor which is r^2. Where is this, let’s look at the first second dimensional stretch. I forced out in the second dimension 100 units. I then defined a line going from the second dimension back to the first. I then defined the area as area between the line and the origin as defined by to radii. So now I have an area centered around a direction vector (45°), so that area is 0 units/height on one end and 141.42 units per height one the other end of each radii. But the radii point along different vectors, the bisector splits the difference. In doing this we create a new basis z for each radii which displaces along z a fraction of the displacement along x (for the reference point). As a consequence that displacement is the cosine of the angle which centers the bisector between to radii that with the chord form the area. The halfchord just happens to be the sine of half the angle of the chord. Thus in this case area along each bisector is (r * sine = halfchord)(r*cosine = bisector) = r^2 sin*cos where angle is 1/2 the chord. As perimeter goes to 2pi, cosine goes to r and area goes to the r^2 * sine of 1/2 the chords angle. In this case To der

BUT MY DOUBT IS STILL NOT CLEARED . I always had the doubt about the discovery of pi . Like you have told , old mathematicians found out that every circles circumference to diameter ratio is a constant , which was later given the notation pi . But we tell pi is an irrational number , and from what i know an irrational number cannot be a ratio . If the old guys were trying to find out a ratio then how did we end up in an irrational number?

Hey, I'm a brazilian fan of your channel, and I loved it last year when you solve our Pinocheo math problem from OBMEP. This year we again had an interesting problem about some flowers which, if you want to, I can translate to you.

This was amazing thanks

What a great channel.

idk i just imagine having 3 squares of side r, and then there's a little bit left to fill once you like cut the corners

Loved it ❤ my fav proof is the 2nd one

Amazing proofs

The symbol pi, has been around THOUSANDS of years. It just was used for the phoneme p.

What do you use to make your animations? They're superb!

I love this channel

From first principles, we define A = 2πr. Now consider the infinitesimal increase in area δA arising from an infinitesimal increase in radius δr. That is very close to 2πr.δr. So δA ≈ 2πrδr which leads to δA/δr ≈ 2πr. In the limit as δr tends to 0, the approximations becomes an equality and we get dA/dr = 2πr. That can be solved for A by taking the integral from 0 to r and we immediately find A = πr^2. It's not particularly rigorous, but once you can see that the rate of increase of the area of a circle wrt its radius is 2πr, it should be obvious what the area is.

That first proof (using the regular n-gon) becomes even nicer if you use τ (tau) as the circle constant (ratio of radius to circumference ≒ 6.28) rather than π. C = τr A = ½Cr = ½τrr = ½τr² Which maintains the ½pa form from the n-gons.

During our school days , we did read about derivation of area of a circle . It was there in famous Geometry book by HALL and STEVENS . Both 1st and 2nd methods were mentioned .

presh checkout the 2024 JEE advanced paper, all the questions are good and u can get to the solution in one step

The third example is almost the radial integral: Let A(x) be the area of a circle of radius x. Then A(x+h) - A(x) is the area of a radial slice of the circle, and it must be limited by h*2pi*x and h*2pi*(x+h), when you try to convert the slice into a rectangle. By the ordinary "sandwich"-method, you get A'(x)=2pi*x, and thus A(x)=pi*x^2, which ends the proof 😊. Note that the derivative of the area is the circumference, which makes perfectly sense if you think about it a few seconds 😎.

Most Excellent!

You can also do it by integrating the circumference by radius

Find the function F(n) = the area of a regular polygon of n sides inscribed in a circle of radius r. Then take the limit of F(n) as n tends to infinity. I did it as an undergraduate taking calculus 2. I found it convincing.

They used pi to represent the value because there was no Greek letter for cake, which as we all know is the superior circular dessert.

It seems like the common insight to all these proofs is: 1) The circular curve can be approximated by a sequence of straight lines placed regularly in some pattern close to the curve (ie a regular piecewise linear approximation). 2) The more lines there are, the more closely the sequence approximates the curve with less distortion to the area. In my opinion, calculating π must have been much more difficult than finding the formula for the area given pi. If I recall correctly, I heard somewhere that Archimedes approximated π by finding the perimeter of an inscribed (inside the circle) regular polygon as the video showed, but also a circumscribed (outside the circle) regular polygon. Then he could measure or calculate the perimeter of the outer (circumscribed) and inner (inscribed) regular polygons and use this to figure out some sort of upper and lower bound on π (I might be citing this story incorrectly).

Wow!..really nice

Pie are not square. Pie are round.

Love this history of math stuff. I know middle school and high school students don't really care about where math comes from (I didn't), but I do think it's vital to teach it before college. You never know whose interest could pique.

While target shooting at 300 yards I found something strange. 1 MOA (1/60 of a degree) at 300 yards is pi, the cord dimension of 1MOA included angle in inches. I checked this on my calculator it rounds to 10 digits so I used EXCEL. (300 yds) 10800" x 1/120° sin x 2 = π and it is correct to 98 digits.

Years and years of math education and I don't recall ever being taught any of these derivations. I was just told how to calculate it.

Could you explain similarly how/why volume formulas work? I know that's kind of broad, but this was so well done I think you could explain it simply

I still Remember i studied about the pie chapter at the Finals of my grade 7 it was fun i was statiesd that i could solve some questions Perimeter=πr^2 and the second Area= 2πr those are the 2 formulas i like it very much.

Draw a infinite number of radii by rotating a single radius around the centre of the circle an infinite small angle theta and then add them all up. Then the area is the double integral from 0 to 2pi of the integral from 0 to r of r dr d-theta.

You might want to re-read Archimedes' "Measurement of a Circle".

nice music as the circle was unwrapped in the 3rd proof. Interesting how each proof used limits to get from a concrete figure towards infinity and a circle.

These are genius, they're so intuitive. I did it myself by summing the areas of n isosceles triangles with apex angle (2π/n) and equal side lengths r, using the fact that sin α = α for small α, and tending n to infinity; but that's hugely complicated compared to any of the wonderful proofs here.

That's why the dollars is going to drop when people ready to tell the true ( actually the diference of value of π is inversely proportional to the difference of radii of the circles. Meaning larger the radius smaller π

Thank u

One thing if you try the second method on a sphere to get the surface area, you will get the wrong answer.... 3blue1brown did a video on why it breaks for a sphere.

I think the question is how to establish a way to prove that this constant ratio of circumference to diameter that we call pi is equal to 3.14...etc . After that areas and volumes follow.

wow... very interesting... haven't thought about how the formula was derived thus far...

I remember at one time thinking π = 22/7 exactly. I missed the point being that it's just a very good approximation that works for a lot of everyday calculations.

That was cool. Those old guys were pretty smart.

Never was a fan of Method #2. Method #1 would be great to teach in schools, while method #3 (via numberphile and 1minutephysics) is my personal favorite

Maravilloso

Stop the vid at 8:44 you will see that Rabi Abraham bar Hiyya's proof resembles a Menorah. What an intriguing coincidence. 🙂

How about the other simple calculation for 3rd method? The area contain circles perimeter with r=0 to r=R so integrain of (2pi*r) and r from (0 to R) will be (2* pi * R^2 / 2) or simply (pi * R^2)

Perhaps you could also expand on how these ancient civilizations came up with the approximation of 3.14###. before the decimal system we use today came into being. I understand that they must have had some way of doing this but nothing I've found gives a satisfactory answer.

At school we were just told the πr² formula. One afternoon I pondered upon the derivation of this formula and came up with the pizza and the onion methods. This shows that it does not take a genius to figure it out. One evening when learning about relativity I wondered how much time would slow down depending upon your velocity, so I drew myself a simple spacetime diagram and came up with an equation, which a few years later found in a physics book to be the Lorentz transformation. My point is that things are often far more simple than you might imagine. To begin with a problem might appear somewhat daunting, yet on reflection having solved it you see how simple it is.

Is there any deeper meaning in the idea that the derivative of the area, with respect to the radius is the circumference?

The last section says at the bottom "A Mechanical Derivation of the Area of the Sphere". Should it be "Circle"?

Hhhh You are the one who impressed me today .... Thanks again

Haven't watched but going with, because that is the definition of pi. Kind of like asking why is the speed of light C and not something else.

The ancients really had the ability of reasoning and mental extrapolation !

I was thinking about cutting the circle to its center and turn it into a strip of paper where the breadth is r and its length is pi r

Ancient People having better imagination than us...

Method4:- Consider a thin ring inside a circle, just like method3, with radius gradient dX and circumference 2piX. Integrate 2piX with respect to X and we get the area.

@1:20 Actually, Liu Hui (劉徽) found the fraction approximation 3927/1250, not the decimal representation.

But how do you prove C/d for any circle is a constant (pi)?

Okay. But how did they even calculate π?

How come i can't find a similar demonstration for surface area or volume of sphere?

Why is there ambiguity going on recently ? Sum of angles inside a circle = 360 OR ♾️ , 0.99999 = 1 or ≠ 1 ( by law of scientific significant figures ) ,

Its how much the radius line rotates thus it equals radius×circumfrence but since the raduis line has two ends so its rotation is half .therefore its =2pieR×R/2 =2pie R^2. Thats how i thought it

Thanks

It’s amazing how much school doesn’t teach you. From the area of a circle to the quadratic formula, they teach a lot of hows, but not a lot of whys. It makes sense for brevity’s sake, but it’s almost like they’re trying to make you hate math by making it a memory game rather than a problem solving game.

A circle, centered around the orgin with radius R can be defined by the equation: x^2+y^2=r^2. But, when does the limit as N -> inf. sided polygon inscribed inside a circle, centered around the Orgin approach the circle equation? Also, if it approaches the shape of a circle, then how come polygons corner points are connected with line segments, but 2 points on a circle's circumference aren't connected with a line segment, being shaped more like a curve between 2 points? Moreover, can it be proved that this area between the polygon and the curve of the circle converges, and approaches ZERO as the number of points on the infinite sided polygon approaches infinity?

8:42 - Ironic that in the Rabi Abraham's solution, it appears as a many-candled Menorah while unwrapping!

Interesting that the ancients used arguments that took a limit to infinity, when the formalism to support the concepts of limits or of infinity wasn’t established.

I had never seen the third method before, it's pretty nice (but was probably a pain to formulate and enunciate clearly back then).

Can anyone tell me when and where the rabbi behind method #3 lived?

Why jacobian at the last method?

Please make a video on Area of sphere = 4 Pi r2

Alternately you can simply roll a wheel of diameter D, x number of revolutions, and distance traveled will be D÷2πr= revolutions. Interestingly enough. If r=h+t hub radius plus tread thickness. And h=t=2"/π What you end up with is a version of the viral incorrect video. 8"÷2(2"+2")=1 revolution. Also, you find that D÷2πr≠D/2πr Illustrating the difference between ÷ and /.

More information saying that traditional Pi = 3.141592653589793 is false part 2 - kloka: Pi is also defined as the ratio of the area of a circle divided by the area of the square that is located on the radius of the circle. If a circle is created with a diameter that is the same measure as the longer edge length of a Square root of the golden ratio √φ = 1.272019649514069 rectangle then one-quarter of the circle’s circumference is the same measure as the shorter edge length of a Square root of the golden ratio √φ = 1.272019649514069 rectangle, plus both the surface area of the circle and the surface area of the Square root of the golden ratio √φ = 1.272019649514069 rectangle have the same surface area. A Square root of the golden ratio √φ = 1.272019649514069 rectangle can be divided into 8 Kepler right triangles and if the shortest edge length of a Kepler right triangle is reduced to 1 then the hypotenuse is equal to the Golden ratio of cosine (36 degrees) multiplied by 2 = (φ) = (√(5) plus 1)/2 = 1.618033988749895, while the second longest edge length of the Kepler right triangle is equal to the Square root of the golden ratio √φ = 1.272019649514069, according to the Pythagorean theorem. A Square root of the golden ratio √φ = 1.272019649514069 rectangle that has been divided into 8 Kepler right triangles has a surface area equal to 4 times √φ = 5.088078598056276. A circle with a diameter that is equal to the longer edge length of a Square root of the golden ratio √φ = 1.272019649514069 rectangle that has been divided into 8 Kepler right triangles also has a surface area equal to 4 times √φ = 5.088078598056276. The longer edge length of the Square root of the golden ratio √φ = 1.272019649514069 rectangle that has been divided into 8 Kepler right triangles has a surface area equal to 4 times √φ = 5.088078598056276 is also equal to 2 times √φ = 2.544039299028138. The shorter edge length of the Square root of the golden ratio √φ = 1.272019649514069 rectangle that has been divided into 8 Kepler right triangles has a surface area equal to 4 times √φ = 5.088078598056276 is also equal to 2. A circle with a diameter that is equal to the longer edge length of a Square root of the golden ratio √φ = 1.272019649514069 rectangle that has been divided into 8 Kepler right triangles also has a radius that is equal to the Square root of the golden ratio √φ = 1.272019649514069. √φ times √φ = the Golden ratio of cosine (36 degrees) multiplied by 2 = (φ) = (√(5) plus 1)/2 = 1.618033988749895. Circumference of the circle = 8. 1-quarter of the circle’s circumference = 2. Diameter of the circle = 2 times √φ = 2.544039299028138. Radius of the circle = the Square root of the golden ratio √φ = 1.272019649514069. The surface area of the circle divided the surface area of the square that is located on the radius of the circle = 4/√φ = 3.144605511029693144, because 4/√φ times √φ times √φ = 4 times √φ/((φ)) = 4/√φ = 3.144605511029693144. Surface area of the circle = 4/√φ times √φ times √φ = 4 times √φ = 5.088078598056276. Radius of the circle = the Square root of the golden ratio √φ = 1.272019649514069. Radius of the circle squared = √φ times √φ = the Golden ratio of cosine (36 degrees) multiplied by 2 = (φ) = (√(5) plus 1)/2 = 1.618033988749895. Pi is also defined as the surface area of the circle divided the surface area of the square that is located on the radius of the circle.

Ok, but how in the first place was the ratio of c/d derived. Was it derived by measuring c and d or what?

Of course it's related with calculus. Just like why deriving the area of a circle is its circumference.

Why is the circumference/diameter a constant ? What proves it ?

So what calculation gets you to pi?

"exactly the approximation"? That ranks up there with some of the silliest expressions I've ever heard - along with the ridiculous "very unique"!

“11 methods for finding the surface area of a circle”: Method 1. The surface area of a circle can be known if the radius of a circle is squared and then multiplied by Golden Pi = 4/√φ = 3.144605511029693144. . Method 2. The surface area of a circle can also be known if half the circumference of the circle is multiplied by the measure for the diameter of the circle and then the result of multiplying half the measure for the circumference of the circle must be divided into 2 resulting in the measure for the surface area of the circle. An isosceles triangle that is made from 2 Kepler right triangles has the same surface area as a circle that has a diameter that is equal in measure to the height of the isosceles triangle that is made from 2 Kepler right triangles. If half the circumference of a circle is divided by the diameter of the circle the result is the half of the ratio Pi. Method 3. The surface area of a circle can also be found if the radius of the circle is multiplied by half the circumference of the circle. If half the circumference of a circle is divided by the radius of a circle the result is the ratio Pi. Method 4. The surface area of a circle can also be discovered if 1 quarter of the circle’s circumference is multiplied by the measure for the diameter of the circle. Method 5. If the surface area of square that has a width equal in measure to 1 quarter of a circle’s circumference is multiplied by the square root of the Golden ratio = √φ = 1.272019649514069 the result is the surface area of the circle. Method 6. If the diameter of a circle is divided by the ratio √√φ = 1.127838485561682 the result is the edge of a square that has the same surface area as the circle. Please remember that the ratio √√φ = 1.127838485561682 is the square root of the ratio √φ = 1.272019649514069 and the ratio 1.272019649514069 is the square root of the Golden ratio of cosine (36 degrees) multiplied by 2 = 1.618033988749895. Method 7. If the surface area of a square that has a width that is equal in measure to the diameter of a circle is divided by the square root of the Golden ratio = √φ = 1.272019649514069 the result is the surface area of the circle. For example: If the shortest edge length of a Kepler right triangle is 36 equal units of measure and also is equal to the surface area of a circle the second longest edge length of the Kepler right triangle can be equal in measure to the surface area of a square with a width that is equal to the diameter of the circle that has a surface area that is the same measure as the shortest edge length of the Kepler right triangle. The mean proportional of a rectangle with its longer edge length being equal to the second longest edge length of the Kepler right triangle while the shorter edge length of the Kepler right triangle is equal to 1 out of 36 equal units if measure from the shortest edge length of the Kepler right triangle is equal in measure to both the width of the square and the diameter of the circle that has a surface area that is equal in measure to the shortest edge length of the Kepler right triangle. 1 time the second longest edge length of the Kepler right triangle is equal to the surface area of the appropriate rectangle that has its longer edge length equal in measure to the second longest edge length of the Kepler right triangle. The mean proportional is the square root of the surface area of a rectangle. The surface area for any rectangle can be derived if the shorter edge length is multiplied by the longer edge length. Method 8. If the surface area of a square that has a diagonal that is equal in measure to the diameter of a circle is multiplied by half of Golden Pi = 4/√φ = 3.144605511029693144 = 2/√φ = 1.572302755514847 the result is the surface area of the circle. Method 9. If the surface area of a square that has a diagonal that is equal in measure to the diameter of a circle is divided by half the square root of the Golden ratio = 0.636009824757035 the result is the surface area of the circle. Method 10: Multiply the diameter of the circle by (2/(√(√(3)) times √√ φ)) = the ratio 1.347419325335723 to get the edge of an equilateral triangle that has the same surface area as the circle. Multiply the edge of the equilateral triangle by half the width of the equilateral triangle times the square root of 3 divided by 2 to get the surface area of the equilateral triangle and confirm that both the equilateral triangle and the circle have the same surface area. The ratio 1.347419325335723 can be derived through the following formulas: (2/(√(√(3)) times √√ φ)) = 1.347419325335723. 2/(√(√3) multiplied by √√ φ = 1.347419325335723. 2/(√(√3) multiplied by 1.127838485561682= 1.347419325335723. 2/(square root (square root 3) multiplied by square root square root Phi = 1.347419325335723. 2(φ/3) ^ (1/4)/ √ φ = 1.347419325335723. 2(1.618033988749895/3) ^ (1/4)/ 1.272019649514069 = 1.34741932533572. 2/(3 X Golden Ratio) ^ (1/4) = 1.347419325335723. 2/(3 times Golden Ratio) ^ (1/4) = 1.347419325335723. 2/(3 x cos (36 degrees) x 2) ^ (1/4) = 1.347419325335723. 2/(3 x sin (54 degrees) x 2) ^ (1/4) = 1.347419325335723. 2/(3 multiplied by φ) ^ (1/4) = 1.347419325335723. 2 divided by the Golden ratio multiplied by 3 ^ (1/4) = 1.347419325335723. 3 times the Golden ratio = 4.854101966249685. (1/2 + √(5)/2 ) = The Golden ratio = 1.618033988749895. Method 11: Divide the diameter of the circle by the ratio = 1.479351567442321 to get the edge of a Pentagon that has the same surface area as the circle. Multiply the edge of the Pentagon by half the edge of the Pentagon times TAN (54 degrees) divided by 2 times 5 to calculate the surface area of the Pentagon and also confirm that both the Pentagon and the circle have the same surface area. The ratio 1.479351567442321 can be derived through the following formulas: (√(34 times 17 times TAN (54 degrees)/2 times 5) times √√φ/34) = the ratio 1.479351567442321. √(34 times 17 times TAN (54 degrees)/2 times 5) times 1.127838485561682/34 = 1.479351567442321. (75/32 + (35 square root (5))/32) ^ (1/4) = 1.479351567442321. 1/2 (5/2 (15 + 7 square root (5))) ^ (1/4) = 1.479351567442321. 1/2 square root (5) (1/2 (1 + 2/square root (5)) (1 + square root (5))) ^ (1/4) = 1.479351567442321. 34 multiplied by 17 multiplied by TAN (54 degrees) divided by 2 multiplied by 5 = 1988.87187508084575. 34 X 17 X TAN (54 degrees)/2 X 5 = 1988.87187508084575. Square root of 1988.87187508084575 multiplied by √√φ = the square root of the square root of Phi = 1.127838485561682 divided by 34 = 1.479351567442321. √1988.87187508084575 multiplied by √√φ/34 = 1.479351567442321. √1988.87187508084575 X √√φ/34 = 1.479351567442321. The true value of Pi = 4/√φ = 3.144605511029 is NOT Transcendental: Pi = 4/√φ = 4 divided by 1.2720196495141 = 3.144605511029. π = 4/√φ = 3.144605511029693144. THE REAL VALUE OF Pi IS NOT TRANSCENDENTAL BECAUSE THE REAL VALUE OF PI = 4/√φ = 3.144605511029693144 IS THE ONLY VALUE OF PI THAT CAN FIT THE FOLLOWING POLYNOMIAL EQUATION: 4th dimensional equation/polynomial for Golden Pi = 4/√φ = 3.144605511029693 Minimal Polynomial: x4 + 16x2 - 256 = 0. www.tiger-algebra.com/drill/x~4-16x~2-256=0/ THE REAL VALUE OF PI = 4/√φ = 3.144605511029693144: Please copy and paste the following link into your web browser if you cannot click onto the following link: www.wolframalpha.com/input/?i=4+divided+by+the+square+root+of+the+golden+ratio PLEASE CLICK ON THE RED DOTS IN THE FOLLOWING LINK TO CONFIRM THAT THE REAL VALUE OF PI = 4/√φ = 3.1446 IS NOT TRANSCENDENTAL. THE REAL VALUE OF PI = 4/√φ = 3.144605511029693144. Minimal polynomial: x4 + 16x2 - 256 = 0 www.wolframalpha.com/input/?i=x4+%2B+16x2+%E2%80%93+256+%3D+0 3D plot of a graph proving that the real value of Pi is NOT transcendental: (Please click on to the following links or copy and them into your web browser): PLEASE DOWNLOAD THE GOOLE DRIVE LINK drive.google.com/file/d/1nT0xGIYHHim49pwjo9oo80MmyWaqOWj1/view?usp=sharing • Panagiotis Stefanides fourth order equation: www.stefanides.gr/Html/piquad.html • Panagiotis Stefanides: Quadrature of circle, theoretical definition: www.stefanides.gr/Html/QuadCirc.html

Very nice - though sadly "rectangle" has been spelt as "rectange" on a few screens.