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# Prove 3<pi<4 watch

1. Can anyone show me prove of 3<pi<4 ??

I can prove pi<4 by
drawing circle then square around it outside then compare area of circle and square.
each side of square will have length=2r
so (2r)^2 > pi r^2
4 r^2 > pi r^2
so 4> pi

I dont know the other part though

Thanks..
2. get a scientific calculator, hit the pie button and then equals to get: 3.14...

Don't believe me? test it out
3. (Original post by ilyking)
get a scientific calculator, hit the pie button and then equals to get: 3.14...

Don't believe me? test it out

Sorry I'm finding people who are smarter than using scientific calculator
4. (Original post by cazzy-joe)
Can anyone show me prove of 3<pi<4 ??

I can prove pi<4 by
drawing circle then square around it outside then compare area of circle and square.
each side of square will have length=2r
so (2r)^2 > pi r^2
4 r^2 > pi r^2
so 4> pi

I dont know the other part though

Thanks..

Spoiler:
Show
Hint: A regular hexagon can be divided into six equilateral triangles
5. I can help.

How about using the Newton Raphson method to show that pi tends to a certain value in the range 3<x<4

Where x is the actual value of pi.
6. (Original post by cazzy-joe)

Sorry I'm finding people who are smarter than using scientific calculator
Take a photo of the result on the calculator, print it off, prit-stick in designated space on answer booklet.
7. (Original post by Ari Ben Canaan)
I can help.

How about using the Newton Raphson method to show that pi tends to a certain value in the range 3<x<4

Where x is the actual value of pi.
what function should I use?????
8. I can only show 2<pi

draw a square within the circle.
The diagonal is then = 2r.
Each side of the square is = x. Then (pythagoras) (2r)^2 = 2x^2
4r^2 = 2x^2
area of square = x^2 = 2r^2

Then you have 2r^2 < pi r^2

and 2 < pi.
9. (Original post by cazzy-joe)
what function should I use?????
Er ... I just told you how to do this geometrically!?
10. (Original post by sudo_euclid)
I can only show 2<pi

draw a square within the circle.
The diagonal is then = 2r.
Each side of the square is = x. Then (pythagoras) (2r)^2 = 2x^2
4r^2 = 2x^2
area of square = x^2 = 2r^2

Then you have 2r^2 < pi r^2

and 2 < pi.
Well perhaps you need a polygon with more sides inside your circle? Say ... 6?
11. (Original post by Mr M)
Well perhaps you need a polygon with more sides inside your circle? Say ... 6?
12. (Original post by Mr M)

Spoiler:
Show
Hint: A regular hexagon can be divided into six equilateral triangles
Thanks..i'm trying with hexagon
13. Do as Mr M said.
14. (Original post by Mr M)
Well perhaps you need a polygon with more sides inside your circle? Say ... 6?
I got 4>pi>

using
15. (Original post by Mr M)

Spoiler:
Show
Hint: A regular hexagon can be divided into six equilateral triangles
Using a hexagon gives

Unless I did the calculation wrong.
16. Not to blow everyone's trumpet but you should start with a geometric definition of pi and prove that it's value is unaltered for a given circle (i.e. it is well defined).

That is pretty important - the other geometry you are using needs no proof (i.e. separate areas combining result in full area, bounded shapes have area less than the figure that bounds them, etc).

Newton Raphson would be unproductive here - not only would you have to define Pi with the cosine function but you would then have a lot of work proving convergence and finding appropriate functions - geometrically this problem is far easier.
17. (Original post by Noble.)
Using a hexagon gives

Unless I did the calculation wrong.
You did the calculation wrong.

Draw a circle enclosing a regular hexagon that touches the circle.

Let the diameter of the circle = 1 unit.

What is the perimeter of the hexagon?
18. (Original post by Mr M)
You did the calculation wrong.

Draw a circle enclosing a regular hexagon that touches the circle.

Let the diameter of the circle = 1 unit.

What is the perimeter of the hexagon?
Oh ok, using the perimeter. I didn't think of that. I was using areas.

Using r as the radius of the circle, it does indeed produce

Thanks
19. (Original post by Mr M)
You did the calculation wrong.

Draw a circle enclosing a regular hexagon that touches the circle.

Let the diameter of the circle = 1 unit.

What is the perimeter of the hexagon?
(Original post by Noble.)
Oh ok, using the perimeter. I didn't think of that. I was using areas.

Using r as the radius of the circle, it does indeed produce

Thanks
Thank you for helping
20. (Original post by DeanK22)
Newton Raphson would be unproductive here - not only would you have to define Pi with the cosine function but you would then have a lot of work proving convergence and finding appropriate functions - geometrically this problem is far easier.
However, that is the standard analytical definition - .

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