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Math Help - Simple derivations on geometrical figures .

  1. #1
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    Simple derivations on geometrical figures .

    Hii , this is the thread for simple derivations on geometrical figures.

    Here I begin with the easiest proof of the area of circle. Although there are many proofs but easiest is that , which I am telling you all ( I think ) .

    1. The proof by Greeks 2000 years ago ! : http://www.slideshare.net/yaherglani...-proof-1789707


    2. The analytical proof :

    Divide the circumference into n equal parts each of length l
    => nl = 2π r => l = 2π r/n
    When n is very large and l very small, each sector of the circle formed by the small arc of length l is like a triangle and
    its area = (1/2) base x altitude
    = (1/2) l * r
    = (1/2) 2π r/n * r
    and area of the circle
    = n * area of each sector
    = n * [ (1/2) 2π r/n * r ]
    = π r^2.

    3. The proof which I discovered !

    View image: proof


    4. Proof of Archimedes by Dr. Math : Math Forum - Ask Dr. Math

    5. Euclid's Proof by Dr. Math : Math Forum - Ask Dr. Math

    6. Archimedes' simplified proof :
    Theorem: The area of a circle is (1/2)circumference * radius

    Proof:

    (1) Let K = (1/2)*C*R where C = circumference and R = radius.

    (2) Let A = the area of the circle.

    (3) Assume that A is greater than K

    (4) We can inscribe a polygon inside A that is greater than K and less than A.

    (5) So, the area of the polygon = (1/2)*Q*h where h is the distance from the center to the base and where Q is the perimeter of the polygon. [See Lemma 1 above]

    (6) But Q is less than C (see Postulate 1 above) and h is less than R.

    (7) So we have Area Polygon = (1/2)Q*h which is less than (1/2)*C*R

    (8) But this contradicts step #4 so we reject step #3.

    (9) Now, let's assume that K is greater than A.

    (10) We can circumscribe a polygon P around A such that P is greater than A but less than K. [See Lemma 3 and Method of Exhaustion here for details.]

    (11) From Lemma 1 again, we know that the area of this polygon is (1/2)*Q*h where Q is the perimeter of the polygon and h is the height.

    (12) In the case of the circumscribed polygon (see diagram for Postulate 2), h = R.

    (13) Using Postulate 2 above, we see that Q is greater than C.

    (14) But then the area of the polygon is greater than K since (1/2)*Q*R is greater than (1/2)*C*R

    (15) But this contradicts step #10 so we reject our assumption at step #9.

    (16) Now, we apply the Law of Trichomoty (see here) and we are done.

    (17) From the theorem, the area of a circle is (1/2)(circumference)(radius)

    (18) From the definition above, π = C/D

    This means that C = D*π = 2*r*π

    (19) Putting this all together gives us:

    area = (1/2)(circumference)(radius) = (1/2)(2*r*π)(r) = πr2
    QED


    Other proofs

    1. Proof of the area of a circle by inscribing a polygon :
    http://www.basic-mathematics.com/pro...-a-circle.html

    2. By inscribing a regular polygon
    Evaluating pi. The area of a circle. Topics in trigonometry

    3. Another proof :
    http://www.maa.org/pubs/Calc_articles/ma018.pdf

    4. By using equation of circle :
    http://iamsuhasm.wordpress.com/2009/...y-integration/

    5. By calculus :
    http://www.vias.org/calculus/06_appl...ral_03_07.html
    http://answers.yahoo.com/question/in...3000747AAjNUsa

    6. Other proofs :
    Pi

    __________________________________________________ _________

    Please share more proofs on geometrical figures , 3rd proof I discovered myself though .


    Thanks !

    Regards - Sankalp

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  2. #2
    Member agentmulder's Avatar
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    Thumbs up Re: Simple derivations on geometrical figures .

    Quote Originally Posted by sankalpmittal View Post

    3. The proof which I discovered !

    I like your proof very much because I never thought about it that way and it shows 'thinking outside the box'
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  3. #3
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    Re: Simple derivations on geometrical figures .

    Quote Originally Posted by agentmulder View Post
    I like your proof very much because I never thought about it that way and it shows 'thinking outside the box'
    Please reply here : http://www.mathhelpforum.com/math-he...tml#post683617
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