# Thread: Simple derivations on geometrical figures .

1. ## 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

6. Other proofs :
Pi

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

Thanks !

Regards - Sankalp

2. ## Re: Simple derivations on geometrical figures .

Originally Posted by sankalpmittal

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'

3. ## Re: Simple derivations on geometrical figures .

Originally Posted by agentmulder
I like your proof very much because I never thought about it that way and it shows 'thinking outside the box'