## Tuesday, May 02, 2006

### Archimedes and the area of a circle

In today's blog, I will go over Archimede's proof that the area of a circle is (1/2)rc. When this proof is combined with Euclid's proof (most likely from Eudoxus), it is possible to show that for all circles, the ratio of C/D is constant.

Postulate 1: For a given chord of a circle, the segment of the circumference is longer than the chord.

In the example above, BDC is longer than BC.

Postulate 2: For a given segment of the circumference, lines connected above it combined are longer.

In the example above, AG + GE is greater than the segment of the circumference AE.

Lemma 1: The area of a regular polygon is (1/2)h*Q where Q is the sum of the perimeter.

Proof:

(1) A regular polygon can be divided up into a sum of triangles.

(2) The area of each triangle is equal to (1/2)*(GH)*(DE) [See here for details if needed]

(3) So, the area of the polygon = (# of sides)*(1/2)*(GH)*(DE)

(4) Let Q = the sum of the bases (for example, for the hexagon above, Q = DE + EF + FA + AB + BC + CD)

(5) Then, the area of the polygon = (1/2)*(GH)*Q

QED

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. [See Lemma 2 and the Method of Exhaustion here for details]

(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.

QED

Corollary 1: For all circles, the ratio of Circumference to Diameter is constant

Proof:

(1) We know from Euclid (see here) that for any two circles C1 and C2 that:

Area1/Area2 = (Diameter1)2/(Diameter2)2

(2) We know from Theorem 1 above that:

(3) Combining step #1 with step #2 and using R=(1/2)D gives us:

[(1/2)*(1/2)D1*C1]/[(1/2)*(1/2)D2*C2] = [D1]2/[D2]2

(4) Canceling out (1/4)/(1/4) gives us:

[D1*C1]/[D2*C2] = [D1]2/[D2]2

(5) Multiplying both sides by (D2/D1) gives us:

C1/C2 = D1/D2

which means that:

C1*D2 = C2*D1

and finally that:

C1/D1 = C2/D2

QED

Definition 1: π

From the Corollary, we know that the ratio of the circumference to the diameter is constant. π is this ratio from circumference to diameter. In other words, π = C/D.

Corollary 2: The area of a circle is πr2.

Proof:

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

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

This means that C = D*π = 2*r*π

(3) Putting this all together gives us:

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

QED

References

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