Now let's find the ratio of the circumference of a circle to its diameter from 1st principles. Let's assume the radius is "r".
At this point in time all that we know is that a circle has a center and the distance from the center to any point on the circle is r. We have to find the length of the circumference.
We can draw 6. (Note: If instead of 6 triangles we had infinite triangles, the sum of the bases of the triangle will be the circumference of the circle.)
Each triangle has two sides which are radii of the circle and a base. If you sum the base of all the 6 triangles, that should be close to the circumference of the circle (not exactly equal but a little less than the circumference)
What is the angle in each triangle between the two radius? Since the total angle around the center is 360 deg and since there are 6 triangles, the angle between the two adjacent radius is 360/6=60 deg. Matter of fact, each angle in the triangle is 60 deg and each of the triangles incidentally is an equilateral triangle (all sides are of same size).
Now in the picture above let's draw a (imaginary) perpendicular from the center of the circle to one of the bases of the triangle. So each equilateral triangle will be bisected into 2 isosceles triangles (with angle between perpendicular and radius =60/2=30 deg. Angle between base and radius = 60 deg as before. And angle between base and perpendicular = 90 deg).
What is sin (30 deg)? It is equal to 0.5 and hence the base of the isosceles triangle is half the radius. Meaning the base of ONE isosceles triangle is half the radius (hypotenuse of the isosceles right angled triangle).
How many isosceles triangles are there in the circle? There are 2 in each equilateral triangle. So there are a total of 12 isosceles triangles.
What is the total length of base of all the 12 isosceles triangles? It is 12 * [0.5 * (radius)] = 6 * radius.
So as a first approximation the ratio of circumference of the circle to the radius is 6. Since radius is half the diameter, hence the ratio of circumference to diameter is 3 (approximate). The actual ratio would be slightly higher.
How do we get the right value of the ratio?
By splitting triangles into many smaller ones with the same radius but a smaller base and with smaller angles between the two sides (that are radii of the circle). Let's try splitting the 6 equilateral triangles into 12 triangles with 30 deg between the radius. Again repeat the process of bisecting each of these triangles into 2 isosceles triangles which will have angle 15 deg between the perpendicular and the hypotenuse (radius). How many such isosceles triangles will be there now? 24. (=360/15). The sum of the bases of all the 24 triangles will now be closer in value to the circumference of the circle.
The base of each of the 24 isosceles triangles will be Sin (15) * Radius.
And the circumference will be 24 * Sin (15) * Radius Equation (0)
How do we get the value of Sin (15)?
Now, let's get some basic trigonometry.
What is Sin (x+y)?
Sin (x+y) = Sin (x) Cos (y) + Cos (x) Sin (y)
and if x = y?
Sin (2x) = 2Sin(x) Cos(x) (A)
And if we square the equation?
Sin2(2x) = 4 Sin2(x) Cos2(x) = 4 (Sin2(x) - Sin4(x)) (B)
( Applying Cos2(x) = 1 - Sin2(x) ) ( C )
If we assume, p=Sin2(x) (note, "p" cannot be negative) (D)
then 1/4 * Sin2(2x) = (p-p2) (E)
Or, p2 - p + 1/4 * Sin2(2x) = 0 (F)
Now we know in the quadratic equation ax2 + bx + c = 0 (G)
rhe solution is: x= (-b± √(b2-4ac)) / (2a) (H)
Now let's assume x=15 deg. 2x = 30 deg. Sin (2x) = Sin (30) = 0.5 (I)
Sin2(2x) = 0.5 * 0.5 = 0.25 (J)
Applying (J) in equation (F) we get
p2 - p + (1/4)*Sin2(2x) = 0 (K)
Applying (D), (H) and (J) in equation (K) we get,
p=Sin2(x) = 0.07 (approx)
Sin (x) = Sin(15) = Sqrt (p) = 0.26 (L)
Now applying (L) in equation (0), we get circumference is 24 * 0.26 = 6.21 times the radius or about 3.11 times the diameter. Applying (L) in (K) recursively (meaning first make 2x=15 deg, then 2x=7.5 deg and so on), we can get values of Sin of smaller angles. Now repeating the above steps few more times. we get the following data.
No. of Isosceles Angle (Q) Between Sin (Q) Circumference / Diameter
Triangles Radii Ratio (=no. of triangles * SinQ)
12 30 0.50 3.00000
24 15 0.26 3.10583
48 7.5 0.13 3.13263
96 3.75 0.07 3.13935
192 1.875 0.03 3.14103
384 0.9375 0.02 3.14145
See the ratio in the last column converging. That is the value of Pi.
As the number of isosceles triangles tend to infinity, the sum of the bases of the isosceles triangles will tend to the circumference of the circle. We can probably get closer to the actual value of Pi if we increase the number of isosceles triangles.
Now let's see how to find the area of a circle. It will be the sum of area of all triangles that we created. As the base of the triangle tends to zero and as the number of triangles tends to infinity, the height of each triangle will be equal to the radius. And the sum of bases of ALL the triangles will be equal to the circumference of the circle.
So the area of a circle is 1/2 * (sum of bases=circumference) * (height = radius) = 1/2 * (2 Pi r) * (r) = Pi r2