1. Day 141 – Arguments about circumference

2. Introduction
We understand that the distance around the circle is given by 2𝜋𝑟 where 𝑟 is the radius and 𝜋 a constant number that is irrational. It is normally approximated using …. There are arguments that try to visualize what actually the circumference is. Since 𝜋 is the only value that is clear, the argument slightly shifts to 𝜋. This could be an interesting concept to try and figure out what this number is in real sense. In this lesson, we are going to give an informal argument for the formulas for the circumference.

3. Vocabulary
Circumference The distance around a circle Radius The line segments from the center of the circle to the circumference

4. Estimation of circumference We would like to come up with an argument (by Archimedes) that would help us estimate the circumference of a circle. Consider a circle of radius 𝑟 and center O. Pick a point, P, on the circle and connect it to the center. Using a compass of radius OP, mark several points on the arc of the circle such the distance from one mark to the other is 𝑟=𝑂𝑃. Connect the points together to get a hexagon.

5. Drawing the diameters from the vertices of the hexagon, we get six equilateral triangles. Estimating the circumference using that of the hexagon, we get 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒≈6𝑟=2×3𝑟. But we know that 𝑐=2𝜋𝑟, thus, 3 is an approximation of 𝜋. Since the sides of the hexagon are less than the length of the arc, 𝜋 must be more than 3. O P Q R S T U 𝑟

6. We can proceed further by coming up with a dodecagon, 12 sided figure
We can proceed further by coming up with a dodecagon, 12 sided figure. M is the intersection of radius ON and chord QP. Since N is chosen so that 𝑄𝑁=𝑁𝑃, which implies that 𝑄𝑀=𝑀𝑃, we have ON being perpendicular to QP. Consider triangle OPM. 𝑂𝑃=𝑟 and 𝑀𝑃= 1 2 𝑟. O P Q R S T U 𝑟 N M

7. By Pythagorean theorem, we have 𝑂𝑀= 𝑟 2 − 1 2 𝑟 2 = 𝑟 2 − 1 4 𝑟 2 = 4−1 4 𝑟 2 = 𝑟 2 3 Thus, 𝑁𝑀=𝑂𝑁−𝑂𝑀=𝑟− 𝑟 2 3 = 2 2 𝑟− 𝑟 2 3 𝑁𝑀= 𝑟 2 2− 3 MP = 1 2 𝑟 Considering triangle MPN, we can get the hypotenuse NP which is the length of the side of the dodecagon. Using the Pythagorean theorem, we get 𝑁𝑃= 1 2 𝑟 2 + 𝑟 2 2− 3 2

8. 𝑁𝑃= 1 4 𝑟 2 + 𝑟 −4 3 = 1 4 𝑟 2 + 𝑟 2 4 7−4 3 = 𝑟 2 4 8−4 3 = 𝑟 2 4 2− 3 =𝑟 2− 3 Thus, the perimeter is given by 12𝑁𝑃 =12𝑟 2− 3 ≅ 𝑟=2× 𝑟 Hence 𝜋 is approximately

9. Example An informal approximation of circumference give the circumference of a circle as 𝑟. What is the approximation for 𝜋. Solution The formula is 𝑐=2𝜋𝑟= 𝑟 Thus, 2𝜋= implying that 𝜋=

10. homework
An informal approximation of circumference give the circumference of a circle as 𝑟. What is the approximation for 𝜋.

11. Answers to homework 𝜋≈

12. THE END