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Two Elementary : Pythagorean & Area of a Two Elementary Area Theorems: Charles Delman &

Charles Delman

August 26, 2016 Why the Pythagorean Theorem is true

Two Elementary Area Theorems: Pythagorean Theorem & Area of a Circle a 2+ b 2 = c 2 Charles Delman

a c a c c b cc b b a b a The area of a circle

Two Elementary Area Theorems: Pythagorean Theorem & Area of a Circle r Charles Delman

The area of a circle is clearly proportional to the of its . That is, A = kr 2. Clearly, k < 4. Why? And k > 2. Why? In fact, dissection of the regular dodecagon shows that k > 3.

Two Elementary Area Theorems: Pythagorean Theorem & Area of a Circle

Charles Delman The area of a circle of radius r is A = πr 2

Two Elementary Area Theorems: In fact, k = π (as you probably remember). Pythagorean Theorem & Area of a Remember that π is defined in terms of linear Circle measurements; it is the of to . Charles Delman Thus, we have another deep relationship between and area!

C A = π = 2r r 2 Why does π, the ratio of circumference to diameter, also out to be the ratio of the area of the circle to the area of a square on the radius? Is it just a miracle, or can we understand the reason? Why A = πr 2

Two Elementary Area Theorems: Pythagorean Theorem & Area of a Circle

Charles Delman h b

As the number of sides, n, increases, the area of the inscribed n-gon approaches the area of the circle. Why A = πr 2, continued

Two Elementary Area Theorems: Pythagorean Theorem & Area of a Circle

Charles Delman h b

1 Each has area bh. 2 Why A = πr 2, continued

Two Elementary Area Theorems: Pythagorean Theorem & Area of a Circle

Charles Delman h b

n So the area of the inscribed is bh. (There are n 2 .) Why A = πr 2, continued

Two Elementary Area Theorems: Pythagorean Theorem & Area of a Circle

Charles h Delman

nb

nb is the of the polygon. As n → ∞, nb → C, the circumference of the circle, and h → r, the radius of the circle. Remember that C = 2πr. Why A = πr 2, conclusion

Two Elementary Area Theorems: Pythagorean Theorem & Area of a Circle

Charles Delman

Thus, as n → ∞, the area of the inscribed polygon, (nb)h 2πr · r , approaches = πr 2. 2 2