1. Develop the formula for the area of a kite and prove that it works. § 11.1 The area of a kite is ½ d 1 d 2. Proof by dissection. By theorem the diagonals of a kite are perpendicular and one of them bisects the other. See the drawing above. We can form a rectangle around the kite that has length d 1 and width d 2. The area of this rectangle is exactly twice the area of the kite. Hence, A = ½ d 1 d 2 d 1 d 2

2. Find the area of an equilateral triangle with side of 1 unit. (A unit triangle.) You need the Pythagorean Theorem to find the altitude. h = √(1 2 - (1/2) 2 ) = √3/2 = So the area is ½ b h = (½) (1) (0.8660) = /2

3. Find the area of a unit regular hexagon. Use the results of problem 2 and the fact that the hexagone consist of six unit triangles. A =

4. Find the area of a unit regular octagon. 1 Calculate x and add the areas of the pieces together. Did you get  2 1 x x 1

5. A can holds three tennis balls. Which is larger the height of the circumference? If r is the radius of a tennis ball then the height of the can is 6r while the circumference is 2πr = r. The circumference is larger.

6. Use the classical area formulas to determine the ratio of the side of a square to the radius of a circle if they have the same area S 2 = π r 2

7. If a wire is stretched around the earth at ground level it would be approximately 24,900 miles long. If you were to increase the length of the wire by 50 feet and uniformly wrap it about the earth at the same height, would a small bug be able to crawl under it? Would a small dog be able to crawl under it? Would you be able to crawl under it? If the circumference of the earth is about 24,900 miles the radius would be about 3963 miles. Why? In feet the radius would be about 20,924,419 feet. If we add 50 feet to the circumference of the earth is the radius would be about 20,924,427 feet. The new radius is about 8 feet longer and you could easily walk under the new fence. Are you surprised.

8. Investigate a Sierpinski Triangle. In stage one what is the ratio the number of triangles shaded to the number of same sized triangles in the entire triangle? In stage two what is the ratio the number of triangles shaded to the number of same sized triangles in the entire triangle? In stage three what is the ratio the number of triangles shaded to the number of same sized triangles in the entire triangle? Make a conjecture about this ratio. STAGE# ∆ S SHADEDTOTAL # SMALL ∆ SPERCENT SHADED N3 N N - 1 (100) (0.75 N – 1 )

9. Investigate Pick’s Formula for area. Show me an example of it with an plane figure of your choice. Pick’s formula concerns regions placed on a grid. Points of the grid that lie on the perimeter are border, b, points. Points that lie on the interior are interior, i, points. Area is then: A = i + b/2 - 1