By Thomas Hull
Venture Origami: actions for Exploring arithmetic, moment variation provides a versatile, discovery-based method of studying origami-math subject matters. It is helping readers see how origami intersects a number of mathematical issues, from the extra visible realm of geometry to the fields of algebra, quantity concept, and combinatorics.
With over a hundred new pages, this up-to-date and multiplied variation now contains 30 actions and provides larger strategies and instructing counsel for all activities.
The ebook comprises unique plans for 30 hands-on, scalable origami actions. each one task lists classes during which the task may possibly healthy, contains handouts for school room use, and offers notes for teachers on options, how the handouts can be utilized, and different pedagogical feedback. The handouts also are to be had at the book’s CRC Press internet page.
Reflecting suggestions from academics and scholars who've used the ebook, this classroom-tested textual content offers a simple and wonderful manner for academics to include origami right into a variety of school and complicated highschool math classes.
Read Online or Download Project Origami Activities for Exploring Mathematics PDF
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Extra resources for Project Origami Activities for Exploring Mathematics
Example text
We need to find a formula for the area A of the equilateral triangle and then try to maximize this formula in terms of θ. ) Since the base of the triangle is x, its height is ( 3/2) x. So A = ( 3/4) x2 , but we wanted it in terms of θ. Well, cos θ = 1/x, so x = 1/ cos θ = sec θ. Thus we have √ 3 sec2 θ. A= 4 Folding Equilateral Triangles in a Square 11 We could take the derivative of this and try to maximize it using calculus, but we don’t really need to. Since cos θ is a decreasing function on the interval 0 ≤ θ ≤ π/12 (we really should be working in radians, after all), we know that sec θ is an increasing function on this interval.
The key is that EDB is similar to ABC since they’re both right triangles and they share the angle at B. Therefore, √ √ √ √ √ 3 3 3 ( 2 − 3) x √ = = ⇒ x= = 2 3 − 3. 1−x 2 1 2+ 3 √ √ Thus our 15◦ -75◦ -90◦ triangle has long leg 3, short leg 2 3 − 2, and hypothenuse √ 2 6 − 3 3. These values are rather awkward, and normalizing the shortest leg to be 1 seems√ somewhat √ standard for √ this kind of thing. Doing this gives us a long leg of length 3/(2 3 − 2) = 2 +√ 3 once we rationalize the denominator.
Since the base of the triangle is x, its height is ( 3/2) x. So A = ( 3/4) x2 , but we wanted it in terms of θ. Well, cos θ = 1/x, so x = 1/ cos θ = sec θ. Thus we have √ 3 sec2 θ. A= 4 Folding Equilateral Triangles in a Square 11 We could take the derivative of this and try to maximize it using calculus, but we don’t really need to. Since cos θ is a decreasing function on the interval 0 ≤ θ ≤ π/12 (we really should be working in radians, after all), we know that sec θ is an increasing function on this interval.