Pythagorean Theorem - March 25-29, 1996
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We have done a fair bit with the Pythagorean Theorem this year, so let's take a look at the theorem itself.One very familiar proof of this theorem is a right triangle with squares on each side. Here's my question: Do we have to use squares? What about hexagons, or other shapes? What would we have to do to use other shapes, if that's at all possible? And why do you suppose squares are usually used for this proof?
Extra: name a U.S. President who discovered a proof of the Pythagorean Theorem.
Annie says: Solutions
Only two folks got this one right! This may partly be a case of my having an answer in mind and not really asking the right questions, so I'll try to do be more explicit next time.
A number of people said no, you don't have to use squares, and suggested that other shapes such as regular polygons would work. A couple of people showed that you could use hexagons and proved it, but their answers weren't general enough - the key here is that you can use any shape you want, as long as the shapes on the three sides are similar ! Thomas and Brian both got that idea in; Thomas actually said similar shapes, and Brian said shapes with proportional lengths. So you could really say the proof says that, given similar shapes on all the edges, the area of the shape on one leg plus the area of the shape on the other leg equals the area of the shape on the hypotenuse. You can check this out in Sketchpad or a similar program - it is pretty cool!
A lot of people said that squares are used because they are easiest, and that's certainly true - the area of a square looks just like the terms we are used to seeing in the Pythagorean Theorem, so this makes it really convenient.
If you come across a problem like this, and you determine that other shapes will work, try to figure out what those other shapes might look like, or what limitations there might be. While you might find a couple that work, see if you can find any that don't, and hence narrow things down a bit.
Thomas S. Kuo Grade: 7 School: Murray Junior High School, Ridgecrest, California (1) We don't have to use squares to prove the Pythagorean Theorem. (2) Hexagons or other shapes could be used too. For a regular hexagon with side a: * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * a * * c * * * * * * * * * * * * * * * * * * * * * * * * * * * * * a b The hexagon is formed by six congruent equilateral triangles. Each triangle has side a and height (sqrt(3)/2) * a. Then the area of a triangle is (1/2)*(a)*(sqrt(3)*a/2) = (a^2) * sqrt(3)/4. The area of the hexagon is 6 * (a^2) * sqrt(3)/4 = (a^2) * 3 * sqrt(3) / 2. Let a, b and c are three side of a right triangle as shown. When we form three hexagons with side a, b, and c, their areas will be (a^2) * 3 * sqrt(3) / 2, (b^2) * 3 * sqrt(3) / 2, and (c^2) * 3 * sqrt(3) / 2, respectively. Now I have to check out whether the sum of areas of the first two is equal to that of the third one. After we cancel out the common factor, we have only a^2, b^2 and c^2 left and the relationship a^2 + b^2 = c^2 still holds. The conclusion is that we can prove the Pythagorean Theorem by using hexagons. It is also true if other shapes are used. The fact is that as long as we use similar shapes with side a, b, and c, their areas will have ratio of a^2, b^2, and c^2. After cancelling out the common factor, it can be simplified to a^2 + b^2 = c^2. (3) I believe that the reason to use squares in the proof is due to the fact that it is the most direct and easiest method.
Brian Gordon Dartmouth '92 I have an answer for the bonus. It's James Garfield, right? I believe the proof is like this: |\ b | \ c |__\ a Paste two copies of this triangle together and connect the other acute vertices to make a trapezoid like this: a ____ | /* b | /c * |/ * |`-c_ * a |____\* b Now.... the area of the trapezoid = 1/2 * height * sum of bases = 1/2 * (a+b) * (a+b) The areas of the triangles are: 1. 1/2 * a * b 2. 1/2 * a * b 3. 1/2 * c * c because they meet at a right angle. This can be shown by noting that the straight angle that is the left side of the trapezoid is made up of two complementary angles (from the two congruent right triangles) and the angle of the two sides of length c (which is then 180 - 90 = 90). Equate the trapezoid formula with the sum of the triangles: Double everything to eliminate those pesky 1/2's: (a+b)(a+b) = ab + ab + cc do the FOIL... a^2 + 2ab + b^2 = 2ab + c^2 subtract the 2ab.... a^2 + b^2 = c^2 I like this one much more than the proof I originally learned, which was based on the altitude-on-hypotenuse and similar triangles. I never actually learned the squares-on-the-sides proof. As for other shapes, I think you could do it, if all the constructions are done right. That's because the areas of the shapes would be proportional to the square of the lengths of the sides. I have a formula for a regular n-gon with side s. It's A=.25 * n * s^2 * cotangent(180/n). Could you guys patent that for me? :) --briA number of you have a poster in your classrooms that answers the question about what U.S. President discovered a proof of the Pythagorean Theorem, and you were smart enough to send it in: James Garfield. Here's a list of the folks who got the extra part:
Granada High School, Livermore, California: Collin Beighley Mike Sue Martin County High School, Stuart, Florida: Ben Ngo Jaime Uhazie Melissa Sloane Sara Holtzman Ridgeview High School, Bakersfield, California: Nathaniel Prevost Anabel Arambula Ebone' King Sandra Rodriguez Jenna Myers and Kim Jackson Justin Poncetta Ryan Duncan Janelle Green Gina Torchia Andrea Brazier and Sandy Blanco Julie Brinkman and Matt Rengers
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