By G. Hardy
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Etc. = r 3((vol. 1st little cube) + (vol. ) = r 3 (volume of original 3-dimensional figure) . Using smaller and smaller cubes, I can get better and better approximations of the original 3-dimensional figure. So: If a 3-dimensional figure is magnified by a factor of its volume is multiplied by r3. r, We can do the same kind of thing for 2-dimensional figures (surfaces): Even if the surface is curved, each little is approximately a square, so again by using the magnification principle for squares, and then adding up all the squares, we get: If a 2-d~ensional figure is magnified by a factor of r, its area is multiplied by r2.
L (Hint: Construct a perpendicular M to L through P, and then at P, construct a perpendicular N to M. Show that L and N contain opposite sides of a rectangle. ) 4. Show that, through a point not on a given line, there passes only one line parallel to the given line. • p • L (Hint: Assume that there are two lines parallel to L that pass through P. z p. I I I ....... ) X Y • L I12e Intuition exercises: 5. el.. ::'x:; N 6. el.. 90· 90· I I12e 37 Intuition: 38 II3 Areas: The principle of parallel slices Suppose I'm trying to find the area of a figure in the plane: Make a lot of cuts parallel to the base: Each horizontal strip is very close to a little rectangle.
The length of the (less and less) bumpy bottom is always C/2. Why? 1 < - - - 1_ C/2 As you cut more and more finely, the figure gets closer and closer to a rectangle of base C/2 and height 1. It still takes 1r cans of paint to paint it. So C/2 = 1r. So: C = (circumference of circle of radius 1) = 21r Now the circle of radius r is obtained by magnifying the circle of radius 1 with magnification factor r. Since length is one-dimensional, the Magnification Principle says that: C = (circumference of circle of radius r) r·21r 21rr Intuition: 52 I18 When are triangles congruent?