Ladnor Geissinger
Posts:
313
From:
University of North Carolina at Chapel Hill
Registered:
12/4/04


[mathlearn] length and area; sidewalks and roads
Posted:
May 5, 2008 12:45 AM



In response to several recent email notes about computing areas of simple geometric regions, I will resend my note to mathlearn from Nov 2003. That original note was in response to comments by Rex Boggs. I hope he forgives me for throwing it out there again. ================================================================ Rex Boggs complained about his students getting mixed up about circumference and area and remembering formulas and knowing which to use when. Perhaps some teachers would find it useful to develop the idea of computing area from two common examples of geometric regions: constant width sidewalks (and borders and ribbon paths) and roads. For each of these the computation of area is simple: multiply the constant width w by the centerline length m.
Note that for any trapezoid with width w (distance between parallel sides) and midline length m, the area is w*m. And borders around parts of buildings, patios, and other objects are often made up of a collection of such trapezoids all of the same width w and with two successive trapezoids meeting at a corner along a line which makes the same angle with each edge at the corner. Sidewalks are usually made this way. So the total area of such a sidewalk or border is the product of the constant width w multiplied by the sum of the midline lengths of each of its segments. Also, suppose we take a long piece of ribbon and tape a string to its centerline and then start pasting in down flat onto a surface. When we want to have it go off in a different direction we cut the ribbon but not the string and flip the ribbon over and continue with the two cut edges realigned. Clearly the total area of our ribbon path is the width multiplied by the total midline length of the string (no matter at what angle the ends are cut). Of course a special case is a single triangle with w the "altitude" from vertex C dropped to side AB (possibly extended), and the area is w*m where m is the length of the midline parallel to AB and halfway between AB and C. So our "formula" works for rectangles, squares, triangles, trapezoids,.. sidewalks and pieces of them.
Roads of constant width w are nearly a series of rectangles of width w connected by sectors of circular rings. In this case again, the total surface area of such a road is the product of w and the centerline length. Does this kind of area computation appear in any texts? Note that special cases include a sector of a ring or the whole ring for a circular race track, and the really special case of a sector of a circle or a whole circle. Note that one could argue that this works for roads by approximating closely a road by using a ribbon path. So we don't have to know about pi*rad^2 before doing this.
On the other hand, if we do know a formula for the area of a circle, we could consider two concentric circles with radii a>b. Then for any sector with angular measure t, the area of the part of the ring between the two circles and within the sector is (t/2pi)*pi*a^2  (t/2pi)*pi*b^2 or (ab)*t*(a+b)/2. And ab is the constant width of that ring while t*(a+b)/2 is the length of the midline, that is, the arc of the circle of radius (a+b)/2 which lies in this sector. So our simple formula for area has appeared again.

