Do Figures with Equal Sides Have the Same Area?Date: 11/21/2008 at 12:19:17 From: Richard Subject: Changing area of quadrilateral If you have a rectangle (figure A), with sides X and Y and Area = X x Y, and you do not change the length of the sides but change the angle formed by sides X and Y (i.e. decrease from 90 to 85 degrees)to make figure B, why is the area of figure B (calculated by the formula 1/2 B x H) now less than figure A? This seems counter intuitive as the length of the sides of both figures A and B are still identical. Date: 11/21/2008 at 16:32:03 From: Doctor Peterson Subject: Re: Changing area of quadrilateral Hi, Richard. It's only counter-intuitive if your "intuition" wrongly assumes that figures with the same sides should have the same area. There are several ways to train your intuition to see that the true result is perfectly natural. First, imagine a rectangle made of jointed pieces of metal, hinged at the corners. It starts out as a rectangle, o------------------o | | | | | | | | o------------------o and then becomes o------------------o / / / / / / o------------------o which is not quite as high but may look to you like about the same area. Push it over farther and keep watching: o------------------o / / / / o------------------o o------------------o / / o------------------o o------------------o o------------------o o------------o-----o------------o Now its area is zero! Is there any doubt that the area has been changing all along? It just wasn't so obvious when you didn't push it so far. This is a technique used by mathematicians: to check whether something is likely to be true in all circumstances (e.g. the idea that the area should not change), push it to the extreme and see if it still makes sense. Another way to make this a little more intuitive is an approach that leads to calculus. Think of a rectangle as a side view of a stack of cards: ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- If you push this over so that the side slants, it keeps the same height, rather than losing height as our other rectangle did; since it is still made of the same cards, this new figure must have the same area: ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- This time, no matter how far you push it, the height remains the same (though the pile would get a little unstable), and the area remains the same. ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- ----------------------- Moreover, the length of the slanting side increases; so you can see that if you push a rectangle over and keep the length of that side the same, you'll be losing area--some cards have to be taken out in order to preserve the length. Does that help? - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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