 mersenneforum.org Area of Triangle, non-obvious case
 Register FAQ Search Today's Posts Mark Forums Read 2012-01-16, 05:12 #1 Unregistered   23·223 Posts Area of Triangle, non-obvious case As is well known, the area of a triangle ABC is 1/2 * base * height. Here, the base is the length of any side (say AB, without loss of generality) and the height is the distance from C to the line containing AB. Of course, we also know that the distance from a point to a line is the length of the segment that is perpendicular to the given line and passes through the given point. In the case where the altitude passes through the line segment AB, this formula is obvious enough. For example, consider the triangle with vertices (0,0), (1,1) and (2,0) and suppose we compute the base using the side with vertices (0,0) and (2,0). In this case, we can justify the area formula by dividing the original triangle into two triangles, one with vertices (0,0), (1,0) and (1,1), and the other with vertices (1,0), (1,1) and (2,0). The first triangle's area is clearly half that of the rectangle with vertices at (0,0), (1,0), (0,1) and (1,1) while the second triangle's area is clearly half that of the rectangle with vertices at (1,0), (2,0), (1,1) and (2,1). Adding that together, the area of the original triangle must be half the area of the rectangle with vertices (0,0), (2,0), (0,1) and (2,1), which is exactly what the area formula predicts. But the area formula is not so clear if the altitude does not pass through AB. For example, consider the triangle with vertices (0,0), (1,0) and (-1,1). If we take (-1,1) as the point from which to drop the altitude, the height will be the length of the segment from (-1,1) to (-1,0), and the area formula will say that the area of the triangle is half the area of the rectangle with vertices (-1,0), (1,0), (-1,1) and (1,1). This is a lot less obvious, and in fact, I'm not sure how to prove it without calculus. Is there a way to prove the triangle area formula is this tougher case using (coordinate-less) geometry alone? Thanks  2012-01-16, 05:49 #2 wblipp   "William" May 2003 New Haven 22×32×5×13 Posts You find the formula trivial for a right triangle with one of the sides as the base. You then handle your obvious case by creating two right triangles that you can manipulate to form the full triangle. That's a good strategy.   2012-01-16, 05:49   #3
LaurV
Romulan Interpreter

Jun 2011
Thailand

2·5·887 Posts Quote:
 Originally Posted by Unregistered ... and the area formula will say that the area of the triangle is half the area of the rectangle with vertices (-1,0), (1,0), (-1,1) and (1,1).
That is not true. The area is 1/2, that is half of the rectangle (-1,0)(0,0)(0,1)(-1,1), or if you like, (0,0)(0,1)(1,1)(1,0). And you prove it exactly in the same way, extending the triangle to the rectangle that goes through the base and the opposite vertex, and showing the triangles that forms are congruent. There is a simple way to show it too, showing that if you move the vertex on a parallel to the base, the area stays the same. That is the same idea used to compute volumes for cones and spheres, see the principle of Cavalieri.

Last fiddled with by LaurV on 2012-01-16 at 06:05   2012-01-16, 09:31   #4
Unregistered

5×1,931 Posts Quote:
 Originally Posted by LaurV That is not true. The area is 1/2, that is half of the rectangle (-1,0)(0,0)(0,1)(-1,1), or if you like, (0,0)(0,1)(1,1)(1,0).
Yes, I should have said said the rectangle with vertices (0,0), (1,0), (1,1) and (1,0). After all, the base is (0,0)(1,0) in this case.

Quote:
 Originally Posted by LaurV And you prove it exactly in the same way, extending the triangle to the rectangle that goes through the base and the opposite vertex, and showing the triangles that forms are congruent.
Sorry, I still don't understand... Suppose we look at the (-1,1)(0,0)(1,0) example again. Do you mean draw the rectangle (-1,0)(1,0)(1,1)(-1,1)? That gives rise to three (apparent) triangles: The original triangle, (-1,0)(0,0)(-1,1) and (-1,1)(0,1)(1,1). Neither of the three is congruent any of the other two.

Quote:
 Originally Posted by LaurV There is a simple way to show it too, showing that if you move the vertex on a parallel to the base, the area stays the same. That is the same idea used to compute volumes for cones and spheres, see the principle of Cavalieri.
Of course, that would solve the problem. But how does one prove this?

Thanks again  2012-01-16, 22:24 #5 science_man_88   "Forget I exist" Jul 2009 Dumbassville 8,369 Posts this is a picture of what I read the problem to be. sorry forgot to upload the first time lol. Attached Images Last fiddled with by science_man_88 on 2012-01-16 at 22:30   2012-01-18, 13:34   #6
Unregistered

22·563 Posts Quote:
 Originally Posted by science_man_88 this is a picture of what I read the problem to be. sorry forgot to upload the first time lol.
Sorry, all I see is a red dot.  2012-01-18, 22:34   #7
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts Quote:
 Originally Posted by Unregistered Sorry, all I see is a red dot.
it's only 7 pixel by 7 pixel.   2012-01-19, 01:43   #8
ewmayer
2ω=0

Sep 2002
República de California

9,791 Posts Quote:
 Originally Posted by science_man_88 it's only 7 pixel by 7 pixel.
So why waste our time posting it and then writing your usual lol-laden drivel about it? Sheesh ... you could've simply used some ascii characters to make your "point", whatever it may have been.

Over 4000 posts and "still 100% content-free!" lol.

To the OP, suggest you use wblipp's suggestion:

Code:
                       A
.|
. /|
.  / |
.   /  |
.    /   |
.     /    |
.      /     |
.       /      |
.        /       |
.         /        |
.          /         |
.           /          |
------------------------
D          C           B
You say you understand the formula for right triangles, so the areas of ABC and ABD make sense, right? Now it's obvious that area(ACD) = area(ABD) - area(ABC), from which the "mystery formula" results, namely that it doesn't matter if the triangle in question encloses the line segment giving the height.

Another way to make intutive sense of this result: Imagine building an approximate triangle by horizontally stacking thin strips of wood of constant width (which accumulate to sum to the final height) and steadily decreasing length. (You can use popsicle sticks, and you don't even need to cut them - just lay them out as an aligned grid and use a ruler and pencil to draw a right triangle on the resulting rectangle). Then line a ruler up with the left or right edges and, keeping its intersect with the bottom stick (the x-axis) fixed, push the stack so as to create a tilted triangle like ACD above. Obviously there will be some small imperfections at the edges of the result, but it is easy to see that those vanish as your strips get thinner and thinner. (Cue music of "infinitesimal calculus", pop in DVD of Newton vs Leibniz in their heavyweight slugout at the London Haymarket in the late 1600s, while fondly gazing on one's tatty-eared original "Haymakers at Haymarket: The Ultimate Intercontinental Maths Smackdown" fight-promo poster.)

Last fiddled with by ewmayer on 2012-01-19 at 02:11   2012-01-19, 01:59   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20B116 Posts Quote:
 Originally Posted by ewmayer So why waste our time posting it and then writing your usual lol-laden drivel about it? Sheesh ... you could've simply used some ascii characters to make your "point", whatever it may have been. Over 4000 posts and "still 100% content-free!" lol.
I made it that small because that's all it needed to be to draw out what the triangle looked like.   2012-01-19, 12:26 #10 Unregistered   2×5×72 Posts That does it. Thanks Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post a1call Puzzles 17 2018-02-23 16:28 fivemack Factoring 0 2014-05-01 07:08 Fusion_power Puzzles 10 2013-09-19 03:41 Mini-Geek Puzzles 29 2009-03-19 22:22 mfgoode Puzzles 6 2006-08-05 04:24

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