20120116, 05:12  #1 
2^{3}·223 Posts 
Area of Triangle, nonobvious 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 (coordinateless) geometry alone? Thanks 
20120116, 05:49  #2 
"William"
May 2003
New Haven
2^{2}×3^{2}×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.

20120116, 05:49  #3 
Romulan Interpreter
Jun 2011
Thailand
2·5·887 Posts 
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 20120116 at 06:05 
20120116, 09:31  #4  
5×1,931 Posts 
Quote:
Quote:
Quote:
Thanks again 

20120116, 22:24  #5 
"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.
Last fiddled with by science_man_88 on 20120116 at 22:30 
20120118, 13:34  #6 
2^{2}·563 Posts 

20120118, 22:34  #7 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 

20120119, 01:43  #8 
∂^{2}ω=0
Sep 2002
República de California
9,791 Posts 
So why waste our time posting it and then writing your usual lolladen 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% contentfree!" lol. To the OP, suggest you use wblipp's suggestion: Code:
A . . / . /  . /  . /  . /  . /  . /  . /  . /  . /  . /   D C B 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 xaxis) 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 tattyeared original "Haymakers at Haymarket: The Ultimate Intercontinental Maths Smackdown" fightpromo poster.) Last fiddled with by ewmayer on 20120119 at 02:11 
20120119, 01:59  #9 
"Forget I exist"
Jul 2009
Dumbassville
20B1_{16} Posts 
I made it that small because that's all it needed to be to draw out what the triangle looked like.

20120119, 12:26  #10 
2×5×7^{2} Posts 
That does it. Thanks

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