mersenneforum.org  

Go Back   mersenneforum.org > New To GIMPS? Start Here! > Homework Help

Reply
 
Thread Tools
Old 2012-01-16, 05:12   #1
Unregistered
 

23·223 Posts
Default 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
  Reply With Quote
Old 2012-01-16, 05:49   #2
wblipp
 
wblipp's Avatar
 
"William"
May 2003
New Haven

22×32×5×13 Posts
Default

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.
wblipp is offline   Reply With Quote
Old 2012-01-16, 05:49   #3
LaurV
Romulan Interpreter
 
LaurV's Avatar
 
Jun 2011
Thailand

2·5·887 Posts
Default

Quote:
Originally Posted by Unregistered View Post
<snip>...
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
LaurV is offline   Reply With Quote
Old 2012-01-16, 09:31   #4
Unregistered
 

5×1,931 Posts
Default

Quote:
Originally Posted by LaurV View Post
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 View Post
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 View Post
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
  Reply With Quote
Old 2012-01-16, 22:24   #5
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts
Default

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
science_man_88 is offline   Reply With Quote
Old 2012-01-18, 13:34   #6
Unregistered
 

22·563 Posts
Default

Quote:
Originally Posted by science_man_88 View Post
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.
  Reply With Quote
Old 2012-01-18, 22:34   #7
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts
Default

Quote:
Originally Posted by Unregistered View Post
Sorry, all I see is a red dot.
it's only 7 pixel by 7 pixel.
science_man_88 is offline   Reply With Quote
Old 2012-01-19, 01:43   #8
ewmayer
2ω=0
 
ewmayer's Avatar
 
Sep 2002
Rep├║blica de California

9,791 Posts
Default

Quote:
Originally Posted by science_man_88 View Post
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
ewmayer is offline   Reply With Quote
Old 2012-01-19, 01:59   #9
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

20B116 Posts
Default

Quote:
Originally Posted by ewmayer View Post
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.
science_man_88 is offline   Reply With Quote
Old 2012-01-19, 12:26   #10
Unregistered
 

2×5×72 Posts
Default

That does it. Thanks
  Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Area of a Triangle as a Function of the Lengths of its 3 Sides a1call Puzzles 17 2018-02-23 16:28
A piece of information obvious in retrospect fivemack Factoring 0 2014-05-01 07:08
Col. Chemistry, General Math & Capt. Obvious Fusion_power Puzzles 10 2013-09-19 03:41
area by percent of sun in sky Mini-Geek Puzzles 29 2009-03-19 22:22
Area mfgoode Puzzles 6 2006-08-05 04:24

All times are UTC. The time now is 09:56.

Mon Oct 26 09:56:18 UTC 2020 up 46 days, 7:07, 0 users, load averages: 1.50, 1.82, 1.86

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.