Area of a triangle.

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Area of a triangle.

Postby wbarnes » Fri Jan 15, 2010 8:32 pm UTC

Ok, so I need to find the area of a triangle with the points A=(-2,1) , B=(4,1) , C=(7,4). Having no clear idea how to do this (I feel like I may have learned this somewhere, then forgotten it.) here is what I did.

1- Line segments.
AB= b
AC= square root of 90 --> Base of Triangle
BC= square root of 18

2- determine height
determine slope of base line segment to be 1/3 , create parallel line running through B, find y-intercept of Line Segment AC and new line running through B, determine distance between the two lines, which gives me a height of 2.

So going into 3, I have base = square root of 90 and height = 2

3- A=(1/2)(b)(h) w/ A=(1/2)(square root of 90)(2) = square root of 90 - reduce to 3(square root of 10).

So, with that as my answer I have to know, is there an easier way to do this that I just forgot? Thanks!
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Re: Area of a triangle.

Postby JBJ » Fri Jan 15, 2010 8:38 pm UTC

So, you sacked the cocky khaki Kicky Sack sock plucker?
The second cocky khaki Kicky Sack sock plucker I've sacked since the sixth sitting sheet slitter got sick.
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Re: Area of a triangle.

Postby wbarnes » Fri Jan 15, 2010 8:42 pm UTC

Yup, that is definitely the easier way to do it.

Thank you for the quick answer and the new resource.
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Re: Area of a triangle.

Postby BlackSails » Fri Jan 15, 2010 10:13 pm UTC

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Re: Area of a triangle.

Postby Tirian » Fri Jan 15, 2010 11:55 pm UTC

Heron's formula is crazy overkill, especially when you have the coordinates of the vertices instead of the lengths of their sides. Just draw a bounding rectangle whose sides are parallel to the axes, find the area of that rectangle, and then subtract the areas of the three (or so) extraneous triangles. When the coordinates are integers, the rectangles sides and the base and altitude of each of the triangles will obviously be integers as well, so the formulas work out without having to factor out all of the square roots that would generally come out of Heron's formula.
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Re: Area of a triangle.

Postby wbarnes » Sat Jan 16, 2010 12:39 am UTC

Is there a circumstance under which Heron's Forumla will not give me the area of a triangle? Basically, is it the end all of triangle area finding?
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Re: Area of a triangle.

Postby duckshirt » Sat Jan 16, 2010 1:14 am UTC

Yes, it will work for any triangle in euclidean geometry. It's just extra work to calculate the sides of the angles in many cases.
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Re: Area of a triangle.

Postby Talith » Sat Jan 16, 2010 1:31 am UTC

If you know what a cross product of vectors is, probably the easiest way is to take the corresponding vectors of two of the sides (make sure the vectors both begin at the same vertex) call these vectors u and v. You then have A=(u x v)/2.

N.B. The cross product in two dimensions is the same as the cross product in three dimensions, just set the k component of the vectors to 0.
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Re: Area of a triangle.

Postby Secateurs » Sat Jan 16, 2010 2:28 am UTC

Just to put how I worked it out in here, I did a quick sketch and realised that points A and B have the same y-coordinate. You can find the length of the base easily and then do perpendicular height by looking at the difference between the y-coordinates of B and C. Then it becomes basic area = 0.5*base*height.

It was in the website posted as well, under "If one side is vertical or horizontal".
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Re: Area of a triangle.

Postby Macbi » Sun Jan 17, 2010 11:29 am UTC

    Indigo is a lie.
    Which idiot decided that websites can't go within 4cm of the edge of the screen?
    There should be a null word, for the question "Is anybody there?" and to see if microphones are on.
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Re: Area of a triangle.

Postby Talith » Sun Jan 17, 2010 1:08 pm UTC

Macbi wrote:Uh, Pick's thereom?

Not a trivial thing to compute if all you're given is 3 vertices.
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Re: Area of a triangle.

Postby Macbi » Mon Jan 18, 2010 11:25 am UTC

Well I'd have thought that drawing a sketch and counting was easier than doing Pythagoras' theorem three times and Herons formula once. Obviously the applet has the nicest formula.
    Indigo is a lie.
    Which idiot decided that websites can't go within 4cm of the edge of the screen?
    There should be a null word, for the question "Is anybody there?" and to see if microphones are on.
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Re: Area of a triangle.

Postby Talith » Mon Jan 18, 2010 7:19 pm UTC

Deciding whether a line lies on a point or passes near it isn't that easy. Take http://www.simeonmagic.com/images/Triangle.gif as an example of how hard it is for us to determine such things without writing down the ratio of the height of the line to the base and playing with the coordinates of the point to see if it's in the same ration to a known point on the line. (note, neither of the images depicts a triangle, the first is an irregular quadrilateral and the second is another irregular quadrilateral with a unit square missing)
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