----- Original Message -----

From: Justin .

To: geometry@yahoogroups.com

Sent: Thursday, November 16, 2006 7:08 AM

Subject: [Norton AntiSpam] Re: Re: [geometry] Re: How to find an angle in a triangle, and how to calculate an irregular area?

If you think about that carefully Carl, that would
give me a triangle with 2 angles being right angles,
right? Don't think so. Or did I miss something?

--- Carl Adkins <ca93901@alltel.net> wrote:

> An angle inscribed in a semicircle is always a right
> angle. You state the small angle is 30 degrees, so
> you have a 30-60 right triangle. The side opposite
> the 30 degree angle is half of the hypotenuse (aka
> diameter), which would be equal to the radius. So,
> you have a right triangle with sides of 2X, X, and
> X*(sqrt of 3), with an area equal to (X/2)*[X(sqrt
> of 3)].
>
> Carl Adkins
> ----- Original Message -----
> From: Justin .
> To: geometry@yahoogroups.com
> Sent: Wednesday, November 15, 2006 8:24 PM
> Subject: [Norton AntiSpam] Re: [geometry] Re: How
> to find an angle in a triangle, and how to calculate
> an irregular area?
>
>
> Hi Everyone
> I have an idea on how to do this calculation of
> irregular area. I will try to explain my idea to
> see
> if you all think it makes good sense and would be
> accurate. As I am not a mathematician, it may be
> wrong, and, sorry I do not know the conventional
> way
> to express it, so forgive me for my misuse/abuse
> of
> terms and symbols, but hopefully you can
> understand
> what I will try to express.
>
> By measuring 6 diameters, I have 12 triangles. The
> small angle of all the triangles are all the same
> - 30
> degrees. I do not know the formula for areas of a
> triangle, but, there must be a very simple one
> when
> the small angle and its 2 lengths are known
> (Anyone
> know the formula?)
>
> So from that it is easy to calculate the area of
> the
> shape.
>
> However, this shape has 12 straight sides, whereas
> the
> actual shape it represents (the real tube
> crossection)
> is more curvey. If this irregular curve were
> approximately a straight line, I would expect the
> curve to be going "over" AND "under" the outer
> side of
> the triangle equally, and thus the result to be
> fair.
> However, as the irregular curve is approximately a
> circle, I would expect it to be going "over" more
> than
> "under", by which I mean, more of the irregular
> curve
> (ie more AREA) would lie OUTSIDE the triangles
> than
> INSIDE. Thus, I would expect the area resulting
> from
> the 12 sided shape to be an underestimate.
>
> I have this suggestion for correction:
>
> - Call the area of the 12 sided shape MUM.
> - Generate a circle where
> radius = length b of isosceles triangle (with
> angle
> inbetween its 2 b sides being 30 degrees and area
> being MUM/12 )
> Call the area of this circle DAD
>
> - DAD minus MUM = BABY
>
> Okay, so, I presume the real area of the irregular
> shape must be greater than MUM, and less than DAD.
> Do
> you think? (It may help to visualize that the
> shape
> looks somwhat like the crossection of a
> treetrunk.)
>
> So I guess that the formula would be
> Area = MUM + (BABY * x)
> where x is less than 1.
>
> Does that sound right? Then, I do not know what x
> should be! Any suggestions?
>
> Also, as the complete formula includes the formula
> of
> the triangles and so on, and it takes all the
> steps
> for the different diameters, if anyone can write
> this
> more clearly, it would be great! Plus, this should
> be
> very simple (for those who know how!) to turn into
> a
> little program (or is it called a routine?). So
> that
> all I have to do is input the diameters, the
> number of
> diameters and the angle (between the diameters,
> which
> anyway is implicite in the number of diameters),
> and
> then it will do all of this calculation.
>
> Thank you all!
> Justin
>
> --- Ben Saucer <bsaucer2@comcast.net> wrote:
>
> > At 11:48 PM 11/10/2006, you wrote:
> >
> > >Hi all
> > >Just to further clarify on the second of my too
> > >questions:
> > >
> > >I am not concerned with volume. I am trying to
> work
> > >out area. The area of a crossection (a number
> of
> > >crossections actually). And it seems to me also
> > that
> > >the average diameter will not have a direct
> > >relationship to the area, surely? For example,
> a
> > line
> > >may have its longest diameter at 20cm and its
> > shortest
> > >diameter at 0cm. Although its average diameter
> is
> > >therefore 10cm, its area is 0. That's why I
> thought
> > >there may be a more accurate way of working it
> out.
> > >For example, wouldn't an elipse use a different
> > >formula for working out the area than a circle?
> > What I
> > >am measuring is closer to an elipse. However it
> is
> > >more irregular. It may aid you if I give you
> the
> > image
> > >of a lake. How would we work out the area of a
> > lake,
> > >when our available data is a number of diameter
> > >measurements of the lake?
> > >By the way I figured if I actually used the
> > diameters
> > >to plot out a map on graph paper, I could count
> the
> > >little squares to find the area!! However, as I
> > have
> > >to do a large number of such calculations, that
> is
> > not
> > >practical. I feel sure there must be a suitable
> > >formula.
> > >
> > >Thank you!
> > >Best wishes
> > >Justin
> >
> > My previous answer was based on knowing the
> radii
> > from some center.
> > However, knowing diameters is more difficult,
> since
> > they may not all
> > meet at a common center.
> >
> > There might be a way to use a tree as a
> "center",
> > but instead of
> > using a common angle, you must measure the
> distance
> > of each point of
> > the lake from the tree, and the distance between
> > consecutive chosen
> > points on the lake. The area of the triangles
> (no
> > longer isosceles)
> > require a different formula. Also, triangles
> formed
> > from points on
> > the near side of the lake must be subtracted
> from
> > the triangles from
> > the far side of the lake. Some triangles have no
> > area at all.
> >
> >
>
=== message truncated ===

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