13 replies to this topic

[Scooby]

Do we have any calculation geniuses around here ?

How does one figure the area of a 4 sided polygon (or quadrilateral) in which all sides are of different lengths ?

For example, if we've a lote size of 10 x 15 x 20 x 25 meters, what are the m2 (the area in square meters) ?

Also, what is an easy method of converting area from meters to standard and visa-versa ?

[Klort]

I think you would also need to add in the degree of the angles for each corner... then someone here can come up with an answer.

Hint: Definitely not coming from me.

[filbert]

You will need at least one angle to do the calculations, or you can measure one of the diagonals.  For instance, do you have at least one right angle in the plot?  If so you should be able to use Heron's formula for the areas of the two triangles and then add them up.  Also, you can use Brotschneider's formula, but I believe you will still need at least one known angle which could be derived from the length of a diagonal (connecting two opposing points crisscrossing the quadrilateral).

Then it would simply be a matter of plugging your square meters into square feet conversion engines that are plentiful on the web.

[CB]

solve first for the square 10x15x10x15 which is 150m(2), then add the remaining triangular area which is 1/2 of 10x10 which is 50m(2). so in your case, its 200 square meters which is roughly 2000 sq. feet.

peace

[sdc]

Nice work CB.....impressive...

[CB]

what did you think? that i was just another face?

[marlin]

If I remember from my math classes, what you have here is an irregular quadrilateral, and the exact area cannot be computed with just the perimeter measurements alone. Because of the four different side lengths, using the rectangle area plus the triangle area is not correct.

[sdc]

The triangles would be rifght triangles making it easy to find the area....

[Scooby]

I'm still confused. I need a formula to follow.

Making two triangles sounds like the correct path.
Whatever it is should also work for any parallelogram (square, rectangle, rhombus).
However, unequal sided quadrilaterals (debatedly called a trapezium) I believe would be different
because any triangles created wouldn't any right angles.

[sdc]

If you make a Square or rectacle from a robus the two touch sides fo the newly formed triangles would be 90 degree angles..

[TJDave]

If you make a Square or rectacle from a robus the two touch sides fo the newly formed triangles would be 90 degree angles..

SDC... I think you may be confused about the question... The shape labeled "trapezium" is what they are asking about... no side is equal (no parallels) You can see in the diagram that all of the other quadrilaterals can fit your definition... the trapezuim (also called an "irregular quadrilateral") cannot.




Quadrilaterals

A quadrilateral is a plane figure bounded by four straight lines. There are several familiar types of quadrilaterals. Trapezoids are quadrilaterals that have two parallel sides of unequal lengths. Parallelograms are quadrilaterals that have opposite sides of equal length. A rhombus is a parallelogram (and therefore also a quadrilateral) whose sides are equal, a rectangle is a parallelogram whose angles are all right angles, and a square is a parallelogram whose angles are right angles and whose sides are of equal length. The diagonals of a parallelogram bisect each other; if the parallelogram is a rectangle, the diagonals are also equal. Irregular quadrilaterals have four unequal and nonparallel sides:




The area of a trapezoid is half the sum of the bases times the altitude, or A = [(b1 + b2)/2]h. For a parallelogram, area equals base times height: A = bh.



For irregular quadrilaterals, a good method for determining the area is to divide the figure into two triangles by means of a diagonal, then find the individual areas of the triangles and add them together.

[TJDave]

Once you divide the "irregular quadrilateral" with a diagonal, the resulting 2 triangles will be "scalene triangles"... again, no sides will be equal. An interior angle will be necessary to find the area by using this equation:




If you bisect this triangle from the top corner down to the base you can create 2 right triangles... however to solve for the area you would still need to know the height of the 2 new right triangles...

[sdc]

You are rigth Dave....thanks...

[Scooby]

..........

Yes, that seems like it would be correct.
However, I don't think I can do any formulas involving sinning. It's against my religion (of basic math).

Interestingly (to me) with some of these quadrilaterals, while changing only the angles but not the segment lengths, the areas will change.
Other quadrilaterals will find the area remaining the same.