Need help calculating area of land with 4 unequal sides

How would I calculate the area of a parcel of land that has four unequal sides?

The four sides are:

104.641 feet   X   33.528 feet   X   18.655 feet   X   110.683 feet

Thanks.
DanielAttardAsked:
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d-glitchCommented:
Sorry.  That is not enough information.  You need to know something else.

Do you know any of the angles?

Or one of the diagnol (corner to corner) measurements.

A four sided figure is not adequately defined by the lengths of its sides.
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d-glitchCommented:
If you know the angle of any one corner, you can solve the problem.

  Let   ABCD be the corner points on your plot

        AB  =  104.641 feet  
        BC  =    33.528 feet
        CD  =    18.655 feet
        DA  =   110.683 feet
=================================================

Assume the angle  ABC = 75 degrees

Now you can use the Law of Cosines to find the diagonal between B and C                  

          http://mathworld.wolfram.com/LawofCosines.html
 
                  AC  =  sqrt( AB²  +  BC²  -  2AB*cos(75))  =   101.281 feet

==================================================

Now you have broken you plot into two triangles.

        AB  =  104.641 feet  
        BC  =    33.528 feet
        AC  =  101.281 feet

        and

        AC  =  101.281 feet
        CD  =    18.655 feet
        DA  =   110.683 feet


You can user Heron's formula to find the area of each triangle, and add them up to get the total area of your plot.

            http://mathworld.wolfram.com/Semiperimeter.html

1
avizitCommented:
I wonder if one angle is enough for this  correct me if I am wrong here


Lets say we have 4 sides AB, BC, CD and DA

and lets say we know the angle B

so we can draw   AB and BC with the angle between them as given

now to obtain side D , we take a compass and draw and arc of length CD with C as centre and a arc of length AD with A as centre ,

now these two arcs can cross each other at two places ,  producing two quadrilaterals , one convex and the other concave

                   *A

*B          *D'    * D"  
                   
                    *C

I hope the above diagram comes to scale (somewhat atleast ).. if it does ....you can see if we have angle B fixed , and all the 4 sides fixed , we still have two poinst satisfying for D , D' and D"  ..such that AD' = AD"  = AD and CD'=CD"=CD

Abhijit
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Sergio_HdezCommented:
Well, one angle (or one diagonal) is not strictly enough as avizit claims, as a diagonal/angle will determine both of the 2 triangles the shape is formed of, but not the layout: the second triangle  could be in two positions: Adding more area to the first triangle (like in a usual rectangle), or subtracting it (like in a arrow-like V).

Foir instance, lenghts of 10m each of the four sides, and an angle of 90 degree, can form both a rectangle or a L shape with no area inside.

It is just a thought, in "real world" ones knows how the shape is, so no more meassuraments are necesaries except one diagonal (or angle).

BTW, meassuring an angle is very difficult in real world, while measuring a diagonal is easy, and calculating area from a diagonal is far more easy that from the angle, so better go for a diagonal.
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d-glitchCommented:
>> avizit  
 >> Sergio_Hdez

I agree with everything you've said.  I was assuming a real world problem,
i.e. a convex polygon, from a surveyed plot plot plan.

If you wind up with a convex polygon, you can sill calculate the area by subtracting
the area of the smaller triangle from the area of the larger.

You have real trouble if you wind up with a degenerate bowtie quadrilateral.


                   *A

*B  
                   
                    *C

             *D'        * D"
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DanielAttardAuthor Commented:
Thank you for all your comments.  I guess I should have provided a bit more information.  I only know the angle of one of the corners, which is a right angle.  Here is a sketch, showing the dimensions of each side:

                                        110.683
                A-----------------------
                 |                             ---------------
                 |                                                 -------------B
                 |                                                                /
                 |                                                               /
      33.528 |                                                              / 18.655
                 |                                                             /
                 |                                                            /
               C| _________________________________/ D  
            (right angle)                104.641

The AB side is a straight line, I just had a difficult time drawing it.  I hope you get the idea of the layout.  Can we calculate the area of this since we know the angle of C is 90 degrees?
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d-glitchCommented:
By the method in my second post

I get a diagonal   AD = 109.8811

And an area of 2778.225 sq ft
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d-glitchCommented:
First the diagonal:

    AD  =  sqrt( AC²  +  CD² )  =   109.8811 feet



You have broken you plot into two triangles.


        CD  =  104.641 feet  
        AC  =    33.528 feet
        AD  =  109.881 feet

        S1  =  124.025 feet  ==>  semiperimeter of  Triangle ACD

       Area of ACD  =  sqrt( S1*(S1-CD)*(S1-AC)*(S1-AD))  =  1754.202 sq ft

      ------------------------------------------------------------------------------------

        AB  =  110.683 feet
        BD  =    18.655 feet
        AD  =   109.881 feet


        S2  =  119.610 feet  ==>  semiperimeter of  Triangle ABD

       Area of ABD  =  sqrt( S2*(S2-AB)*(S2-BD)*(S2-AD))  =  1024.023 sq ft

     ------------------------------------------------------------------------------------

    ACD  +  ABD  =  2778.225 sq ft


I did the math with Excel, so it should be okay.  But do check it.









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DanielAttardAuthor Commented:
That's great!  Very understandable.  I really appreciate your help with this.  Thank you so much!
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