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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.

The four sides are:

104.641 feet X 33.528 feet X 18.655 feet X 110.683 feet

Thanks.

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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

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

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.

>> 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"

110.683

A-----------------------

| ---------------

| -------------B

| /

| /

33.528 | / 18.655

| /

| /

C| __________________________

(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?

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)

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

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)

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

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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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.