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promark
07-28-2006, 06:08 AM
Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

Here are the dimensions.

238' 224' 179' 241'.

I am trying to sell the property and there seems to be conflict with the size?

Thanks:eek:

Robert Fogt
07-28-2006, 10:19 AM
That is an irregular polygon and you cant calculate the area with the given information.

You will also need to know the length of one of the diagonals. That will enable us to split the odd shape into two triangles which we can then calculate the areas of.

grimesm
08-10-2006, 02:09 PM
Wouldn't you just take the average of the sides and the average of the front and back then multiply?

Robert Fogt
08-10-2006, 06:17 PM
No that would be terrible inaccurate.

Take for example a perfect square that is 1 foot wide by 1 foot long. It has an area of 1 square foot. Taking that same square and push the top towards the right making it a parallelogram that is now 1/2 as tall. The sides are still 1 foot by 1 foot, but the area is now only 0.5 square foot.

It's similar to shapes with all sides unequal. Imagine each side being a stick of a certain length, these sticks connected on the ends by rope. You can move the sticks around making different shapes with different areas, but the length of the sticks never change.

So an irregular polygon could have an infinite number of possible areas.

Unregistered
08-14-2006, 09:33 PM
I think Fogt is correct in his explanation. But I believe if you actually can plot the four corners of the land area going by the deed description on paper you can calculate the area in question by subtracting the two irregular triangles remaining from the total area of the square area of the longest leg of your example (241x241/43560=1.33a) The trick is knowing the correct angles of the two irregular triangles. I hope this is not confusing and may be totally incorrect but I am just trying to respond.

Unregistered2
11-17-2006, 11:10 AM
"Unregistered" is correct: you need to know the exact angles, as from the deed description. A property with more than 4 sides needs to be calculated by dividing the shape into squares or rectangles, and triangles, and then figuring the total area of each part then adding them all up. (OR maybe some calculus formula will do it) Our local courthouse Mapping Dept. has some program on their computer that does the math from the property description. MY problem was that the two ends of the property line didn't meet, and slight adjustments had to be made in the numbers so that the two ends would meet (in the computer) and then it could do the math.

Unregistered
12-21-2006, 01:08 PM
Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

Here are the dimensions.

238' 224' 179' 241'.

I am trying to sell the property and there seems to be conflict with the size?

Thanks:eek:
= 1.0564 Acres

Unregistered
04-03-2007, 06:20 AM
Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

Here are the dimensions.

238' 224' 179' 241'.

I am trying to sell the property and there seems to be conflict with the size?

Thanks:eek:
238+179/2=208.5
224+241/2=232.5
208.5*232.5=48476.25

chaitanya shirole
01-06-2008, 06:25 AM
Dear Sir,

Please confirm the area of the plot having the following dimensions by triangular method as we are unable to calculate the correct area and request you to confirm with the sketch.

Plot Size 105.13 L x 52.98W x 108.96 L x 47.86 W

Thanks and Best Regards,

Chaitanya Shirole

JohnS
01-06-2008, 06:17 PM
For a four-sided figure, just the four sides are insufficient to determine the area, as the figure is not rigid. It can be flexed at the corners.

At least one more piece of information is required:
*Either one of the diagonals (and its relation to the sides.
*Or one of the angles (and what sides bound it)

The diagonal divides it into two rigid triangles, or the angle allows the diagonal to be calculated.

If the sides are known accurately, they have probably been taken from a plat which also specifies the bearing angles of each of the lotlines. That also provides the necessary information.

Without that additional information, there is no way to proceed with any real accuracy. There is a theorem that gives an upper bound from the four sides, but the real area could be ANY figure between that upper bound and zero.

Dekerivers101@hotmail.com
09-05-2008, 12:40 PM
Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

Here are the dimensions.

238' 224' 179' 241'.

I am trying to sell the property and there seems to be conflict with the size?

Thanks:eek:

If you use the following formula you can determine the acreage.

a x b x 1/2 =e
c x d x 1/2 =f

add e and f then divide by 43,560'. 43,560 is the square foot area of an acre.

ex: 238 x 224 x 1/2 = 26,656
179 x 241 x 1/2 = 21,569.5

now divide by an acre square foot. 48,225.5 / 43,560 = 1.107 acres.

JohnS
09-05-2008, 02:48 PM
If you use the following formula you can determine the acreage.

a x b x 1/2 =e
c x d x 1/2 =f

add e and f then divide by 43,560'. 43,560 is the square foot area of an acre.

ex: 238 x 224 x 1/2 = 26,656
179 x 241 x 1/2 = 21,569.5

now divide by an acre square foot. 48,225.5 / 43,560 = 1.107 acres.

That would only be correct if ab and cd each formed right triangles. If they did, the figure would be a rectangle so the four sides would be two equal pairs. Since this is false, the method does NOT give the correct area.

As mentioned earlier, you need one angle, or one diagonal, or the bearings of the four sides in addition to length.

Edit: If two opposite angles are known, it can be solved by Bretschneider's formula. Let A and C be the known, opposite angles, and s = 0.5(a+b+c+d), then
area = SQRT((s-a)*(s-b)*(s-c)*(s-d) - a*b*c*d*cos²((A+C)/2))

This is a maximum when A+C = 180°, making the cosine term zero, and for the given problem is less than or equal to 48044.5 ft². (It could be considerably less, depending on the angles, in fact, nearly zero for "pathological" angles.)

Edit 2: Without being completely bizarre, it could be as small as 20198 ft². The angles are essential to any accurate estimate.

01-17-2010, 03:27 AM
[QUOTE=promark;5231]Hi,I
Dear Sir,

Please confirm the area of the plot in square feet having the following dimensions by triangular method as we are unable to calculate the correct area and request you to confirm with the sketch.

Plot Size 79L x 48.4W x 74L x 49.4 W

Thanks and Best Regards,

JohnS
01-17-2010, 05:10 AM
[QUOTE=promark;5231]Hi,I
Dear Sir,

Please confirm the area of the plot in square feet having the following dimensions by triangular method as we are unable to calculate the correct area and request you to confirm with the sketch.

Plot Size 79L x 48.4W x 74L x 49.4 W

Thanks and Best Regards,

There is no triangular method without more info to define the triangles. Please review posts 2, 4, 10, 12 for an explanation of what more we need and why.

Without it, I can say the property is greater than or equal to ZERO square feet, and less than or equal to 3735.88 sq ft.

Edit: Since none of the sides are equal, the area can't actually collapse to zero. The minimum area is 441.2 ft². The maximum as given above. The shape info is critical, as this range is too wide to be useful.

lovey
02-25-2010, 06:37 AM
please tell me the conversion of these measurements in order for me to know the exact size of this lot 192.02 , 243.36 ,40.00 270.96 thank you

JohnS
02-25-2010, 07:34 AM
please tell me the conversion of these measurements in order for me to know the exact size of this lot 192.02 , 243.36 ,40.00 270.96 thank you

Unregistered
03-02-2010, 08:05 PM
how bout if you add all four numbers together and divide by 4 to get an average then multiply length and width average numbers and divide by 43560?

JohnS
03-03-2010, 02:05 AM
how bout if you add all four numbers together and divide by 4 to get an average then multiply length and width average numbers and divide by 43560?

Sure, you can do that if you like playing with your calculator, but it won't be the area.

Consider a rectangle, where we know the area is length * width. Now deform one of the corner angles (all the others vary too) to make a parallelogram. The angle becomes the original length*width multiplied by the sine of the corner angle, and the angle can be deformed to 0° resulting in zero area.

Four unequal sides is slightly more complicated, but the angles matter. A four sided object is not rigid, it can flex at the corners, unless braced with a diagonal, and the area changes while the four sides stay the same.

Unregistered
05-06-2010, 08:11 AM
Area = 5395.5632, Perimeter = 315.4394 in mm plot area

Unregistered
05-06-2010, 02:28 PM
Using Trigonometry to work out the angles, I created two attached triangles. Next, using Heron's Formula, I determined the area of each triangle. Finally I added them together. Here is the solution:

The area of the lot is 5,236.61207 square feet, or 0.12122 acres. This is a very small lot.

Xax

dnrcnnmtb@yahoo.com
09-01-2010, 12:47 AM
sir please measure the area of 4 unequal sides A=10 B=68 C=61 D=44

JohnS
09-01-2010, 02:29 AM
sir please measure the area of 4 unequal sides A=10 B=68 C=61 D=44

Please review posts 2,10,12,14,18 for why you are not giving us enough info to answer your question.

Edit: Lets mark the two-year anniversary of post 12, which gives Breitschneider's formula. If some angular information were known, the area could be determined. As no angular information is given, the first term of the formula gives an upper bound. The maximum possible area is 1665.75 ft².

The quadrilateral is not rigid and may be formed into various shapes, with the ultimate limits being four different triangles formed by two sides collapsing into a straight line. The minimum area of these four triangles is the one with sides, 68, 61, 34, having an area of 1035.07 ft². This is the minimum possible area of the quadrilateral.

Several people have proposed averaging opposite sides and multiplying. This gives 1988 ft² which significantly overestimates the area. It can be proven if the four sides are unequal, this method always overestimates.

Averaging the four sides and squaring is even worse at 2093 ft². Breitschneider's formula gives the only accurate upper bound. With the two required angles, it will give the actual area, or the triangles can be analyzed for minimum and maximum diagonals, and bounds computed for various shapes. The shape with the smallest area is the lower bound.

With either one angle or one diagonal (and where it is located in the diagram) the figure is determined. It can be broken into two triangles and solved. For an n-sided polygon, you need to know the n sides and at least n-3 additional pieces of information (diagonals or sides) to solve. Only a triangle may be solved with just the three sides.

hari_1985
11-05-2010, 02:43 AM
i have a land with four different dimenion now i want to know about that how to Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

Here are the dimensions.

238' 224' 179' 241'.

I am trying to sell the property and there seems to be conflict with the size?

Thanks:eek:

JohnS
11-05-2010, 11:07 AM
i have a land with four different dimenion now i want to know about that how to

Please review posts 2,10,12,14,18, 22 for why you are not giving us enough info to answer your question.

kr_sonawane@yahoo.co.in
12-15-2010, 12:17 AM
380' 35' 374' 245'.

JohnS
12-15-2010, 01:34 AM
380' 35' 374' 245'.

Please review posts 2,10,12,14,18, 22 for why you are not giving us enough info to answer your question.

However, using the upper bound from posts 12 and 22, it is guarenteed to be smaller than 50680 sq ft.

Unregistered
07-09-2011, 08:00 PM
calculate area with sides 76,82,106,112

07-24-2011, 07:07 AM
The four different feets for my land
48-35-49-36.5 ft how to calculate total squrefeet
pls helpme

Robert Fogt
07-24-2011, 02:02 PM
The four different feets for my land
48-35-49-36.5 ft how to calculate total squrefeet
pls helpme

You cant really do that without knowing one of the corner angles or the length of a diagonal.

The sides are very close, but don't be tempted to estimate by multiplying together unless the corner angles are 90°

HerrWarum
07-24-2011, 03:58 PM
Today I have learned something.
http://en.wikipedia.org/wiki/Bretschneider's_formula
Thanks.
:)

JohnS
07-24-2011, 04:25 PM
Today I have learned something.
http://en.wikipedia.org/wiki/Bretschneider's_formula
Thanks.
:)

It's a neat formula and gives an upper bound with no angular information.

However, it requires one more angle than is technically needed. With four sides and ANY angle given, the diagonal opposite that angle can be found by law of cosines(and the area of that triangle given by two sides and the angle). With the diagonal, the other triangle can be solved by Heron's formula.

HerrWarum
07-24-2011, 04:44 PM
I begin to think I learned just enough geometry to get by. . .:(

Unregistered
10-21-2011, 08:40 AM
i have a piece of land cleared and need to know how many acres it is, here is the measurements
a-238'
b-60'
c-260'
d-210'

and the measurement from 1 corner to the other is 260'

JohnS
10-21-2011, 09:22 AM
i have a piece of land cleared and need to know how many acres it is, here is the measurements
a-238'
b-60'
c-260'
d-210'

and the measurement from 1 corner to the other is 260'

The diagonal is a common leg to two triangles. For each of the triangles which two legs plus the diagonal make a triangle
(I'm basically trying to figure out which of the two diagonals you have lised).

Unregistered
10-21-2011, 10:07 AM
the 260 is from the end of c to the end of a that connects to b

Unregistered
10-21-2011, 10:12 AM
a and b makes 1 triangle

JohnS
10-21-2011, 10:28 AM
Once it is divided into two triangles, it can be calculated by Heron's formula. We have a calculator for that:
http://www.onlineconversion.com/shape_area_scalene_triangle.htm

The 238, 60, 260 triangle is 6899.1 ft² and the 260, 210, 260 triangle is 24974.8 ft². totalling 31873.9 ft².

Note: The calculator demands (wrongly) that all three sides be different. For the two 260 sides, I used 260.0001 and 259.9999

Unregistered
10-21-2011, 11:09 AM
please correct me if i'm wrong so would it be 1.46 acres?

JohnS
10-21-2011, 11:16 AM
An acre is 43560 ft², so that is 0.732 acre.

Unregistered
12-16-2011, 08:20 PM
Sir, I have a site measuring total area of 337 sq. yards. length on one side is 68 ft. breadth on one side is 40. Kindly calculate for me the length and breadth of the other 2 sides.

Unregistered
01-01-2012, 11:03 AM
Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

Here are the dimensions.

238' 224' 179' 241'.

I am trying to sell the property and there seems to be conflict with the size?

Thanks:eek:

One Acre, 2 kanal, 15.6 marlas

Unregistered
02-15-2012, 07:57 AM
Plz hellp ..
east side 80 feet
west= 86 feet
north= 84 feet
south = 88 feet

calculate square yards

Unregistered
07-25-2012, 09:56 PM
Hi, I have the same problem and am not too good in math, so here are my measurements of my property. Can someone PLEASE convert them into acres for me?
The property is: 930 x 292 x 1150 x 556ft Thanks!

Unregistered
08-01-2012, 08:27 AM
If you use the following formula you can determine the acreage.

a x b x 1/2 =e
c x d x 1/2 =f

add e and f then divide by 43,560'. 43,560 is the square foot area of an acre.

ex: 930 x 292 x 1/2 = 135,780
1150 x 556 x 1/2 = 319,700

now divide by an acre square foot. 455,480/ 43,560 = 10.456382 acres

Hope this helps.
Also remember earth is not flat and circles will not give you an exact measurement. (Longitude, latitude and land elevations rolling hills etc)
It will always be (+/-) on the number I gave you per the structure of the land (not being flat) Unless your teacher wants the numbers to be rounded to a whole number

Brett

Unregistered
09-02-2012, 09:17 PM
?formula to measure acre, when all 4 sides are different size

Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

Here are the dimensions.

700'*580'*600'*200'

I am trying to purchase the property and there seems to be conflict with the size?

Thanks?

JohnS
09-03-2012, 02:25 AM
?formula to measure acre, when all 4 sides are different size

Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

Here are the dimensions.

700'*580'*600'*200'

I am trying to purchase the property and there seems to be conflict with the size?

Thanks?

The area of a four-sided figure is not fixed by just the four sides. You need additional information which can be:
*the diagonal
*the corner angles
*the bearing angles of the four sides.

Note that depending on angles, a figure with two pair of equal and parallel opposite sides can be either a rectangle or a parallelogram. If a rectangle, the area is L*W. If a parallelogram, L*W*sin(theta) where theta is a corner angle, and the area can be as small as zero. Similar considerations apply when all sides are unequal.

Sierra
01-23-2013, 01:32 PM
I need help it says 6 7/8 in. On the top and 8 1/2 on the side and 5 3/4 in the middle and the want me to find out the area of each parallelogram. Round each decimal answer to the nearest tenth! I need help on it someone help Me please

Robert Fogt
01-23-2013, 05:49 PM
I need help it says 6 7/8 in. On the top and 8 1/2 on the side and 5 3/4 in the middle and the want me to find out the area of each parallelogram. Round each decimal answer to the nearest tenth! I need help on it someone help Me please

Not exactly sure of what you mean by "each" parallelogram, and the measurements don't seem like that of a parallelogram.

http://www.onlineconversion.com/shape_area_parallelogram.htm

rubeshares
01-23-2013, 06:46 PM
Given information pertains to irregular polygon and cannot measure.

Robert Fogt
01-23-2013, 07:16 PM
Given information pertains to irregular polygon and cannot measure.

Possible its a trapezoid.
http://www.onlineconversion.com/shape_area_trapezoid.htm

Unregistered
03-21-2013, 06:23 AM
238+179/2=208.5
224+241/2=232.5
208.5*232.5=48476.25

Unregistered is right !!

JohnS
03-21-2013, 12:02 PM
Unregistered is right !!

This method is an estimate, just not a good one. It ALWAYS over-estimates. See post 12 and/or 22 for details

Unregistered
04-07-2013, 06:08 PM
This method is an estimate, just not a good one. It ALWAYS over-estimates. See post 12 and/or 22 for details

Y'all are all wrong, even if each corner is a right angle, you can't add two sides and times them and add the other two side together and times them and add them all together and divide them by two, it sill will be wrong.

JohnS
04-08-2013, 02:54 AM
Y'all are all wrong, even if each corner is a right angle, you can't add two sides and times them and add the other two side together and times them and add them all together and divide them by two, it sill will be wrong.

Actually, if all four corners are 90°, the figure is a rectangle and opposite sides are equal. The method does work in this special case.

If one angle differs from 90°, at least one more must also differ as the four add to 360°. If any angles differ from 90°, the method fails and as previously mentioned it ALWAYS fails in the direction of overestimating.

lalit
06-21-2013, 03:58 AM
plz hellp ..
East side 80 feet
west= 86 feet
north= 84 feet
south = 88 feet

calculate square yards

Unregistered
07-16-2013, 08:07 PM
[QUOTE=promark;5231]Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

from feet to acre.

Here are the dimensions.

757' 638' 444' 485'.

Thanks:rose

JohnS
07-17-2013, 02:00 AM
[QUOTE=promark;5231]Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

from feet to acre.

Here are the dimensions.

757' 638' 444' 485'.

Thanks:rose

Not enough info to calculate. Please review post #22 (and others referenced there) for details.

Unregistered
08-12-2013, 01:33 AM
Use brahmagupta formula.

Add all the four sides and divide by 2, which is called as S.

Now go to formula.

Area=SQ root of(S-A)*(S-B)*(S-C)*(S-D),

Where a,b,c,d are the different sides of the land.

Sethu.M

JohnS
08-12-2013, 03:06 AM
Use brahmagupta formula.

Add all the four sides and divide by 2, which is called as S.

Now go to formula.

Area=SQ root of(S-A)*(S-B)*(S-C)*(S-D),

Where a,b,c,d are the different sides of the land.

Sethu.M

This formula is only valid for a cyclic quadrilateral (all four vertices lie on the circumference of a circle. Bretschneider's formula contains an additional term which included the cosine of the sum of two opposite angles. When that sum is 180°, the area is given by Brahmagupta, otherwise the area is less.

Unregistered
08-29-2013, 09:59 AM
Wow, one offered up over 10 acres? Best way is to know the angles also. If you have the lot survey then you can add up the sqare ares plus the triangles if you at least scale the straight cross measurements. All else fails call a surveyor for a one hour job. I have worked with some chaining and closed two huge multi acre boundaries flat. Not in Dayton Ohio though where now they only take Kettering High students and Work House inmates for instrument class at the local Jr College and rehab center.

Unregistered
09-19-2013, 02:44 AM
Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

Here are the dimensions.

side A-B 43", side A-C 43.2" , side B-D 32.6" , side C-D 43.5" .

I am trying to buy the property and there seems to be conflict with the size?

Thanks

JohnS
09-19-2013, 02:54 AM
Hi,I hope someone can help me.I am trying to figure out how to measure a odd parcel of land all 4 sides are different dimensions.

Here are the dimensions.

side A-B 43", side A-C 43.2" , side B-D 32.6" , side C-D 43.5" .

I am trying to buy the property and there seems to be conflict with the size?