How to calculate lot size

[pomchop]

If a lot is 50 feet x 94 x 140 x 100 how many square meters would that be? Anyone know how to calculate?

[mikekim1219]

8,159.298 sqf or 758 sqm

[toybits]

What you need to do is draw the lot on paper - I assume you already have done that, the break it up into either Triangles and rectangles. Getting the area of a rectangle is relatively easy. What about the triangle? If it is a right Triangle (meaning, one of the angles is 90 degrees, simply imagine a duplicate mirror image of that triangle to double the area of the right triangle. What you end up with is another rectangle. Calculate for area - the divide by two to get area of right triangle. If the triangle is not a right triangle, it is a bit more complicated... But what you should try to do is redivide that triangle to get two right triangle.

Am I triangling you out???

[mikekim1219]

Simple method is using Brahmaguta's formula.

Area is equal to square root of (s-a)(s-(s-c)(s-d)

a,b,c,d are the length of the four size and where:

s = (a+b+c+d)/2

Hence you get square feet of 8,159 sqt. convert it to meter, it is 758 sqm minus the decimals.

[johnlandy]

8,159.298 sqf or 758 sqm

8,159.298 sqf or 758 sqm

The question was does "anyone know how to calculate" (the land size) not what the land size is

The response "simple method is using Brahmaguta's formula" I don't think he was expecting an answer from Einstein

The OP has made the question unnecessarily confusing by putting in extraneous dimensions when he could have (should have)

just said, e.g. 50' x 140' which would be converted into metric and that would give the area.

But "stupid is, as stupid does"

Edited July 18, 2013 by johnlandy

[pomchop]

8,159.298 sqf or 758 sqm

thank you

[oneday]

How can you know this formula mikekin1219 and NOT know this only works for a "cyclic quadrilateral"? In other words it must be able to be inscribed inside a circle. It's unlikely this one can be inscribed within a circle.

OP you need to follow what "toybits" said.

[pomchop]

8,159.298 sqf or 758 sqm

8,159.298 sqf or 758 sqm

The question was does "anyone know how to calculate" (the land size) not what the land size is

The response "simple method is using Brahmaguta's formula" I don't think he was expecting an answer from Einstein alt=clap2.gif width=31 height=25>

The OP has made the question unnecessarily confusing by putting in extraneous dimensions when he could have (should have)

just said, e.g. 50' x 140' which would be converted into metric and that would give the area.

But "stupid is, as stupid does" alt=coffee1.gif width=32 height=24>

actually the main questions was: If a lot is 50 feet x 94 x 140 x 100 how many square meters would that be?

And though I may be "stupid" I don't think your simple 50 x140 converted to metric is correct. The lot is not a perfect rectangle.

So perhaps you are correct that" stupid is as stupid does."

[pomchop]

8,159.298 sqf or 758 sqm

8,159.298 sqf or 758 sqm

The question was does "anyone know how to calculate" (the land size) not what the land size is

The response "simple method is using Brahmaguta's formula" I don't think he was expecting an answer from Einstein alt=clap2.gif width=31 height=25>

The OP has made the question unnecessarily confusing by putting in extraneous dimensions when he could have (should have)

just said, e.g. 50' x 140' which would be converted into metric and that would give the area.

But "stupid is, as stupid does" alt=coffee1.gif width=32 height=24>

actually the main questions was: If a lot is 50 feet x 94 x 140 x 100 how many square meters would that be?

And though I may be "stupid" I don't think your simple 50 x140 converted to metric is correct. The lot is not a perfect rectangle.

So perhaps you are correct that" stupid is as stupid does."

[pomchop]

sorry for double post...

i don't begin to claim to understand all the math equations you guys are talking about...i do appreciate the positive answers and calculations...so is the correct answer approx. 758 square meters?

[sunsamourai]

draw a polyline in autocad or equivalent, click properties, you got your area. Change units to metrics if still in imperial.

done.

[ShopBoy]

draw a polyline in autocad or equivalent, click properties, you got your area. Change units to metrics if still in imperial.

done.

you expect layman to have autocad or equivalent installed to just calculate the area size ?

[Fiddlesticks]

sorry for double post...

i don't begin to claim to understand all the math equations you guys are talking about...i do appreciate the positive answers and calculations...so is the correct answer approx. 758 square meters?

Without knowing two of the angles from the lot, it will difficult to calculate the exact area but 758 square meters is a close approximation (probably as close as you will get !

[Lalo27]

Well, at the end we still don't know how to calculate the area of an totally irregular parallelepiped. ..! You do have to divide into triangles but if a right triangle is super easy to calculate, others are not that easy and especially when you are talking about plots of land with trees, bushes and water around.

I am interested to find an easy way as well...

[Shiver]

If you google "calculate area of irregular polygon", it quickly becomes clear that there is no simple way to do that without getting into serious maths. If you don't have access to an app like Acad, then what I would do is get a simple drawing app (like MSPaint that is standard with windows, or even a pencil and paper with grid lines on it) and decide on units for each square, Say a big square = 10ft, and a small square =1ft, you can get an approximate answer.

If you know how to do vertices, then you can do it online as shown here http://www.wikihow.com/Calculate-the-Area-of-a-Polygon

Edited July 18, 2013 by Shiver

[mikekim1219]

.so is the correct answer approx. 758 square meters?

Yes, that's correct

[mikekim1219]

If you google "calculate area of irregular polygon", it quickly becomes clear that there is no simple way to do that without getting into serious maths. If you don't have access to an app like Acad, then what I would do is get a simple drawing app (like MSPaint that is standard with windows, or even a pencil and paper with grid lines on it) and decide on units for each square, Say a big square = 10ft, and a small square =1ft, you can get an approximate answer.

If you know how to do vertices, then you can do it online as shown here http://www.wikihow.com/Calculate-the-Area-of-a-Polygon

All you need is calculator with square root feature and use Brahmaguta's formula as stated in my reply to Pomchop.

[lineofentry]

Draw it on graph paper, count the squares. Yawn.

[pomchop]

Thanks guys...i'll let you math gurus figure the finer details but for my purposes I will just go with 758 square meters.

[HerbalEd]

Since the area is not a square or rectangle, you must know the degree of at least one angle to calculate the area's size. At least, that's how I'd do it. Maybe there's another way?

Edited July 18, 2013 by HerbalEd

[petertucker48]

How has any rectangle got 4 dimensions ?

[petertucker48]

Sorry 4 Different dimensions.

[Moouan]

Add all the dimensions together and take that new dimension as the circumference of a circle then apply formula below:

C = 2pir ( 2 x pi x r) So knowing the circumference, you can find the radius and this is all you need to find the area.

Area = pir^2 ( pi x radius^2)

Edited July 19, 2013 by Moouan

[oxymoron]

Add all the dimensions together and take that new dimension as the circumference of a circle then apply formula below:

C = 2pir ( 2 x pi x r) So knowing the circumference, you can find the radius and this is all you need to find the area.

Area = pir^2 ( pi x radius^2)

Rubbish, try testing your hypothesis

[DGIE]

....?????!!!!!

[Sophon]

sorry for double post...

i don't begin to claim to understand all the math equations you guys are talking about...i do appreciate the positive answers and calculations...so is the correct answer approx. 758 square meters?

No, the correct answer is that with the information you have given it's impossible to calculate the area of your land. With an irregular polygon it's not enough to know the length of the (in this case four) sides, you also need to know the angels at some of the corners.

There are calculators on the net e.g. this one, that will calculate the area for you given enough information. Here are a couple of examples, just to show you how different the results can be depending on the angels.

First example, a somewhat "regular" shape where the angels are close to 90 degrees:

As you can see the area of this land is 8,159 sq.feet or 758 sq.m.

Second example, where the sum of the opposing angels are changed from 180 degrees to a somewhat extreme 30 degrees:

Now your land (without changing the length of any of the sides) is only 2,276 sq.feet or 211 sq.m.

Sophon

Edited July 19, 2013 by Sophon

[wayned]

Here's the area if one of the angles is 90 degrees;

Use the law of cosines to find the diagonal length AC:
SQRT( 50² + 94² - 2(50)(94)cos(90) )
= 106.47.

With the diagonal, the lot is divided into two triangles and I know all the lengths of the sides of these triangles, so I can use Heron's formula to find the area of each triangle:
S_ABC = (50 + 94 + 106.47) / 2 = 125.24
A_ABC = SQRT( (125.24)(125.24 - 50)(125.24 - 94)(125.24 - 106.47) )
= 2350.0.

S_ACD = (106.47 + 140 + 100) / 2 = 173.24
A_ACD = SQRT( (173.24)(173.24 - 106.47)(173.24 - 140)(173.24 - 100) )
= 5305.8.

That makes a total for the two triangles (which is the area of your property) of:

2350.0 + 5305.8

= 7655.8 square feet

= 850.65 square yards

= 711.25 square meters

= 0.17575 acres

= 0.071125 hectares

Go to this website, enter the four sides and one of the angles and it will calculate it for you:

http://mathcentral.uregina.ca/QQ/database/QQ.09.06/s/foursidedlot.php

Edited July 19, 2013 by wayned

[Sophon]

Here's the area if one of the angles is 90 degrees;

Use the law of cosines to find the diagonal length AC:

SQRT( 50² + 94² - 2(50)(94)cos(90) )

= 106.47.

With the diagonal, the lot is divided into two triangles and I know all the lengths of the sides of these triangles, so I can use Heron's formula to find the area of each triangle:

S_ABC = (50 + 94 + 106.47) / 2 = 125.24

A_ABC = SQRT( (125.24)(125.24 - 50)(125.24 - 94)(125.24 - 106.47) )

= 2350.0.

S_ACD = (106.47 + 140 + 100) / 2 = 173.24

A_ACD = SQRT( (173.24)(173.24 - 106.47)(173.24 - 140)(173.24 - 100) )

= 5305.8.

That makes a total for the two triangles (which is the area of your property) of:

2350.0 + 5305.8

= 7655.8 square feet

= 850.65 square yards

= 711.25 square meters

= 0.17575 acres

= 0.071125 hectares

Go to this website, enter the four sides and one of the angles and it will calculate it for you:

http://mathcentral.uregina.ca/QQ/database/QQ.09.06/s/foursidedlot.php

Note that you cannot really calculate the area with just one angle, that is why the website you link to is making the assumption that the shape of the land is convex to get around needing the second angle.

Also, you write "Here's the area if one of the angles is 90 degrees", but the area will actually differ depending on which of the angles it is that is 90 degrees.

Sophon

[pomchop]

Ok for all you math gurus here is a pic of the actual lot...not sure if you can calculate the angles you guys are talking about from this or not....but this is all the info I have....the measurements are all shown in feet as the lot is in the USA. I'm ok with the previous 758 sm but if anyone wants to keep calculating go for it.

[mikekim1219]

Ok for all you math gurus here is a pic of the actual lot...not sure if you can calculate the angles you guys are talking about from this or not....but this is all the info I have....the measurements are all shown in feet as the lot is in the USA. I'm ok with the previous 758 sm but if anyone wants to keep calculating go for it.

Pomchop,

758 sqm is area if it's cyclic quadrilateral shape. If it's not, you need to know at least one of the diagonal length or angel of any one of the corners. Looking at your picture it may not be cyclic quadrilateral meaning all four vertices must lie in a circle. Best way to be sure is measure the distance of any two points.