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### Land Plot Area

```Date: 08/28/2002 at 12:10:16
From: Tom Guess
Subject: Geometry (land plat area)

I have a parcel of land that is a trapezoid. The four sides are not
equal.

The road frontage is 363.04 ft in a northeasterly direction.

One side is 1950.16 ft in a southeasterly direction.

The back is 338.49 ft in a southwesterly direction.

The final side is 2000.75 ft in a northwesterly direction.

The surveyor has determined it to be 624,630.88 square ft and 14.340
acres overall. The legal description is 15.8 acres.

How can I figure out the square footage?

Thanks!
Tom
```

```
Date: 08/29/2002 at 08:56:04
From: Doctor Peterson
Subject: Re: Geometry (land plat area)

Hi, Tom.

You call this a trapezoid, but I suspect you do not mean that two
sides are exactly parallel, or you would probably have said so.
Without knowing at least one exact angle (better than "northeast"), I
can't find the exact area. Do you know the bearing along each side?

The square feet you give do convert to 14.340 acres; and 15.8 acres
is very close to what you get by just multiplying the average "length"
by the average "width," which may suggest that the area was originally
found by a simplistic and inaccurate method, or it may be correct if
the angles are just right.

Please send me some angle data, and I can give you a better answer.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/29/2002 at 18:04:46
From: Tom Guess
Subject: Geometry (land plat area)

Doctor Peterson;

The angles and measurements that are on the plat are as follows:

363.04 ft -- N 38 deg 15' 00" E
1950.16 ft -- plat reads: S 80 deg 36' 34" E I would assume it should
actually be the reciprocal of: S 260 deg 36' 34" W
338.47 ft -- plat reads: S 30 deg 59' 49" W  I would assume this
should also be the reciprocal of: S 210 deg 59' 49" W
200.75 fr -- N 80 deg 42' 11" W

Hope this helps; that's all the plat has on it.

Thanks!
Tom
```

```
Date: 08/29/2002 at 23:27:33
From: Doctor Peterson
Subject: Re: Geometry (land plat area)

Hi, Tom.

Okay, let's first convert the angles from surveyors' terms to mine.
I'll measure angles clockwise from north, and put them in decimal form
(the form "N x E" means we start at north and go x degrees toward the
east):

AB: N 38 deg 15' 00" E --> 38.25 deg
BC: S 80 deg 36' 34" E --> 180-80.61 = 99.391 deg
CD: S 30 deg 59' 49" W --> 180+30.997 = 210.997 deg
DA: N 80 deg 42' 11" W --> 360-80.703 = 279.297 deg

This should look something like

B----_____ 1950
363/          -----_____
/                     ----C
A----_____                /
-----_____     /338
2000      ----D

Now we can turn each side into a vector, indicating how far N and
E it goes; the north component is L cos(A) where L is the length of
the side and A is the angle from north, and the east component is
L sin(A).

Side   Length   Angle     North     East
----   ------   -----     -----     ----
AB    363.04   38.250   285.101   224.756
BC   1950.16   99.391  -318.195  1924.026
CD    338.47  210.997  -290.135  -174.309
DA   2000.75  279.297   323.224 -1974.469

From here we can find the coordinates of each vertex relative to A:

North     East
-----     ----
A     0         0
B   285.101   224.756
C   -33.094  2148.782
D  -323.229  1974.473
(A)    -.005      .004

I included the location we get back to when we come to A, to check
the accuracy of the calculations, and it looks pretty good.

Now the last step is to find the area using this formula from our FAQ
on analytic geometry:

http://mathforum.org/dr.math/faq/formulas/faq.ag2.html#twopolygons

K = [(x1y2 + x2y3 + x3y4 + ... + xny1) -
(x2y1 + x3y2 + x4y3 + ... + x1yn)]/2

This makes the area

K = [(0*224.756 + 285.101*2148.782 + -33.094*1974.473 + -323.229*0)
- (285.101*0 + -33.094*224.756 + -323.229*2148.782 + 0*1974.473)]
/2

= [547276.687 - -701986.732]/2 = 1249263.419/2
= 624631.709 ft^2

This, divided by 43560 ft^2 per acre, gives 14.34 acre.

Not surprisingly, the surveyor appears to have used his own data
correctly.

I've been very happy to go through this, because people are always
asking us to find the area of their land given the lengths of the
four sides, but I don't recall that anyone has ever given us the
information necessary to actually do it. This will therefore go into
our archives as an example.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/30/2002 at 12:22:53
From: Tom Guess
Subject: Thank you (Geometry (land plat area))

Doctor Peterson,

Thank you very much for your time.  This has helped me tremendously!

Tom Guess
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Practical Geometry
High School Trigonometry

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