Triangulation and Representing Area
Given a land area $ABCD$:
You can easily measure the lengths and the perimeter of the land simply by using tapes, but you can't easily determine the internal angles without the use of equipment like theodolites. Since theodolites are somewhat pricey, you can use triangulation to determine the internal angles of the land.
Triangulation is a method used in surveying, wherein a land area is split into a number of triangles. You measure the sides of the triangles and then you can use the measurements to determine the internal angles and the land area by means of mathematical formula.
Some formula that are very resourceful for triangulation are:
Sine Formula.
Cosine Formula.
Heron's Formula
Area of a Triangle General Formula
Using the given land $ABCD$, we first measure the lengths of the different sides. After measurement, we got $60.80m (199ft)$, $30.40m (203ft)$, $61.80m (99.74ft)$ and $31.30m (103ft)$.
Since we do not have the grid coordinates of the different points of the land, we would have to use triangulation, thus, we measure any of the diagonals of the land. Here the measured triangle is $67.42m (221ft)$. [If we had the grid coordinates, we would simply use the Shoelace formula to get the area of the land.]
Now, we have two triangles: $\triangle ABD$ and $\triangle BCD$. We can then use the Heron's formula to get the area of the two triangular portions of the land.
Side AD ($b$): $31.30\text{ m}$
Diagonal BD ($c$): $67.42\text{ m}$ $$s_{ABD} = \frac{60.80 + 31.30 + 67.42}{2} = \frac{159.52}{2} = 79.76\text{ m}$$ $$\text{Area}_{ABD} = \sqrt{79.76 \times (79.76 - 60.80) \times (79.76 - 31.30) \times (79.76 - 67.42)}$$ $$A_{ABD} = \sqrt{79.76 \times 18.96 \times 48.46 \times 12.34}$$ $$A_{ABD} = \sqrt{904,785.4988}$$ $$A_{ABD} \approx \mathbf{951.20\text{ m}^2}$$
Side BC ($b$): $30.40\text{ m}$
Diagonal BD ($c$): $67.42\text{ m}$ $$s_{BCD} = \frac{61.80 + 30.40 + 67.42}{2} = \frac{159.62}{2} = 79.81\text{ m}$$ $$\text{Area}_{BCD} = \sqrt{79.81 \times (79.81 - 61.80) \times (79.81 - 30.40) \times (79.81 - 67.42)}$$ $$A_{BCD} = \sqrt{79.81 \times 18.01 \times 49.41 \times 12.39}$$ $$A_{BCD} = \sqrt{880,958.0772}$$ $$A_{BCD} \approx \mathbf{938.59\text{ m}^2}$$
To determine the internal angles of the land, we use the Cosine formula:
we would have to make the angle the subject of the formula:
Still considering our given quadrilateral land space $ABCD$. We have already splitted the land area into two triangles: $\triangle ABD$ and $\triangle BCD$:
Mark the internal angles and label them:
Since we are having six (6) angles we would solve for all on a spreadsheet:
| $\angle$ | $a$ | $b$ | $c$ | $a^2 +b^2-c^2$ | $2ab$ | $\frac{a^2 +b^2-c^2}{2ab}$ | $\cos^{-1}{(\frac{a^2 +b^2-c^2}{2ab})}$ |
| a | $60.80$ | $31.30$ | $67.42$ | $130.87$ | $3806.08$ | $0.034385$ | $88.03$ |
| b | $67.42$ | $60.80$ | $31.30$ | $7262.41$ | $8198.27$ | $0.885846$ | $27.64$ |
| c | $31.30$ | $67.42$ | $60.80$ | $1828.51$ | $4220.49$ | $0.433244$ | $64.33$ |
| d | $67.42$ | $61.80$ | $30.40$ | $7440.54$ | $8333.11$ | $0.892888$ | $26.76$ |
| e | $67.42$ | $30.40$ | $61.80$ | $1650.38$ | $4099.14$ | $0.402616$ | $66.26$ |
| f | $61.80$ | $30.40$ | $67.42$ | $197.94$ | $3757.44$ | $0.052680$ | $86.98$ |
With the internal angles updated, we have:
To simply represent a land area, one would have to choose a suitable scale (e.g., 1:500, 1:1000, 1:2500) that would allow the parcel of land to fit comfortably on the paper size (A3, A4, etc.). When a suitable scale is chosen, it is recommended to use both a numerical scale (e.g., 1:1,000) and a graphical scale. Here is an example of a scale:
Next, you draw the land area according to the scale to represent the actual measured land.
In a standard land area representation, you would have to show the following in addition to the scale: North arrow, Coordinate grid (Easting/Northing), Datum specification, Survey beacons, Bearings and distances, Area calculation (e.g., 960sqm), Project metadata (e.g. name of owner, name of project, location, etc.), Surveyor's identity, Legend, etc.
- Triangulation is the splitting of a land area into measureable triangles such that the area and internal angles of the splitted land area can be determined.
- Some formulae used in trianguulation are:
Sine formula: $\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}$
Cosine formula: $c = \sqrt{a^2 + b^2 - 2ab\cos{C}}$
Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$; $s = \frac{a + b + c}{2}$
Area of a Triangle: $A = \frac{1}{2}bc\sin{A}$ - To determine a land area using triangulation, first, we split the land area to measureable triangles. Then, we calculate the areas of the different triangles and then sum them to find the area of the land area.
- To determine the internal angles using triangulation, after splitting the land area to measureable triangles, we use the adjusted Cosine Rule to get different internal angles of the triangles, which hints all the internal angles in the land space:
$C= \cos^{-1}{(\frac{a^2 + b^2 - c^2}{2ab})}$ - To representa a land area, we consider a suitable scale (e.g. 1:500, 1:1000). Thereafter, we draw the land area according to indicate. We also indicate the scale on the top or bottom side of the drawing.