UTM coordinates are given as a zone number together with an easting-northing pair. Given two points (X1, Y1), (X2, Y2) in the same zone.
Simply distance “over the Earth” between the two points is well approximated by the euclidean distance between the easting-northing coordinates using the following formula.
Now calculate the distance b/w stations using the above formula,
\[
\begin{aligned}
\text { Distance between } 1 \text { to } 2 \text { station } & =\sqrt{(744523.2121-744492.8768)^{2}+(2031527.8810-2031516.2320)^{2}} \\
& =32.49 \text { Meters }
\end{aligned}
\]
\[
\begin{aligned}
\text { Distance between } 2 \text { to } 3 \text { station } & =\sqrt{(744523.2121-744518.1044)^{2}+(2031527.8810-2031549.6850)^{2}} \\
& =22.39 \text { Metres }
\end{aligned}
\]
\[
\begin{aligned}
\text { Distance between } 3 \text { to } 4 \text { station } & =\sqrt{(744518.1044-744517.3945)^{2}+(2031549.6850-2031566.7380)^{2}} \\
& =17.06 \text { Metres }
\end{aligned}
\]
\[
\begin{aligned}
\text { Distance between } 4 \text { to } 5 \text { station } & =\sqrt{(744517.3945-744484.8212)^{2}+(2031566.7380-2031556.6270)^{2}} \\
& =34.10 \text { Metres }
\end{aligned}
\]
\[
\begin{aligned}
\text { Distance between } 5 \text { to } 6 \text { station } & =\sqrt{(744484.8212-744484.8769)^{2}+(2031556.6270-2031544.9390)^{2}} \\
& =11.68 \text { Metres }
\end{aligned}
\]
\[
\begin{aligned}
\text { Distance between } 6 \text { to } 1 \text { station } & =\sqrt{(744484.8769-744492.8768)^{2}+(2031544.9390-2031516.2320)^{2}} \\
& =29.80 \text { Metres }
\end{aligned}
\]
Using the triangulation principle, Divides the entire field into triangles.
Now calculate the distance of diagonals 1 to 4, 1 to 5, and 2 to 4 Using the Euclidean formula.
\[
\begin{aligned}
\text { Distance between } 1 \text { to } 4 \text { station } & =\sqrt{(744517.3945-744492.8768)^{2}+(2031566.7380-2031516.2320)^{2}} \\
& =56.14 \text { Meters }
\end{aligned}
\]
\[
\begin{aligned}
\text { Distance between } 1 \text { to } 5 \text { station } & =\sqrt{(744484.8212-744492.8768)^{2}+(2031556.6270-2031516.2320)^{2}} \\
& =41.19 \text { Metres }
\end{aligned}
\]
\[
\begin{aligned}
\text { Distance between } 2 \text { to } 4 \text { station } & =\sqrt{(744523.2121-744517.3945)^{2}+(2031527.8810-2031566.7380)^{2}} \\
& =39.29 \text { Metres }
\end{aligned}
\]
\[
\begin{aligned}
\text { Distance between } 1 \text { to } 4 \text { station } & =\sqrt{(744517.3945-744492.8768)^{2}+(2031566.7380-2031516.2320)^{2}} \\
& =56.14 \text { Meters }
\end{aligned}
\]
\[
\begin{aligned}
\text { Distance between } 1 \text { to } 5 \text { station } & =\sqrt{(744484.8212-744492.8768)^{2}+(2031556.6270-2031516.2320)^{2}} \\
& =41.19 \text { Metres }
\end{aligned}
\]
\[
\begin{aligned}
\text { Distance between } 2 \text { to } 4 \text { station } & =\sqrt{(744523.2121-744517.3945)^{2}+(2031527.8810-2031566.7380)^{2}} \\
& =39.29 \text { Metres }
\end{aligned}
\]
Now, using the NOS problem we can calculate the area of the field as follows
First of all, Calculate offset cutting point distance to stations 2 and 5 from 1.
OFFSET DISTANCE
\[
\begin{aligned}
1-A=\frac{1-4^{2}+1-5^{2}-5-4^{2}}{2 \times 14} & =\frac{56.14^{2}+41.19^{2}-34.10^{2}}{2 \times 56.14} \\
& =32.83 \text{ Meters}
\end{aligned}
\]
\[
\begin{aligned}
1 -B =\frac{1-4^{2}+1-2^{2}-2-4^{2}}{2 \times 14} & =\frac{56.14^{2}+32.49^{2}-39.29^{2}}{2 \times 56.14} \\
& =23.73 \text{ Meters}
\end{aligned}
\]
\[
\begin{aligned}
5-A & =\sqrt{(41.19)^{2}-(32.83)^{2}} \\
& =24.87 \text{ Metres}
\end{aligned}
\]
\[
\begin{aligned}
2-B & =\sqrt{(32.49)^{2}-(23.73)^{2}} \\
& =22.19 \text{ Metres}
\end{aligned}
\]
From Trapezium 1-2-4-5,
\[
\begin{aligned}
\text{Area} & =\frac{1}{2} *(1-4) *[(5-A)+(2-B)] \\
& =\frac{1}{2} * 56.14 *[24.87+22.19] \\
& =1320.97 \text{ Sq.metres}
\end{aligned}
\]
From Triangle 1-5-6
OFFSET CUTTING POINT DISTANCE
\[
\begin{aligned}
1-C & =\frac{(1-5)^{2}+(1-6)^{2}-(5-6)^{2}}{2 *(1-5)} \\
& =\frac{41.19^{2}+29.80^{2}-11.68^{2}}{2 * 41.19}=29.71 \text{ Metres}
\end{aligned}
\]
OFFSET DISTANCE
\[
\begin{aligned}
6-C & =\sqrt{29.80^{2}-29.71^{2}} \\
& =2.31 \text{ Metres} \\
\text{Area} & =\frac{1}{2} * 41.19 * 2.31 \\
& =47.57 \text{ Sq.metres}
\end{aligned}
\]
From triangle 2-3-4,
OFFSET CUTTING POINT DISTANCE
\[
\begin{aligned}
2-D & =\frac{(2-4)^{2}+(2-3)^{2}-(3-4)^{2}}{2 *(2-4)} \\
& =\frac{39.29^{2}+22.39^{2}-17.06^{2}}{2 * 39.29} \\
& =22.32 \text{ Metres}
\end{aligned}
\]
OFFSET DISTANCE
$$
\begin{aligned}
3-D & =\sqrt{22.39^{2}-22.32^{2}} \\
& =1.76 \text{ Metres} \\
\text{Area} & =\frac{1}{2} * 39.29 * 1.76 \\
& =34.57 \text{ Sq.metres}
\end{aligned}
$$
Therefore,
Total area of field \( = 1320.97 + 47.57 – 34.57 = 1333.97 \) Sq. metres (0.329 Acres)
