GEFS-CE11 - (Fundamentals of Surveying)
INSTR. MMMayos UNIVERSITY OF THE CORDILLERAS MODULE 3
METHODS OF DETERMINING AREA OF CLOSED TRAVERSE
A. AREA BY TRIANGLES
1. Using the definition of the area of a triangle
When the sides of the traverse have few sides and have a small area, this method
can be used. This method is based on the area of a triangle which is equal to one-
half of the product of its base and altitude.
P
θ
β 90 0
Q
N
M
Fig. A
1
𝐴𝑟𝑒𝑎 = (𝐵𝑎𝑠𝑒)(𝐴𝑙𝑡𝑖𝑡𝑢𝑑𝑒)
2
1
𝐴 = 𝐵𝐻
2
But from right triangle MNP, solving for H
𝐻
𝑠𝑖𝑛 𝛽 =
𝑋
𝐻 = 𝑋𝑠𝑖𝑛 𝛽
Thus
1
𝐴 = 𝐵𝐻
2
𝐴 = 𝐵(𝑋𝑠𝑖𝑛 𝛽)
Or from right triangle QNP, solving for H
𝐻
sin 𝜃 =
𝑌
𝐻 = 𝑌𝑠𝑖𝑛 𝜃
1
𝐴 = 𝐵𝐻
2
1
𝐴= 𝐵(𝑌𝑠𝑖𝑛 𝜃)
2
, GEFS-CE11 - (Fundamentals of Surveying)
INSTR. MMMayos UNIVERSITY OF THE CORDILLERAS MODULE 3
Therefore, the area of a triangle can be determined as equal to one-half the
product of two adjacent sides multiplied by the sine of the included angle. In
Fig. A above, if the adjacent sides are X and B, the included angle is β, and if the
two adjacent sides are Y and B, the included angle is θ. This can be use when two
adjacent sides and the included angle are known.
2. Using Heron’s Formula
Heron’s formula can be use when the three sides of the triangle are known. In
the figure above (Fig. A), sides X, Y, and B are the sides, the area of the triangle
is,
𝐴= 𝑆(𝑆 − 𝑋)(𝑆 − 𝑌)(𝑆 − 𝐵)
Where:
1
𝑆= (𝐴 + 𝐵 + 𝑌)
2
The two methods mentioned above can be used to determine the area of a closed
traverse by dividing the closed traverse into a series of triangles.
Sample Problem no. 1
Given in the table below is the technical descriptions of a transit-tape survey of a
closed traverse. Determine the area of the closed traverse using triangles.
Course Distance Bearing Latitudes (m) Departures (m)
(m)
1–2 𝟐𝟗𝟖. 𝟒𝟒𝟔 S 36057’ E −238.506 +179.401
2–3 𝟒𝟎𝟑. 𝟒𝟕𝟒 S 43038’ W −292.022 – 278.413
3–4 𝟐𝟓𝟐. 𝟒𝟏𝟎 N 73016’W +72.674 – 241.721
4–5 𝟑𝟒𝟖. 𝟓𝟓𝟎 N 12010’ E +340.721 +73.459
5–1 𝟐𝟗𝟏. 𝟖𝟎𝟎 N 66020’E +117.133 +267.259
EL = 0.00 m ED = - 0.015 m
N
Final Figure 1
0
66 20’
5 36057’
2
0
12 10’ 0
43 38’
4 0
73 16’
3
Solution:
INSTR. MMMayos UNIVERSITY OF THE CORDILLERAS MODULE 3
METHODS OF DETERMINING AREA OF CLOSED TRAVERSE
A. AREA BY TRIANGLES
1. Using the definition of the area of a triangle
When the sides of the traverse have few sides and have a small area, this method
can be used. This method is based on the area of a triangle which is equal to one-
half of the product of its base and altitude.
P
θ
β 90 0
Q
N
M
Fig. A
1
𝐴𝑟𝑒𝑎 = (𝐵𝑎𝑠𝑒)(𝐴𝑙𝑡𝑖𝑡𝑢𝑑𝑒)
2
1
𝐴 = 𝐵𝐻
2
But from right triangle MNP, solving for H
𝐻
𝑠𝑖𝑛 𝛽 =
𝑋
𝐻 = 𝑋𝑠𝑖𝑛 𝛽
Thus
1
𝐴 = 𝐵𝐻
2
𝐴 = 𝐵(𝑋𝑠𝑖𝑛 𝛽)
Or from right triangle QNP, solving for H
𝐻
sin 𝜃 =
𝑌
𝐻 = 𝑌𝑠𝑖𝑛 𝜃
1
𝐴 = 𝐵𝐻
2
1
𝐴= 𝐵(𝑌𝑠𝑖𝑛 𝜃)
2
, GEFS-CE11 - (Fundamentals of Surveying)
INSTR. MMMayos UNIVERSITY OF THE CORDILLERAS MODULE 3
Therefore, the area of a triangle can be determined as equal to one-half the
product of two adjacent sides multiplied by the sine of the included angle. In
Fig. A above, if the adjacent sides are X and B, the included angle is β, and if the
two adjacent sides are Y and B, the included angle is θ. This can be use when two
adjacent sides and the included angle are known.
2. Using Heron’s Formula
Heron’s formula can be use when the three sides of the triangle are known. In
the figure above (Fig. A), sides X, Y, and B are the sides, the area of the triangle
is,
𝐴= 𝑆(𝑆 − 𝑋)(𝑆 − 𝑌)(𝑆 − 𝐵)
Where:
1
𝑆= (𝐴 + 𝐵 + 𝑌)
2
The two methods mentioned above can be used to determine the area of a closed
traverse by dividing the closed traverse into a series of triangles.
Sample Problem no. 1
Given in the table below is the technical descriptions of a transit-tape survey of a
closed traverse. Determine the area of the closed traverse using triangles.
Course Distance Bearing Latitudes (m) Departures (m)
(m)
1–2 𝟐𝟗𝟖. 𝟒𝟒𝟔 S 36057’ E −238.506 +179.401
2–3 𝟒𝟎𝟑. 𝟒𝟕𝟒 S 43038’ W −292.022 – 278.413
3–4 𝟐𝟓𝟐. 𝟒𝟏𝟎 N 73016’W +72.674 – 241.721
4–5 𝟑𝟒𝟖. 𝟓𝟓𝟎 N 12010’ E +340.721 +73.459
5–1 𝟐𝟗𝟏. 𝟖𝟎𝟎 N 66020’E +117.133 +267.259
EL = 0.00 m ED = - 0.015 m
N
Final Figure 1
0
66 20’
5 36057’
2
0
12 10’ 0
43 38’
4 0
73 16’
3
Solution:
