Property/Boundary Surveys and Lot Computation - Detailed Explanation
A property or boundary survey is the cornerstone of Philippine land titling. It physically establishes or re-establishes the corners of a parcel of land and produces the technical description — the set of bearings, distances, and computed area that appears on the Certificate of Title issued under Presidential Decree 1529 (Property Registration Decree). Under Republic Act 8560 (Geodetic Engineering Act of 1998), only licensed geodetic engineers may conduct cadastral, boundary, and subdivision surveys in the Philippines. This chapter covers the full workflow: from recovering control points and monument corners, through computing coordinates of lot corners using the shoelace (coordinate) method, to deriving bearing-and-distance lines for the technical description, and finally checking closure and adjusting misclosure. Mastery of these skills is essential for the PRC Geodetic Engineer Licensure Examination, where lot computation problems appear in multiple choice, identification, and problem-solving formats.
Boundary and Relocation Surveys: Definitions and Legal Framework
Examples
A landowner in Bulacan holds a TCT whose technical description cites four corners tied to BLLM No. 1, Bocaue Cadastre. The concrete monuments have been removed during road construction. What type of survey is needed and what is the legal basis?
Solution
A relocation survey is required. The geodetic engineer recovers BLLM No. 1 from the NAMRIA/LMB control network and uses the bearing and distance ties in the LDC sheet to re-establish the four corner monuments.
This is a relocation survey because the corners existed previously and are being re-established from the record. Legal basis: PD 1529 Section 48 (reestablishment of lost monuments); RA 8560 Section 22 (only licensed GEs may conduct surveys).
A boundary survey establishes or confirms the corners of a parcel of land from the evidence contained in the title, approved plan (Lot Data Computation / LDC sheet), and on-the-ground monuments. It resolves overlaps (encroachments) and gaps with adjoining lots. A relocation survey re-establishes lost, obliterated, or disputed corners using the recorded technical description and tie lines to known control monuments. Both types of surveys are regulated by the Land Registration Authority (LRA) and the Department of Environment and Natural Resources (DENR) through the Land Management Bureau (LMB). Under RA 8560, surveys must be conducted by a licensed geodetic engineer. Under PD 1529, the approved plan and technical description are attached to the decree of registration and are binding on all parties. Under CA 141 (Public Land Act), surveys of public agricultural lands follow rules prescribed by the DENR.
Applications
- Titling of untitled agricultural land under CA 141 free patent applications.
- Subdivision of a mother lot into individual lots before sale (requires DENR-approved subdivision plan).
- Resolution of encroachment disputes filed before the LRA or regular courts.
- Preparation of verified survey returns submitted to the LMB for approval.
Common misconceptions
- A boundary survey is NOT the same as a cadastral survey. A cadastral survey is a systematic, government-initiated survey of all lands in a municipality; a boundary survey is parcel-specific and privately commissioned.
- Relocation does not change the technical description — it merely recovers the physical position of corners from the existing record.
- RA 4374 created the system of cadastral surveys but the actual survey operations are now administered by DENR-LMB, not a separate cadastral bureau.
Related concepts
- Lot Data Computation (LDC) Sheet
- Technical Description
- Land Registration Authority (LRA)
- NAMRIA Control Network / PRS92
- Cadastral Survey (RA 4374)
Common exam questions
Example
Q: A survey that re-establishes the lost corners of a titled lot from its approved plan and tie lines is called a ___. A: Relocation survey.
Approach
Know the distinction between boundary survey (establishing corners from title/plan for the first time or confirming existing corners) vs. relocation survey (recovering lost corners).
Example
Q: Under which law may a court order the reestablishment of lost or obliterated land survey monuments? A: PD 1529, Section 48.
Approach
Memorize which law governs which aspect: RA 8560 = profession, PD 1529 = registration, CA 141 = public land, RA 4374 = cadastral.
Key points to remember
- Only licensed geodetic engineers may sign and seal survey plans under RA 8560, Section 22.
- PD 1529 requires an approved survey plan (LRC/DENR-approved) before a decree of registration can be issued.
- A boundary survey works from title/plan to the ground; a relocation survey recovers lost corners from the title/plan record.
- Overlaps and gaps with adjacent lots must be reconciled and reported to LRA/DENR.
- CA 141 governs surveys of public lands; RA 4374 created the Cadastral Survey system; RA 8560 regulates the geodetic engineering profession.
- Control monuments used are BM (benchmarks) and triangulation/traverse control points of the NAMRIA national network, which is referenced to PRS92 (tied to WGS84).
Coordinate System and Lot Corner Coordinates
Examples
A lot has corners listed in PPCS Zone IV: A(500010.00 E, 1620030.00 N), B(500060.00 E, 1620030.00 N), C(500070.00 E, 1620070.00 N), D(500020.00 E, 1620060.00 N). Translate to local coordinates for computation.
Solution
Subtract the smallest easting (500010) and smallest northing (1620030) from all points: A(0,0), B(50,0), C(60,40), D(10,30). Area computation proceeds identically.
Translating to a local origin does not change distances or areas. This simplifies arithmetic and reduces risk of transcription errors — a practical technique for board exams.
All Philippine surveys are now referenced to the Philippine Reference System 1992 (PRS92), which is aligned to WGS84. Field coordinates are expressed in the Philippine Plane Coordinate System (PPCS), a Transverse Mercator projection divided into five zones (Zones I–V), equivalent to UTM zones but with Philippine-specific parameters. Corner coordinates are given as Easting (E) and Northing (N) in metres. For lot computation purposes — especially in board-exam problems — a local Cartesian system with an assumed origin may be used. The important rule is consistency: all corners must be in the same system, with the same origin, and traversed in a consistent direction (either all clockwise or all counter-clockwise) around the lot boundary.
Applications
- Setting up the coordinate table for the shoelace (DMD) area computation.
- Transforming PPCS coordinates to local grid for simpler arithmetic in field computation.
- Checking consistency of coordinates against the approved plan scale and dimensions.
Common misconceptions
- PRS92 is NOT the same as the old Luzon Datum of 1911. They differ significantly — using Luzon 1911 coordinates with PRS92 formulas introduces errors of tens of metres.
- PPCS and UTM use the same projection mathematics but different zone boundaries and false origins — they are not interchangeable without transformation.
Related concepts
- WGS84
- PRS92
- PPCS / UTM
- Transverse Mercator Projection
- Easting and Northing
Common exam questions
Example
Q: The official geodetic reference system used in Philippine land surveys since 1992 is ___. A: Philippine Reference System 1992 (PRS92).
Approach
Know that PRS92 is referenced to WGS84 and that PPCS is its map projection.
Key points to remember
- PRS92 is the official horizontal datum of the Philippines, referenced to WGS84 with a 3-parameter transformation.
- PPCS uses Transverse Mercator projection; false easting = 500,000 m; scale factor k₀ = 0.99995 for most zones.
- For lot area computation, the shoelace formula works in any consistent Cartesian coordinate system.
- Corners must be listed in order around the boundary — either all clockwise or all counter-clockwise. The absolute value in the shoelace formula handles the sign.
- In board exams, coordinates may be given as local grid (arbitrary origin) — treat them identically to PPCS/UTM coordinates.
Shoelace Formula for Lot Area
Examples
Board-style Problem: A lot has corners (E, N) in metres: A(0, 0), B(50, 0), C(60, 40), D(10, 30). Find the area in m² and in hectares.
Solution
Step 1 — Set up the shoelace table:
| Corner | E | N | E·N(next) | E(next)·N |
|---|---|---|---|---|
| A | 0 | 0 | 0×0=0 | 50×0=0 |
| B | 50 | 0 | 50×40=2000 | 60×0=0 |
| C | 60 | 40 | 60×30=1800 | 10×40=400 |
| D | 10 | 30 | 10×0=0 | 0×30=0 |
(back to A)
Step 2 — Sum each column:
Σ(E·N_next) = 0 + 2000 + 1800 + 0 = 3800
Σ(E_next·N) = 0 + 0 + 400 + 0 = 400
Step 3 — Apply formula:
A = ½|3800 − 400| = ½ × 3400 = 1700 m²
Step 4 — Convert:
1700 m² ÷ 10,000 = 0.17 ha
The wrap-around is critical: after corner D(10,30), the 'next' is A(0,0). The formula pairs each corner's E with the NEXT corner's N, and vice versa. Missing the wrap-around (a very common error) gives a wrong answer of 900 m² instead of 1700 m².
Exercise 1 from chapter: Find the area of a lot with corners (0,0), (80,0), (90,50), (20,40).
Solution
Shoelace table (corners listed as A,B,C,D → back to A):
A(0,0)→B(80,0): 0×0−80×0 = 0
B(80,0)→C(90,50): 80×50−90×0 = 4000
C(90,50)→D(20,40): 90×40−20×50 = 3600−1000 = 2600
D(20,40)→A(0,0): 20×0−0×40 = 0
Sum = 0 + 4000 + 2600 + 0 = 6600
A = ½|6600| = 3300 m²
In hectares: 3300 ÷ 10,000 = 0.33 ha
Each product EᵢNᵢ₊₁ − Eᵢ₊₁Nᵢ is computed for each pair of consecutive corners. Summing all products and taking half the absolute value gives the area.
The shoelace formula (also called the Coordinate Method or DMD Method in Philippine survey practice) computes the area of any polygon from the Cartesian coordinates of its vertices taken in order. Given n corners with coordinates (E₁,N₁), (E₂,N₂), …, (Eₙ,Nₙ), the signed area is: A = ½|Σᵢ(Eᵢ·Nᵢ₊₁ − Eᵢ₊₁·Nᵢ)| where the index wraps: Eₙ₊₁ = E₁ and Nₙ₊₁ = N₁. The absolute value ensures a positive area regardless of whether corners are listed clockwise or counter-clockwise. An equivalent tabular method used in Philippine practice lists two columns: (Eᵢ·Nᵢ₊₁) in one column and (Eᵢ₊₁·Nᵢ) in the other, then subtracts column totals and halves the absolute difference. The result is in square metres (m²); divide by 10,000 to convert to hectares.
Applications
- Computing the area stated on the Certificate of Title and LDC sheet.
- Verifying that a surveyed lot area matches the approved plan area within tolerance.
- Checking computations after adjusting misclosure — recompute area from adjusted coordinates.
- Area computation for irregular (multi-sided) lots in subdivision surveys.
Common misconceptions
- CRITICAL: Forgetting the wrap-around — the last corner MUST connect back to the first corner in the shoelace table.
- Using latitude/departure sums instead of shoelace products — the DMD method requires a separate procedure; do not mix methods.
- The shoelace formula requires corners to be in ORDER around the boundary. Random listing gives a wrong answer.
- Area in m² must be divided by 10,000 (not 1,000) to get hectares.
Related concepts
- DMD (Double Meridian Distance) Method
- Departure and Latitude
- Polygon Area
- Unit Conversion (m² to ha)
Common exam questions
Example
Q: A triangular lot has corners A(0,0), B(30,0), C(15,20). What is its area? Sol: ½|(0×0−30×0)+(30×20−15×0)+(15×0−0×20)| = ½|0+600+0| = 300 m².
Approach
Set up the two-column shoelace table. Always include the wrap-around row. Sum columns, subtract, take absolute value, divide by 2.
Example
Q: Convert 4,500 m² to hectares and ares. A: 0.45 ha = 45 ares.
Approach
Know: 1 ha = 10,000 m²; 1 are = 100 m²; 1 ha = 100 ares.
Key points to remember
- Shoelace formula: A = ½|Σ(EᵢNᵢ₊₁ − Eᵢ₊₁Nᵢ)|; indices wrap (last corner connects to first).
- Works for any simple (non-self-intersecting) polygon — triangles, quadrilaterals, irregular lots.
- 1 hectare = 10,000 m²; 1 are = 100 m².
- The DMD (Double Meridian Distance) method gives the same result as the shoelace — they are mathematically equivalent.
- Always list corners in consistent order around the boundary before applying the formula.
- The formula gives ZERO if any three consecutive corners are collinear (no area contribution from that triangle).
Technical Description: Bearing and Distance of Boundary Lines
Examples
Board-style Problem: Find the distance and bearing of line D→A, where D(10, 30) and A(0, 0).
Solution
Step 1: ΔE = 0 − 10 = −10 m (West), ΔN = 0 − 30 = −30 m (South)
Step 2: d = √((-30)² + (-10)²) = √(900 + 100) = √1000 = 31.62 m
Step 3: θ = arctan(|−10| / |−30|) = arctan(10/30) = arctan(0.3333) = 18.4349° = 18°26'06"
Step 4: ΔN < 0 → S prefix; ΔE < 0 → W suffix
Bearing: S 18°26'06" W
Distance: 31.62 m
The quadrant is determined by the SIGNS of ΔN and ΔE, not by the calculator output. arctan always gives a positive acute angle; you add the compass prefix and suffix from the signs of ΔN and ΔE.
Exercise from chapter: Find the distance and bearing of line B(50,0)→C(60,40).
Solution
ΔE = 60 − 50 = +10 m (East)
ΔN = 40 − 0 = +40 m (North)
d = √(40² + 10²) = √(1600 + 100) = √1700 = 41.23 m
θ = arctan(10/40) = arctan(0.25) = 14.0362° = 14°02'10"
ΔN > 0 → N prefix; ΔE > 0 → E suffix
Bearing: N 14°02'10" E, Distance: 41.23 m
When both ΔN and ΔE are positive, the line points into the NE quadrant.
Each side of the lot boundary is described by its bearing and distance. These form the technical description of the parcel, which is the legal description attached to the title. Given two consecutive corners Pᵢ(Eᵢ, Nᵢ) and Pᵢ₊₁(Eᵢ₊₁, Nᵢ₊₁), compute: Departure ΔE = Eᵢ₊₁ − Eᵢ (positive = East, negative = West) and Latitude ΔN = Nᵢ₊₁ − Nᵢ (positive = North, negative = South). Then: Distance d = √(ΔN² + ΔE²) and Reference bearing angle θ = arctan(|ΔE| / |ΔN|). The full bearing is expressed in the Philippine quadrant system: N θ E, N θ W, S θ E, or S θ W, where the leading letter (N or S) is from the sign of ΔN and the trailing letter (E or W) is from the sign of ΔE. The angle θ is always between 0° and 90°. If ΔN = 0 (line is due East or West), the bearing is 'Due East' or 'Due West'; if ΔE = 0, the bearing is 'Due North' or 'Due South'.
Applications
- Writing the technical description in the approved survey plan and title.
- Staking boundary lines in the field using a total station or theodolite set up at a corner monument.
- Checking whether boundary lines from the title match the as-surveyed lines.
- Computing back-bearing for the return direction (reverse the prefix and suffix).
Common misconceptions
- Do NOT use arctan(ΔE / ΔN) with signed values directly — the calculator may place the angle in the wrong quadrant. Always use arctan(|ΔE| / |ΔN|) and assign the quadrant separately from the signs.
- The bearing angle is measured from North or South toward East or West — it is NOT an azimuth (which is always measured from North, 0°–360°).
- Distance is always positive — it is a magnitude, not signed.
- When ΔN = 0, do NOT attempt arctan(|ΔE|/0) — the bearing is simply Due East or Due West.
Related concepts
- Azimuth vs Bearing
- Departure and Latitude
- Traverse Computation
- Technical Description Format
- DMS Conversion
Common exam questions
Example
Q: Corner P(100,200) to Q(50,350). Find bearing of PQ. ΔE=−50, ΔN=+150. θ=arctan(50/150)=18°26'06". Bearing: N 18°26'06" W.
Approach
Compute ΔE and ΔN. Apply arctan(|ΔE|/|ΔN|). Assign quadrant from signs. Express in DMS.
Example
Q: If bearing A→B is N 45°00'00" E, what is the bearing B→A? A: S 45°00'00" W.
Approach
The back bearing of line A→B is the bearing of B→A. Simply reverse the N/S prefix and E/W suffix.
Key points to remember
- ΔE = E(next) − E(current); ΔN = N(next) − N(current).
- Distance d = √(ΔN² + ΔE²) — always positive.
- Bearing angle θ = arctan(|ΔE| / |ΔN|) — always between 0° and 90°.
- Quadrant determination: ΔN > 0 → N prefix; ΔN < 0 → S prefix; ΔE > 0 → E suffix; ΔE < 0 → W suffix.
- Express bearing in degrees-minutes-seconds (DMS) for technical descriptions: e.g., S 18°26'00" W.
- arctan(1/3) = 18.4349° = 18°26'06" — common in board-exam problems.
Closure Check and Misclosure Adjustment
Examples
A four-sided traverse has: AB: ΔN=+0.00, ΔE=+50.00; BC: ΔN=+40.00, ΔE=+10.00; CD: ΔN=−30.00, ΔE=−50.00; DA: ΔN=−30.00, ΔE=−10.00. Check closure.
Solution
ΣΔN = 0.00 + 40.00 − 30.00 − 30.00 = −20.00 ≠ 0 — ERROR IN PROBLEM
Corrected for the lot A(0,0),B(50,0),C(60,40),D(10,30):
AB: ΔN=0, ΔE=50; BC: ΔN=40, ΔE=10; CD: ΔN=−10, ΔE=−50; DA: ΔN=−30, ΔE=−10
ΣΔN = 0+40−10−30 = 0 ✓
ΣΔE = 50+10−50−10 = 0 ✓ — PERFECT CLOSURE (as expected for coordinate-derived lot).
When coordinates are given exactly (as in board exam problems), closure is mathematically perfect. In real field surveys, small measurement errors produce non-zero sums that require Bowditch adjustment.
A closed polygon boundary must return to its starting point. This means the algebraic sum of all latitudes (ΔN values) must equal zero, and the algebraic sum of all departures (ΔE values) must also equal zero. In practice, measurement errors produce a small non-zero result called the linear misclosure (or error of closure). The ratio of linear misclosure to total perimeter is the relative error (precision ratio). For DENR-approved surveys, acceptable precision depends on survey class (Class 1: 1:10,000 minimum; Class 2: 1:5,000; Class 3: 1:3,000). If within tolerance, the misclosure is distributed among the traverse legs by the Bowditch (Compass) Rule: correction to latitude of leg i = −(total latitude misclosure) × (length of leg i / total perimeter); similarly for departure. After adjustment, the corrected coordinates are used for all subsequent computations including area.
Applications
- Verifying field traverse data before computing lot area.
- Applying Bowditch adjustment to distribute misclosure before computing final coordinates.
- Reporting precision ratio to DENR-LMB as part of survey returns.
- Detecting blunders: a very large misclosure signals a measurement error, not just random error.
Common misconceptions
- A zero sum of ΔN and ΔE does NOT mean the survey is error-free — it only means the mathematical closure is satisfied (common in problems derived from exact coordinates).
- Misclosure is NOT distributed equally — Bowditch distributes it proportionally to leg length.
- The precision ratio denominator is rounded DOWN to the nearest hundred or thousand, NOT up.
Related concepts
- Bowditch / Compass Rule
- Transit Rule
- Traverse Adjustment
- Error of Closure
- Relative Precision
Common exam questions
Example
Q: A traverse has eN=0.05 m, eE=0.03 m, total perimeter=400 m. Find relative error. Sol: e=√(0.0025+0.0009)=√0.0034=0.0583 m; ratio=0.0583/400=1:6,861.
Approach
Sum all ΔN and all ΔE. Compute e = √(eN²+eE²). Compute ratio = e/P.
Example
Q: eN=+0.06 m, leg length=100 m, P=300 m. Bowditch correction to leg latitude = −0.06×(100/300) = −0.02 m.
Approach
Apply cΔN,i = −eN×(dᵢ/P) for each leg. Show corrected ΔN and ΔE.
Key points to remember
- Closure condition: Σ(ΔN) = 0 and Σ(ΔE) = 0 for a perfectly closed boundary.
- Linear misclosure e = √(eN² + eE²), where eN = Σ(ΔN) and eE = Σ(ΔE).
- Relative error = e / perimeter (e.g., 1:5000 means 1 m error per 5000 m surveyed).
- Bowditch correction for each leg: cΔN,i = −eN × (dᵢ/P); cΔE,i = −eE × (dᵢ/P), where P = total perimeter.
- Recompute corner coordinates after adjustment: adjusted E and N = running sum of corrected departures and latitudes from the starting point.
- Recompute area from adjusted coordinates for the final answer.
Practice problems
Problem 1 (Shoelace Area): A residential lot in Quezon City has corners with PPCS Zone IV coordinates shifted to a local grid: A(0.00, 0.00), B(80.00, 0.00), C(90.00, 50.00), D(20.00, 40.00), all in metres. (a) Compute the area in m². (b) Convert to hectares. (c) Convert to ares.
Solution
(a) Shoelace table (list corners in order, wrap last to first):
| Pair | Eᵢ·Nᵢ₊₁ | Eᵢ₊₁·Nᵢ |
|---|---|---|
| A→B (to B,C) | 0×0=0 | 80×0=0 |
| B→C | 80×50=4000 | 90×0=0 |
| C→D | 90×40=3600 | 20×50=1000 |
| D→A | 20×0=0 | 0×40=0 |
Σ(Eᵢ·Nᵢ₊₁) = 0+4000+3600+0 = 7600
Σ(Eᵢ₊₁·Nᵢ) = 0+0+1000+0 = 1000
A = ½|7600−1000| = ½×6600 = 3300 m²
(b) 3300 ÷ 10,000 = 0.33 ha
(c) 3300 ÷ 100 = 33 ares
The wrap-around (D back to A) produces the term 20×0−0×40 = 0, which does not affect the result here but must always be included in the table. Missing it would give 3300 m² in this case since the D→A term is 0, but in other problems the wrap-around term is non-zero and critical.
Problem 2 (Bearing and Distance): Using the same lot corners A(0,0), B(80,0), C(90,50), D(20,40), compute the bearing and distance of each of the four sides: AB, BC, CD, and DA.
Solution
Line AB: A(0,0)→B(80,0)
ΔE=+80, ΔN=0 → Due East
d=80.00 m, Bearing: Due East (or N 90°00'00" E)
Line BC: B(80,0)→C(90,50)
ΔE=+10, ΔN=+50
d=√(50²+10²)=√2600=50.99 m
θ=arctan(10/50)=arctan(0.2)=11.310°=11°18'36"
ΔN>0, ΔE>0 → N 11°18'36" E
Line CD: C(90,50)→D(20,40)
ΔE=−70, ΔN=−10
d=√(10²+70²)=√5000=70.71 m
θ=arctan(70/10)=arctan(7)=81.870°=81°52'12"
ΔN<0, ΔE<0 → S 81°52'12" W
Line DA: D(20,40)→A(0,0)
ΔE=−20, ΔN=−40
d=√(40²+20²)=√2000=44.72 m
θ=arctan(20/40)=arctan(0.5)=26.565°=26°33'54"
ΔN<0, ΔE<0 → S 26°33'54" W
For line AB: ΔN=0 means the line is horizontal — bearing is Due East because ΔE is positive. For CD and DA: both ΔN and ΔE are negative → SW quadrant. The bearing angle is always arctan(|ΔE|/|ΔN|), always positive and between 0° and 90°.
Problem 3 (Closure and Bowditch): A four-leg traverse has the following computed latitudes and departures:
Leg 1: ΔN=+120.00, ΔE=+80.00, d=144.22 m
Leg 2: ΔN=+60.00, ΔE=−100.00, d=116.62 m
Leg 3: ΔN=−90.00, ΔE=−70.00, d=114.02 m
Leg 4: ΔN=−91.00, ΔE=+91.00, d=128.69 m
(a) Find eN and eE. (b) Find linear misclosure e. (c) Find the relative error. (d) Apply Bowditch correction to Leg 1 latitudes and departures.
Solution
(a) eN = 120+60−90−91 = −1.00 m; eE = 80−100−70+91 = +1.00 m
(b) e = √(1.00²+1.00²) = √2 = 1.414 m
(c) Total perimeter P = 144.22+116.62+114.02+128.69 = 503.55 m
Relative error = 1.414/503.55 = 1:356 — this is poor precision; would NOT pass DENR Class 3 (requires 1:3000).
(d) Bowditch correction for Leg 1 (d₁=144.22 m):
cΔN,1 = −(−1.00)×(144.22/503.55) = +1.00×0.2865 = +0.29 m (rounded to 2 dec)
cΔE,1 = −(+1.00)×(144.22/503.55) = −0.29 m
Corrected Leg 1: ΔN=120.00+0.29=120.29 m; ΔE=80.00−0.29=79.71 m
The large relative error (1:356) is exaggerated for pedagogical purposes. A real DENR Class 1 survey requires 1:10,000 or better. The Bowditch correction is proportional to leg length — longer legs receive larger corrections. After applying corrections to all four legs, verify that Σ(corrected ΔN)=0 and Σ(corrected ΔE)=0.
Problem 4 (Technical Description): Write the technical description (bearing and distance) for the lot in Problem 1 (corners A,B,C,D), starting from corner A and proceeding clockwise. Express all bearings in DMS (degrees-minutes-seconds).
Solution
Technical Description:
From corner A(0.00,0.00):
Thence N 90°00'00" E (Due East), 80.00 m to corner B;
Thence N 11°18'36" E, 50.99 m to corner C;
Thence S 81°52'12" W, 70.71 m to corner D;
Thence S 26°33'54" W, 44.72 m back to corner A (point of beginning).
Total Perimeter = 80.00+50.99+70.71+44.72 = 246.42 m
Total Area = 3,300 m² (0.33 ha or 33 ares)
The technical description always starts and ends at the same point (point of beginning, POB). It lists each line by its direction (bearing) and length (distance), traversing the lot perimeter in a consistent direction. The total perimeter and area are stated at the end. This is the format used in DENR-approved survey plans and attached to Certificates of Title under PD 1529.
Problem 5 (Triangular Lot): A triangular lot has corners P(0,0), Q(60,0), R(30,50) all in metres. (a) Find area by shoelace. (b) Verify using the base-height formula. (c) Find the bearing and distance of side RP.
Solution
(a) Shoelace:
P→Q: 0×0−60×0=0
Q→R: 60×50−30×0=3000
R→P: 30×0−0×50=0
A=½|0+3000+0|=1500 m²
(b) Base PQ=60 m (horizontal). Height = northing of R = 50 m.
A=½×60×50=1500 m² ✓
(c) Line R(30,50)→P(0,0):
ΔE=0−30=−30 m (W)
ΔN=0−50=−50 m (S)
d=√(50²+30²)=√(2500+900)=√3400=58.31 m
θ=arctan(30/50)=arctan(0.6)=30.964°=30°57'50"
Bearing: S 30°57'50" W
The base-height check confirms the shoelace result — use this whenever the lot has a horizontal or vertical base, as a quick verification. The shoelace formula is general and works for any polygon.
Exam preparation tips
- MASTER THE SHOELACE TABLE: Set up the two-column table (Eᵢ·Nᵢ₊₁ and Eᵢ₊₁·Nᵢ) for every area problem. Always write the wrap-around row explicitly. This single technique covers almost all area computation problems on the board exam.
- BEARING QUADRANT RULE: Memorize — ΔN sign gives the N/S prefix; ΔE sign gives the E/W suffix. Bearing angle = arctan(|ΔE|/|ΔN|), ALWAYS between 0° and 90°. Never forget the absolute values.
- DMS CONVERSION: Practice converting decimal degrees to DMS. Multiply the decimal part by 60 to get minutes, then multiply the decimal part of the minutes by 60 to get seconds. e.g., 18.4349° = 18° + 0.4349×60' = 18°26.094' = 18°26'05.6" ≈ 18°26'06".
- UNIT CONVERSIONS: Know these cold — 1 ha = 10,000 m²; 1 are = 100 m²; 1 ha = 100 ares. Board exam answer choices are often in hectares while your computation gives m².
- CLOSURE CHECK: Always verify Σ(ΔN)=0 and Σ(ΔE)=0 for problems with field measurements. If the problem gives you coordinates directly (no traverse data), closure is automatic — do NOT waste time checking.
- BOWDITCH RULE FORMULA: Correction = −(total misclosure) × (leg length / total perimeter). Apply to each leg. After adjustment, Σ(corrected ΔN) must = 0 and Σ(corrected ΔE) must = 0.
- LEGAL FRAMEWORK QUESTIONS: Know RA 8560 (geodetic engineering profession — only licensed GEs may sign surveys), PD 1529 (registration decree — survey plan required for title), CA 141 (public land surveys), RA 4374 (cadastral survey system). These appear as 1–2 item questions in every board exam.
- PRS92 vs WGS84: PRS92 is aligned to WGS84 with 3-parameter transformation; differences are small (<3 m). PPCS = Philippine Plane Coordinate System (Transverse Mercator, k₀=0.99995, false E=500,000 m). These distinctions appear in the geodetic engineering and advanced surveying portions of the exam.
- PRACTICE SPEED: Most board exam computations must be done in 2–3 minutes per item. Practice the shoelace table until you can complete a 4-corner lot in under 2 minutes. Use a scientific calculator's memory functions to store running sums.
- COMMON TRAP — ARCTAN WHEN ΔN=0: If ΔN=0, the bearing is Due East (ΔE>0) or Due West (ΔE<0). Do NOT attempt arctan(|ΔE|/0) = undefined. Similarly, if ΔE=0, bearing is Due North or Due South.
- REVIEW DENR SURVEY PRECISION CLASSES: Class 1 = urban/titled land (1:10,000); Class 2 = agricultural (1:5,000); Class 3 = forest/remote (1:3,000). These values may appear in closure and precision problems.
- TECHNICAL DESCRIPTION FORMAT: A typical board exam fill-in question asks you to write a technical description line. Format: 'Thence [Bearing], [Distance] m to corner [Label].' Always start from the point of beginning and return to it.
Diagrams
Property and Boundary Surveys: Chapter Concept Map
A comprehensive mind map showing all major topics in the chapter — survey types, legal framework, coordinate systems, and lot computation methods — and their relationships. Use this as a study organizer before drilling into individual concepts.
Shoelace Formula Step-by-Step Procedure
A step-by-step flowchart for applying the shoelace (coordinate) method to compute lot area. The critical wrap-around step is highlighted early in the process to prevent the most common board-exam error. Unit conversion branches are shown at the end.
Bearing and Distance Computation: Quadrant Decision Tree
A decision tree flowchart for determining the correct bearing quadrant (NE, NW, SE, SW) from the signs of Delta-E and Delta-N. The special case where Delta-N = 0 (Due East or Due West) is handled at the start to prevent division-by-zero errors in the arctan formula.
Traverse Closure Check and Bowditch Adjustment Workflow
A complete workflow diagram for the traverse closure and Bowditch adjustment process. Shows the decision point (tolerance check), the Bowditch correction step, verification of adjusted sums, and the final connection to the shoelace area computation. DENR precision class tolerances are included in the tolerance decision node.
Key Philippine Land Survey Legislation Timeline
A chronological timeline of the key laws and geodetic milestones governing property surveys in the Philippines. This helps reviewees place each law in historical context and remember which law governs which aspect of land survey practice.
Typical Board Exam Topic Distribution — Cadastral Surveys and Lot Computation
Estimated percentage distribution of board exam items across the major sub-topics of property surveys and lot computation, based on review of past PRC Geodetic Engineer board exam syllabi. Area computation dominates, followed by bearing-distance and closure problems. Legal framework questions are consistently present and easy marks for well-prepared reviewees.
In summary
Property and boundary surveys form the technical backbone of Philippine land titling and cadastral administration. From the field recovery of control monuments to the final area figure on the Certificate of Title, the geodetic engineer must apply a precise, legally defensible workflow. The shoelace formula — A = ½|Σ(EᵢNᵢ₊₁ − Eᵢ₊₁Nᵢ)| — is the single most important computational tool in this chapter, and its correct application (especially the wrap-around rule) is tested in nearly every board examination. Bearing computation from coordinates, with correct quadrant assignment from the signs of ΔN and ΔE, defines the technical description that appears on the title. Closure checking and Bowditch adjustment ensure that field measurements are internally consistent and within DENR-prescribed precision limits before the final area is computed. The legal framework — RA 8560 (professional authority), PD 1529 (registration), CA 141 (public lands), and RA 4374 (cadastral surveys) — underpins the entire practice and provides easy marks for well-prepared reviewees. Together, these skills and knowledge areas are the core of the Cadastral Surveys and Land Laws component of the PRC Geodetic Engineer Licensure Examination. Review the worked examples systematically, practice the shoelace table until it is automatic, master the bearing-quadrant decision tree, and internalize the key legal provisions — and you will be well prepared to pass this portion of the board exam.