CELE Surveying (Geomatics) — Area and Volume Computation (Earthworks)Exam Answer Templates

Marks

1

Topic

Area by Coordinates (Shoelace Formula)

Difficulty

easy

Template Id

T1

Examiner Tip

The absolute value sign is the most-penalized omission on 1-mark definition questions. Write it explicitly.

Model Answer

A = ½ |Σ(xᵢ yᵢ₊₁ − xᵢ₊₁ yᵢ)|, where the subscripts are taken modulo n (i.e., vertex n+1 = vertex 1).

Question Type

very_short_answer

Answer Structure

• Write the complete Shoelace formula with absolute value signs [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting the ½ factor
• Omitting the absolute value bars — sign errors can give a negative area
• Writing the formula for only a triangle instead of a general polygon

Key Phrases To Include

• A = ½|Σ(xᵢyᵢ₊₁ − xᵢ₊₁yᵢ)|
• absolute value
• closed polygon
• subscripts modulo n

Marks

1

Topic

Earthwork Volumes — End-Area vs Prismoidal

Difficulty

easy

Template Id

T2

Examiner Tip

Examiners reward the word 'mid-section' and the phrase 'more accurate' — use both.

Model Answer

The End-Area method averages only the two end cross-section areas [V = (L/2)(A₁ + A₂)], making it simple but slightly over-estimating for tapered prismoids. The Prismoidal method includes the mid-section area Aₘ [V = (L/6)(A₁ + 4Aₘ + A₂)], giving a more accurate result for any prismoid.

Question Type

very_short_answer

Answer Structure

• Define End-Area formula and note its tendency to over-estimate [0.5 mark]
• Define Prismoidal formula and note the role of Aₘ [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating only one formula without comparison
• Confusing Aₘ with the average (A₁+A₂)/2

Key Phrases To Include

• End-Area: V = (L/2)(A₁ + A₂)
• Prismoidal: V = (L/6)(A₁ + 4Aₘ + A₂)
• mid-section area Aₘ
• over-estimates
• more accurate

Marks

2

Topic

Irregular Boundaries — Simpson's One-Third Rule

Difficulty

medium

Template Id

T3

Examiner Tip

Always verify: count of offsets = count of intervals + 1. If you have 6 offsets, you have 5 intervals (odd) — you CANNOT apply Simpson's to all strips directly.

Model Answer

Simpson's One-Third Rule: A = (h/3)[(y₀ + yₙ) + 4(y₁ + y₃ + … + yₙ₋₁) + 2(y₂ + y₄ + … + yₙ₋₂)] where h = equal interval spacing and y₀, y₁, …, yₙ are the measured offsets. Condition: The number of intervals n must be EVEN (equivalently, the number of offsets must be ODD). If n is odd, apply Simpson's rule to all but the last strip, then add the last strip using the Trapezoidal rule.

Question Type

short_answer

Answer Structure

• Write the complete Simpson's 1/3 formula with correct grouping [1 mark]
• State the even-interval (odd-offset) condition and the workaround for odd intervals [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting the factor 4 or 2 in the multiplier groups
• Failing to state the even-intervals condition — this is a dedicated mark
• Using n for number of offsets instead of number of intervals (off-by-one error)

Key Phrases To Include

• h/3
• 4 × (odd-positioned offsets)
• 2 × (even-positioned interior offsets)
• even number of intervals
• odd number of offsets
• Trapezoidal rule for the remaining strip

Marks

2

Topic

Irregular Boundaries — Trapezoidal Rule

Difficulty

easy

Template Id

T4

Examiner Tip

The classic error is treating the Trapezoidal rule as a simple average. Emphasize (y₀ + yₙ)/2 — the end terms get half weight.

Model Answer

Given: h = 8 m; offsets: y₀ = 3.0, y₁ = 5.5, y₂ = 7.2, y₃ = 4.8, y₄ = 2.1 m Trapezoidal Rule: A = h[(y₀ + y₄)/2 + y₁ + y₂ + y₃] A = 8[(3.0 + 2.1)/2 + 5.5 + 7.2 + 4.8] A = 8[2.55 + 17.5] A = 8 × 20.05 A = 160.4 m²

Question Type

numerical

Answer Structure

• Identify given data (h and all offsets) [0 marks — setup; establishes clarity]
• Write the Trapezoidal Rule formula [0.5 mark]
• Correct substitution and grouping [0.5 mark]
• Correct arithmetic with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Adding all offsets equally (forgetting to halve the end offsets)
• Omitting m² in the final answer
• Arithmetic error in summing interior offsets

Key Phrases To Include

• Trapezoidal Rule
• A = h[(y₀ + yₙ)/2 + y₁ + y₂ + … + yₙ₋₁]
• 160.4 m²

Marks

3

Topic

Irregular Boundaries — Simpson's One-Third Rule

Difficulty

medium

Template Id

T5

Examiner Tip

Board exam questions frequently pair Trapezoidal and Simpson's for the same data set. Always do the interval-count check first — it demonstrates methodical thinking.

Model Answer

Given: h = 8 m; y₀ = 3.0, y₁ = 5.5, y₂ = 7.2, y₃ = 4.8, y₄ = 2.1 m Number of intervals n = 4 (EVEN) → Simpson's rule is applicable. Simpson's One-Third Rule: A = (h/3)[(y₀ + y₄) + 4(y₁ + y₃) + 2(y₂)] A = (8/3)[(3.0 + 2.1) + 4(5.5 + 4.8) + 2(7.2)] A = (8/3)[5.1 + 4(10.3) + 14.4] A = (8/3)[5.1 + 41.2 + 14.4] A = (8/3)[60.7] A = 8 × 20.233 A = 161.9 m² Comment: Simpson's rule (161.9 m²) is more accurate than the Trapezoidal rule (160.4 m²) because it fits parabolic arcs to the boundary rather than straight-line chords, thus better approximating most natural terrain profiles.

Question Type

numerical

Answer Structure

• Verify that n = 4 is even — state explicitly [0.5 mark]
• Write Simpson's formula correctly [0.5 mark]
• Substitute and group: end terms, 4× odd, 2× even interior [1 mark]
• Correct numerical result with units [0.5 mark]
• Comparison statement favoring Simpson's with reason [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Not verifying even-interval condition (costs the first half-mark)
• Misidentifying odd-index vs. even-index offsets for the 4× and 2× groups
• Omitting the comparison comment (costs the last mark)

Key Phrases To Include

• n = 4 (even) — Simpson's is applicable
• A = (h/3)[(y₀+y₄) + 4(y₁+y₃) + 2(y₂)]
• 161.9 m²
• parabolic arc
• more accurate than Trapezoidal

Marks

3

Topic

Area by Coordinates (Shoelace Formula)

Difficulty

medium

Template Id

T6

Examiner Tip

Add an extra row at the bottom of your table repeating the first vertex's coordinates. This makes the polygon closure visible to the examiner and prevents the most common error.

Model Answer

Vertices in order: A(0,0), B(50,0), C(60,40), D(10,30) Shoelace Formula: A = ½|Σ(xᵢyᵢ₊₁ − xᵢ₊₁yᵢ)| Tabulation (closing back to A): xᵢ yᵢ xᵢ₊₁ yᵢ₊₁ (xᵢyᵢ₊₁) (xᵢ₊₁yᵢ) 0 0 50 0 0×0=0 50×0=0 50 0 60 40 50×40=2000 60×0=0 60 40 10 30 60×30=1800 10×40=400 10 30 0 0 10×0=0 0×30=0 Σ(xᵢyᵢ₊₁) = 0 + 2000 + 1800 + 0 = 3800 Σ(xᵢ₊₁yᵢ) = 0 + 0 + 400 + 0 = 400 A = ½|3800 − 400| = ½(3400) = 1700 m²

Question Type

numerical

Answer Structure

• Write the Shoelace formula [0.5 mark]
• Set up and correctly complete the tabulation (all 4 pairs) [1 mark]
• Compute Σ(xᵢyᵢ₊₁) and Σ(xᵢ₊₁yᵢ) separately [0.5 mark]
• Correct final answer: A = 1700 m² with absolute value applied [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to close the polygon (connect last vertex back to first vertex) — biggest single error
• Subtracting in the wrong order (backward − forward) without taking absolute value
• Arithmetic error in one cross-product term
• Omitting m² unit

Key Phrases To Include

• A = ½|Σ(xᵢyᵢ₊₁ − xᵢ₊₁yᵢ)|
• tabulation
• close the polygon (last vertex back to first)
• absolute value
• 1700 m²

Marks

3

Topic

Earthwork Volumes — Prismoidal Method

Difficulty

medium

Template Id

T7

Examiner Tip

Aₘ is the cross-section area AT the midpoint of the segment — it is measured or calculated from mid-section dimensions. It is NOT (A₁+A₂)/2. This distinction is tested directly.

Model Answer

Given: A₁ = 35.0 m², Aₘ = 28.0 m², A₂ = 22.0 m², L = 50 m (a) End-Area Method: V_EA = (L/2)(A₁ + A₂) V_EA = (50/2)(35.0 + 22.0) V_EA = 25 × 57.0 V_EA = 1425 m³ (b) Prismoidal Method: V_P = (L/6)(A₁ + 4Aₘ + A₂) V_P = (50/6)(35.0 + 4×28.0 + 22.0) V_P = (50/6)(35.0 + 112.0 + 22.0) V_P = (50/6)(169.0) V_P = 8.3333 × 169.0 V_P = 1408.3 m³ Conclusion: The Prismoidal method (1408.3 m³) is more accurate than the End-Area method (1425 m³) because it accounts for the actual variation of cross-section shape along the length using the mid-section area.

Question Type

numerical

Answer Structure

• List all given data [setup — no dedicated mark but required for clarity]
• Apply End-Area formula and compute V_EA [1 mark]
• Apply Prismoidal formula with correct Aₘ and compute V_P [1.5 marks]
• Conclusion statement identifying the more accurate method with reason [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using Aₘ = (A₁+A₂)/2 = 28.5 m² instead of the given 28.0 m²
• Using L = 25 m (half-length) instead of 50 m in the Prismoidal formula
• Omitting m³ units
• Concluding End-Area is more accurate

Key Phrases To Include

• V_EA = (L/2)(A₁ + A₂)
• V_P = (L/6)(A₁ + 4Aₘ + A₂)
• Aₘ is the mid-section area (not the average)
• 1425 m³
• 1408.3 m³
• Prismoidal is more accurate

Marks

2

Topic

Mass Diagram

Difficulty

easy

Template Id

T8

Examiner Tip

The word 'cumulative' is the core of the definition. Without it, the definition is incomplete and loses the mark.

Model Answer

Mass Diagram: A graph that plots the cumulative algebraic sum of cut (positive) and fill (negative) volumes against the station (chainage) along a highway alignment. The vertical ordinate at any station represents the net accumulated earthwork volume from the starting point. Practical Uses: 1. Determining the economic haul distance — identifying where material from cut sections can be economically moved to fill sections without incurring excess hauling costs. 2. Locating borrow pits and waste areas — stations where the mass diagram curve falls below the balance line indicate required borrow (import of fill material); stations where it rises above the balance line indicate waste (excess cut to be disposed of).

Question Type

short_answer

Answer Structure

• Define the mass diagram: cumulative cut−fill vs. station plot [1 mark]
• State two distinct practical uses (haul planning + borrow/waste identification) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining it as a simple plot of cut-fill without the word 'cumulative'
• Stating only one use
• Confusing 'free haul' with 'overhaul' without defining terms

Key Phrases To Include

• cumulative algebraic sum
• cut (positive) and fill (negative)
• chainage / station
• balance line
• haul distance
• borrow
• waste

Marks

2

Topic

Irregular Boundaries — Trapezoidal Rule

Difficulty

easy

Template Id

T9

Examiner Tip

Note: n = 5 is ODD so Simpson's 1/3 cannot be applied directly to all strips. The Trapezoidal rule has no such restriction — use it confidently for odd n.

Model Answer

Given: h = 5 m; y₀=0, y₁=3.2, y₂=5.8, y₃=4.1, y₄=6.3, y₅=2.0 m Number of intervals n = 5 Trapezoidal Rule: A = h[(y₀ + y₅)/2 + y₁ + y₂ + y₃ + y₄] A = 5[(0 + 2.0)/2 + 3.2 + 5.8 + 4.1 + 6.3] A = 5[1.0 + 19.4] A = 5 × 20.4 A = 102.0 m²

Question Type

numerical

Answer Structure

• Identify h, all offsets, and note n = 5 intervals [setup]
• Write Trapezoidal formula and substitute [1 mark]
• Correct arithmetic: A = 102.0 m² [1 mark]

Scoring Breakdown

Common Mark Deductions

• Including y₀=0 in the full sum without halving
• Arithmetic mistake in interior sum (3.2+5.8+4.1+6.3 = 19.4)
• Omitting unit m²

Key Phrases To Include

• A = h[(y₀+y₅)/2 + y₁+y₂+y₃+y₄]
• end offsets halved
• 102.0 m²

Marks

3

Topic

Irregular Boundaries — Simpson's Rule with Odd Intervals

Difficulty

hard

Template Id

T10

Examiner Tip

This is a high-value trap question. Examiners specifically set up odd-interval data to test whether you know the restriction. Always count intervals before choosing the method.

Model Answer

Given: n = 5 intervals (ODD) — Simpson's rule CANNOT be applied to all 5 strips directly. Strategy: Apply Simpson's rule to the first 4 intervals (y₀ to y₄), then add the last strip (y₄ to y₅) using the Trapezoidal rule. Step 1 — Simpson's for strips 1–4 (n=4, even; offsets y₀ to y₄): A₁ = (h/3)[(y₀ + y₄) + 4(y₁ + y₃) + 2(y₂)] A₁ = (5/3)[(0 + 6.3) + 4(3.2 + 4.1) + 2(5.8)] A₁ = (5/3)[6.3 + 4(7.3) + 11.6] A₁ = (5/3)[6.3 + 29.2 + 11.6] A₁ = (5/3)(47.1) = 78.5 m² Step 2 — Trapezoidal for the last strip (y₄ to y₅): A₂ = h(y₄ + y₅)/2 = 5(6.3 + 2.0)/2 = 5(4.15) = 20.75 m² Total Area: A = A₁ + A₂ = 78.5 + 20.75 = 99.25 m² Note: Compare with Trapezoidal result (102.0 m²). The Simpson's-based composite result (99.25 m²) is generally more accurate for the parabolic portion.

Question Type

numerical

Answer Structure

• Identify that n = 5 is odd and state that direct Simpson's application is invalid [0.5 mark]
• Split: Simpson's on first 4 strips, Trapezoidal on last strip [0.5 mark]
• Correct Simpson's computation for A₁ = 78.5 m² [1 mark]
• Correct Trapezoidal computation for A₂ = 20.75 m² [0.5 mark]
• Total A = 99.25 m² [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Applying Simpson's to all 5 strips ignoring the odd-n condition (most common error)
• Applying the last-strip correction incorrectly (using Simpson's 3/8 rule instead of simple Trapezoidal)
• Not closing with the note that composite result is preferred

Key Phrases To Include

• n = 5 is odd — direct Simpson's not applicable
• apply Simpson's to first 4 intervals
• apply Trapezoidal to last strip
• A₁ = 78.5 m²
• A₂ = 20.75 m²
• Total = 99.25 m²

Marks

2

Topic

Area by Coordinates (Shoelace Formula)

Difficulty

easy

Template Id

T11

Examiner Tip

Verification questions reward checking work. Write 'verified' or use a checkmark ✓ — this signals to the examiner that you understand the result is consistent.

Model Answer

Method 1 — Shoelace Formula: A = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| A = ½|0(0−6) + 8(6−0) + 0(0−0)| A = ½|0 + 48 + 0| A = ½(48) = 24 m² Verification — Basic Triangle Formula: Base = 8 m (along x-axis from P₁ to P₂) Height = 6 m (along y-axis from P₁ to P₃) A = ½ × base × height = ½ × 8 × 6 = 24 m² ✓ Both methods give A = 24 m² — result confirmed.

Question Type

numerical

Answer Structure

• Apply Shoelace formula correctly with all three vertex substitutions [1 mark]
• Verify using A = ½bh with correct base and height identification [0.5 mark]
• State that both methods agree [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Not taking absolute value in Shoelace (gives −24 if vertices are in CW order)
• Skipping the verification step when the question explicitly asks for it

Key Phrases To Include

• Shoelace formula
• A = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|
• 24 m²
• verification
• A = ½ × base × height

Marks

5

Topic

Earthwork Volumes — Prismoidal Correction

Difficulty

hard

Template Id

T12

Examiner Tip

A 5-mark numerical question expects a structured, multi-part solution. Losing one sub-part still lets you earn marks on the others — show all working even if you are unsure of one step.

Model Answer

Given: A₁ = 40 m², Aₘ = 32 m², A₂ = 18 m²; L = 40 m (Sta. 5+000 to 5+040) (a) Prismoidal Volume: V_P = (L/6)(A₁ + 4Aₘ + A₂) V_P = (40/6)(40 + 4×32 + 18) V_P = (40/6)(40 + 128 + 18) V_P = (40/6)(186) V_P = 6.6667 × 186 V_P = 1240.0 m³ (b) End-Area Volume: V_EA = (L/2)(A₁ + A₂) V_EA = (40/2)(40 + 18) V_EA = 20 × 58 V_EA = 1160.0 m³ Wait — recheck: V_EA = 20 × 58 = 1160 m³ (c) Prismoidal Correction: Cp = V_EA − V_P = 1160 − 1240 = −80 m³ Interpretation: The negative correction indicates that the prismoidal volume (1240 m³) is LARGER than the end-area volume (1160 m³) for this case. This can occur when the mid-section area is significantly larger than the average of the end areas — a situation sometimes found in certain concave cross-sections. Apply correction as: V_P = V_EA − Cp = 1160 − (−80) = 1240 m³ ✓

Question Type

numerical

Answer Structure

• List all given: A₁, Aₘ, A₂, L [0 marks — setup]
• Compute V_P correctly [1.5 marks]
• Compute V_EA correctly [1.5 marks]
• Compute Cp = V_EA − V_P [1 mark]
• Interpret the sign of Cp and verify consistency [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using L = 20 m (half the station interval) instead of L = 40 m
• Substituting Aₘ = (40+18)/2 = 29 m² instead of the given 32 m²
• Neglecting to state the sign of the prismoidal correction
• Not verifying V_EA ± Cp = V_P at the end
• Omitting m³ from all three volume answers

Key Phrases To Include

• V_P = (L/6)(A₁ + 4Aₘ + A₂)
• V_EA = (L/2)(A₁ + A₂)
• Cp = V_EA − V_P
• 1240 m³
• 1160 m³
• Cp = −80 m³
• prismoidal correction

Marks

2

Topic

All Topics — Common Pitfalls

Difficulty

easy

Template Id

T13

Examiner Tip

Enumeration questions are free marks — structure as a numbered list with a one-line explanation per item. No need for paragraphs.

Model Answer

1. Simpson's even-interval condition ignored — applying Simpson's 1/3 rule to an odd number of intervals without splitting off the last strip leads to an incorrect formula application. 2. Mid-section area Aₘ confused with arithmetic mean — Aₘ in the prismoidal formula is the measured cross-section at the midpoint, not (A₁+A₂)/2; using the average underestimates volume. 3. Shoelace polygon not closed — failing to connect the last vertex back to the first in the tabulation causes a systematic error across all cross-product terms. 4. Units omitted — answers in m³ written without the unit lose the mark in most rubrics regardless of numerical correctness.

Question Type

short_answer

Answer Structure

• Pitfall 1 with explanation [0.5 mark]
• Pitfall 2 with explanation [0.5 mark]
• Pitfall 3 with explanation [0.5 mark]
• Pitfall 4 with explanation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Repeating the same pitfall in different words (counts as one)
• Listing errors without brief explanations when the question says 'explain'

Key Phrases To Include

• even number of intervals
• mid-section area
• close the polygon
• units (m, m², m³)

Marks

5

Topic

Earthwork Volumes — Applied Prismoidal

Difficulty

hard

Template Id

T14

Examiner Tip

When equally spaced cross-sections are given, always check whether the middle station naturally serves as the prismoidal mid-section. This avoids unnecessary complexity.

Model Answer

Given cross-sections: Sta. 0+000: A₁ = 50 m² Sta. 0+050: A₂ = 30 m² (this serves as Aₘ for Segment 1 and as A₁ for Segment 2) Sta. 0+100: A₃ = 10 m² Note: With three evenly spaced stations at L = 50 m spacing, the prismoidal formula can be applied over the full 100 m length using Sta. 0+050 as the mid-section: Single Prismoidal Application (L = 100 m, Aₘ = 30 m²): V_P = (L/6)(A₁ + 4Aₘ + A₂) V_P = (100/6)(50 + 4×30 + 10) V_P = (100/6)(50 + 120 + 10) V_P = (100/6)(180) V_P = 16.667 × 180 V_P = 3000 m³ Alternative — Two-segment End-Area check: Segment 1 (0+000 to 0+050): V₁ = (50/2)(50+30) = 2000 m³ Segment 2 (0+050 to 0+100): V₂ = (50/2)(30+10) = 1000 m³ V_EA(total) = 3000 m³ In this case both methods coincide because the areas vary linearly, confirming V_P = 3000 m³. Final Answer: Total volume of cut = 3000 m³

Question Type

numerical

Answer Structure

• Identify three stations, their areas, and recognize Sta. 0+050 as mid-section [0.5 mark]
• Apply Prismoidal formula over full 100 m span [2 marks]
• Correct arithmetic: V_P = 3000 m³ [1 mark]
• Two-segment End-Area check to verify [1 mark]
• Clear conclusion with units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using L = 50 m for the full-span Prismoidal formula
• Applying two separate Prismoidal formulas for two 50-m segments (no mid-section data for each sub-segment)
• Omitting verification step for a 5-mark problem
• Failing to label result as 'cut' volume

Key Phrases To Include

• mid-section area at Sta. 0+050
• L = 100 m
• V_P = (L/6)(A₁ + 4Aₘ + A₂)
• 3000 m³
• verification by two-segment end-area

Marks

2

Topic

Earthwork Volumes — Prismoidal Correction

Difficulty

medium

Template Id

T15

Examiner Tip

If the exact Cp formula is forgotten, you can still earn partial marks by writing V_P = V_EA − Cp and explaining what Cp corrects for — always show what you know.

Model Answer

Prismoidal Correction: Cp = (L/12)(c₁ − c₂)(d₁ − d₂) where c₁, c₂ = center heights (cut depths) at the two end sections; d₁, d₂ = corresponding widths at the two end sections. Alternatively expressed as: V_Prismoidal = V_End-Area − Cp When most important: The prismoidal correction is most significant (and must be applied) when: 1. Cross-section dimensions change substantially between consecutive stations (large differences in width or depth), and 2. High-accuracy quantity estimates are required for payment or contract purposes, where over-estimation by the end-area method would lead to overpayment.

Question Type

short_answer

Answer Structure

• State the Prismoidal Correction formula Cp = (L/12)(c₁−c₂)(d₁−d₂) [1 mark]
• State two conditions warranting its use [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing the correction formula with the prismoidal formula itself
• Stating only one condition for application
• Not defining c and d in the formula

Key Phrases To Include

• Cp = (L/12)(c₁−c₂)(d₁−d₂)
• V_P = V_EA − Cp
• center heights
• widths
• over-estimation
• contract payment