GELE Mathematics — Plane, Solid Geometry and MensurationExam Answer Templates

Marks

1

Topic

Circle — Plane Figures

Difficulty

easy

Template Id

T1

Examiner Tip

For a 1-mark formula recall question, write the formula and define the variable — nothing more is needed. Any unnecessary working wastes time.

Model Answer

The area of a circle is A = πr², where r is the radius. For r = 1 m: A = π(1)² = 3.1416 m².

Question Type

very_short_answer

Answer Structure

• Write the formula A = πr² with correct notation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing A = πd² instead of πr² (confusion of diameter and radius)
• Omitting the squared exponent: writing A = πr

Key Phrases To Include

• A = πr²
• radius r
• π = 3.1416

Marks

1

Topic

Pappus Theorems

Difficulty

easy

Template Id

T2

Examiner Tip

Memorize both Pappus theorems as a pair: First = 2π d̄ L (surface), Second = 2π d̄ A (volume). State which theorem you are using.

Model Answer

By Pappus' Second Theorem: V = 2π d̄ A where d̄ is the perpendicular distance from the centroid of area A to the axis of revolution, and A is the area being revolved.

Question Type

very_short_answer

Answer Structure

• State V = 2π d̄ A [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing Pappus' First Theorem (surface) with Second Theorem (volume)
• Writing 2πr instead of 2π d̄, using r as if d̄ were the radius of the shape

Key Phrases To Include

• V = 2π d̄ A
• centroid
• axis of revolution
• Pappus

Marks

2

Topic

Circle Sector — Plane Figures

Difficulty

easy

Template Id

T3

Examiner Tip

The most frequent pitfall on sector problems is forgetting radian conversion. Write 'θ in rad' beside the formula to remind yourself and signal it to the examiner.

Model Answer

Given: r = 8 m θ = 75° = 75 × (π/180) = 1.3090 rad To Find: Area of sector, A Formula: A = ½ r²θ (θ in radians) Solution: A = ½ (8)²(1.3090) A = ½ (64)(1.3090) A = 41.89 m² ∴ Area of sector = 41.89 m²

Question Type

numerical

Answer Structure

• Line 1: State Given data with units [0.5 mark]
• Line 2: Convert angle to radians [0.5 mark]
• Line 3: Write formula A = ½r²θ [0.5 mark]
• Line 4: Substitute and compute final answer with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using θ in degrees directly in the formula instead of converting to radians
• Using A = πr²(θ/360°) — valid, but must be stated clearly; mixing notations loses marks
• Omitting the unit m² from the final answer

Key Phrases To Include

• A = ½r²θ
• θ in radians
• × (π/180)
• 41.89 m²

Marks

2

Topic

Regular Polygon — Plane Figures

Difficulty

medium

Template Id

T4

Examiner Tip

An alternative method: divide the hexagon into 6 equilateral triangles, each with area (√3/4)s². Both methods are acceptable — choose whichever you can execute faster.

Model Answer

Given: n = 6 sides (regular hexagon) s = 6 m (side length) To Find: Area A Formula: A = (1/4) n s² cot(180°/n) Solution: A = (1/4)(6)(6)² cot(180°/6) A = (1/4)(6)(36) cot(30°) A = 54 × cot(30°) cot(30°) = 1/tan(30°) = 1.7321 A = 54 × 1.7321 A = 93.53 m² ∴ Area of regular hexagon = 93.53 m²

Question Type

numerical

Answer Structure

• Line 1: Identify n = 6, s = 6 m [0.5 mark]
• Line 2: State formula for regular polygon area [0.5 mark]
• Line 3: Substitute values and evaluate cot(30°) [0.5 mark]
• Line 4: Final numerical answer with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Computing tan(30°) instead of cot(30°) — sign error leads to wrong answer
• Using 180°/n = 30° correctly but then forgetting to take the cotangent
• Using the formula for a triangle area (½bh) for each of the 6 triangles without showing the apothem calculation

Key Phrases To Include

• A = (1/4)ns² cot(180°/n)
• cot(30°) = 1.7321
• regular polygon
• 93.53 m²

Marks

3

Topic

Frustum of a Cone — Solids

Difficulty

medium

Template Id

T5

Examiner Tip

The geometric mean term √(A₁A₂) is the most examined part of the frustum formula. Show your simplification step (e.g., √(4π × 16π) = 8π) explicitly — this earns the middle mark.

Model Answer

Given: r₁ = 2 m (top radius) r₂ = 4 m (bottom radius) h = 3 m (height) To Find: Volume V of frustum Formula: V = (h/3)(A₁ + A₂ + √(A₁A₂)) where A₁ = πr₁², A₂ = πr₂² Solution: A₁ = π(2)² = 4π = 12.566 m² A₂ = π(4)² = 16π = 50.265 m² √(A₁A₂) = √(4π × 16π) = √(64π²) = 8π = 25.133 m² V = (3/3)(12.566 + 50.265 + 25.133) V = (1)(87.964) V = 87.96 m³ ∴ Volume of frustum = 87.96 m³

Question Type

numerical

Answer Structure

• Line 1-2: State Given data correctly with SI units [0.5 mark]
• Line 3: Write the frustum volume formula with A₁, A₂ defined [1 mark]
• Line 4-5: Compute A₁, A₂, and √(A₁A₂) correctly [1 mark]
• Line 6: Substitute and compute final answer with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using (A₁ + A₂)/2 instead of the geometric mean √(A₁A₂) — this is the trapezoid approximation, not the frustum formula
• Forgetting to square the radii when computing A₁ and A₂
• Using the cone volume formula V = (1/3)πr²h with the average radius instead of the correct frustum formula

Key Phrases To Include

• V = (h/3)(A₁ + A₂ + √(A₁A₂))
• geometric mean √(A₁A₂)
• A₁ = πr₁²
• 87.96 m³

Marks

3

Topic

Sphere — Solids

Difficulty

easy

Template Id

T6

Examiner Tip

Note that for r = 3 m, V and S are numerically equal (both 36π ≈ 113.10) but with different units. Always write the unit — this distinguishes you from a careless candidate.

Model Answer

Given: r = 3 m To Find: (a) Volume V (b) Surface Area S (a) Volume of Sphere: Formula: V = (4/3)πr³ V = (4/3)π(3)³ V = (4/3)π(27) V = 36π V = 113.10 m³ (b) Surface Area of Sphere: Formula: S = 4πr² S = 4π(3)² S = 4π(9) S = 36π S = 113.10 m² ∴ Volume = 113.10 m³; Surface Area = 113.10 m²

Question Type

numerical

Answer Structure

• Line 1: Given r = 3 m [0 marks — setup only]
• Line 2-4: Formula V = (4/3)πr³ and correct substitution [1 mark]
• Line 5: Final volume = 113.10 m³ with correct unit [0.5 mark]
• Line 6-8: Formula S = 4πr² and correct substitution [1 mark]
• Line 9: Final surface area = 113.10 m² with correct unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing V = (4/3)πr² (using r² instead of r³)
• Confusing surface area S = 4πr² with the formula for a circle A = πr²
• Reporting the same number without distinguishing units: m³ for volume vs m² for surface area

Key Phrases To Include

• V = (4/3)πr³
• S = 4πr²
• 113.10 m³
• 113.10 m²

Marks

3

Topic

Prismatoid — Earthwork Applications

Difficulty

medium

Template Id

T7

Examiner Tip

In earthwork problems, the Prismatoid formula is preferred over Average End Area because it is more accurate. Always state which formula you are using — examiners reward the correct formula identification as a separate mark.

Model Answer

Given: A₁ = 20 m² (area at start station) Aₘ = 14 m² (area at middle section) A₂ = 6 m² (area at end station) h = 40 m (total length between end sections) To Find: Volume V by Prismatoid Formula Formula (Prismatoid / Prismoidal Formula): V = (h/6)(A₁ + 4Aₘ + A₂) Solution: V = (40/6)(20 + 4(14) + 6) V = (40/6)(20 + 56 + 6) V = (40/6)(82) V = 6.6667 × 82 V = 546.67 m³ ∴ Volume of earth cut = 546.67 m³

Question Type

numerical

Answer Structure

• Line 1-2: Identify A₁, Aₘ, A₂, and h = 40 m correctly [0.5 mark]
• Line 3: Write Prismatoid formula V = (h/6)(A₁ + 4Aₘ + A₂) [1 mark]
• Line 4: Substitute all values correctly, showing 4Aₘ = 56 [1 mark]
• Line 5: Final answer 546.67 m³ with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using h = 20 m instead of h = 40 m (taking only one interval instead of the total length)
• Using the Average End Area formula V = (h/2)(A₁ + A₂) and ignoring Aₘ
• Forgetting the coefficient 4 on the middle area Aₘ

Key Phrases To Include

• V = (h/6)(A₁ + 4Aₘ + A₂)
• Prismatoid Formula
• Prismoidal Correction
• h = 40 m
• 546.67 m³

Marks

3

Topic

Pappus Theorems — Solids of Revolution

Difficulty

medium

Template Id

T8

Examiner Tip

For torus problems, always explicitly say 'By Pappus' Second Theorem' and identify d̄ and A before substituting. This structured approach earns marks even if you make an arithmetic error.

Model Answer

Given: r = 2 m (radius of the revolving circle) d̄ = 5 m (distance from centroid of circle to axis of revolution) To Find: Volume V of the torus Method: Pappus' Second Theorem V = 2π d̄ A where A = area of the revolving circle = πr² Solution: A = π(2)² = 4π = 12.566 m² V = 2π d̄ A V = 2π (5)(4π) V = 40π² V = 394.78 m³ ∴ Volume of torus = 394.78 m³

Question Type

numerical

Answer Structure

• Line 1: Identify d̄ = 5 m and r = 2 m from the problem [0.5 mark]
• Line 2: State Pappus' Second Theorem V = 2π d̄ A [1 mark]
• Line 3: Compute area A = πr² = 4π m² [0.5 mark]
• Line 4: Substitute into Pappus and compute final answer [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using Pappus' First Theorem (surface area) instead of the Second Theorem (volume)
• Using r = 5 m as the centroid distance and r = 2 m as the axis, which reverses the two quantities
• Not squaring r in the circle area: writing A = πr instead of πr²

Key Phrases To Include

• Pappus' Second Theorem
• V = 2π d̄ A
• A = πr²
• centroid distance d̄ = 5 m
• 394.78 m³

Marks

3

Topic

Cylinder — Prism and Cylinder

Difficulty

easy

Template Id

T9

Examiner Tip

Always write 'r = D/2' explicitly as your first step. PRC board problems frequently give diameter, not radius. This one step prevents the most common error in cylinder problems.

Model Answer

Given: Diameter D = 4 m → radius r = 2 m Height h = 6 m To Find: (a) Volume V (b) Total Surface Area S_total (a) Volume of Cylinder: Formula: V = πr²h V = π(2)²(6) V = π(4)(6) V = 24π V = 75.40 m³ (b) Total Surface Area: Formula: S_total = 2πr² + 2πrh (two circular bases + lateral surface) S_total = 2π(2)² + 2π(2)(6) S_total = 8π + 24π S_total = 32π S_total = 100.53 m² ∴ Volume = 75.40 m³; Total Surface Area = 100.53 m²

Question Type

numerical

Answer Structure

• Line 1: Convert diameter to radius r = 2 m [0.5 mark]
• Line 2-3: Volume formula V = πr²h, substitution, and result [1 mark]
• Line 4-5: Surface area formula S = 2πr² + 2πrh, substitution, and result [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using diameter D = 4 m directly in πr²h without halving — gives V = 301.59 m³, four times the correct value
• Computing only the lateral surface area 2πrh and forgetting to add the two circular ends 2πr²
• Confusing lateral surface area (open tank, no top) with total surface area (both caps included)

Key Phrases To Include

• r = D/2 = 2 m
• V = πr²h
• S_total = 2πr² + 2πrh
• 75.40 m³
• 100.53 m²

Marks

5

Topic

Cone — Volume and Lateral Surface Area

Difficulty

medium

Template Id

T10

Examiner Tip

In a 5-mark cone problem, each part is worth roughly 1–1.5 marks. Even if you cannot complete part (c), earn maximum marks on (a) and (b). The slant height from (b) feeds into (c) — label your answers clearly so the examiner can follow.

Model Answer

Given: r = 4 m (base radius) h = 9 m (vertical height) To Find: (a) Volume V (b) Slant height L (c) Lateral surface area S_lat (a) Volume of Cone: Formula: V = (1/3)πr²h V = (1/3)π(4)²(9) V = (1/3)π(16)(9) V = (1/3)(144π) V = 48π V = 150.80 m³ (b) Slant Height: By Pythagorean theorem: L = √(r² + h²) L = √(4² + 9²) L = √(16 + 81) L = √97 L = 9.849 m (c) Lateral Surface Area: Formula: S_lat = πrL S_lat = π(4)(9.849) S_lat = 39.396π S_lat = 123.74 m² ∴ (a) V = 150.80 m³ (b) L = 9.849 m (c) S_lat = 123.74 m²

Question Type

numerical

Answer Structure

• Block 1: Given data — r, h identified with units [0.5 mark]
• Block 2 (Part a): Write V = (1/3)πr²h, substitution, V = 150.80 m³ [1.5 marks]
• Block 3 (Part b): Write L = √(r² + h²), substitution, L = 9.849 m [1.5 marks]
• Block 4 (Part c): Write S_lat = πrL, substitute L from part (b), S_lat = 123.74 m² [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using vertical height h instead of slant height L in the lateral surface area formula — S_lat = πrh is wrong
• Forgetting the (1/3) factor in the cone volume formula
• Not carrying enough decimal places in L, causing an inaccurate final surface area

Key Phrases To Include

• V = (1/3)πr²h
• L = √(r² + h²)
• S_lat = πrL
• slant height L not vertical height h
• 150.80 m³
• 9.849 m
• 123.74 m²

Marks

5

Topic

Frustum of a Pyramid — Solids

Difficulty

medium

Template Id

T11

Examiner Tip

The geometric mean term √(A₁A₂) is the defining feature of the frustum formula. In this example, √(16 × 4) = √64 = 8 — note that 8 is the area of a 2√2 m × 2√2 m square, the 'mean' section. Understanding this geometry reinforces the formula.

Model Answer

Given: Bottom base: square of side a₁ = 4 m Top base: square of side a₂ = 2 m Height: h = 6 m To Find: Volume V of frustum of a pyramid Step 1 — Compute base areas: A₁ = a₁² = (4)² = 16 m² A₂ = a₂² = (2)² = 4 m² Step 2 — Compute geometric mean area: √(A₁A₂) = √(16 × 4) = √64 = 8 m² Step 3 — Apply Frustum Volume Formula: V = (h/3)(A₁ + A₂ + √(A₁A₂)) V = (6/3)(16 + 4 + 8) V = (2)(28) V = 56 m³ ∴ Volume of frustum = 56 m³

Question Type

numerical

Answer Structure

• Line 1-2: Given data — a₁, a₂, h with units [0.5 mark]
• Line 3-4: Compute A₁ = 16 m², A₂ = 4 m² [1 mark]
• Line 5: Compute geometric mean √(A₁A₂) = 8 m² [1.5 marks]
• Line 6-7: Apply V = (h/3)(A₁ + A₂ + √(A₁A₂)) with full substitution [1.5 marks]
• Line 8: Final answer 56 m³ with unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using arithmetic mean (A₁ + A₂)/2 in place of the geometric mean √(A₁A₂)
• Using the side lengths directly without squaring to get area
• Confusing the frustum of a pyramid with the frustum of a cone (using πr² instead of a²)

Key Phrases To Include

• A₁ = a₁²
• A₂ = a₂²
• V = (h/3)(A₁ + A₂ + √(A₁A₂))
• geometric mean
• √(A₁A₂) = 8 m²
• 56 m³

Marks

5

Topic

Cone vs Cylinder — Volume Relationship

Difficulty

medium

Template Id

T12

Examiner Tip

Derive-then-verify problems reward structured thinking. Write the conceptual statement first, then the formula, then the numbers. The ratio check at the end demonstrates rigor and earns the verification mark.

Model Answer

Part 1 — Conceptual Relationship: A cone and a cylinder with identical base radius r and height h are related by: V_cone = (1/3) V_cylinder This means the volume of a cone is exactly one-third the volume of the enclosing cylinder. Physically, you can fill a cone three times to completely fill the same-sized cylinder. Part 2 — Formulae: Cylinder: V_cyl = πr²h Cone: V_cone = (1/3)πr²h Ratio: V_cone / V_cyl = (1/3) Part 3 — Numerical Verification (r = 3 m, h = 5 m): V_cyl = πr²h = π(3)²(5) = 45π = 141.37 m³ V_cone = (1/3)πr²h = (1/3)π(3)²(5) = 15π = 47.12 m³ Ratio check: V_cone / V_cyl = 47.12 / 141.37 = 0.3333 = 1/3 ✓ ∴ Volume of cone = (1/3) × Volume of corresponding cylinder For r = 3 m, h = 5 m: V_cone = 47.12 m³

Question Type

long_answer

Answer Structure

• Block 1: State the (1/3) relationship in words and formula [1.5 marks]
• Block 2: Write both formulas V_cyl = πr²h and V_cone = (1/3)πr²h [1 mark]
• Block 3: Substitute r = 3, h = 5 into both formulas [1.5 marks]
• Block 4: Compute ratio and verify = 1/3 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating the relationship as V_cone = (1/2)V_cylinder — a common confusion with the area of a triangle
• Not performing the ratio check — the problem asks to 'verify', so the check must be shown
• Omitting units on the final volumes

Key Phrases To Include

• V_cone = (1/3) V_cylinder
• same base and height
• V_cyl = πr²h
• V_cone = (1/3)πr²h
• ratio = 1/3
• 47.12 m³

Marks

5

Topic

Trapezoid — Plane Figures

Difficulty

hard

Template Id

T13

Examiner Tip

For the diagonal part, drawing a quick labeled sketch of the right-angled trapezoid takes 15 seconds and eliminates confusion about which sides form the right triangle. In a 5-mark problem, this diagram can save 2 marks.

Model Answer

Given: b₁ = 12 m (shorter parallel side) b₂ = 18 m (longer parallel side) h = 8 m (perpendicular distance / height) Right angle at one end of the longer base b₂ To Find: (a) Area A (b) Diagonal d (a) Area of Trapezoid: Formula: A = (1/2)(b₁ + b₂)h A = (1/2)(12 + 18)(8) A = (1/2)(30)(8) A = 120 m² (b) Diagonal of Right-Angled Trapezoid: At the right-angle end, the diagonal connects the top corner (end of b₁) to the far bottom corner (end of b₂). The horizontal span = b₂ = 18 m (from the right-angle corner) The offset = b₂ − b₁ = 18 − 12 = 6 m (non-right side horizontal projection) Since right-angled at the longer base, the right leg is: Vertical leg = h = 8 m Horizontal leg = b₂ = 18 m (full base to opposite top corner) Actually, taking the diagonal from the top-left corner to the bottom-right corner: Horizontal component = b₂ = 18 m (because top-left is directly above left end of b₂) But the right angle is at the bottom-right, so: Horizontal component of diagonal from top-left to bottom-right: = b₂ + (b₂ − b₁)/... Simplified approach (right angle at the right end of b₂): The right-angled trapezoid has one vertical leg of height h = 8 m. Diagonal from the top of the vertical leg to the far end of the longer base: Horizontal distance = b₂ − b₁... Correct setup: Right angle at the right end of b₂ means the right side is perpendicular (= h = 8 m). Diagonal = from top-right corner to bottom-left corner: Horizontal span = b₂ = 18 m, Vertical span = h = 8 m d = √(b₂² + h²) = √(18² + 8²) = √(324 + 64) = √388 = 19.70 m ∴ (a) Area = 120 m² (b) Diagonal = 19.70 m

Question Type

numerical

Answer Structure

• Line 1-2: List all given data correctly with units [0.5 mark]
• Block (a): A = (1/2)(b₁+b₂)h with substitution → 120 m² [2 marks]
• Block (b): Identify horizontal and vertical components of diagonal [1.5 marks]
• Line final: d = √(b₂² + h²) = 19.70 m [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using A = b × h (rectangle formula) instead of the trapezoid formula
• Not identifying the correct horizontal and vertical legs for the diagonal — a diagram would prevent this
• Using b₁ instead of b₂ as the horizontal component of the diagonal

Key Phrases To Include

• A = (1/2)(b₁ + b₂)h
• 120 m²
• right-angled trapezoid
• d = √(b₂² + h²)
• 19.70 m
• Pythagorean theorem

Marks

5

Topic

Square Pyramid — Volume and Surface Area

Difficulty

hard

Template Id

T14

Examiner Tip

The slant height of the triangular face uses the apothem (perpendicular distance from center to mid-side = a/2), NOT the half-diagonal (a√2/2). This is the most commonly lost mark in pyramid surface area problems. Draw the right triangle: apex → center of base → midpoint of a side.

Model Answer

Given: Square base side a = 6 m Height h = 8 m To Find: (a) Volume V (b) Total Surface Area S_total (a) Volume of Pyramid: A_base = a² = (6)² = 36 m² Formula: V = (1/3) A_base × h V = (1/3)(36)(8) V = (1/3)(288) V = 96 m³ (b) Total Surface Area: Step 1 — Slant height of triangular face (l): The apothem of the base (distance from center to midpoint of a side) = a/2 = 6/2 = 3 m l = √(h² + (a/2)²) = √(8² + 3²) = √(64 + 9) = √73 = 8.544 m Step 2 — Area of one triangular face: A_tri = (1/2) × base × l = (1/2)(6)(8.544) = 25.63 m² Step 3 — Total Surface Area: S_total = A_base + 4 × A_tri S_total = 36 + 4(25.63) S_total = 36 + 102.52 S_total = 138.52 m² ∴ (a) V = 96 m³ (b) S_total = 138.52 m²

Question Type

numerical

Answer Structure

• Line 1-2: Given data a = 6 m, h = 8 m [0.5 mark]
• Block (a): A_base = 36 m², V = (1/3)(36)(8) = 96 m³ [1.5 marks]
• Block (b) Step 1: Correct slant height l = √(h² + (a/2)²) = 8.544 m [1.5 marks]
• Block (b) Step 2-3: A_tri × 4 + A_base = 138.52 m² [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Computing slant height using the full diagonal of the base (a√2/2) instead of the apothem (a/2)
• Forgetting to include the base area in the total surface area
• Using h directly as the slant height of the triangular face

Key Phrases To Include

• V = (1/3)A_base × h
• slant height l = √(h² + (a/2)²)
• apothem = a/2
• A_tri = (1/2) × a × l
• S_total = A_base + 4A_tri
• 96 m³
• 138.52 m²

Marks

5

Topic

Sphere — Volume and Surface Area from Circumference

Difficulty

medium

Template Id

T15

Examiner Tip

When given the circumference of a great circle, the first step is always to extract r using C = 2πr. In the PRC board exam, this type of multi-step sphere problem (extract → volume → surface) appears frequently. Set up part (a) cleanly since r feeds all subsequent parts.

Model Answer

Given: Circumference of great circle C = 18.85 m To Find: (a) r (b) V (c) S (a) Radius: The great circle has circumference C = 2πr r = C / (2π) = 18.85 / (2π) = 18.85 / 6.2832 = 3.00 m (b) Volume of Sphere: Formula: V = (4/3)πr³ V = (4/3)π(3.00)³ V = (4/3)π(27) V = 36π V = 113.10 m³ (c) Surface Area: Formula: S = 4πr² S = 4π(3.00)² S = 4π(9) S = 36π S = 113.10 m² ∴ (a) r = 3.00 m (b) V = 113.10 m³ (c) S = 113.10 m²

Question Type

numerical

Answer Structure

• Block (a): C = 2πr → r = C/(2π) = 3.00 m [1.5 marks]
• Block (b): V = (4/3)πr³ → 113.10 m³ [1.5 marks]
• Block (c): S = 4πr² → 113.10 m² [1.5 marks]
• Unit annotation: m vs m² vs m³ clearly distinguished [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Treating the circumference as the surface area of the sphere — these are completely different quantities
• Computing r = C/π (forgetting the factor 2): r = 6.00 m, which doubles the computed radius
• Reporting V and S with the same unit — must distinguish m³ and m²

Key Phrases To Include

• great circle
• C = 2πr
• r = C/(2π) = 3.00 m
• V = (4/3)πr³
• S = 4πr²
• 113.10 m³
• 113.10 m²