GELE Mathematics — Plane, Solid Geometry and MensurationDetailed Explanation
Plane, Solid Geometry and Mensuration has a reputation among GELE reviewers for being deceptively tricky in the Mathematics subtest. PRC likes to hide the hard part in the phrasing rather than the concept. This long-form explanation untangles the phrasing traps and takes you through the concept the way someone who scored at the top of the GELE papers would.
Exam context
For the Geodetic Engineer Licensure Examination, Professional Regulation Commission (PRC) — Board of Geodetic Engineering tests Mathematics under a "Core" label, with Plane, Solid Geometry and Mensuration in the 3rd slot across 10 chapters. GELE candidates must clear the 70% weighted average, no sub-test below 50% cut on the 2026 paper, which draws about a meaningful share of Mathematics questions. Date to watch: September 2026.
Plane, Solid Geometry and Mensuration - Detailed Explanation
Plane and Solid Geometry with Mensuration is one of the most consistently tested topics in the PRC Civil Engineer Licensure Examination. Mastery of this chapter is not merely academic — it has direct engineering applications in earthwork volume computation, concrete quantity takeoffs, reservoir and tank sizing, and structural cross-section analysis. In the board exam, expect 5–10 problems per sitting drawn from areas of polygons, circles, and volumes of prisms, pyramids, cones, spheres, frustums, and prismatoids. The Pappus theorems frequently appear as higher-order problems. This chapter provides a systematic review of all critical formulas, worked board-style problems in SI units, and targeted guidance on the specific pitfalls that cost examinees points.
Concepts
Plane Figures — Triangles and Quadrilaterals
A plane figure is any two-dimensional shape. For the board exam, you must know area formulas by heart and recognize which formula applies given the available data. **Triangle:** The most versatile area formulas are: • A = (1/2)bh — when base b and perpendicular height h are known • A = √[s(s−a)(s−b)(s−c)] — Heron's Formula, when all three sides a, b, c are known; s = (a+b+c)/2 is the semi-perimeter • A = (1/2)ab·sin C — when two sides and the included angle are known • A = abc/(4R) — when circumradius R is given • A = rs — when inradius r is given; r = Area/s **Rectangle:** A = length × width = LW **Square:** A = s²; diagonal = s√2 **Parallelogram:** A = bh = ab·sin θ (a, b are adjacent sides, θ is included angle) **Rhombus:** A = (1/2)d₁d₂ (d₁, d₂ are diagonals) **Trapezoid:** A = (1/2)(b₁ + b₂)h where b₁ and b₂ are the parallel sides and h is the perpendicular distance between them. For a general quadrilateral with diagonals d₁ and d₂ intersecting at angle θ: A = (1/2)d₁d₂·sin θ. For a cyclic quadrilateral (inscribed in a circle) with sides a, b, c, d: A = √[(s−a)(s−b)(s−c)(s−d)] — Brahmagupta's Formula, where s = (a+b+c+d)/2.
Examples
This is a classic board-exam triangle. The semi-perimeter is computed first, then substituted into Heron's Formula. Note the 13-14-15 triangle has area 84 m² — a value worth memorizing as it appears frequently.
Scenario
A triangular plot of land has sides a = 13 m, b = 14 m, and c = 15 m. Find its area.
Solution
Step 1 — Compute semi-perimeter: s = (13 + 14 + 15)/2 = 42/2 = 21 m Step 2 — Apply Heron's Formula: A = √[s(s−a)(s−b)(s−c)] A = √[21(21−13)(21−14)(21−15)] A = √[21 × 8 × 7 × 6] A = √[7,056] A = 84 m²
This mirrors a hydraulics problem where the trapezoidal channel cross-section area is needed for flow rate computation (Q = Av). The perpendicular depth is always h — not the slant side of the channel bank.
Scenario
A trapezoidal channel cross-section has parallel widths of 4.0 m and 6.0 m and a perpendicular depth of 2.5 m. Find the cross-sectional area.
Solution
Step 1 — Identify the parallel sides and height: b₁ = 4.0 m, b₂ = 6.0 m, h = 2.5 m Step 2 — Apply the trapezoid formula: A = (1/2)(b₁ + b₂)h A = (1/2)(4.0 + 6.0)(2.5) A = (1/2)(10.0)(2.5) A = 12.5 m²
Applications
- Land area computation and lot surveys (RA 8550, land registration)
- Cross-sectional area of irrigation canals and drainage channels
- Reinforcing bar cut-off layouts using triangular geometry
- Roof plan area calculations for roofing material estimates
Misconceptions
- Using the slant side of a trapezoid as the height h — height must be PERPENDICULAR to the parallel sides
- Confusing perimeter with semi-perimeter in Heron's Formula — s is HALF the perimeter
- Applying Brahmagupta's Formula to non-cyclic quadrilaterals — it is valid ONLY for cyclic (inscribed) quadrilaterals
- Using diameter instead of radius in formulas — a very common 'careless' error in boards
Related Concepts
- Regular Polygons
- Circle Geometry
- Coordinate Geometry (Shoelace Formula for irregular polygons)
Common Exam Questions
Example
Sides 5 m, 12 m, 13 m — note this is a right triangle (5²+12²=13²), so A = (1/2)(5)(12) = 30 m². Always check for right-triangle shortcuts.
Approach
Compute semi-perimeter s first, then apply Heron's Formula. Check: verify the triangle inequality (sum of any two sides > third side).
Question Type
Find area given three sides
Example
Two sides 8 m and 10 m, included angle 30°: A = (1/2)(8)(10)sin30° = (1/2)(8)(10)(0.5) = 20 m²
Approach
Apply A = (1/2)ab·sin C directly. Convert angle to degrees if given in DMS first.
Question Type
Find area given two sides and included angle
Key Points To Remember
- Heron's Formula requires the semi-perimeter s = (a+b+c)/2 first — don't skip this step
- Trapezoid height h is the PERPENDICULAR distance between parallel sides, not the slant leg
- For a rhombus, A = (1/2)d₁d₂ uses the TWO diagonals, not the side
- The included angle in A = (1/2)ab·sin C must be between sides a and b
- Brahmagupta's Formula reduces to Heron's Formula when d = 0 (triangle)
Circles — Area, Circumference, Sectors, and Segments
Circle geometry is heavily tested because it connects to arc lengths, sector areas, and circular segments — all of which appear in structural and hydraulic applications. **Fundamental formulas:** • Area: A = πr² • Circumference: C = 2πr = πd • Diameter: d = 2r **Circular Sector** (a 'pie slice'): • Arc length: s = rθ (θ must be in RADIANS) • Sector area: A_sector = (1/2)r²θ = (θ/2π)·πr² = (θ/360°)·πr² [if θ in degrees] **Circular Segment** (region between a chord and its arc): A_segment = A_sector − A_triangle A_segment = (1/2)r²(θ − sin θ) [θ in radians] The triangle area within the sector: A_triangle = (1/2)r²·sin θ **Circular Ring (Annulus):** A = π(R² − r²) where R = outer radius, r = inner radius **Arc length in degrees:** s = (θ/360°)·2πr **Key relationship:** To convert degrees to radians: θ_rad = θ_deg × (π/180) **Inscribed angle theorem:** An inscribed angle is half the central angle subtending the same arc. **Tangent-chord angle:** equals half the intercepted arc.
Examples
Always convert to radians when using s = rθ or A = (1/2)r²θ. The alternative degree-based check confirms the answer. Note that a 120° sector is exactly 1/3 of the full circle.
Scenario
A circular sector has a radius of 6 m and a central angle of 120°. Find (a) the arc length and (b) the sector area.
Solution
Step 1 — Convert 120° to radians: θ = 120° × (π/180) = (2π/3) rad ≈ 2.0944 rad Step 2 — Arc length: s = rθ = 6 × (2π/3) = 4π ≈ 12.566 m Step 3 — Sector area: A_sector = (1/2)r²θ = (1/2)(6²)(2π/3) A_sector = (1/2)(36)(2π/3) = 12π ≈ 37.699 m² Alternative check using degrees: A_sector = (120/360) × π(6²) = (1/3)π(36) = 12π ✓
The segment is ALWAYS smaller than the sector. The formula A = (1/2)r²(θ − sin θ) in one step: A = (1/2)(100)(π/3 − sin 60°) = 50(1.0472 − 0.8660) = 50(0.1812) = 9.06 m² ✓
Scenario
Find the area of a circular segment formed by a chord that subtends a central angle of 60° in a circle of radius 10 m.
Solution
Step 1 — Convert 60° to radians: θ = 60° × (π/180) = π/3 rad Step 2 — Sector area: A_sector = (1/2)r²θ = (1/2)(10²)(π/3) = (1/2)(100)(π/3) = 50π/3 ≈ 52.36 m² Step 3 — Triangle area (isosceles triangle with two sides = r = 10 m): A_triangle = (1/2)r²·sin θ = (1/2)(10²)·sin 60° A_triangle = (1/2)(100)(√3/2) = 25√3 ≈ 43.30 m² Step 4 — Segment area: A_segment = A_sector − A_triangle A_segment = 52.36 − 43.30 = 9.06 m²
Applications
- Sizing of circular water tanks and pipes (volume and cross-section area)
- Earthwork and grade separation: circular tunnel cross-sections
- Reinforced concrete circular columns — gross area = πr² for axial load capacity
- Roadway curve geometry (arc length of horizontal curves in highway design)
Misconceptions
- Using degrees directly in s = rθ without conversion to radians — will give wrong arc length
- Confusing segment (bounded by chord and arc) with sector (bounded by two radii and arc)
- Forgetting to subtract the triangle from the sector when computing segment area
- Using circumference formula for arc length without applying the angle fraction
Related Concepts
- Regular Polygons (circle as limiting case of infinite-sided polygon)
- Pappus Theorems (revolving circular areas)
- Cone and Cylinder (circular bases)
Common Exam Questions
Example
Radius 5 m, θ = 72°: A = (72/360)·π(25) = (1/5)(25π) = 5π ≈ 15.71 m²
Approach
Either convert to radians and use (1/2)r²θ, OR use (θ/360°)·πr². Both give the same answer. Pick the method you are more comfortable with and stick to it.
Question Type
Sector area with angle in degrees
Example
For major segment with central angle 240°: A_major = πr² − A_minor_segment (where minor has θ = 120°)
Approach
Always find sector area first, then subtract triangle area. For major segment (θ > 180°), find the minor segment area and subtract from total circle area.
Question Type
Area of circular segment (minor or major)
Key Points To Remember
- Sector area formula (1/2)r²θ REQUIRES θ in RADIANS — this is the #1 source of errors
- Segment area = Sector area MINUS Triangle area; NEVER add them
- For a semicircle: A = πr²/2, perimeter = πr + 2r (arc + diameter)
- The chord length for a central angle θ: chord = 2r·sin(θ/2)
- Annulus area: π(R+r)(R−r) — factor of sum and difference of radii
Regular Polygons
A regular polygon has all sides equal and all interior angles equal. These appear in board exams in problems about inscribed/circumscribed circles and tile/section areas. **Key formulas for a regular polygon with n sides, each of length s:** • Interior angle per vertex: α = (n−2)·180°/n • Exterior angle: β = 360°/n • Sum of interior angles: S = (n−2)·180° • Area: A = (1/4)ns²·cot(π/n) = (1/4)ns²·cot(180°/n) • Perimeter: P = ns **Using apothem a** (perpendicular distance from center to a side): • A = (1/2)·P·a = (1/2)·(ns)·a • a = (s/2)·cot(π/n) **Circumscribed circle (circumradius R):** • R = (s/2)·csc(π/n) = (s/2)/sin(π/n) • A_circle = πR² **Inscribed circle (inradius r = apothem a):** • r = a = (s/2)·cot(π/n) • A_circle = πr² **Common polygons:** • n=3 (equilateral triangle): A = (√3/4)s² • n=4 (square): A = s² • n=5 (regular pentagon): A = (s²/4)√(25+10√5) ≈ 1.720s² • n=6 (regular hexagon): A = (3√3/2)s² ≈ 2.598s² • n=8 (regular octagon): A = 2(1+√2)s² ≈ 4.828s²
Examples
Three methods give the same answer — this cross-verification skill is valuable in exams. Method 3 works because a regular hexagon decomposes into 6 congruent equilateral triangles, all with side = s.
Scenario
Find the area of a regular hexagon with side length 6 m.
Solution
Method 1 — Using the general formula: A = (1/4)ns²·cot(180°/n) n = 6, s = 6 m A = (1/4)(6)(6²)·cot(180°/6) A = (1/4)(6)(36)·cot(30°) A = (1/4)(216)·(√3) A = 54√3 ≈ 93.53 m² Method 2 — Using the hexagon-specific formula: A = (3√3/2)s² = (3√3/2)(36) = 54√3 ≈ 93.53 m² ✓ Method 3 — Hexagon = 6 equilateral triangles, each with side s: A_triangle = (√3/4)s² = (√3/4)(36) = 9√3 m² A_hexagon = 6 × 9√3 = 54√3 ≈ 93.53 m² ✓
The apothem-based formula A = (1/2)Pa is often the most intuitive: imagine the polygon divided into n isosceles triangles each with base s and height = apothem a. Their total area is n × (1/2)(s)(a) = (1/2)(ns)(a) = (1/2)Pa.
Scenario
A regular pentagon has a side length of 10 m. Find (a) the apothem and (b) the area.
Solution
Step 1 — Apothem: a = (s/2)·cot(π/n) = (10/2)·cot(180°/5) a = 5·cot(36°) = 5 × (cos36°/sin36°) a = 5 × (0.8090/0.5878) = 5 × 1.3764 = 6.882 m Step 2 — Area using apothem: A = (1/2)·P·a = (1/2)·(5 × 10)·6.882 A = (1/2)(50)(6.882) = 172.05 m² Alternative check: A = (1/4)(5)(100)·cot(36°) = (125)(1.3764) = 172.05 m² ✓
Applications
- Computation of areas of polygonal land parcels and reservoirs
- Hexagonal steel bar cross-sections and bolt head geometry
- Floor tile areas and tessellation patterns in architectural finishes
- Cross-sectional areas of polygonal columns and structural sections
Misconceptions
- Confusing apothem (inradius) with circumradius — for regular hexagon only are they NOT equal; R = s but a = s·cos30° = (√3/2)s
- Using perimeter instead of side length in the area formula
- Forgetting that cot(180°/n) must be computed with the angle in the correct unit (degrees on a calculator)
Related Concepts
- Circles (inscribed and circumscribed)
- Triangles (decomposition into triangles)
- Prisms (regular polygon as the base)
Common Exam Questions
Example
Regular octagon, s = 5 m: A = (1/4)(8)(25)·cot(22.5°) = 50 × 2.4142 = 120.71 m²
Approach
Use A = (1/4)ns²·cot(180°/n). For hexagons, use A = (3√3/2)s². Always evaluate cot carefully using your calculator.
Question Type
Area of regular polygon given side length
Example
Equilateral triangle with A = 25√3 m²: A = (√3/4)s², so s² = 4A/√3 = 4(25√3)/√3 = 100, s = 10 m
Approach
Rearrange the area formula to solve for s: s = √[4A / (n·cot(180°/n))]
Question Type
Find side length given area of regular polygon
Key Points To Remember
- The apothem is the INRADIUS of the polygon — the radius of the inscribed circle
- Area = (1/2) × Perimeter × Apothem — applies to ALL regular polygons
- For a regular hexagon: the circumradius R equals the side length s
- The cot(π/n) in the area formula uses π/n in RADIANS (or equivalently 180°/n in degrees)
- For the board exam, memorize the hexagon formula A = (3√3/2)s² directly
Prisms and Cylinders
A prism is a solid with two parallel, congruent polygonal bases connected by rectangular lateral faces. A cylinder is the curved-surface analogue with circular bases. **RIGHT PRISM:** • Volume: V = A_base × h (h = perpendicular height between bases) • Lateral Surface Area: LSA = Perimeter_base × h • Total Surface Area: TSA = LSA + 2·A_base **OBLIQUE PRISM:** • Volume: V = A_base × h (h = perpendicular height, NOT slant height) • Note: An oblique prism has the same volume as a right prism with equal base area and equal perpendicular height **RIGHT CIRCULAR CYLINDER:** • Volume: V = πr²h • Lateral Surface Area: LSA = 2πrh • Total Surface Area: TSA = 2πrh + 2πr² = 2πr(h + r) **HOLLOW CYLINDER (pipe or annular prism):** • Volume: V = π(R² − r²)h = π(R+r)(R−r)h • where R = outer radius, r = inner radius **Cavalieri's Principle:** Two solids with equal cross-sectional areas at every height have equal volumes. This is why oblique and right prisms with equal bases and heights have equal volumes. **TRUNCATED PRISM** (cut by an oblique plane): • V = A_base × h_avg (where h_avg = average of all vertical edge lengths) • For a triangular truncated prism: h_avg = (h₁ + h₂ + h₃)/3
Examples
This is a standard civil engineering problem for sizing water storage tanks. Note that the volume = 75.40 m³ ≈ 75,400 liters (1 m³ = 1,000 L). For open-top tanks, TSA uses only one circular base: TSA_open = 2πrh + πr² = 24π + 4π = 28π ≈ 87.96 m².
Scenario
A cylindrical water tank has a diameter of 4.0 m and a height of 6.0 m. Find (a) its volume and (b) its total surface area.
Solution
Given: d = 4.0 m → r = 2.0 m, h = 6.0 m (a) Volume: V = πr²h = π(2.0)²(6.0) = π(4)(6) = 24π ≈ 75.40 m³ (b) Total Surface Area: LSA = 2πrh = 2π(2.0)(6.0) = 24π ≈ 75.40 m² 2·A_base = 2·πr² = 2·π(2.0)² = 8π ≈ 25.13 m² TSA = 24π + 8π = 32π ≈ 100.53 m²
The formula V = A_base × h is universal for all prisms. The critical step is computing the base area of the polygon correctly. Hexagonal prisms appear in honeycomb structures and certain structural precast members.
Scenario
A hexagonal prism has a regular hexagon base with side 3 m and a height of 8 m. Find its volume.
Solution
Step 1 — Area of regular hexagonal base: A_base = (3√3/2)s² = (3√3/2)(3²) = (3√3/2)(9) = (27√3)/2 ≈ 23.38 m² Step 2 — Volume: V = A_base × h = 23.38 × 8 = 187.06 m³ Exact form: V = (27√3/2)(8) = 108√3 ≈ 187.06 m³
Applications
- Concrete volume computation for columns, walls, and beams (prismatic members)
- Water tank and reservoir sizing (cylindrical and rectangular prisms)
- Pipe volume and flow capacity calculations
- Earthwork cut and fill computations using prismatic approximations
Misconceptions
- Using the slant height (lateral edge) instead of perpendicular height for oblique prisms — always use perpendicular h
- Forgetting to halve the diameter to get radius for cylinder formulas
- Computing only LSA when the problem asks for TSA — always check if bases are included
- For a hollow cylinder, using outer radius only for volume — must subtract inner cylinder volume
Related Concepts
- Pyramids and Cones (1/3 of prism/cylinder volume)
- Frustums (truncated cones/pyramids)
- Prismatoid Formula (earthwork application)
Common Exam Questions
Example
Right triangular prism, base legs 3 m and 4 m, h = 10 m: A_base = (1/2)(3)(4) = 6 m², V = 6 × 10 = 60 m³
Approach
Identify the base shape and compute A_base first. Then multiply by perpendicular height. Watch out for whether diameter or radius is given for cylinders.
Question Type
Find volume given dimensions
Example
Cylinder r = 3 m, h = 5 m: TSA = 2π(3)(5) + 2π(9) = 30π + 18π = 48π ≈ 150.80 m²
Approach
TSA = 2πrh + 2πr². Remember to add BOTH bases unless the problem specifies an open-top tank.
Question Type
Find total surface area of cylinder
Key Points To Remember
- Volume of prism = BASE AREA × PERPENDICULAR HEIGHT — not slant height
- For a cylinder, LSA is the lateral surface area only (the curved part); TSA adds both circular ends
- Unwrap the cylinder lateral surface to get a rectangle: width = circumference = 2πr, height = h
- For a hollow pipe: V = π(R² − r²)h; identify inner and outer radii clearly
- A truncated prism volume = A_base × (average of the parallel edge heights)
Pyramids and Cones
A pyramid has a polygonal base with all lateral faces meeting at a single apex. A cone is the curved-surface version with a circular base. **KEY RULE:** Volume of any pyramid or cone = (1/3) × Volume of corresponding prism or cylinder with the same base and height. **PYRAMID:** • Volume: V = (1/3)·A_base·h • Lateral Surface Area: LSA = (1/2)·P_base·l (where l = slant height of lateral face) • Slant height of lateral face: l = √(h² + a²) where a = apothem of base • Lateral edge length (from apex to base corner): e = √(h² + R²) where R = circumradius of base **RIGHT CIRCULAR CONE:** • Volume: V = (1/3)πr²h • Slant height: L = √(r² + h²) • Lateral Surface Area: LSA = πrL • Total Surface Area: TSA = πrL + πr² = πr(L + r) **IMPORTANT DISTINCTION:** • h = vertical (perpendicular) height from apex to base plane • L = slant height (along the lateral surface) = √(r² + h²) • These are DIFFERENT; the lateral area uses L, the volume uses h **OBLIQUE PYRAMID/CONE:** V = (1/3)·A_base·h (still valid where h = perpendicular height to base) **RELATIONSHIP:** For the same base and height: V_pyramid = (1/3)·V_prism V_cone = (1/3)·V_cylinder
Examples
This is a complete cone problem as commonly set in boards. Notice the slant height L ≠ h. If you used h = 9 instead of L ≈ 9.849 for the LSA, you'd get πrh = 4π(9) = 113.10 m² — a classic wrong answer choice in multiple-choice exams.
Scenario
A cone has a base diameter of 8 m and a vertical height of 9 m. Find (a) volume, (b) slant height, (c) lateral surface area, and (d) total surface area.
Solution
Given: d = 8 m → r = 4 m, h = 9 m (a) Volume: V = (1/3)πr²h = (1/3)π(4²)(9) = (1/3)π(16)(9) = 48π ≈ 150.80 m³ (b) Slant height: L = √(r² + h²) = √(4² + 9²) = √(16 + 81) = √97 ≈ 9.849 m (c) Lateral Surface Area: LSA = πrL = π(4)(√97) = 4π√97 ≈ 123.82 m² (d) Total Surface Area: TSA = πrL + πr² = 4π√97 + π(16) = π(4√97 + 16) ≈ 123.82 + 50.27 = 174.09 m²
For a square pyramid, the apothem of the base is half the side length (3 m for a 6 m square). This gives the slant height l of the lateral FACE. The lateral EDGE (from apex to corner) would be √(8² + (3√2)²) = √(64 + 18) = √82 ≈ 9.055 m — different from the slant height of the face.
Scenario
A square pyramid has a 6 m × 6 m base and a height of 8 m. Find its volume and lateral surface area.
Solution
Given: base = 6 m × 6 m, h = 8 m A_base = 6 × 6 = 36 m² (a) Volume: V = (1/3)·A_base·h = (1/3)(36)(8) = 96 m³ (b) Slant height of lateral face: Apothem of square base = s/2 = 6/2 = 3 m l = √(h² + apothem²) = √(8² + 3²) = √(64 + 9) = √73 ≈ 8.544 m (c) Lateral Surface Area: LSA = (1/2)·P_base·l = (1/2)(4 × 6)(√73) = (1/2)(24)(√73) = 12√73 ≈ 102.54 m²
Applications
- Stockpile volume computation (gravel, sand piles approximate a cone)
- Volume of soil in conical spoil heaps in earthwork
- Roof pyramid volume and surface area for roofing material estimates
- Conical strainer and funnel sizing in plumbing and process engineering
Misconceptions
- Using h (vertical height) in the lateral area formula instead of L (slant height) — yields a systematically low LSA
- Forgetting the 1/3 factor in the volume formula — gives volume 3× too large
- For pyramid lateral area, using circumradius R instead of apothem a for the slant height of the face
- Confusing lateral edge length (apex to corner) with slant height of face (apex to midpoint of base edge)
Related Concepts
- Frustum of Cone or Pyramid
- Prisms and Cylinders (pyramid = 1/3 of prism)
- Pappus Theorems (revolution of triangular area generates a cone)
Common Exam Questions
Example
Slant height L = 13 m, r = 5 m: h = √(13² − 5²) = √(169 − 25) = √144 = 12 m; V = (1/3)π(25)(12) = 100π ≈ 314.16 m³
Approach
Apply V = (1/3)πr²h directly. If diameter is given, halve it first. If slant height L is given instead of h, find h = √(L² − r²).
Question Type
Find volume of cone given radius and height
Example
Cone r = 6 m, h = 8 m: L = √(36+64) = 10 m; LSA = π(6)(10) = 60π ≈ 188.50 m²
Approach
LSA = πrL. If h is given instead of L, compute L = √(r² + h²) first.
Question Type
Find lateral surface area given radius and slant height
Key Points To Remember
- Volume of pyramid/cone = (1/3) of the corresponding prism/cylinder — the 1/3 factor is ALWAYS there
- Cone lateral area = πrL uses SLANT height L, NOT vertical height h
- Slant height L = √(r² + h²) — this is the Pythagorean theorem on the right triangle formed by r, h, and L
- For a pyramid, the slant height l of a lateral face uses the APOTHEM of the base, not the circumradius
- LSA of cone is the area of a sector when the cone is 'unrolled': sector radius = L, arc = 2πr
Sphere and Spherical Solids
The sphere is arguably the most tested solid in the PRC CE board exam for its clean, elegant formulas. **SPHERE:** • Volume: V = (4/3)πr³ • Surface Area: SA = 4πr² **HEMISPHERE:** • Volume: V = (2/3)πr³ • Curved Surface Area: CSA = 2πr² • Total Surface Area (including flat base): TSA = 2πr² + πr² = 3πr² **SPHERICAL CAP** (a slice cut by a plane at distance d from center, where the cap has height h): • Volume: V_cap = (πh²/3)(3r − h) where h = height of cap, r = sphere radius • Lateral Surface Area: A_cap = 2πrh • Relationship: h = r − d (where d = distance from center to cutting plane) **SPHERICAL ZONE** (region between two parallel planes cutting the sphere): • Volume: V_zone = (πh/6)(3a² + 3b² + h²) where h = height of zone, a = radius of base circle 1, b = radius of base circle 2 • Lateral Surface Area: A_zone = 2πrh (same formula as cap, depending only on sphere radius r and zone height h) **SPHERICAL SECTOR** (cone + cap): • Volume: V_sector = (2/3)πr²h (h = height of the corresponding cap) **SPHERICAL WEDGE** (like an orange slice from pole to pole): • Volume: V_wedge = (θ/2π)·(4/3)πr³ = (2/3)θr³ (θ in radians) **SPHERICAL LUNE** (surface area between two great circles with dihedral angle θ): • A_lune = 2r²θ (θ in radians)
Examples
This demonstrates an elegant coincidence: for r = 3, V and SA are numerically equal but have different units. Water tank sizing: 113,097 liters is equivalent to about 113.1 kL of storage capacity — useful for community water supply planning.
Scenario
A spherical water tank has an inner radius of 3 m. Find (a) its volume in m³ and in liters, and (b) its surface area.
Solution
(a) Volume: V = (4/3)πr³ = (4/3)π(3)³ = (4/3)π(27) = 36π ≈ 113.10 m³ In liters: 113.10 m³ × 1,000 L/m³ = 113,097 liters ≈ 113.1 kL (b) Surface Area: SA = 4πr² = 4π(3)² = 4π(9) = 36π ≈ 113.10 m² Note: For this sphere, V (in m³) numerically equals SA (in m²) — this is a mathematical coincidence for r = 3.
The cap height h = r − d assumes the plane cuts on the same side as the cap. If the plane is at distance d from center and we want the MAJOR cap (the larger piece), use h = r + d. Always draw a cross-section diagram to confirm which cap you are computing.
Scenario
A spherical cap is cut from a sphere of radius 10 m by a plane 6 m from the center. Find the volume and lateral surface area of the cap.
Solution
Given: r = 10 m, distance from center to plane d = 6 m Step 1 — Cap height: h = r − d = 10 − 6 = 4 m Step 2 — Cap volume: V_cap = (πh²/3)(3r − h) V_cap = (π(4²)/3)(3(10) − 4) V_cap = (16π/3)(30 − 4) V_cap = (16π/3)(26) = 416π/3 ≈ 435.47 m³ Step 3 — Lateral surface area of cap: A_cap = 2πrh = 2π(10)(4) = 80π ≈ 251.33 m²
Applications
- Spherical water towers and elevated storage tanks
- Dome structures (geodesic and spherical shell analysis)
- Ball valve and spherical bearing geometry
- Estimation of stockpile volume for aggregate and earth (hemisphere approximation)
Misconceptions
- Using d (diameter) instead of r (radius) in the sphere formulas — gives volume 8× too large, area 4× too large
- For a hemisphere, computing TSA as just 2πr² — must add the flat circular base πr² for total surface area
- Computing cap height h incorrectly — draw the cross-section and use h = r − d carefully
- Confusing spherical cap (one cut) with spherical zone (two parallel cuts)
Related Concepts
- Pappus Theorems (semicircle area revolved = sphere volume)
- Circles (great circle of sphere has area πr²)
- Frustum (spherical zone is the curved analogue)
Common Exam Questions
Example
Diameter 12 m: r = 6 m; V = (4/3)π(216) = 288π ≈ 904.78 m³; SA = 4π(36) = 144π ≈ 452.39 m²
Approach
Identify r (check if diameter is given — halve it). Apply V = (4/3)πr³ and SA = 4πr².
Question Type
Find volume and surface area of sphere
Example
Sphere r = 5 m, plane at d = 3 m from center: h = 5 − 3 = 2 m; V = (π·4/3)(15−2) = (4π/3)(13) = 52π/3 ≈ 54.45 m³
Approach
Draw the cross-section. Identify r (sphere radius) and h (cap height). Use V = (πh²/3)(3r−h) and A = 2πrh.
Question Type
Spherical cap problems
Key Points To Remember
- Sphere surface area = 4πr² = 4 times the area of a great circle (πr²)
- V_sphere = (4/3)πr³; V_hemisphere = (2/3)πr³ — hemisphere is exactly HALF the sphere
- Spherical cap volume = (πh²/3)(3r − h) — depends on sphere radius r AND cap height h
- Spherical zone lateral area = 2πrh — same as a cylindrical band of the same height and radius
- For diameter d: V = (4/3)π(d/2)³ = πd³/6; SA = 4π(d/2)² = πd²
Frustum of a Cone or Pyramid
A frustum is what remains after cutting a cone or pyramid with a plane parallel to the base, removing the smaller top portion. It is one of the most important solids in civil engineering because earthwork volumes in road construction frequently use the frustum formula or the more general prismatoid formula. **FRUSTUM OF A CONE (Truncated Cone):** • Volume: V = (πh/3)(R² + Rr + r²) where R = radius of larger base, r = radius of smaller base, h = perpendicular height • Alternatively: V = (h/3)(A₁ + A₂ + √(A₁A₂)) where A₁ = πR², A₂ = πr² • Slant height: L = √(h² + (R−r)²) • Lateral Surface Area: LSA = π(R + r)L • Total Surface Area: TSA = π(R + r)L + πR² + πr² **FRUSTUM OF A PYRAMID:** • Volume: V = (h/3)(A₁ + A₂ + √(A₁A₂)) This formula works for ANY frustum regardless of base shape • where A₁ = area of larger base, A₂ = area of smaller base, h = perpendicular height **KEY INSIGHT — Why √(A₁A₂)?** The middle term √(A₁A₂) is the geometric mean of the two base areas. It represents the area of a 'middle' cross-section between the two bases. For similar bases (pyramid/cone), √(A₁A₂) is the area at the average height. **LIMITING CASES:** • When A₂ = 0: frustum becomes a cone/pyramid, V = (h/3)(A₁ + 0 + 0) = (h/3)A₁ = (1/3)A₁h ✓ • When A₂ = A₁: frustum becomes a prism/cylinder, V = (h/3)(A₁ + A₁ + A₁) = (h/3)(3A₁) = A₁h ✓
Examples
Note that √(A₁A₂) = √(16π)(4π) = √(64π²) = 8π, which equals π(Rr) = π(4)(2) = 8π. This confirms that √(A₁A₂) = πRr for circular frustums. The formula reduces to V = (πh/3)(R² + Rr + r²) — always a useful cross-check.
Scenario
A frustum of a cone has a bottom base radius of 4 m, a top base radius of 2 m, and a height of 3 m. Find the volume.
Solution
Given: R = 4 m, r = 2 m, h = 3 m A₁ = πR² = π(4²) = 16π m² A₂ = πr² = π(2²) = 4π m² √(A₁A₂) = √(16π × 4π) = √(64π²) = 8π m² V = (h/3)(A₁ + A₂ + √(A₁A₂)) V = (3/3)(16π + 4π + 8π) V = (1)(28π) V = 28π ≈ 87.96 m³ Verification using cone formula: V = (πh/3)(R² + Rr + r²) V = (π·3/3)(16 + 8 + 4) V = π(28) = 28π ✓
For square bases: A₁ = s₁², A₂ = s₂², √(A₁A₂) = s₁s₂. Here, √(36 × 9) = 6 × 3 = 18. This shortcut works for similar polygonal bases — the geometric mean of areas equals the product of the corresponding sides.
Scenario
A frustum of a square pyramid has a 6 m × 6 m bottom base, a 3 m × 3 m top base, and a height of 4 m. Find the volume.
Solution
A₁ = 6² = 36 m² A₂ = 3² = 9 m² √(A₁A₂) = √(36 × 9) = √324 = 18 m² V = (h/3)(A₁ + A₂ + √(A₁A₂)) V = (4/3)(36 + 9 + 18) V = (4/3)(63) V = 84 m³
Applications
- Earthwork volume computation for road cuts and fills using end-area and average-end-area method
- Retaining wall volume (trapezoidal cross-section frustum)
- Concrete volume for tapered columns and bridge piers
- Volume of grain storage silos with conical lower section
Misconceptions
- Using A₁A₂ instead of √(A₁A₂) in the frustum formula — this is the most common error in board exams
- Using the arithmetic mean (A₁ + A₂)/2 instead of the geometric mean √(A₁A₂) for the middle term
- Using the vertical height h instead of slant height L in the lateral area formula LSA = π(R+r)L
- Forgetting the 1/3 factor — the frustum volume denominator is 3, not 2 or 6
Related Concepts
- Prismatoid Formula (generalization of frustum formula)
- Pyramid and Cone (frustum with upper base → 0)
- Earthwork Volume Computation
Common Exam Questions
Example
Conical frustum: R = 5 m, r = 3 m, h = 6 m: V = (6π/3)(25 + 15 + 9) = 2π(49) = 98π ≈ 307.88 m³
Approach
Compute A₁, A₂, then √(A₁A₂). Apply V = (h/3)(A₁ + A₂ + √(A₁A₂)). For circular frustums, also check: does √(A₁A₂) = πRr?
Question Type
Frustum volume given two radii and height
Example
R = 4, r = 2, h = 3: L = √(9 + 4) = √13; LSA = π(6)(√13) = 6π√13 ≈ 67.94 m²
Approach
Compute slant height L = √[h² + (R−r)²], then LSA = π(R+r)L.
Question Type
Find the lateral surface area of a frustum
Key Points To Remember
- Frustum volume uses SQUARE ROOT of product of base areas: √(A₁A₂), NOT A₁A₂
- The frustum formula works for ANY frustum (cone, pyramid, any polygonal base) — universal formula
- Slant height of frustum cone: L = √[h² + (R−r)²] — use this for lateral area
- The 1/3 factor in V = (h/3)(A₁ + A₂ + √(A₁A₂)) is still there from the derivation
- Verify with limiting cases: A₂ = 0 gives pyramid; A₁ = A₂ gives prism
Prismatoid and Prismatoid Formula
The Prismatoid Formula (also called Simpson's Rule for Volumes or the Prismoidal Formula) is the most powerful single formula in earthwork computation. **DEFINITION:** A prismatoid is a polyhedron with all vertices lying in one of two parallel planes. It includes prisms, pyramids, frustums, wedges, and obelisks as special cases. **PRISMATOID FORMULA (Prismoidal Formula):** $$V = \frac{h}{6}(A_1 + 4A_m + A_2)$$ where: • h = perpendicular distance between the two parallel base planes • A₁ = area of base 1 (one parallel plane) • A₂ = area of base 2 (other parallel plane) • Aₘ = area of the cross-section at MID-HEIGHT (h/2 from each base) **DERIVATION BASIS:** This is Simpson's 1/3 Rule applied to the area function A(x) integrated over height h. It is exact (not approximate) for any prismatoid. **APPLICATIONS TO COMMON SOLIDS:** • Prism: A₁ = A₂ = Aₘ = A: V = (h/6)(A + 4A + A) = (h/6)(6A) = Ah ✓ • Pyramid: A₂ = 0, Aₘ = A₁/4: V = (h/6)(A₁ + 4(A₁/4) + 0) = (h/6)(2A₁) = A₁h/3 ✓ • Sphere: h = 2r, A₁ = A₂ = 0, Aₘ = πr²: V = (2r/6)(0 + 4πr² + 0) = (4πr³)/3 ✓ • Cone: A₁ = πR², A₂ = 0, Aₘ = π(R/2)²= πR²/4: V = (h/6)(πR² + 4·πR²/4 + 0) = (h/6)(2πR²) = πR²h/3 ✓ **EARTHWORK APPLICATION:** In road earthwork, each end area method uses: • A₁ = cross-sectional area at one station • A₂ = cross-sectional area at the next station • Aₘ = cross-sectional area at the mid-station • h = distance between stations (typically 20 m) The prismatoid formula gives more accurate earthwork volumes than the simple average-end-area method: V_avg = h(A₁+A₂)/2.
Examples
This demonstrates that the prismatoid formula gives the exact answer for a wedge (it is a prismatoid). The mid-height cross-section is found by linear interpolation of dimensions — the width at mid-height is 2 m (half of 4 m), giving Aₘ = 6 × 2 = 12 m².
Scenario
A wedge-shaped solid (triangular prism with one edge collapsed to a point) has a rectangular base 6 m × 4 m, a triangular top edge of length 6 m, and a height of 5 m. Apply the prismatoid formula.
Solution
A₁ = 6 × 4 = 24 m² (rectangular base) A₂ = (1/2)(6)(0) = 0 m² (top edge collapsed — length 6 m, width 0 — it's a line, so A₂ = 0) Actually, this is a triangular wedge. Let's reconsider: Base: 6 m × 4 m rectangle → A₁ = 24 m² Top: a line (ridge) parallel to the 6 m side → A₂ = 0 Mid-height cross-section: a rectangle 6 m × 2 m = 12 m² (linear interpolation) Aₘ = 6 × 2 = 12 m² V = (h/6)(A₁ + 4Aₘ + A₂) V = (5/6)(24 + 4(12) + 0) V = (5/6)(24 + 48) V = (5/6)(72) = 60 m³ Verification: Wedge V = (1/2)(base area)(height) = (1/2)(24)(5) = 60 m³ ✓
The prismatoid formula gives the more accurate earthwork volume. The average-end-area method overpredicts by about 5.4% in this case. In large earthwork contracts, this difference can represent significant cost discrepancies.
Scenario
In road earthwork, a cut section has the following cross-sectional areas: at Station 10+000: A₁ = 45 m², at Station 10+020 (mid): Aₘ = 30 m², at Station 10+040: A₂ = 20 m². The total distance between end stations is h = 40 m. Find the volume using the prismatoid formula.
Solution
V = (h/6)(A₁ + 4Aₘ + A₂) V = (40/6)(45 + 4(30) + 20) V = (40/6)(45 + 120 + 20) V = (40/6)(185) V = 7400/6 ≈ 1,233.33 m³ Comparison with average-end-area method: V_avg = h(A₁ + A₂)/2 = 40(45 + 20)/2 = 40(32.5) = 1,300 m³ Prismoidal correction = V_avg − V_prismatoid = 1,300 − 1,233.33 = 66.67 m³ overestimate by average-end-area method
Applications
- Earthwork volume computation in road, railway, and canal construction
- Concrete volume for irregular structural members
- Volume of transition sections in culverts and spillways
- Verification formula for volumes of any prismatoid shape
Misconceptions
- Using (A₁ + A₂)/2 for Aₘ — the mid-height area must be computed geometrically, not averaged
- Using h as the HALF-height instead of the FULL height between the two end planes
- Coefficient confusion: the middle area gets coefficient 4 and the ends get 1 each, not the other way around
- Thinking the prismatoid formula is approximate — it is EXACT for all prismatoids
Related Concepts
- Frustum (special case of prismatoid)
- Average-End-Area Method (less accurate alternative for earthwork)
- Simpson's Rule (mathematical basis of prismatoid formula)
Common Exam Questions
Example
A₁ = 50 m², Aₘ = 35 m², A₂ = 25 m², h = 12 m: V = (12/6)(50 + 140 + 25) = 2(215) = 430 m³
Approach
Substitute directly into V = (h/6)(A₁ + 4Aₘ + A₂). Ensure Aₘ is the area at mid-height, not the average.
Question Type
Apply prismatoid formula given three cross-section areas
Example
Sphere: h=2r, A₁=A₂=0, Aₘ=πr²: V=(2r/6)(0+4πr²+0)=(2r)(4πr²)/6=8πr³/6=4πr³/3 ✓
Approach
Substitute specific area expressions for sphere, cone, pyramid and confirm they reduce to correct formulas. Good exam essay question.
Question Type
Verify that prismatoid formula reduces to known solid formulas
Key Points To Remember
- V = (h/6)(A₁ + 4Aₘ + A₂) — the MIDDLE area has coefficient 4, the two ends have coefficient 1 each
- Aₘ is the area at EXACTLY mid-height — NOT the average of A₁ and A₂
- The prismatoid formula is EXACT for polynomials of degree ≤ 3 (covering all common solids)
- For earthworks: prismatoid formula gives a prismoidal correction to the average-end-area method
- Memorize: '1–4–1' coefficient pattern for the three cross-sections
Pappus Theorems — Surfaces and Volumes of Revolution
The two Pappus Theorems (Pappus-Guldinus Theorems) are elegant tools for finding the surface area or volume of a solid of revolution without integration. They are high-value board exam topics because they can solve complex-looking problems very quickly. **THEOREM 1 — Surface of Revolution (First Theorem of Pappus):** The lateral surface area generated by revolving a plane curve of length L about an external axis equals the product of the arc length L and the circumference of the circle traced by the centroid of the arc: $$S = 2\pi\bar{d}\cdot L$$ where: • S = surface area of revolution • $\bar{d}$ = perpendicular distance from the centroid of the ARC to the axis of revolution • L = length of the arc (curve) **THEOREM 2 — Volume of Revolution (Second Theorem of Pappus):** The volume generated by revolving a plane area A about an external axis equals the product of the area A and the circumference of the circle traced by the centroid of the area: $$V = 2\pi\bar{d}\cdot A$$ where: • V = volume of revolution • $\bar{d}$ = perpendicular distance from the centroid of the AREA to the axis of revolution • A = area of the plane figure **CONDITION:** The axis of revolution must be EXTERNAL to the figure (must not intersect the area or curve). **APPLICATIONS:** • Circle revolved about external axis → TORUS • Rectangle revolved about external axis → Hollow cylinder • Triangle revolved about its base → Cone • Semicircle revolved about diameter → Sphere (from area standpoint) **Centroid locations for common shapes:** • Semi-circle arc: $\bar{d}$ from diameter = 2r/π • Semi-circular area: $\bar{d}$ from diameter = 4r/(3π) • Triangle: $\bar{d}$ from base = h/3; from apex = 2h/3 • Rectangle: $\bar{d}$ from edge = h/2
Examples
For a torus: V = 2π²Rr² and SA = 4π²Rr. Note that in this specific problem (r = 2, R = 5), V and SA are numerically equal by coincidence. The torus is a common board exam question. The centroid of a circle (both arc and area) is its geometric center.
Scenario
A circle of radius 2 m has its center located 5 m from an axis of revolution. The circle is revolved about this axis to form a torus. Find (a) the volume and (b) the surface area of the torus.
Solution
Given: r = 2 m (circle radius), R = 5 m (distance from circle center = centroid to axis) (a) Volume — use Second Theorem: Area of circle A = πr² = π(2²) = 4π m² $\bar{d}$ = R = 5 m (centroid of circle = its center) V = 2π$\bar{d}$·A = 2π(5)(4π) = 40π² ≈ 394.78 m³ Direct formula: V_torus = 2π²Rr² = 2π²(5)(4) = 40π² ≈ 394.78 m³ ✓ (b) Surface Area — use First Theorem: Arc length of circle (circumference) L = 2πr = 2π(2) = 4π m $\bar{d}$ for arc = R = 5 m (centroid of the circle arc = center) S = 2π$\bar{d}$·L = 2π(5)(4π) = 40π² ≈ 394.78 m² Direct formula: SA_torus = 4π²Rr = 4π²(5)(2) = 40π² ≈ 394.78 m²
Pappus gives the same answer as the direct cone formula — confirming both. The centroid of a right triangle is at 1/3 of the base from the right angle vertex. When the triangle revolves about the vertical leg, it sweeps out a cone.
Scenario
A right triangle with legs 3 m and 4 m is revolved about its 4 m leg (the vertical leg). Find the volume of the resulting solid.
Solution
When revolved about the 4 m leg, the triangle generates a CONE with: • Height h = 4 m (the axis of revolution = the 4 m leg) • Radius = 3 m (the other leg, at maximum distance from axis) Using Pappus Second Theorem: A = (1/2)(3)(4) = 6 m² $\bar{d}$ = centroid distance from the 4 m leg = (1/3)(3) = 1 m (centroid of right triangle is 1/3 of base from any side) V = 2π$\bar{d}$·A = 2π(1)(6) = 12π ≈ 37.70 m³ Verification (cone formula): V = (1/3)πr²h = (1/3)π(3²)(4) = (1/3)π(9)(4) = 12π ≈ 37.70 m³ ✓
Applications
- Volume and surface area of toroidal structures (donut-shaped tanks, pipe bends)
- Generating surfaces of revolution for machine parts and structural members
- Computing volumes of solids of revolution without calculus integration
- Finding centroids of complex shapes using reverse Pappus theorem
Misconceptions
- Confusing First Theorem (arc length, centroid of arc) with Second Theorem (area, centroid of area) — they use DIFFERENT centroids
- Using the wrong centroid distance — for a semicircle arc, $\bar{d}$ = 2r/π; for semicircular AREA, $\bar{d}$ = 4r/(3π)
- Applying Pappus when the axis intersects the figure — the theorem requires an external axis
- For a triangle: using 2h/3 from base when it should be h/3 (centroid is at h/3 from base, 2h/3 from apex)
Related Concepts
- Centroids of plane figures
- Sphere (Pappus derivation)
- Torus geometry
- Calculus of solids of revolution (Pappus without integration)
Common Exam Questions
Example
Circle radius r = 3 m, axis distance R = 8 m: V = 2π²(8)(9) = 144π² ≈ 1,421.22 m³
Approach
Identify the circle radius r and the distance R from circle center to revolution axis. Apply V = 2π²Rr² or V = 2πR·(πr²).
Question Type
Volume of a torus
Example
A semicircle of radius r has area πr²/2; it generates a sphere of volume 4πr³/3 when revolved about diameter. Verify: $\bar{d}$ = V/(2πA) = (4πr³/3)/(2π·πr²/2) = (4πr³/3)/(π²r²) = 4r/(3π) ✓ (matches known centroid of semicircle)
Approach
If volume V and area A are known, find centroid: $\bar{d}$ = V/(2πA). If surface S and arc length L are known: $\bar{d}$ = S/(2πL).
Question Type
Find centroid using reverse Pappus
Key Points To Remember
- First Theorem uses the arc LENGTH and centroid of the ARC (curve), not area
- Second Theorem uses the AREA and centroid of the AREA, not arc
- The centroid distance is from the CENTROID to the AXIS OF REVOLUTION
- The axis must be EXTERNAL — if it intersects the figure, Pappus does not apply directly
- Torus: circle of radius r with centroid at distance R from axis; V = 2πR·(πr²) = 2π²Rr²
Practice Problems
The regular hexagon is unique: its circumradius R = s = side length. This means a regular hexagon can always be divided into 6 equilateral triangles with side = R = s. This fact is frequently tested.
Problem
Problem 1: A regular hexagon has an area of 93.53 m². Find its side length and the radius of its circumscribed circle.
Solution
Step 1 — Find side length from area: A = (3√3/2)s² 93.53 = (3√3/2)s² s² = 93.53 × 2/(3√3) = 187.06/(5.1962) = 36.00 s = 6.00 m Step 2 — Circumscribed circle radius: For a regular hexagon, R = s (the circumradius equals the side length) R = 6.00 m Verification: A = (3√3/2)(36) = 54√3 = 93.53 m² ✓
The concrete occupies the annular ring on the sides and the full bottom disc. The key is correctly identifying the inner radius (outer radius minus side thickness) and inner height (total depth minus bottom thickness). This type of problem appears in concrete quantity takeoff calculations.
Problem
Problem 2: A cylindrical cistern 2.5 m in diameter and 3.0 m deep is to be lined with concrete 15 cm thick on the sides and 20 cm thick on the bottom. Find the volume of concrete used.
Solution
Outer cylinder (total excavation): R = 2.5/2 = 1.25 m, H = 3.0 m V_outer = πR²H = π(1.25²)(3.0) = π(1.5625)(3) = 4.6875π m³ Inner cylinder (usable cistern space): r = 1.25 − 0.15 = 1.10 m (sides: 15 cm = 0.15 m thick) h = 3.0 − 0.20 = 2.80 m (bottom: 20 cm = 0.20 m thick) V_inner = πr²h = π(1.10²)(2.80) = π(1.21)(2.80) = 3.388π m³ Volume of concrete: V_concrete = V_outer − V_inner V_concrete = 4.6875π − 3.388π = 1.2995π ≈ 4.08 m³
The key insight is using similar triangles to find the height of the cut. The similarity ratio of radii = 1:2 means the similarity ratio of heights = 1:2 (all linear dimensions scale equally). Volume scales as the cube: V_small/V_large = (1/2)³ = 1/8. Check: V_small = 6π = (1/8)(48π) = 6π ✓
Problem
Problem 3: A conical pile of gravel has a base diameter of 12 m and a height of 4 m. If the gravel is scooped from the top to form a frustum with a top radius of 3 m and the same original base, find the volume of gravel removed.
Solution
Volume of original cone: r = 6 m, h = 4 m V_cone = (1/3)πr²h = (1/3)π(36)(4) = 48π m³ Volume of frustum (remaining pile): R = 6 m, r_top = 3 m, h_frustum = ? The smaller cone cut off from the top is similar to the original cone. Similarity ratio = r_top/R = 3/6 = 1/2 Therefore, the small cone height = (1/2) × original height = (1/2)(4) = 2 m Frustum height h = 4 − 2 = 2 m V_frustum = (πh/3)(R² + Rr + r²) = (π·2/3)(36 + 18 + 9) = (2π/3)(63) = 42π m³ Alternatively: V_small_cone = (1/3)π(3²)(2) = 6π m³ V_frustum = V_cone − V_small_cone = 48π − 6π = 42π m³ ✓ Volume removed = V_cone − V_frustum = 48π − 42π = 6π ≈ 18.85 m³
The curved surface area of a spherical cap 2πrh is elegantly simple — it depends only on the sphere radius r and cap height h, not on the cap base radius. This is the curved (domed) portion only. If the problem asks for total surface area of the cap as a solid, add the base circle area: Total SA = 24π + 20π = 44π ≈ 138.23 m².
Problem
Problem 4: A sphere of radius 6 m is cut by a plane 4 m from the center. Find the volume of the smaller cap and the curved surface area of the cap.
Solution
Given: sphere radius r = 6 m, distance from center to plane d = 4 m Step 1 — Cap height: h = r − d = 6 − 4 = 2 m Step 2 — Volume of spherical cap: V_cap = (πh²/3)(3r − h) V_cap = (π(2²)/3)(3(6) − 2) V_cap = (4π/3)(18 − 2) V_cap = (4π/3)(16) V_cap = 64π/3 ≈ 67.02 m³ Step 3 — Curved surface area of cap: A_cap = 2πrh = 2π(6)(2) = 24π ≈ 75.40 m² Note: The base circle of the cap has radius: a = √(r² − d²) = √(36 − 16) = √20 = 2√5 ≈ 4.47 m Base circle area = πa² = 20π ≈ 62.83 m²
This confirms that the prismatoid formula and the frustum formula are equivalent for a frustum — as expected, since a frustum is a special case of a prismatoid. The mid-section side = (8+4)/2 = 6 m by linear interpolation (valid because the frustum cross-section varies linearly with height). Note: √(A₁A₂) = 32 from the frustum formula matches the check: s₁·s₂ = 8×4 = 32 (geometric mean of sides for square frustum).
Problem
Problem 5: A frustum has a square base of 8 m × 8 m, a square top of 4 m × 4 m, and a height of 6 m. Using the prismatoid formula, verify the frustum volume formula result.
Solution
Method 1 — Frustum Formula: A₁ = 8² = 64 m², A₂ = 4² = 16 m², h = 6 m √(A₁A₂) = √(64 × 16) = √1024 = 32 m² V_frustum = (h/3)(A₁ + A₂ + √(A₁A₂)) V_frustum = (6/3)(64 + 16 + 32) = 2(112) = 224 m³ Method 2 — Prismatoid Formula: At mid-height (h/2 = 3 m from each base): By linear interpolation, the middle cross-section side: s_m = (8 + 4)/2 = 6 m (linear interpolation between 8 and 4) Aₘ = 6² = 36 m² V = (h/6)(A₁ + 4Aₘ + A₂) V = (6/6)(64 + 4(36) + 16) V = (1)(64 + 144 + 16) V = 224 m³ ✓ Both methods give 224 m³.
The centroid distance must be measured from the axis to the centroid of the area. If the flat diameter is 3 m from the axis and the curved part faces away, the centroid (at 4r/3π above the diameter) is on the far side, giving d_total = 3 + 4r/(3π). Draw a clear diagram showing the axis, diameter, and centroid position to avoid sign errors.
Problem
Problem 6: A semicircular area of radius 4 m is revolved about a line parallel to its diameter and 3 m away from it (measured from the diameter). Find the volume of the resulting solid.
Solution
Using Pappus' Second Theorem: Area of semicircle: A = (1/2)πr² = (1/2)π(4²) = 8π m² Centroid of semicircular area from its diameter: $\bar{y}$ = 4r/(3π) = 4(4)/(3π) = 16/(3π) m ≈ 1.698 m The axis of revolution is 3 m from the diameter (external to the semicircle since 3 m > 0 and on the other side). Total centroid distance from axis: $\bar{d}$ = 3 + 16/(3π) [if the semicircle is on the same side as the axis and the diameter is 3 m from axis] Assume the flat side (diameter) faces the axis, with the axis 3 m beyond the diameter: $\bar{d}$ = 3 + 4r/(3π) = 3 + 16/(3π) ≈ 3 + 1.698 = 4.698 m Volume: V = 2π$\bar{d}$·A = 2π(4.698)(8π) V = 2π × 4.698 × 8π V = 16π² × 4.698 ≈ 16(9.8696)(4.698) ≈ 741.68 m³ Exact form: V = 2π(3 + 16/(3π))(8π) = 16π²(3 + 16/(3π)) = 48π² + 256π/3 ≈ 473.74 + 267.94 = 741.68 m³
Step (a) uses s = rθ in reverse to find r — always ensure θ is in radians. Step (c) notes that sin(150°) = sin(30°) = 0.5 (reference angle). Since 150° > 120°, the triangle is obtuse and the segment area is larger than the triangle area. The segment 'cuts off' 152.50 m² from the 60π ≈ 188.50 m² sector.
Problem
Problem 7 (Board-Style): A circular sector has a central angle of 150° and an arc length of 10π m. Find: (a) the radius, (b) the area of the sector, and (c) the area of the minor segment cut off by the chord.
Solution
(a) Radius: θ = 150° × (π/180) = 5π/6 rad s = rθ → r = s/θ = 10π/(5π/6) = 10π × (6/(5π)) = 12 m (b) Sector area: A_sector = (1/2)r²θ = (1/2)(144)(5π/6) = (72)(5π/6) = 60π ≈ 188.50 m² Alternative: A_sector = (150/360)·πr² = (5/12)·π(144) = 60π ✓ (c) Area of segment: A_segment = A_sector − A_triangle A_triangle = (1/2)r²·sin θ = (1/2)(144)·sin(150°) A_triangle = (72)(0.5) = 36 m² A_segment = 60π − 36 ≈ 188.50 − 36 = 152.50 m² Note: This is a MAJOR segment (150° > 90°), so the segment is large.
This engineering-scale problem is realistic — the Philippine Arena (Bulacan) has a dome span of ~170 m. The hemisphere volume (2/3)πr³ is exactly half the sphere volume (4/3)πr³. The CSA (curved surface) = 2πr² is double the floor area πr². These ratios are useful for rapid mental checks.
Problem
Problem 8: The Great Dome of a sports complex is a hemisphere of inner radius 50 m. Determine: (a) the floor area covered, (b) the interior curved surface area, and (c) the volume of air enclosed.
Solution
(a) Floor area (circle): A_floor = πr² = π(50²) = 2,500π ≈ 7,853.98 m² (b) Interior curved surface area (hemisphere): CSA = 2πr² = 2π(50²) = 5,000π ≈ 15,707.96 m² (c) Volume of air (hemisphere): V = (2/3)πr³ = (2/3)π(50³) = (2/3)π(125,000) V = (250,000/3)π ≈ 261,799.39 m³ ≈ 261,799 m³ ≈ 261,800 kL of air
Exam Preparation Tips
- MEMORIZE THE CRITICAL FORMULAS COLD: V_sphere = (4/3)πr³, SA_sphere = 4πr², V_cone = (1/3)πr²h, V_frustum = (h/3)(A₁+A₂+√(A₁A₂)), V_prismatoid = (h/6)(A₁+4Aₘ+A₂). You cannot look these up in the exam.
- ALWAYS IDENTIFY r vs d (radius vs diameter): When a problem gives 'diameter', halve it immediately before writing any formula. Diameter errors account for a significant percentage of mensuration mistakes in board exams.
- DRAW CROSS-SECTION DIAGRAMS: For sphere cap, frustum, and spherical zone problems, always sketch the cross-section showing the relevant dimensions (r, h, d) before computing. This prevents substitution errors.
- CHECK UNITS CONSISTENTLY: Volume answers are in m³ (or cm³), area answers in m² (or cm²). If the problem gives mixed units (e.g., diameter in cm, height in m), convert all to the same unit before computing.
- KNOW THE 1/3 PATTERN: Pyramid volume = (1/3) × prism volume; cone volume = (1/3) × cylinder volume. These comparisons are often the basis of 'ratio' type board questions.
- SECTOR ANGLES IN RADIANS: Any formula with θ (sector area, arc length, segment area) requires θ in RADIANS. The formula A = (1/2)r²θ only works in radians. If degrees are given, convert first: θ_rad = θ_deg × π/180.
- FRUSTUM FORMULA CHECK: The middle term is √(A₁A₂), not (A₁+A₂)/2 or A₁A₂. Practice computing √(A₁A₂) quickly: for circular frustums, √(πR²·πr²) = πRr. For square frustums, √(a²·b²) = ab.
- PAPPUS CENTROID LOCATIONS: Memorize the key centroids — triangle: h/3 from base; semicircular area: 4r/(3π) from diameter; semicircular arc: 2r/π from diameter; quarter-circle: 4r/(3π) from each straight side.
- USE ANSWER CHOICES STRATEGICALLY: In multiple-choice exams, if your answer contains π (like 28π), evaluate it numerically only if needed. Keep exact forms as long as possible to avoid rounding errors.
- VERIFY WITH LIMITING CASES: When checking prismatoid or frustum formulas, set A₂ = 0 (should give pyramid/cone) or A₁ = A₂ (should give prism/cylinder). This is a rapid self-check.
- PRACTICE UNIT CONVERSIONS: 1 m³ = 1,000 liters = 1 kL. For concrete quantities: 1 m³ of concrete requires approximately 7-8 bags of cement (40 kg each). Knowing these conversions helps verify the reasonableness of answers.
- TIME MANAGEMENT IN BOARDS: Mensuration problems that require only formula substitution should take 2-3 minutes. Complex multi-step problems (like frustum + Pappus) may take 5-7 minutes. Skip and return if stuck — do not spend more than 8 minutes on any single problem.
- RECOGNIZE TRICK QUESTIONS: Watch for problems where slant height is given but vertical height is needed (or vice versa). For cones: if L and r are given, compute h = √(L²−r²) before finding volume.
- LEARN COMMON PYTHAGOREAN TRIPLES: 3-4-5, 5-12-13, 8-15-17, 7-24-25. These frequently appear in cone/pyramid slant height calculations and save computation time.
- PRISMATOID vs FRUSTUM FORMULA: Both give the same answer for frustums. However, the prismatoid formula (h/6)(A₁+4Aₘ+A₂) requires computing Aₘ at mid-height — use this when Aₘ is given directly in the problem. The frustum formula (h/3)(A₁+A₂+√(A₁A₂)) is faster when only the two end areas are given.
In summary
Plane, Solid Geometry and Mensuration is a high-yield chapter in the PRC Civil Engineer Licensure Examination that rewards systematic preparation. The key to mastery is not passive memorization of formulas but active pattern recognition — knowing WHEN to apply each formula and recognizing the subtle distinctions that separate correct from incorrect solutions. The core hierarchy to internalize: (1) Plane figure areas form the foundation — all solid volume formulas reference a base area. (2) Prisms and cylinders have volume = A_base × h. (3) Pyramids and cones have volume = (1/3) × A_base × h — the 1/3 factor is always present. (4) The frustum formula V = (h/3)(A₁ + A₂ + √(A₁A₂)) generalizes the pyramid formula and includes a geometric mean term. (5) The prismatoid formula V = (h/6)(A₁ + 4Aₘ + A₂) is the most general and serves as the basis for earthwork volume computation in practice. The Pappus Theorems provide elegant, calculus-free methods for surfaces and volumes of revolution — recognize them as high-value problems in the exam because few examinees master them, giving those who do a distinct advantage. For the board exam, build your problem-solving discipline around three practices: (1) Always identify what is given and what is asked before writing a single formula. (2) Draw a cross-section sketch for any sphere, cone, or frustum problem. (3) Verify your answer using a limiting case, a ratio check (cone = 1/3 of cylinder), or a units check. With consistent practice of the worked examples in this chapter and deliberate review of the common pitfalls, you will approach the mensuration section of the board exam with confidence and accuracy. Magsipag at maniwala sa sarili — you can ace this topic!
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