CELE Engineering Mathematics — Plane, Solid Geometry and MensurationRevision Notes

Formulas

Exam Tips

• Memorise cot(30°) = √3, cot(45°) = 1, cot(60°) = 1/√3 for quick regular polygon area calculations (n = 6, 4, 3).
• For a right triangle with legs a and b, area = ½ab — no need to find height separately.
• In multiple-choice items, verify your answer by dimensional analysis: area must be in m², volume in m³.
• Heron's formula is slower; use the sine formula A = ½ab sin C whenever two sides and included angle are given.

Key Points

• Triangle area can be computed by three methods: (1) base-height A = ½bh, (2) Heron's formula A = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2, and (3) two-sides-and-included-angle A = ½ab sin C.
• The trapezoid formula A = ½(b₁ + b₂)h applies to any quadrilateral with exactly one pair of parallel sides.
• For a regular polygon with n sides of length s, the area is A = (1/4)ns² cot(180°/n), derived from dividing the polygon into n isosceles triangles.
• Circle area is A = πr²; circumference C = 2πr = πd.
• A circular sector with central angle θ (in radians) has area A_sector = ½r²θ and arc length L = rθ.
• A circular segment area = A_sector − A_triangle = ½r²(θ − sin θ), with θ in radians.
• An annulus (ring) has area A = π(R² − r²) where R is the outer and r is the inner radius.
• An ellipse has area A = πab where a and b are the semi-major and semi-minor axes.

Definitions

Section Title

Plane Figures — Areas and Perimeters

Common Mistakes

• Using diameter instead of radius in area formulas — always halve the given diameter before substituting into A = πr².
• Forgetting to convert the central angle to radians before using A = ½r²θ or L = rθ. Multiply degrees by π/180.
• Confusing sector area with segment area — the segment requires subtracting the triangle (½r² sin θ) from the sector.
• Applying A = ½bh to non-right triangles using the hypotenuse as 'b' without finding the true perpendicular height.

Formulas

Exam Tips

• Cylinder vs Cone memory aid: 'The cone is a pyramid — 1/3 of the cylinder.' A full cylinder becomes a cone when you take 1/3.
• When a board problem gives a cone inscribed in a cylinder, their volumes are in ratio 1:3.
• Always compute slant height L first from a right triangle (r, h, L) before computing lateral area of a cone or pyramid.
• For a box problem, if they ask for the amount of material (sheet metal), they want total surface area.

Key Points

• Any prism or cylinder has volume V = A_base × h, where h is the perpendicular height between the two parallel bases.
• Lateral surface area of a right prism = Perimeter_base × h; for a right cylinder = 2πrh.
• Total surface area of a right cylinder = 2πrh + 2πr² (lateral + two circular ends).
• Any pyramid or cone has volume V = (1/3) A_base × h — exactly one-third of the corresponding prism or cylinder.
• Lateral surface area of a right circular cone = πrL, where L = √(r² + h²) is the slant height.
• Total surface area of a right circular cone = πrL + πr² (lateral + base).
• A rectangular parallelepiped (box): V = lwh, total surface area = 2(lw + lh + wh).
• An oblique prism still has V = A_base × h_perpendicular, not the slant height.

Definitions

Section Title

Solid Geometry — Prisms, Cylinders, Pyramids, Cones

Common Mistakes

• Using the vertical height h instead of slant height L in the cone lateral area formula πrL — the most common error.
• Forgetting to add the base areas when total surface area is required (not just lateral).
• Applying V = (1/3)A_base × h to a prism — pyramids and cones are 1/3, prisms and cylinders are full height.
• For an oblique prism, using the slant side length as h — always use the perpendicular distance between bases.

Formulas

Exam Tips

• Memory device for frustum: 'A1 plus A2 plus geometric mean, divided by 3, times h.' Write it out once — never approximate √(A₁A₂) as (A₁+A₂)/2.
• For a cone frustum with base radii r₁ and r₂: √(A₁A₂) = √(πr₁²)(πr₂²) = πr₁r₂. So V = (πh/3)(r₁² + r₂² + r₁r₂).
• Quick check on prismatoid formula: apply it to a 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 ✓
• Sphere volume and surface area are numerically equal when r = 3 (both equal 36π ≈ 113.10) — useful as a self-check.

Key Points

• Sphere: V = (4/3)πr³ and total surface area S = 4πr².
• A hemisphere has V = (2/3)πr³ and curved surface area = 2πr² (flat base adds another πr²).
• A spherical cap (zone) of height h cut from a sphere of radius R: V = (πh²/3)(3R − h), curved area = 2πRh.
• A frustum of a cone or pyramid (a solid with two parallel bases A₁ and A₂ separated by height h): V = (h/3)(A₁ + A₂ + √(A₁A₂)).
• The prismatoid formula (also called the Prismoidal formula or Simpson's Rule for volumes): V = (h/6)(A₁ + 4Aₘ + A₂), where Aₘ is the area of the midsection at h/2. This is the basis of earthwork volume computations.
• The prismatoid formula is exact (not an approximation) for prisms, pyramids, cones, wedges, and frustums.
• For earthwork computations using the average end-area method: V = (L/2)(A₁ + A₂) — gives a slight overestimate compared to the prismatoid formula.
• A torus (donut shape) formed by revolving a circle of radius r about an axis d from its centre has V = 2π²r²d and S = 4π²rd.

Definitions

Section Title

Sphere, Frustum, and Prismatoid

Common Mistakes

• Writing the frustum formula as V = (h/3)(A₁ + A₂ + A₁A₂) — the geometric mean term is √(A₁A₂), not the product A₁A₂.
• Confusing the sphere surface area formula (4πr²) with the circle area formula (πr²) — the sphere has a coefficient of 4.
• Computing Aₘ at the wrong elevation — it must be at exactly mid-height h/2, not at the average dimension.
• Using the average end-area formula for earthworks when the prismatoid formula is specifically required — they give different answers and both appear on boards.

Formulas

Exam Tips

• Pappus mnemonic: 'The solid knows how far its centroid travels — multiply that travel distance (2π d̄) by what it carries (A or L).'
• Board exam torus problem shortcut: V_torus = 2π²r²d. Memorise this directly rather than deriving from Pappus each time.
• Verification: Apply Pappus Second Theorem to revolve a semicircle of radius r about its diameter to get a sphere — d̄ = 4r/(3π), A = πr²/2; V = 2π(4r/3π)(πr²/2) = (4/3)πr³ ✓
• If a board problem asks for the volume of a solid of revolution of any standard shape, identify d̄ first using centroid tables, then apply V = 2π d̄ A — faster than triple integration.

Key Points

• Pappus' First Theorem (Surface of Revolution): The surface area S generated by revolving a plane curve of arc length L about an external axis equals 2π times the distance from the centroid of the arc to the axis, multiplied by L. S = 2π d̄ L.
• Pappus' Second Theorem (Volume of Revolution): The volume V generated by revolving a plane area A about an external axis equals 2π times the distance from the centroid of the area to the axis, multiplied by A. V = 2π d̄ A.
• The axis of revolution must not intersect the figure (it must be external) for both theorems to apply.
• d̄ in both theorems is the distance from the centroid (of the arc or area) to the axis of revolution.
• A torus generated by revolving a circle of radius r with centroid at distance d from the axis: V = 2πd(πr²) = 2π²r²d and S = 2πd(2πr) = 4π²rd.
• The centroid of a semicircle arc is at 2r/π from the diameter; the centroid of a semicircular area is at 4r/(3π) from the diameter.

Definitions

Section Title

Pappus Theorems — Solids of Revolution

Common Mistakes

• Using the centroid of the area when arc length is involved (First Theorem) — the two centroids are different: 2r/π for arc, 4r/(3π) for area.
• Applying Pappus theorems when the axis intersects the figure — both theorems require an external axis.
• Forgetting the 2π factor — the formula is V = 2π d̄ A, not V = π d̄ A.
• Confusing arc length L with the chord length in First Theorem — L = rθ (arc), not the straight chord.

Formulas

Exam Tips

• Write the formula before substituting numbers — this prevents substitution errors and gives partial credit in recall.
• For a cylindrical tank problem, always check whether the tank is open-top or closed — this affects surface area but not volume.
• Percentage increase/decrease in volume when a dimension changes: for a sphere, V ∝ r³, so a 10% increase in r gives V increases by (1.10)³ − 1 = 33.1%.
• For the frustum of a cone with known radii r₁ and r₂ and slant height L (not height h): h = √(L² − (r₁−r₂)²) — find h first, then apply the frustum formula.

Key Points

• Problem Type 1: Direct formula application — given dimensions, find area or volume. Strategy: identify the shape, write the formula, substitute, compute.
• Problem Type 2: Reverse problems — given volume or area, find a missing dimension. Strategy: write the formula, isolate the unknown, solve.
• Problem Type 3: Composite solids — a solid composed of two or more basic shapes. Strategy: decompose, compute each part separately, add or subtract.
• Problem Type 4: Rate problems — volume filled at a given flow rate, find time. Strategy: find total volume first, then t = V/Q.
• Problem Type 5: Solids of revolution — identify the generating figure, find its centroid, apply Pappus.
• Problem Type 6: Inscribed/circumscribed figures — use geometric relationships (e.g., sphere inscribed in cone: r_sphere = r_cone × h / (√(r²+h²) + r)).
• Always verify units — areas in m², volumes in m³, lengths in m. The PRC board exam is entirely SI.
• Draw a clear sketch for every problem — label all given dimensions before writing any equation.

Definitions

Section Title

Board-Exam Problem-Solving Strategy and Common Question Types

Common Mistakes

• For composite solids, forgetting to subtract the hollow or void portion (e.g., computing total cylinder volume without removing the hollow core).
• Mixing up the inscribed and circumscribed conditions — a sphere inscribed in a cube has diameter equal to the cube's edge, not the diagonal.
• In rate problems, not converting units consistently (e.g., mixing litres and cubic metres — 1 m³ = 1000 L).
• Using diameter where radius is needed after reading a problem that states 'diameter = 6 m' — write r = 3 m immediately.