Mensuration: Areas and Volumes

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Mensuration is the branch of geometry that quantifies the size of plane figures (2-D) and solids (3-D). Three measurements dominate: perimeter (sum of side lengths around a shape), area (space enclosed, in square units), and volume (space occupied by a solid, in cubic units).

Must-know formulas:

Triangle area: A = ½ × base × height
Circle: A = πr², circumference C = 2πr
Trapezium: A = ½(a + b)h
Cylinder: V = πr²h, total surface area (TSA) = 2πr(r + h)
Cone: V = ⅓πr²h, curved surface area (CSA) = πrl where l = √(r² + h²)
Sphere: V = 4/3 πr³, surface area = 4πr²

Take π = 22/7 when r is a multiple of 7, otherwise π = 3.142. Always confirm whether the question gives diameter or radius — halve the diameter first.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Plane Figures (2-D)

Triangle — Area is half the base times the perpendicular height. For a right triangle the two legs act as base and height. Heron’s formula A = √[s(s−a)(s−b)(s−c)], where s = (a+b+c)/2, handles sides-only triangles and appears in NECO Paper II as a 5-mark item.

Trapezium — A quadrilateral with one pair of parallel sides a and b, separated by perpendicular height h. Area A = ½(a + b)h.

Circle, sector and segment — A full circle gives A = πr². A sector (pie-slice) of angle θ takes A = (θ/360) × πr² in degrees, or ½r²θ in radians. A segment is a sector minus the triangle formed by the two radii; its area is A_segment = A_sector − A_triangle.

Solids (3-D)

Cylinder — Two parallel circles joined by a curved surface. Volume V = πr²h; the curved surface alone (no ends) is CSA = 2πrh; the closed can including both lids is TSA = 2πr(r + h).

Cone — A circular base tapering to an apex. Volume V = ⅓πr²h (one-third of the enclosing cylinder). The slant height l = √(r² + h²) feeds CSA = πrl and TSA = πr(r + l).

Sphere — Every point on the surface is equidistant from the centre. Surface area = 4πr², volume = 4/3 πr³.

Composite Shapes

Split the figure into recognisable parts (e.g. a cylinder topped by a hemisphere), calculate each part, then add or subtract as the geometry dictates. Track units carefully: 1 m² = 10 000 cm²; 1 m³ = 1 000 000 cm³.

Typical NECO Question Patterns

Direct substitution (e.g. “find the volume of a cylinder of radius 7 cm and height 10 cm” → 1540 cm³).
Converting the answer from cm³ to litres (1 litre = 1000 cm³).
Multi-step word problems on tanks, frustums of cones, or metallic casts.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Frustum of a Cone

A frustum is what remains when the top of a cone is sliced off parallel to the base. With lower radius R, upper radius r, and perpendicular height h, the volume is V = ⅓πh(R² + Rr + r²) and the slant height ℓ = √[(R − r)² + h²]. CSA = π(R + r)ℓ. NECO frequently frames this as a bucket or a water-storage tank.

Hemisphere

Half a sphere: curved surface area = 2πr², total surface area (curved + flat circle) = 3πr², volume = ⅔πr³. Solid shapes combining a cylinder with a hemispherical end use these directly.

Edge Cases and Pitfalls

Slant vs vertical height. A cone question that quotes “height 12 cm and slant height 13 cm” expects you to verify l² = r² + h² and back-solve r before any area step.
Diameter trap. “A circular tank of diameter 14 m” means r = 7 m; substituting d = 14 into πr² inflates the answer by a factor of 4.
Unit drift. A volume in cm³ must be divided by 1000 to become litres; surface area in cm² never becomes litres.
Composite subtraction. When a cylindrical hole is drilled through a solid, the removed volume is subtracted, not added.

Connection to Other Topics

Mensuration links directly to similar shapes and scale factors (areas scale as k², volumes as k³), trigonometry of angles in a circle (sector angles), and algebraic word problems that set up equations in r or h.

Worked Micro-Example

A cone has slant height 13 cm and radius 5 cm. Height h = √(13² − 5²) = √144 = 12 cm.

Volume = ⅓π(5²)(12) = 100π ≈ 314.16 cm³.
CSA = π(5)(13) = 65π ≈ 204.20 cm².
TSA = π(5)(5 + 13) = 90π ≈ 282.74 cm².

Practice Prompts

A frustum has R = 10 cm, r = 4 cm, h = 12 cm. Find its volume in cm³ and capacity in litres.
A solid is a cylinder of radius 7 cm and height 20 cm with a hemisphere of the same radius removed from one end. Calculate the remaining volume, correct to two significant figures.

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Mensuration: Areas and Volumes