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3.2 Solid Mensuration, Measurement, and Unit Conversion

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1 Introduction & Exam Overview

1.1 AdUCET Overview, Format, and Admission Process 1.2 Test-Taking and Time-Management Strategies

2 Chapter 2: Mathematics I — Numbers, Algebra, and Functions

2.1 Number Sense, Operations, and Fraction–Decimal–Percent Conversions 2.2 Algebra, Polynomials, and Linear/Quadratic Equations 2.3 Inequalities, Absolute Value, and Systems of Equations 2.4 Radicals, Rational Expressions, and Scientific Notation 2.5 Coordinate Geometry, Lines, and Functions

3 Chapter 3: Mathematics II — Geometry, Measurement, and Trigonometry

3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles 3.2 Solid Mensuration, Measurement, and Unit Conversion 3.3 Right-Triangle Trigonometry and Applications

4 Chapter 4: Mathematics III — Sequences, Proportion, and Data

4.1 Arithmetic and Geometric Sequences and Series 4.2 Ratio, Proportion, Percentage, Work, and Mixture Problems 4.3 Descriptive Statistics and Data Interpretation 4.4 Probability: Rules, Compound Events, and Conditional Probability 4.5 Counting Principles: Permutations and Combinations

5 Chapter 5: English I — Reading Comprehension and Vocabulary

5.1 Reading Comprehension: Main Idea, Purpose, and Passage Structure 5.2 Inference, Supporting Detail, and Author's Tone 5.3 Word Roots, Prefixes, and Suffixes 5.4 Vocabulary in Context, Synonyms, Antonyms, and Confusables

6 Chapter 6: English II — Grammar, Usage, and Written Expression

6.1 Subject-Verb Agreement and Verb Tenses 6.2 Pronouns, Modifiers, and Parallel Structure 6.3 Sentence Boundaries, Clauses, and Punctuation 6.4 Sentence Correction, Redundancy, and Idiomatic Usage 6.5 Verbal Analogies and Relationship Bridges 6.6 Paragraph Organization, Coherence, and Transitions

7 Chapter 7: General Science I — Biology and Chemistry

7.1 Cell Biology, Transport, and Cellular Energetics 7.2 Genetics, Heredity, and Molecular Biology 7.3 Human Body Systems, Ecology, and Evolution 7.4 Matter, Phase Changes, and Atomic Structure 7.5 Periodic Trends, Chemical Bonding, Reactions, Acids and Bases

8 Chapter 8: General Science II — Physics and Earth Science

8.1 Mechanics: Motion, Newton's Laws, Work, and Energy 8.2 Waves, Sound, Light, Heat, Electricity, and Magnetism 8.3 Earth's Structure, Plate Tectonics, and the Rock Cycle 8.4 The Atmosphere, Weather, the Water Cycle, and Astronomy

9 Chapter 9: Aptitude and Abstract Reasoning

9.1 Figure Series and Pattern Progression 9.2 Matrix Reasoning and Figural Analogies 9.3 Spatial Reasoning: Rotation, Reflection, and Chirality 9.4 Paper Folding, Cube Nets, and Orthographic Views 9.5 Deductive Reasoning, Syllogisms, and Venn Diagrams 9.6 Number, Letter, and Alphanumeric Series 9.7 Odd-One-Out, Classification, and Logic Puzzles

3.2 Solid Mensuration, Measurement, and Unit Conversion

Key Takeaways Explain This Topic AI

Cone lateral surface area is πrℓ where the slant height ℓ = √(r² + h²); volume formulas use the vertical height h, never the slant height.

Multiplying every linear dimension by k multiplies surface areas by k² and volumes by k³, so doubling all dimensions gives eight times the capacity.

1 m² = 10,000 cm² and 1 m³ = 1,000,000 cm³ because the linear factor of 100 is squared or cubed with the unit.

1 L = 1,000 cm³ = 1 dm³, so 1 m³ holds exactly 1,000 litres — a cylindrical tank 2.8 m across and 3 m tall holds about 18,480 L using π ≈ 22/7.

A hemisphere has volume (2/3)πr³, curved area 2πr² and total surface area 3πr² once the flat circular face is counted.

Last updated: August 2026

3.2 Solid Mensuration, Measurement, and Unit Conversion

Mensuration items are pure formula recognition plus disciplined bookkeeping of units. Nothing here needs a calculator if you keep π \pi π symbolic or use the friendly approximation π ≈ 22 7 \pi \approx \frac{22}{7} π ≈ 7 22 ​ when the radius is a multiple of 7 7 7 . This section is deliberately about solids and units ; the perimeter and area of flat figures belong to the plane-geometry section.

1. The core solids

  • Solid Volume V V V Lateral / curved area Total surface area
  • Cube (edge s s s ) s 3 s^3 s 3 4 s 2 4s^2 4 s 2 6 s 2 6s^2 6 s 2

Rectangular prism l w h lwh lw h 2 h ( l + w ) 2h(l+w) 2 h ( l + w ) 2 ( l w + l h + w h ) 2(lw+lh+wh) 2 ( lw + l h + w h )

Any right prism B h B h B h P h P h P h 2 B + P h 2B + Ph 2 B + P h

Cylinder π r 2 h \pi r^2 h π r 2 h 2 π r h 2\pi r h 2 π r h 2 π r 2 + 2 π r h 2\pi r^2 + 2\pi r h 2 π r 2 + 2 π r h

Cone 1 3 π r 2 h \frac{1}{3}\pi r^2 h 3 1 ​ π r 2 h π r ℓ \pi r \ell π r ℓ π r 2 + π r ℓ \pi r^2 + \pi r \ell π r 2 + π r ℓ

Sphere 4 3 π r 3 \frac{4}{3}\pi r^3 3 4 ​ π r 3 4 π r 2 4\pi r^2 4 π r 2 4 π r 2 4\pi r^2 4 π r 2

Hemisphere 2 3 π r 3 \frac{2}{3}\pi r^3 3 2 ​ π r 3 2 π r 2 2\pi r^2 2 π r 2 3 π r 2 3\pi r^2 3 π r 2 (curved + flat face)

Pyramid (base area B B B ) 1 3 B h \frac{1}{3}B h 3 1 ​ B h 1 2 P ℓ \frac{1}{2}P\ell 2 1 ​ P ℓ B + 1 2 P ℓ B + \frac{1}{2}P\ell B + 2 1 ​ P ℓ

Square pyramid (edge s s s ) 1 3 s 2 h \frac{1}{3}s^2 h 3 1 ​ s 2 h 2 s ℓ 2s\ell 2 s ℓ s 2 + 2 s ℓ s^2 + 2s\ell s 2 + 2 s ℓ

Here B B B is the base area, P P P the base perimeter, h h h the vertical height and ℓ \ell ℓ the slant height .

Slant height, derived not given

The slant height is the hypotenuse of a right triangle standing inside the solid:

Cone : the legs are the radius and the vertical height, so ℓ = r 2 + h 2 \ell = \sqrt{r^2 + h^2} ℓ = r 2 + h 2 ​ . A cone with r = 6 r = 6 r = 6 and h = 8 h = 8 h = 8 has ℓ = 36 + 64 = 10 \ell = \sqrt{36+64} = 10 ℓ = 36 + 64 ​ = 10 .

Square pyramid : the legs are the vertical height and half the base edge, so ℓ = h 2 + ( s 2 ) 2 \ell = \sqrt{h^2 + \left(\frac{s}{2}\right)^2} ℓ = h 2 + ( 2 s ​ ) 2 ​ . A pyramid with s = 10 s = 10 s = 10 and h = 12 h = 12 h = 12 has ℓ = 144 + 25 = 13 \ell = \sqrt{144 + 25} = 13 ℓ = 144 + 25 ​ = 13 .

Curved-surface formulas ( π r ℓ \pi r\ell π r ℓ , 1 2 P ℓ \frac{1}{2}P\ell 2 1 ​ P ℓ ) always want ℓ \ell ℓ ; volume formulas always want h h h . Swapping them is the single most common mensuration error.

  • Space diagonals +-----------+ D = sqrt(l^2 + w^2 + h^2)
  • /| /|
  • / | / | Cube: face diagonal = s*sqrt(2)
  • +-----------+ | space diagonal = s*sqrt(3)
  • | +--------|--+
  • | / D | /
  • |/ |/
  • +-----------+

A crate measuring 3 × 4 × 12 3 \times 4 \times 12 3 × 4 × 12 m accepts a rod of length 9 + 16 + 144 = 169 = 13 \sqrt{9+16+144} = \sqrt{169} = 13 9 + 16 + 144 ​ = 169 ​ = 13 m — the longest straight object that fits.

Composite solids

Split, compute, then add or subtract. A grain silo formed by a cylinder of radius 3 3 3 m and height 10 10 10 m capped by a hemisphere of the same radius holds V = π ( 3 ) 2 ( 10 ) + 2 3 π ( 3 ) 3 = 90 π + 18 π = 108 π ≈ 339.3 m 3 . V = \pi(3)^2(10) + \tfrac{2}{3}\pi(3)^3 = 90\pi + 18\pi = 108\pi \approx 339.3\ \text{m}^3. V = π ( 3 ) 2 ( 10 ) + 3 2 ​ π ( 3 ) 3 = 90 π + 18 π = 108 π ≈ 339.3 m 3 .

Note the hemisphere's flat face disappears inside the joint, so it is not counted in surface area.

2. The scaling law — the classic trap

Multiply every linear dimension of a figure by k k k and lengths scale by k k k , areas by k 2 k^2 k 2 , volumes by k 3 k^3 k 3 .

  • k k k Length Surface area Volume
  • 2 2 2 × 2 \times 2 × 2 × 4 \times 4 × 4 × 8 \times 8 × 8
  • 3 3 3 × 3 \times 3 × 3 × 9 \times 9 × 9 × 27 \times 27 × 27
  • 1 2 \tfrac{1}{2} 2 1 ​ × 0.5 \times 0.5 × 0.5 × 0.25 \times 0.25 × 0.25 × 0.125 \times 0.125 × 0.125
  • 1.2 1.2 1.2 × 1.2 \times 1.2 × 1.2 × 1.44 \times 1.44 × 1.44 × 1.728 \times 1.728 × 1.728

"Doubling the tank doubles the water" is false: doubling all dimensions gives eight times the capacity. Read the wording precisely — if only the radius of a cylinder doubles while the height is unchanged, volume rises by 2 2 = 4 2^2 = 4 2 2 = 4 , because r r r appears squared and h h h does not change at all.

3. Metric prefixes and conversion

  • Prefix Symbol Factor
  • mega M 10 6 10^{6} 1 0 6
  • kilo k 10 3 10^{3} 1 0 3
  • hecto h 10 2 10^{2} 1 0 2
  • deka da 10 1 10^{1} 1 0 1
  • (base unit) — 10 0 10^{0} 1 0 0
  • deci d 10 − 1 10^{-1} 1 0 − 1
  • centi c 10 − 2 10^{-2} 1 0 − 2
  • milli m 10 − 3 10^{-3} 1 0 − 3
  • micro μ \mu μ 10 − 6 10^{-6} 1 0 − 6

Moving down the ladder multiplies by 10 10 10 each step; moving up divides by 10 10 10 . The order kilo–hecto–deka–base–deci–centi–milli is worth reciting once before the exam.

Common metric–English equivalents (memorize the exact ones):

1 1 1 inch = 2.54 = 2.54 = 2.54 cm exactly; 1 1 1 foot = 30.48 = 30.48 = 30.48 cm; 1 1 1 mile ≈ 1.609 \approx 1.609 ≈ 1.609 km

  • 1 1 1 kg ≈ 2.205 \approx 2.205 ≈ 2.205 lb; 1 1 1 lb ≈ 453.6 \approx 453.6 ≈ 453.6 g
  • 1 1 1 US gallon ≈ 3.785 \approx 3.785 ≈ 3.785 L; 1 1 1 L ≈ 0.264 \approx 0.264 ≈ 0.264 US gallon
  • 1 1 1 hectare = 10,000 = 10{,}000 = 10 , 000 m 2 ^2 2 ; 1 1 1 km 2 = 100 ^2 = 100 2 = 100 hectares
  • Why 1 1 1 m 2 = 10,000 ^2 = 10{,}000 2 = 10 , 000 cm 2 ^2 2

Because the conversion factor is raised to the same power as the unit. 1 m = 100 cm 1\ \text{m} = 100\ \text{cm} 1 m = 100 cm , so 1 m 2 = ( 100 cm ) 2 = 10,000 cm 2 , 1 m 3 = ( 100 cm ) 3 = 1,000,000 cm 3 . 1\ \text{m}^2 = (100\ \text{cm})^2 = 10{,}000\ \text{cm}^2,\qquad 1\ \text{m}^3 = (100\ \text{cm})^3 = 1{,}000{,}000\ \text{cm}^3. 1 m 2 = ( 100 cm ) 2 = 10 , 000 cm 2 , 1 m 3 = ( 100 cm ) 3 = 1 , 000 , 000 cm 3 .

Also fix in memory: 1 cm 3 = 1 mL 1\ \text{cm}^3 = 1\ \text{mL} 1 cm 3 = 1 mL , 1 L = 1,000 cm 3 = 1 dm 3 1\ \text{L} = 1{,}000\ \text{cm}^3 = 1\ \text{dm}^3 1 L = 1 , 000 cm 3 = 1 dm 3 , and therefore 1 m 3 = 1,000 1\ \text{m}^3 = 1{,}000 1 m 3 = 1 , 000 litres.

4. Density and rates

Density is mass per unit volume, ρ = m V \rho = \dfrac{m}{V} ρ = V m ​ , rearranged as m = ρ V m = \rho V m = ρ V and V = m ρ V = \dfrac{m}{\rho} V = ρ m ​ . Fresh water is 1 1 1 g/cm 3 ^3 3 , which is the same as 1,000 1{,}000 1 , 000 kg/m 3 ^3 3 — the factor between those two units is exactly 1,000 1{,}000 1 , 000 .

Rates convert by chaining fractions so unwanted units cancel: 72 km h × 1,000 m 1 km × 1 h 3,600 s = 20 m s . 72\ \frac{\text{km}}{\text{h}} \times \frac{1{,}000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3{,}600\ \text{s}} = 20\ \frac{\text{m}}{\text{s}}. 72 h km ​ × 1 km 1 , 000 m ​ × 3 , 600 s 1 h ​ = 20 s m ​ .

The shortcut worth memorizing: divide km/h by 3.6 3.6 3.6 to get m/s, multiply by 3.6 3.6 3.6 to go back. Peso rates behave the same way — diesel at ₱62.50 per litre is 62.50 × 3.785 ≈ 62.50 \times 3.785 \approx 62.50 × 3.785 ≈ ₱236.56 per US gallon.

Round only at the very end, and round up whenever the answer counts whole purchased items (bags of cement, cans of paint, jeepney trips).

5. Worked example 1 — barangay water tank

Problem. A cylindrical tank has diameter 2.8 2.8 2.8 m and height 3 3 3 m. Using π ≈ 22 7 \pi \approx \frac{22}{7} π ≈ 7 22 ​ : (a) find its volume in cubic metres, (b) convert to litres, (c) if 20 20 20 households each draw 220 220 220 L per day, how long does a full tank last, and (d) how many 1 1 1 -litre cans of paint cover the curved outside wall if one can covers 8 8 8 m 2 ^2 2 ?

Radius r = 2.8 2 = 1.4 r = \frac{2.8}{2} = 1.4 r = 2 2.8 ​ = 1.4 m, so r 2 = 1.96 r^2 = 1.96 r 2 = 1.96 m 2 ^2 2 .

V = π r 2 h = 22 7 ( 1.96 ) ( 3 ) V = \pi r^2 h = \frac{22}{7}(1.96)(3) V = π r 2 h = 7 22 ​ ( 1.96 ) ( 3 ) . Since 1.96 7 = 0.28 \frac{1.96}{7} = 0.28 7 1.96 ​ = 0.28 , this is 22 × 0.28 × 3 = 18.48 22 \times 0.28 \times 3 = 18.48 22 × 0.28 × 3 = 18.48 m 3 ^3 3 .

18.48 m 3 × 1,000 = 18,480 18.48\ \text{m}^3 \times 1{,}000 = 18{,}480 18.48 m 3 × 1 , 000 = 18 , 480 litres.

Daily draw = 20 × 220 = 4,400 = 20 \times 220 = 4{,}400 = 20 × 220 = 4 , 400 L, so the tank lasts 18,480 4,400 = 4.2 \frac{18{,}480}{4{,}400} = 4.2 4 , 400 18 , 480 ​ = 4.2 days.

Curved wall = 2 π r h = 2 × 22 7 × 1.4 × 3 = 2 × 4.4 × 3 = 26.4 = 2\pi r h = 2 \times \frac{22}{7} \times 1.4 \times 3 = 2 \times 4.4 \times 3 = 26.4 = 2 π r h = 2 × 7 22 ​ × 1.4 × 3 = 2 × 4.4 × 3 = 26.4 m 2 ^2 2 , needing 26.4 8 = 3.3 \frac{26.4}{8} = 3.3 8 26.4 ​ = 3.3 cans — buy 4 , because paint is sold whole.

6. Worked example 2 — concrete footing and the scaling trap

Problem. A rectangular concrete slab measures 4 4 4 m by 3 3 3 m by 150 150 150 mm thick. (a) Find its volume in m 3 ^3 3 and cm 3 ^3 3 . (b) At an assumed 9 9 9 bags of cement per cubic metre, how many bags must be bought? (c) If the client later doubles the length, width and thickness, what volume is poured?

Convert the thickness first: 150 150 150 mm = 0.15 = 0.15 = 0.15 m. Mixing millimetres with metres is what wrecks this item.

V = 4 × 3 × 0.15 = 1.8 V = 4 \times 3 \times 0.15 = 1.8 V = 4 × 3 × 0.15 = 1.8 m 3 ^3 3 .

In cubic centimetres, 1.8 × 1,000,000 = 1,800,000 1.8 \times 1{,}000{,}000 = 1{,}800{,}000 1.8 × 1 , 000 , 000 = 1 , 800 , 000 cm 3 ^3 3 (equivalently 1,800 1{,}800 1 , 800 litres).

Cement: 1.8 × 9 = 16.2 1.8 \times 9 = 16.2 1.8 × 9 = 16.2 bags, so 17 bags are purchased.

Doubling every linear dimension multiplies volume by 2 3 = 8 2^3 = 8 2 3 = 8 : 1.8 × 8 = 14.4 1.8 \times 8 = 14.4 1.8 × 8 = 14.4 m 3 ^3 3 . Answering 3.6 3.6 3.6 m 3 ^3 3 is the trap.

7. Worked example 3 — density of a steel ball

A solid steel ball has radius 3 3 3 cm and steel has density 7.8 7.8 7.8 g/cm 3 ^3 3 . Its volume is V = 4 3 π ( 3 ) 3 = 36 π ≈ 113.1 V = \frac{4}{3}\pi(3)^3 = 36\pi \approx 113.1 V = 3 4 ​ π ( 3 ) 3 = 36 π ≈ 113.1 cm 3 ^3 3 , so its mass is m = ρ V = 7.8 × 113.1 ≈ 882 m = \rho V = 7.8 \times 113.1 \approx 882 m = ρ V = 7.8 × 113.1 ≈ 882 g, or about 0.88 0.88 0.88 kg.

8. Common traps

Using vertical height inside π r ℓ \pi r \ell π r ℓ , or slant height inside 1 3 π r 2 h \frac{1}{3}\pi r^2 h 3 1 ​ π r 2 h .

Believing 1 1 1 m 2 = 100 ^2 = 100 2 = 100 cm 2 ^2 2 ; it is 10,000 10{,}000 10 , 000 cm 2 ^2 2 .

Forgetting the 1 3 \frac{1}{3} 3 1 ​ in cone and pyramid volumes, which inflates the answer threefold.

Taking the total surface area of a hemisphere as 2 π r 2 2\pi r^2 2 π r 2 when the flat face is included — it is 3 π r 2 3\pi r^2 3 π r 2 .

Being handed a diameter and using it as the radius.

Assuming doubled dimensions double the volume.

Leaving one measurement in mm or cm while the rest are in m.