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3.4 Geometry, Measurement, and Data Interpretation

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Philippines Computer-Based National Career Assessment Examination CB-NCAE

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1 Chapter 1: Overview, Architecture, and Testing Policies

1.1 CB-NCAE Overview, Governance, and Purpose 1.2 Computer-Based Platform, Delivery, and Accommodations 1.3 Senior High School Track Integration and Career Guidance

2 Chapter 2: GSA — Verbal Reasoning (English & Filipino)

2.1 Reading Comprehension and Passage Analysis 2.2 Vocabulary in Context, Context Clues, and Idiomatic Usage 2.3 Verbal Analogies and Word Relationships 2.4 Critical Evaluation, Tone, and Argument Synthesis

3 Chapter 3: GSA — Numerical Reasoning

3.1 Arithmetic Operations, Fractions, Decimals, and Percentages 3.2 Ratios, Proportions, and Rates 3.3 Basic Algebra, Sequences, and Patterns 3.4 Geometry, Measurement, and Data Interpretation

4 Chapter 4: GSA — Logical Reasoning

4.1 Categorical Syllogisms and Deductive Reasoning 4.2 Conditional Logic, Truth Tables, and Contrapositives 4.3 Logical Fallacies and Argument Flaws 4.4 Analytical Puzzles, Linear Ordering, and Grouping

5 Chapter 5: GSA — Abstract Reasoning

5.1 Linear Figure Series and Sequential Progression 5.2 Matrix Completion and Multi-Variable Grids 5.3 Figure Classification and Odd-One-Out 5.4 Rule Induction, Pattern Analogy, and Complex Transformations

6 Chapter 6: GSA — Spatial Reasoning

6.1 2D to 3D Mental Folding and Surface Nets 6.2 Orthographic Projection and Multiview Visualization 6.3 3D Spatial Rotation and Block Visualization 6.4 Mirror Images, Reflections, and Spatial Symmetry

7 Chapter 7: GSA — Mechanical Reasoning

7.1 Simple Machines: Levers, Moments, and Mechanical Advantage 7.2 Pulleys, Wheels and Axles, and Inclined Planes 7.3 Gears, Belt Drives, and Rotational Mechanics 7.4 Fluid Mechanics, Pressure, and Basic Electrical Circuits

8 Chapter 8: GSA — Perceptual Accuracy and Response Efficiency (PARE)

8.1 Alphanumeric Code and String Matching 8.2 Clerical Filing, Indexing, and Alphabetical Sorting 8.3 Symbol Counting, Visual Inspection, and Transposition Detection

9 Chapter 9: CORE — Career and Occupational Roles Explorer & Track Guidance

9.1 CORE-A: Holland's RIASEC Typology 9.2 CORE-B: Thurstone Model of Occupational Interests 9.3 Profile Synthesis and Senior High School Track Selection

3.4 Geometry, Measurement, and Data Interpretation

Key Takeaways Explain This Topic AI

Perimeter measures boundary length, area measures enclosed two-dimensional surface, and volume measures three-dimensional capacity; linear scaling by k k k scales area by k 2 k^2 k 2 and volume by k 3 k^3 k 3 .

The Pythagorean theorem ( a 2 + b 2 = c 2 a^2 + b^2 = c^2 a 2 + b 2 = c 2 ) and primitive triples ( ( 3 , 4 , 5 ) (3, 4, 5) ( 3 , 4 , 5 ) , ( 5 , 12 , 13 ) (5, 12, 13) ( 5 , 12 , 13 ) , ( 8 , 15 , 17 ) (8, 15, 17) ( 8 , 15 , 17 ) ) allow rapid calculation of distances and hypotenuses without squaring large numbers.

Angle relationships on parallel lines cut by a transversal establish that alternate interior, alternate exterior, and corresponding angles are equal, while consecutive interior angles are supplementary ( 180 ∘ 180^\circ 18 0 ∘ ).

Data interpretation relies on careful unit inspection, axis verification, and percentage-to-degree conversions for pie charts ( 1 % = 3.6 ∘ 1\% = 3.6^\circ 1% = 3. 6 ∘ , 100 % = 360 ∘ 100\% = 360^\circ 100% = 36 0 ∘ ), while the median is the most robust average in skewed data.

Last updated: October 2026

3.4 Geometry, Measurement, and Data Interpretation

The final module of Numerical Reasoning synthesizes spatial-quantitative measurement with data literacy. Geometry items evaluate your grasp of plane figures, geometric scaling, angle theorems, and solids, while data interpretation tests your capacity to extract facts, discern trends, and compute summary statistics from graphs and tables under timed conditions.

Plane Geometry: Perimeter and Area

Perimeter is the one-dimensional total distance around the boundary of a closed figure, whereas area measures the two-dimensional surface enclosed within that boundary.

Primary Plane Figures and Area Formulations

Geometric Figure Perimeter / Circumference Area Formula Key Notes & Non-Calculator Tips

Square P = 4 s P = 4s P = 4 s A = s 2 = 1 2 d 2 A = s^2 = \frac{1}{2} d^2 A = s 2 = 2 1 ​ d 2 Diagonal d = s 2 d = s\sqrt{2} d = s 2 ​

Rectangle P = 2 ( l + w ) P = 2(l + w) P = 2 ( l + w ) A = l × w A = l \times w A = l × w Diagonal d = l 2 + w 2 d = \sqrt{l^2 + w^2} d = l 2 + w 2 ​

Triangle P = a + b + c P = a + b + c P = a + b + c A = 1 2 b h A = \frac{1}{2} b h A = 2 1 ​ bh Altitude h h h must be perpendicular to base b b b

Parallelogram P = 2 ( a + b ) P = 2(a + b) P = 2 ( a + b ) A = b × h A = b \times h A = b × h Use vertical height h h h , never slant side

Trapezoid P = a + b 1 + c + b 2 P = a + b_1 + c + b_2 P = a + b 1 ​ + c + b 2 ​ A = b 1 + b 2 2 × h A = \frac{b_1 + b_2}{2} \times h A = 2 b 1 ​ + b 2 ​ ​ × h Area equals median × \times × height

Circle C = 2 π r = π d C = 2\pi r = \pi d C = 2 π r = π d A = π r 2 A = \pi r^2 A = π r 2 Use π ≈ 22 7 \pi \approx \frac{22}{7} π ≈ 7 22 ​ when r r r is a multiple of 7; else 3.14 3.14 3.14

Special Right Triangles

Memorizing the exact side ratios of special right triangles eliminates the need for manual square root extraction:

45 ∘ − 45 ∘ − 90 ∘ 45^\circ - 45^\circ - 90^\circ 4 5 ∘ − 4 5 ∘ − 9 0 ∘ Triangle (Isosceles Right): Side ratio is 1 : 1 : 2 1 : 1 : \sqrt{2} 1 : 1 : 2 ​ . If legs are length x x x , the hypotenuse is x 2 x\sqrt{2} x 2 ​ .

30 ∘ − 60 ∘ − 90 ∘ 30^\circ - 60^\circ - 90^\circ 3 0 ∘ − 6 0 ∘ − 9 0 ∘ Triangle: Side ratio is 1 : 3 : 2 1 : \sqrt{3} : 2 1 : 3 ​ : 2 . Opposite the 30 ∘ 30^\circ 3 0 ∘ angle is x x x (shortest leg), opposite 60 ∘ 60^\circ 6 0 ∘ is x 3 x\sqrt{3} x 3 ​ , and opposite 90 ∘ 90^\circ 9 0 ∘ is 2 x 2x 2 x (hypotenuse).

Composite Plane Figures

To compute the area of composite figures (e.g., an L-shaped room, an athletic track, or a shaded border), decompose the figure into standard shapes. Either:

Partition into non-overlapping sub-rectangles/triangles and sum their individual areas; or

Compute the bounding area and subtract the unshaded cutout portion ( A shaded = A outer − A inner A_{\text{shaded}} = A_{\text{outer}} - A_{\text{inner}} A shaded ​ = A outer ​ − A inner ​ ).

Solid Geometry: Volume and Geometric Scaling Laws

Volume quantifies three-dimensional capacity, measured in cubic units ( m 3 m^3 m 3 , c m 3 cm^3 c m 3 , or liters, where 1 liter = 1 , 000 cm 3 1\text{ liter} = 1,000\text{ cm}^3 1 liter = 1 , 000 cm 3 ).

Standard 3D Solids

Rectangular Prism (Box): V = l × w × h V = l \times w \times h V = l × w × h Total Surface Area ( T S A ) = 2 ( l w + l h + w h ) \text{Total Surface Area } (TSA) = 2(lw + lh + wh) Total Surface Area ( TS A ) = 2 ( lw + l h + w h ) Space Diagonal = l 2 + w 2 + h 2 \text{Space Diagonal} = \sqrt{l^2 + w^2 + h^2} Space Diagonal = l 2 + w 2 + h 2 ​

Right Circular Cylinder: V = π r 2 h V = \pi r^2 h V = π r 2 h Lateral Surface Area ( L S A ) = 2 π r h \text{Lateral Surface Area } (LSA) = 2\pi r h Lateral Surface Area ( L S A ) = 2 π r h Total Surface Area ( T S A ) = 2 π r h + 2 π r 2 \text{Total Surface Area } (TSA) = 2\pi r h + 2\pi r^2 Total Surface Area ( TS A ) = 2 π r h + 2 π r 2

The Fundamental Geometric Scaling Laws

Questions frequently test what occurs when dimensions are scaled by a constant factor k k k :

All linear dimensions (perimeter, circumference, radius, height, diagonal) scale by k 1 k^1 k 1 .

All surface areas (lateral area, base area, total surface area) scale by k 2 k^2 k 2 .

All volumes and capacities scale by k 3 k^3 k 3 .

Non-Calculator Example: If the radius of a cylindrical water tank is doubled while its height remains constant, how does its volume change? The radius is squared in the volume formula ( V = π r 2 h V = \pi r^2 h V = π r 2 h ), so doubling the radius ( k = 2 k = 2 k = 2 ) multiplies the volume by 2 2 = 4 2^2 = 4 2 2 = 4 .

  • Angle Theorems and the Pythagorean Theorem
  • Angle Fundamentals

Complementary Angles: Two angles whose measures sum to 90 ∘ 90^\circ 9 0 ∘ .

Supplementary Angles: Two angles whose measures sum to 180 ∘ 180^\circ 18 0 ∘ .

Vertical Angles: Non-adjacent angles formed by two intersecting lines; vertical angles are always congruent (equal in measure).

Triangle Angle Sum Theorem: The interior angles of any planar triangle sum to 180 ∘ 180^\circ 18 0 ∘ .

Exterior Angle Theorem: The measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.

Parallel Lines Cut by a Transversal

When two parallel lines are intersected by a transversal line, eight angles are created, forming exactly two equal measure groups:

Congruent pairs: Alternate interior angles, alternate exterior angles, and corresponding angles are equal.

Supplementary pairs: Consecutive interior angles (co-interior) on the same side of the transversal sum to 180 ∘ 180^\circ 18 0 ∘ .

The Pythagorean Theorem and Primitive Triples

In every right triangle with legs a a a and b b b and hypotenuse c c c : a 2 + b 2 = c 2 a^2 + b^2 = c^2 a 2 + b 2 = c 2

To save time on the non-calculator exam, memorize the essential primitive Pythagorean triples and their common scalar multiples ( k × { a , b , c } k \times \{a, b, c\} k × { a , b , c } ):

Family { 3 , 4 , 5 } \{3, 4, 5\} { 3 , 4 , 5 } : Multiples include { 6 , 8 , 10 } \{6, 8, 10\} { 6 , 8 , 10 } , { 9 , 12 , 15 } \{9, 12, 15\} { 9 , 12 , 15 } , { 12 , 16 , 20 } \{12, 16, 20\} { 12 , 16 , 20 } , { 15 , 20 , 25 } \{15, 20, 25\} { 15 , 20 , 25 } .

Family { 5 , 12 , 13 } \{5, 12, 13\} { 5 , 12 , 13 } : Multiples include { 10 , 24 , 26 } \{10, 24, 26\} { 10 , 24 , 26 } .

Family { 8 , 15 , 17 } \{8, 15, 17\} { 8 , 15 , 17 } : Multiples include { 16 , 30 , 34 } \{16, 30, 34\} { 16 , 30 , 34 } .

Family { 7 , 24 , 25 } \{7, 24, 25\} { 7 , 24 , 25 } : Primitive set.

If legs are 9 and 12, recognize the ratio 3 : 4 3 : 4 3 : 4 , indicating the hypotenuse must be 5 × 3 = 15 5 \times 3 = 15 5 × 3 = 15 , avoiding squaring and root operations.

Data Interpretation: Reading Tables, Charts, and Graphs

Data interpretation items evaluate your ability to synthesize information presented in visual and tabular formats.

Formats and Scanning Rules

Data Tables: Always verify row and column headers and note measurement units (e.g., "in thousands", "in millions of PHP").

Bar Charts: Used for discrete categorical comparisons. Check the baseline: if the vertical axis begins at a non-zero value, bar heights exaggerate relative differences.

Line Graphs: Used for continuous time-series data. The slope between two points indicates the rate of change; steeper lines represent faster growth or decline.

Pie Charts (Circle Graphs): Sectors represent parts of a whole ( 100 % 100\% 100% or 360 ∘ 360^\circ 36 0 ∘ ).

Conversion factor: 1 % = 3.6 ∘ 1\% = 3.6^\circ 1% = 3. 6 ∘ , and 10 % = 36 ∘ 10\% = 36^\circ 10% = 3 6 ∘ .

To convert degrees ( θ \theta θ ) into percentage: P = θ 360 ∘ × 100 % = θ 3.6 ∘ P = \frac{\theta}{360^\circ} \times 100\% = \frac{\theta}{3.6^\circ} P = 36 0 ∘ θ ​ × 100% = 3. 6 ∘ θ ​ .

To convert percentage into degrees: θ = P × 3.6 ∘ \theta = P \times 3.6^\circ θ = P × 3. 6 ∘ .

  • Descriptive Statistics: Measures of Central Tendency and Dispersion
  • 1. Arithmetic Mean (Average)

x ˉ = ∑ x n = x 1 + x 2 + ⋯ + x n n \bar{x} = \frac{\sum x}{n} = \frac{x_1 + x_2 + \dots + x_n}{n} x ˉ = n ∑ x ​ = n x 1 ​ + x 2 ​ + ⋯ + x n ​ ​

Tip

The Assumed Mean (Deviation) Shortcut: When computing the mean of numbers close to each other (e.g., 83, 87, 88, 92, 85), do not sum the full values. Pick an assumed mean of 85 and sum the signed deviations: ( 83 − 85 ) + ( 87 − 85 ) + ( 88 − 85 ) + ( 92 − 85 ) + ( 85 − 85 ) = − 2 + 2 + 3 + 7 + 0 = + 10 (83-85) + (87-85) + (88-85) + (92-85) + (85-85) = -2 + 2 + 3 + 7 + 0 = +10 ( 83 − 85 ) + ( 87 − 85 ) + ( 88 − 85 ) + ( 92 − 85 ) + ( 85 − 85 ) = − 2 + 2 + 3 + 7 + 0 = + 10 . Divide the net deviation by n = 5 n=5 n = 5 : + 10 5 = + 2 \frac{+10}{5} = +2 5 + 10 ​ = + 2 . True mean = 85 + 2 = 87 = 85 + 2 = 87 = 85 + 2 = 87 .

2. Median

The middle value when data points are arranged in ascending order:

If n n n is odd, the median is the single middle element at position n + 1 2 \frac{n + 1}{2} 2 n + 1 ​ .

If n n n is even, the median is the arithmetic mean of the two middle elements at positions n 2 \frac{n}{2} 2 n ​ and n 2 + 1 \frac{n}{2} + 1 2 n ​ + 1 .

The median is resistant to extreme outliers , making it superior to the mean when distributions are heavily skewed (such as household income statistics).

3. Mode and Range

Mode: The value that appears with the greatest frequency. A distribution may have one mode (unimodal), two modes (bimodal), multiple modes, or no mode if all values appear with equal frequency.

Range: The simplest measure of dispersion: Range = x max − x min \text{Range} = x_{\text{max}} - x_{\text{min}} Range = x max ​ − x min ​ .

  • Step-by-Step Worked Problems
  • Worked Example 1: Composite Area and the Pythagorean Theorem

Problem: A school community garden in Laguna is designed in the shape of a right-angled triangle. Its two perpendicular legs measure 12 meters and 16 meters. A circular concrete composting station with a radius of 2 meters is constructed inside the garden. What is the length of the perimeter boundary fence along the hypotenuse, and what is the remaining plantable soil area of the garden? (Use π ≈ 3.14 \pi \approx 3.14 π ≈ 3.14 ).

Step-by-Step Solution:

Determine the hypotenuse using Pythagorean triples:

The legs are 12 and 16. Divide both by their greatest common factor, 4: 12 4 = 3 \frac{12}{4} = 3 4 12 ​ = 3 and 16 4 = 4 \frac{16}{4} = 4 4 16 ​ = 4 .

This is a scaled { 3 , 4 , 5 } \{3, 4, 5\} { 3 , 4 , 5 } triple with scale factor k = 4 k = 4 k = 4 .

Hypotenuse = 5 × 4 = 20 meters = 5 \times 4 = 20\text{ meters} = 5 × 4 = 20 meters .

Calculate the total area of the triangular garden: A triangle = 1 2 × base × height = 1 2 × 12 × 16 = 6 × 16 = 96 m 2 A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 16 = 6 \times 16 = 96\text{ m}^2 A triangle ​ = 2 1 ​ × base × height = 2 1 ​ × 12 × 16 = 6 × 16 = 96 m 2

Calculate the area occupied by the circular composting station: A circle = π r 2 = 3.14 × 2 2 = 3.14 × 4 = 12.56 m 2 A_{\text{circle}} = \pi r^2 = 3.14 \times 2^2 = 3.14 \times 4 = 12.56\text{ m}^2 A circle ​ = π r 2 = 3.14 × 2 2 = 3.14 × 4 = 12.56 m 2

Calculate the remaining plantable soil area: A plantable = A triangle − A circle = 96 − 12.56 = 83.44 m 2 A_{\text{plantable}} = A_{\text{triangle}} - A_{\text{circle}} = 96 - 12.56 = 83.44\text{ m}^2 A plantable ​ = A triangle ​ − A circle ​ = 96 − 12.56 = 83.44 m 2

Worked Example 2: Target Score Using the Deviation Method

Problem: A student completes four periodic assessment quizzes in Numerical Reasoning, scoring 82, 87, 89, and 84. What score must the student achieve on the fifth and final quiz to obtain an overall average score of exactly 86 across all five assessments?

Step-by-Step Solution:

Method A — Total Points Architecture:

Target total points required: 5 × 86 = 430 points 5 \times 86 = 430\text{ points} 5 × 86 = 430 points .

Current total points earned: 82 + 87 + 89 + 84 = 342 points 82 + 87 + 89 + 84 = 342\text{ points} 82 + 87 + 89 + 84 = 342 points .

Required fifth score: 430 − 342 = 88 points 430 - 342 = 88\text{ points} 430 − 342 = 88 points .

Method B — Deviation Architecture (Mental Shortcut):

Calculate deviations of each known quiz from target 86:

  • Quiz 1: 82 − 86 = − 4 82 - 86 = -4 82 − 86 = − 4
  • Quiz 2: 87 − 86 = + 1 87 - 86 = +1 87 − 86 = + 1
  • Quiz 3: 89 − 86 = + 3 89 - 86 = +3 89 − 86 = + 3
  • Quiz 4: 84 − 86 = − 2 84 - 86 = -2 84 − 86 = − 2

Sum the deviations: ( − 4 ) + ( + 1 ) + ( + 3 ) + ( − 2 ) = − 2 (-4) + (+1) + (+3) + (-2) = -2 ( − 4 ) + ( + 1 ) + ( + 3 ) + ( − 2 ) = − 2 .

To achieve an overall net deviation of 0, Quiz 5 must provide a balance of + 2 +2 + 2 .

Required fifth score: 86 + 2 = 88 points 86 + 2 = 88\text{ points} 86 + 2 = 88 points .