1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area

Ateneo College Entrance Test ACET

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1 Chapter 1: Mathematics

1.1 Real Numbers, Ratios, Rates & Percentages 1.2 Algebraic Expressions, Polynomials & Linear Equations 1.3 Functions, Domain/Range, Graphs & Transformations 1.4 Quadratic, Polynomial, Exponential & Logarithmic Equations 1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area 1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections 1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities 1.8 Arithmetic & Geometric Sequences, Series & Word Patterns 1.9 Permutations, Combinations, Probability & Statistics

2 Chapter 2: English Proficiency

2.1 Subject-Verb Agreement, Verb Tenses & Aspect 2.2 Pronoun-Antecedent Agreement, Case, Misplaced & Dangling Modifiers 2.3 Sentence Structure, Run-Ons, Fragments & Parallelism 2.4 Contextual Vocabulary, Word Roots, Prefixes & Suffixes 2.5 Advanced Synonyms, Antonyms & Nuanced Vocabulary 2.6 Verbal Analogies: Relationship Types & Solving Strategies 2.7 Punctuation, Capitalization, Mechanics & Common Errors 2.8 Diction, Idiomatic Expressions, Prepositions & Frequently Confused Words

3 Chapter 3: Reading Comprehension

3.1 Identifying Main Ideas, Central Themes & Paragraph Summaries 3.2 Locating Explicit Details, Supporting Evidence & Data 3.3 Drawing Logical Inferences & Unstated Conclusions 3.4 Analyzing Author's Purpose, Point of View, Tone & Attitude 3.5 Interpreting Figurative Language, Tone Words & Context Clues 3.6 Text Structure, Rhetorical Devices & Evaluating Arguments

4 Chapter 4: Logical and Abstract Reasoning

4.1 Number & Letter Series Patterns & Formulas 4.2 Abstract Reasoning: Matrix Patterns & Figure Rotations 4.3 Formal Syllogisms, Categorical Logic & Deductive Proofs 4.4 Coding-Decoding, Cipher Patterns & Symbol Operations 4.5 Spatial Visualization, Paper Folding & Venn Diagram Logic

5 Chapter 5: General Knowledge

5.1 Philippine History, Government, Constitution & Civics 5.2 Integrated Science: Biology, Chemistry, Physics & Earth Science 5.3 Philippine & World Geography, Basic Economics & Cultural Facts 5.4 Current Events, World Landmarks, Literature & General Trivia

6 Chapter 6: ACET Essay Writing & Exam Strategies

6.1 ACET Essay: Thesis Development, Structuring & Argumentation 6.2 Time Allocation, Pacing Strategies & Subtest Tactics 6.3 Guessing Strategies (No Penalty), Test Anxiety & Exam Day Readiness

1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area

Key Takeaways Explain This Topic AI

Memorize common Pythagorean triples and special right triangle ratios (30-60-90 and 45-45-90) to solve geometric lengths without full calculations.

Apply similarity properties: linear scale factor k yields area scale factor k^2 and volume scale factor k^3.

Utilize circle theorems regarding inscribed angles, central angles, tangent lines, and secant lengths.

Calculate regular polygon interior angles using (n-2)*180/n and exterior angles using 360/n.

Last updated: July 2026

Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area

Geometry on the ACET emphasizes fast shape identification, spatial reasoning, area/volume ratios, and circle properties. Calculators are unavailable, so memorizing key geometric ratios saves time.

1. Angle Relationships & Parallel Lines

When two parallel lines L 1 ∥ L 2 L_1 \parallel L_2 L 1 ∥ L 2 are cut by a transversal T T T :

Corresponding Angles are equal ( 1 = 5 1 = 5 1 = 5 ).

Alternate Interior Angles are equal ( 3 = 6 3 = 6 3 = 6 ).

Alternate Exterior Angles are equal ( 2 = 7 2 = 7 2 = 7 ).

Consecutive Interior Angles are supplementary ( 3 + 5 = 180 ∘ 3 + 5 = 180^\circ 3 + 5 = 18 0 ∘ ).

2. Triangles: Properties, Special Triangles & Area Formulas
Fundamental Triangle Inequalities & Angle Theorems

Angle Sum: θ 1 + θ 2 + θ 3 = 180 ∘ \theta_1 + \theta_2 + \theta_3 = 180^\circ θ 1 + θ 2 + θ 3 = 18 0 ∘ .

Exterior Angle Theorem: The measure of an exterior angle equals the sum of its two remote interior angles.

Triangle Inequality Theorem: For sides a , b , c a, b, c a , b , c : ∣ a − b ∣ < c < a + b |a - b| < c < a + b ∣ a − b ∣ < c < a + b .

Special Right Triangles & Pythagorean Triples 45 ∘ − 45 ∘ − 90 ∘ Triangle: Legs x , x ⟹ Hypotenuse x 2 30 ∘ − 60 ∘ − 90 ∘ Triangle: Short leg x ⟹ Long leg x 3 , Hypotenuse 2 x \begin{aligned}

\mathbf{45^\circ-45^\circ-90^\circ \text{ Triangle:}} & \quad \text{Legs } x, x \implies \text{Hypotenuse } x\sqrt{2}

\\ \mathbf{30^\circ-60^\circ-90^\circ \text{ Triangle:}} & \quad \text{Short leg } x \implies \text{Long leg } x\sqrt{3}, \text{ Hypotenuse } 2x

\end{aligned} 4 5 ∘ − 4 5 ∘ − 9 0 ∘ Triangle: 3 0 ∘ − 6 0 ∘ − 9 0 ∘ Triangle: Legs x , x ⟹ Hypotenuse x 2 Short leg x ⟹ Long leg x 3 , Hypotenuse 2 x

High-Frequency ACET Pythagorean Triples:

3 : 4 : 5 ( 6 : 8 : 10 , 9 : 12 : 15 , 12 : 16 : 20 ) 3 : 4 : 5 \quad (6:8:10, \; 9:12:15, \; 12:16:20) 3 : 4 : 5 ( 6 : 8 : 10 , 9 : 12 : 15 , 12 : 16 : 20 )

5 : 12 : 13 ( 10 : 24 : 26 ) 5 : 12 : 13 \quad (10:24:26) 5 : 12 : 13 ( 10 : 24 : 26 )
8 : 15 : 17 8 : 15 : 17 8 : 15 : 17
7 : 24 : 25 7 : 24 : 25 7 : 24 : 25
Comprehensive Area Formulas for Triangles
Standard Formula: A = 1 2 b h A = \frac{1}{2} b h A = 2 1 bh
SAS Trigonometric Formula: A = 1 2 a b sin ⁡ C A = \frac{1}{2} a b \sin C A = 2 1 ab sin C
Equilateral Triangle Area: A = s 2 3 4 A = \frac{s^2 \sqrt{3}}{4} A = 4 s 2 3

Heron's Formula (Given 3 sides a , b , c a, b, c a , b , c ): A = s ( s − a ) ( s − b ) ( s − c ) , s = a + b + c 2 A = \sqrt{s(s - a)(s - b)(s - c)}, \quad s = \frac{a + b + c}{2} A = s ( s − a ) ( s − b ) ( s − c ) , s = 2 a + b + c

3. Polygon Properties & Interior/Exterior Angles

For a convex polygon with n n n sides:

Sum of Interior Angles: S int = ( n − 2 ) × 180 ∘ S_{\text{int}} = (n - 2) \times 180^\circ S int = ( n − 2 ) × 18 0 ∘

Interior Angle of Regular Polygon: I = ( n − 2 ) × 180 ∘ n I = \frac{(n - 2) \times 180^\circ}{n} I = n ( n − 2 ) × 18 0 ∘

Sum of Exterior Angles: S ext = 360 ∘ S_{\text{ext}} = 360^\circ S ext = 36 0 ∘
Exterior Angle of Regular Polygon: E = 360 ∘ n E = \frac{360^\circ}{n} E = n 36 0 ∘
Number of Diagonals: D = n ( n − 3 ) 2 D = \frac{n(n - 3)}{2} D = 2 n ( n − 3 )
4. Circle Theorems & Sector Measures
Primary Circle Formulas
Circumference: C = 2 π r = π d C = 2\pi r = \pi d C = 2 π r = π d
Area: A = π r 2 A = \pi r^2 A = π r 2

Arc Length ( L L L for central angle θ ∘ \theta^\circ θ ∘ ): L = θ 360 ∘ × 2 π r L = \frac{\theta}{360^\circ} \times 2\pi r L = 36 0 ∘ θ × 2 π r

Sector Area ( A sector A_{\text{sector}} A sector ): A sector = θ 360 ∘ × π r 2 A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 A sector = 36 0 ∘ θ × π r 2

Core Circle Theorems

Inscribed Angle Theorem: An inscribed angle is half the central angle intercepting the same arc: θ inscribed = 1 2 θ central \theta_{\text{inscribed}} = \frac{1}{2} \theta_{\text{central}} θ inscribed = 2 1 θ central .

Thales' Theorem: An angle inscribed in a semicircle is always a right angle ( 90 ∘ 90^\circ 9 0 ∘ ).

Tangent Line Theorem: A line tangent to a circle is perpendicular to the radius at the point of tangency.

Intersecting Chords Theorem: P A ⋅ P B = P C ⋅ P D PA \cdot PB = PC \cdot PD P A ⋅ PB = PC ⋅ P D .

Secant-Tangent Theorem: P T 2 = P A ⋅ P B PT^2 = PA \cdot PB P T 2 = P A ⋅ PB .

Worked Step-by-Step ACET Exam Problems
Problem 1: Sector Area and Shaded Segment Calculation

A circle has a radius of 6 cm 6\text{ cm} 6 cm . A central angle of 60 ∘ 60^\circ 6 0 ∘ intercepts an arc. What is the area of the shaded segment formed between the chord and the arc?

Step 1: Calculate total sector area. A sector = 60 ∘ 360 ∘ × π ( 6 ) 2 = 1 6 × 36 π = 6 π cm 2 A_{\text{sector}} = \frac{60^\circ}{360^\circ} \times \pi (6)^2 = \frac{1}{6} \times 36 \pi = 6\pi \text{ cm}^2 A sector = 36 0 ∘ 6 0 ∘ × π ( 6 ) 2 = 6 1 × 36 π = 6 π cm 2

Step 2: Calculate area of the subtended triangle. Since the central angle is 60 ∘ 60^\circ 6 0 ∘ and two sides are radii ( r = 6 r=6 r = 6 ), the triangle is equilateral with side length s = 6 s = 6 s = 6 . A triangle = s 2 3 4 = 6 2 3 4 = 36 3 4 = 9 3 cm 2 A_{\text{triangle}} = \frac{s^2 \sqrt{3}}{4} = \frac{6^2 \sqrt{3}}{4} = \frac{36 \sqrt{3}}{4} = 9\sqrt{3} \text{ cm}^2 A triangle = 4 s 2 3 = 4 6 2 3 = 4 36 3 = 9 3 cm 2

Step 3: Subtract triangle area from sector area. A segment = A sector − A triangle = ( 6 π − 9 3 ) cm 2 A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}} = (6\pi - 9\sqrt{3}) \text{ cm}^2 A segment = A sector − A triangle = ( 6 π − 9 3 ) cm 2

Problem 2: Similar Triangles and Area Scale Factors

In △ A B C \triangle ABC △ A BC , a line parallel to base B C BC BC intersects side A B AB A B at D D D and side A C AC A C at E E E . If A D = 3 cm AD = 3\text{ cm} A D = 3 cm and D B = 5 cm DB = 5\text{ cm} D B = 5 cm , find the ratio of the area of △ A D E \triangle ADE △ A D E to the area of trapezoid D B C E DBCE D BCE .

Step 1: Determine side ratio between similar triangles. △ A D E ∼ △ A B C \triangle ADE \sim \triangle ABC △ A D E ∼ △ A BC . Side length A B = A D + D B = 3 + 5 = 8 cm AB = AD + DB = 3 + 5 = 8\text{ cm} A B = A D + D B = 3 + 5 = 8 cm .

Scale factor k = A D A B = 3 8 k = \frac{AD}{AB} = \frac{3}{8} k = A B A D = 8 3 .

Step 2: Determine area ratio. Area ( △ A D E ) Area ( △ A B C ) = k 2 = ( 3 8 ) 2 = 9 64 \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = k^2 = \left(\frac{3}{8}\right)^2 = \frac{9}{64} Area ( △ A BC ) Area ( △ A D E ) = k 2 = ( 8 3 ) 2 = 64 9

Step 3: Calculate area ratio relative to trapezoid. Let Area ( △ A B C ) = 64 x \text{Area}(\triangle ABC) = 64x Area ( △ A BC ) = 64 x , so Area ( △ A D E ) = 9 x \text{Area}(\triangle ADE) = 9x Area ( △ A D E ) = 9 x . Area ( Trapezoid D B C E ) = 64 x − 9 x = 55 x \text{Area}(\text{Trapezoid } DBCE) = 64x - 9x = 55x Area ( Trapezoid D BCE ) = 64 x − 9 x = 55 x Area ( △ A D E ) Area ( Trapezoid D B C E ) = 9 x 55 x = 9 55 \frac{\text{Area}(\triangle ADE)}{\text{Area}(\text{Trapezoid } DBCE)} = \frac{9x}{55x} = \frac{9}{55} Area ( Trapezoid D BCE ) Area ( △ A D E ) = 55 x 9 x = 55 9

ACET Speed Tactics & Traps Summary

Problem Type Common Student Trap ACET Speed Tactic
Triangles Using Pythagorean theorem on non-right triangles Check angle measures or use Law of Cosines

Similar Figures Assuming area scales linearly with side length k k k Area scales as k 2 k^2 k 2 ; volume scales as k 3 k^3 k 3

Inscribed Angles Setting inscribed angle equal to arc measure Inscribed angle is half the intercepted arc

Regular Polygons Confusing interior angle sum with single angle Divide sum ( n − 2 ) 180 ∘ (n-2)180^\circ ( n − 2 ) 18 0 ∘ by n n n for regular shapes