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3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles

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Adamson University College Entrance Test AdUCET

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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

3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles

Key Takeaways Explain This Topic AI

Co-interior (same-side interior) angles formed by a transversal across parallel lines are supplementary, while corresponding, alternate interior and alternate exterior pairs are congruent.

With two sides a and b fixed, the third side of a triangle must satisfy |a - b| < c < a + b, so 5, 7 and 12 form a straight segment rather than a triangle.

SSS, SAS, ASA, AAS and HL prove congruence; AA, SAS and SSS prove similarity; SSA and AAA prove neither.

For similar figures with linear scale factor k, perimeters are in ratio k and areas in ratio k squared.

A sector with central angle θ has arc length (θ/360)·2πr and area (θ/360)·πr², and an inscribed angle is exactly half the central angle subtending the same arc.

Last updated: August 2026

3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles

Plane geometry on the AdUCET rewards recognition , not derivation. A typical item hands you a figure or a one-sentence description and expects you to name the single relationship that unlocks it, quickly and without a diagram of your own. Every problem in this section is designed to be finished by hand, so work in exact form: leave π \pi π and surds such as 3 \sqrt{3} 3 ​ unresolved unless the answer choices are decimals.

1. Angles, linear pairs, and the transversal family

Complementary angles sum to 90 ∘ 90^\circ 9 0 ∘ ; supplementary angles sum to 180 ∘ 180^\circ 18 0 ∘ . An angle of 38 ∘ 38^\circ 3 8 ∘ has complement 52 ∘ 52^\circ 5 2 ∘ and supplement 142 ∘ 142^\circ 14 2 ∘ .

A linear pair — two adjacent angles whose outer sides form a straight line — is always supplementary.

Vertical angles , the opposite pair produced when two lines cross, are always congruent.

Angle bisector : a ray that splits an angle into two congruent halves.

When a transversal (a line crossing two others) cuts a pair of parallel lines, eight angles appear but only two distinct measures exist: some x x x and its supplement 180 ∘ − x 180^\circ - x 18 0 ∘ − x . \ transversal T

  • 1 \ 2
  • L1 -------------------\-----------------
  • 3 \ 4
  • \
  • \
  • 5 \ 6
  • L2 ------------------------\------------
  • 7 \ 8
  • \
  • Pair name Example from the diagram Relationship
  • Corresponding ∠ 1 \angle 1 ∠1 and ∠ 5 \angle 5 ∠5 Congruent
  • Alternate interior ∠ 3 \angle 3 ∠3 and ∠ 6 \angle 6 ∠6 Congruent
  • Alternate exterior ∠ 1 \angle 1 ∠1 and ∠ 8 \angle 8 ∠8 Congruent
  • Co-interior (same-side interior) ∠ 3 \angle 3 ∠3 and ∠ 5 \angle 5 ∠5 Supplementary
  • Vertical ∠ 1 \angle 1 ∠1 and ∠ 4 \angle 4 ∠4 Congruent

The converse is examinable too: if one pair of corresponding angles is congruent, the two lines must be parallel. That is how items ask you to justify parallelism rather than compute a number.

2. Triangles

Angle Sum Theorem : the three interior angles of any triangle total 180 ∘ 180^\circ 18 0 ∘ .

Exterior Angle Theorem : an exterior angle equals the sum of the two remote (non-adjacent) interior angles, so it is always larger than either one.

Triangle Inequality : the sum of any two sides must exceed the third. Equivalently, with sides a a a and b b b fixed, the third side c c c satisfies ∣ a − b ∣ < c < a + b |a-b| < c < a+b ∣ a − b ∣ < c < a + b . Sides 7 7 7 and 10 10 10 admit any third side strictly between 3 3 3 and 17 17 17 .

Side–angle correspondence : the longest side lies opposite the largest angle.

An isosceles triangle has congruent base angles, and the converse holds; an equilateral triangle is equiangular at 60 ∘ 60^\circ 6 0 ∘ each, with area A = 3 4 s 2 A = \frac{\sqrt{3}}{4}s^2 A = 4 3 ​ ​ s 2 .

Congruence and similarity criteria

Congruent triangles are identical in size and shape. The valid shortcuts are SSS , SAS , ASA , AAS , plus HL (hypotenuse–leg) for right triangles only. SSA is ambiguous and AAA fixes only shape, so neither proves congruence.

Similar triangles have equal angles and proportional sides. Prove similarity by AA (two pairs of equal angles), SAS similarity (one equal angle between proportional sides), or SSS similarity (all three ratios equal). If the linear scale factor is k k k , then corresponding perimeters are in ratio k k k but corresponding areas are in ratio k 2 k^2 k 2 .

The Basic Proportionality (Thales) Theorem is the workhorse: a line parallel to one side of a triangle cuts the other two sides proportionally.

Special right triangles and Pythagorean triples a 2 + b 2 = c 2 a^2 + b^2 = c^2 a 2 + b 2 = c 2

Triangle Side ratio Reading

45 ∘ 45^\circ 4 5 ∘ – 45 ∘ 45^\circ 4 5 ∘ – 90 ∘ 90^\circ 9 0 ∘ 1 : 1 : 2 1 : 1 : \sqrt{2} 1 : 1 : 2 ​ legs equal; hypotenuse is leg × 2 \times\sqrt{2} × 2 ​

30 ∘ 30^\circ 3 0 ∘ – 60 ∘ 60^\circ 6 0 ∘ – 90 ∘ 90^\circ 9 0 ∘ 1 : 3 : 2 1 : \sqrt{3} : 2 1 : 3 ​ : 2 short leg faces 30 ∘ 30^\circ 3 0 ∘ ; hypotenuse is twice it

Memorize the triples 3 3 3 - 4 4 4 - 5 5 5 , 5 5 5 - 12 12 12 - 13 13 13 , 8 8 8 - 15 15 15 - 17 17 17 , 7 7 7 - 24 24 24 - 25 25 25 , 9 9 9 - 40 40 40 - 41 41 41 and their multiples ( 6 6 6 - 8 8 8 - 10 10 10 , 9 9 9 - 12 12 12 - 15 15 15 , 10 10 10 - 24 24 24 - 26 26 26 , 15 15 15 - 20 20 20 - 25 25 25 , 21 21 21 - 28 28 28 - 35 35 35 ). Spotting one saves you a square-root extraction.

3. Quadrilaterals and polygons

Trapezoid : exactly one pair of parallel sides; the median joining the leg midpoints equals 1 2 ( b 1 + b 2 ) \frac{1}{2}(b_1+b_2) 2 1 ​ ( b 1 ​ + b 2 ​ ) . An isosceles trapezoid has congruent legs and congruent base angles.

Parallelogram : both pairs of opposite sides parallel and congruent, opposite angles congruent, consecutive angles supplementary, diagonals bisect each other.

Rectangle : a parallelogram with four right angles; its diagonals are congruent.

Rhombus : a parallelogram with four congruent sides; its diagonals are perpendicular and bisect the vertex angles.

Square : simultaneously a rectangle and a rhombus.

For a convex polygon with n n n sides: interior angles total ( n − 2 ) × 180 ∘ (n-2)\times 180^\circ ( n − 2 ) × 18 0 ∘ ; exterior angles always total 360 ∘ 360^\circ 36 0 ∘ no matter how many sides; a regular n n n -gon has each interior angle ( n − 2 ) 180 ∘ n \frac{(n-2)180^\circ}{n} n ( n − 2 ) 18 0 ∘ ​ and each exterior angle 360 ∘ n \frac{360^\circ}{n} n 36 0 ∘ ​ ; the number of diagonals is n ( n − 3 ) 2 \frac{n(n-3)}{2} 2 n ( n − 3 ) ​ .

  • Figure Perimeter Area Note
  • Square 4 s 4s 4 s s 2 s^2 s 2 diagonal s 2 s\sqrt{2} s 2 ​
  • Rectangle 2 ( l + w ) 2(l+w) 2 ( l + w ) l w lw lw diagonal l 2 + w 2 \sqrt{l^2+w^2} l 2 + w 2 ​
  • Triangle a + b + c a+b+c a + b + c 1 2 b h \frac{1}{2}bh 2 1 ​ bh h h h perpendicular to b b b
  • Parallelogram 2 ( a + b ) 2(a+b) 2 ( a + b ) b h bh bh h h h is the height, not the slant side

Trapezoid sum of 4 sides 1 2 ( b 1 + b 2 ) h \frac{1}{2}(b_1+b_2)h 2 1 ​ ( b 1 ​ + b 2 ​ ) h average of the parallel sides

  • Rhombus / kite 4 s 4s 4 s (rhombus) 1 2 d 1 d 2 \frac{1}{2}d_1d_2 2 1 ​ d 1 ​ d 2 ​ diagonals perpendicular
  • Regular polygon n s ns n s 1 2 P a \frac{1}{2}Pa 2 1 ​ P a a a a = apothem
  • Circle C = 2 π r C = 2\pi r C = 2 π r π r 2 \pi r^2 π r 2 d = 2 r d = 2r d = 2 r
  • 4. Circles

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 for central angle θ \theta θ in degrees: L = θ 360 ∘ ⋅ 2 π r L = \frac{\theta}{360^\circ}\cdot 2\pi r L = 36 0 ∘ θ ​ ⋅ 2 π r .

Sector area : A sec = θ 360 ∘ ⋅ π r 2 A_{\text{sec}} = \frac{\theta}{360^\circ}\cdot \pi r^2 A sec ​ = 36 0 ∘ θ ​ ⋅ π r 2 . A segment is the sector minus the triangle formed by the two radii.

Inscribed Angle Theorem : an inscribed angle is half the central angle subtending the same arc; consequently an angle inscribed in a semicircle is exactly 90 ∘ 90^\circ 9 0 ∘ , inscribed angles on the same arc are congruent, and opposite angles of a cyclic quadrilateral are supplementary.

Tangent–radius : a tangent is perpendicular to the radius at the point of tangency, and the two tangent segments drawn from one external point are congruent.

A radius perpendicular to a chord bisects that chord.

5. Worked example 1 — parallel cut inside a triangle

Problem. In △ A B C \triangle ABC △ A BC , point D D D lies on A B ‾ \overline{AB} A B and E E E on A C ‾ \overline{AC} A C with D E ‾ ∥ B C ‾ \overline{DE}\parallel\overline{BC} D E ∥ BC . Given A D = 6 AD = 6 A D = 6 cm, D B = 4 DB = 4 D B = 4 cm, A E = 9 AE = 9 A E = 9 cm and B C = 15 BC = 15 BC = 15 cm, find (a) E C EC EC , (b) D E DE D E , (c) 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 .

Proportionality. A D D B = A E E C ⇒ 6 4 = 9 E C ⇒ E C = 4 × 9 6 = 6 \dfrac{AD}{DB} = \dfrac{AE}{EC} \Rightarrow \dfrac{6}{4} = \dfrac{9}{EC} \Rightarrow EC = \dfrac{4\times 9}{6} = 6 D B A D ​ = EC A E ​ ⇒ 4 6 ​ = EC 9 ​ ⇒ EC = 6 4 × 9 ​ = 6 cm.

Similarity. ∠ A \angle A ∠ A is shared and ∠ A D E = ∠ A B C \angle ADE = \angle ABC ∠ A D E = ∠ A BC (corresponding angles), so △ A D E ∼ △ A B C \triangle ADE \sim \triangle ABC △ A D E ∼ △ A BC by AA. The scale factor uses whole sides: k = A D A B = 6 6 + 4 = 3 5 k = \dfrac{AD}{AB} = \dfrac{6}{6+4} = \dfrac{3}{5} k = A B A D ​ = 6 + 4 6 ​ = 5 3 ​ .

D E = k ⋅ B C = 3 5 ( 15 ) = 9 DE = k\cdot BC = \dfrac{3}{5}(15) = 9 D E = k ⋅ BC = 5 3 ​ ( 15 ) = 9 cm.

Areas. [ A D E ] [ A B C ] = k 2 = 9 25 \dfrac{[ADE]}{[ABC]} = k^2 = \dfrac{9}{25} [ A BC ] [ A D E ] ​ = k 2 = 25 9 ​ , so the trapezoid takes the remaining 16 16 16 parts and [ A D E ] : [ D B C E ] = 9 : 16 [ADE] : [DBCE] = 9 : 16 [ A D E ] : [ D BCE ] = 9 : 16 .

6. Worked example 2 — rotunda: arc, sector, tangent, inscribed angle

Problem. A circular rotunda in Ermita has radius 21 21 21 m. A landscaped sector is bounded by two radii meeting at a central angle of 120 ∘ 120^\circ 12 0 ∘ . Using π ≈ 22 7 \pi \approx \frac{22}{7} π ≈ 7 22 ​ : (a) how much edging borders the curved side? (b) what is the sector's area? (c) a straight walkway is tangent to the rotunda at T T T , and a lamp post P P P on that walkway is 28 28 28 m from T T T — how far is P P P from the centre O O O ? (d) a statue S S S stands on the major arc; what angle do the two sector endpoints subtend at S S S ?

Fraction of the circle: 120 ∘ 360 ∘ = 1 3 \frac{120^\circ}{360^\circ} = \frac{1}{3} 36 0 ∘ 12 0 ∘ ​ = 3 1 ​ .

L = 1 3 ⋅ 2 π ( 21 ) = 14 π ≈ 14 × 22 7 = 44 L = \frac{1}{3}\cdot 2\pi(21) = 14\pi \approx 14\times\frac{22}{7} = 44 L = 3 1 ​ ⋅ 2 π ( 21 ) = 14 π ≈ 14 × 7 22 ​ = 44 m of edging.

A sec = 1 3 π ( 21 ) 2 = 147 π ≈ 147 × 22 7 = 462 A_{\text{sec}} = \frac{1}{3}\pi(21)^2 = 147\pi \approx 147\times\frac{22}{7} = 462 A sec ​ = 3 1 ​ π ( 21 ) 2 = 147 π ≈ 147 × 7 22 ​ = 462 m 2 ^2 2 .

Tangent–radius gives right angle ∠ O T P \angle OTP ∠ OTP , so O P = 21 2 + 28 2 = 441 + 784 = 1225 = 35 OP = \sqrt{21^2+28^2} = \sqrt{441+784} = \sqrt{1225} = 35 OP = 2 1 2 + 2 8 2 ​ = 441 + 784 ​ = 1225 ​ = 35 m — this is the 3 3 3 - 4 4 4 - 5 5 5 triple scaled by 7 7 7 .

The inscribed angle at S S S intercepts the 120 ∘ 120^\circ 12 0 ∘ minor arc, so it measures 1 2 ( 120 ∘ ) = 60 ∘ \frac{1}{2}(120^\circ) = 60^\circ 2 1 ​ ( 12 0 ∘ ) = 6 0 ∘ .

7. Common traps

Co-interior angles are supplementary , not congruent — the most reversed rule in the whole topic.

360 ∘ 360^\circ 36 0 ∘ is the total of the exterior angles; dividing by n n n gives one exterior angle only when the polygon is regular.

The triangle inequality is strict: 5 , 7 , 12 5, 7, 12 5 , 7 , 12 collapses into a straight segment, not a triangle.

SSA and AAA prove nothing about congruence.

Arc length uses 2 π r 2\pi r 2 π r ; sector area uses π r 2 \pi r^2 π r 2 . Mixing them is the standard circle distractor.

Substituting the diameter where the formula wants r r r quadruples an area.

Trapezoid and parallelogram areas need the perpendicular height, never the slanted side.

For similar figures the area ratio is k 2 k^2 k 2 , not k k k .