Quadrilaterals and Polygons

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Quadrilaterals and Polygons

1. Properties of Quadrilaterals — Comparison Table

A quadrilateral is any polygon with exactly 4 sides. The six special quadrilaterals you need to know for Sec 2 E-Math are listed below.

Quick Reference Table

Property Parallelogram Rhombus Rectangle Square Trapezium Kite
Sides Opp. sides equal & parallel All 4 sides equal; opp. sides parallel Opp. sides equal & parallel All 4 sides equal; opp. sides parallel One pair of parallel sides Two pairs of adjacent sides equal
Angles Opp. angles equal; co-int. angles supp. Opp. angles equal; co-int. angles supp. All 4 angles = 90° All 4 angles = 90° Co-interior angles (between parallels) are supp. One pair of opp. angles equal (between unequal sides)
Diagonals Bisect each other Bisect each other at 90°; bisect vertex angles Bisect each other; equal in length Bisect each other at 90°; equal in length; bisect vertex angles Generally unequal; do not bisect each other Perpendicular; one bisects the other; one is an axis of symmetry
Lines of Symmetry 0 2 2 4 0 (isosceles: 1) 1
Rotational Symmetry Order 2 Order 2 Order 2 Order 4 Order 1 (none) Order 1 (none)

Memory tip: Square ⊂ Rectangle ⊂ Parallelogram, and Square ⊂ Rhombus ⊂ Parallelogram. A square is the most special — it satisfies every property in the table.

Detailed Properties

Parallelogram

  • AB ∥ DC and AD ∥ BC
  • AB = DC and AD = BC
  • ∠A = ∠C; ∠B = ∠D
  • ∠A + ∠B = 180° (co-interior angles)
  • Diagonals bisect each other (AO = OC, BO = OD)

Rhombus

  • All sides equal: AB = BC = CD = DA
  • Opposite sides parallel
  • Opposite angles equal
  • Diagonals bisect each other at right angles
  • Each diagonal bisects the vertex angles
  • 2 lines of symmetry (the diagonals themselves)

Rectangle

  • All angles = 90°
  • Opposite sides equal and parallel
  • Diagonals are equal in length and bisect each other
  • 2 lines of symmetry (through midpoints of opposite sides)

Square

  • All sides equal; all angles = 90°
  • Diagonals equal, bisect each other at 90°, and bisect vertex angles (each = 45°)
  • 4 lines of symmetry; rotational symmetry of order 4

Trapezium

  • Exactly one pair of parallel sides (AB ∥ DC)
  • Co-interior angles between the parallel sides are supplementary
  • Isosceles trapezium: non-parallel sides equal; base angles equal; diagonals equal; 1 line of symmetry

Kite

  • Two pairs of adjacent (not opposite) sides equal: AB = AD and CB = CD
  • One pair of opposite angles equal: ∠B = ∠D
  • Diagonals are perpendicular
  • The longer diagonal (axis of symmetry) bisects the shorter diagonal
  • 1 line of symmetry (along the longer diagonal)

2. SVG Diagrams of Each Quadrilateral

Parallelogram

<svg width="260" height="160" xmlns="http://www.w3.org/2000/svg">
  <polygon points="50,130 200,130 230,30 80,30" fill="#e8f4f8" stroke="#2c5f8a" stroke-width="2"/>
  <!-- Tick marks: AB = DC -->
  <line x1="120" y1="130" x2="126" y2="124" stroke="#2c5f8a" stroke-width="2"/>
  <line x1="126" y1="130" x2="132" y2="124" stroke="#2c5f8a" stroke-width="2"/>
  <line x1="150" y1="30" x2="156" y2="24" stroke="#2c5f8a" stroke-width="2"/>
  <line x1="156" y1="30" x2="162" y2="24" stroke="#2c5f8a" stroke-width="2"/>
  <!-- Tick marks: AD = BC -->
  <line x1="62" y1="82" x2="70" y2="80" stroke="#c0392b" stroke-width="2"/>
  <line x1="208" y1="82" x2="216" y2="80" stroke="#c0392b" stroke-width="2"/>
  <!-- Arrows for parallel sides -->
  <text x="120" y="148" font-size="11" fill="#2c5f8a" text-anchor="middle">AB ∥ DC</text>
  <text x="140" y="90" font-size="11" fill="#c0392b" text-anchor="middle">AD ∥ BC</text>
  <!-- Labels -->
  <text x="44" y="135" font-size="13" fill="#333" font-weight="bold">A</text>
  <text x="202" y="145" font-size="13" fill="#333" font-weight="bold">B</text>
  <text x="232" y="26" font-size="13" fill="#333" font-weight="bold">C</text>
  <text x="72" y="26" font-size="13" fill="#333" font-weight="bold">D</text>
  <text x="5" y="20" font-size="12" fill="#555" font-style="italic">Parallelogram</text>
</svg>

Rhombus

<svg width="220" height="180" xmlns="http://www.w3.org/2000/svg">
  <polygon points="110,15 200,90 110,165 20,90" fill="#f0e8f8" stroke="#6c3483" stroke-width="2"/>
  <!-- Diagonals -->
  <line x1="110" y1="15" x2="110" y2="165" stroke="#6c3483" stroke-width="1" stroke-dasharray="5,4"/>
  <line x1="20" y1="90" x2="200" y2="90" stroke="#6c3483" stroke-width="1" stroke-dasharray="5,4"/>
  <!-- Right angle marker at centre -->
  <rect x="110" y="90" width="8" height="8" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <!-- Tick marks on all sides -->
  <line x1="59" y1="51" x2="65" y2="55" stroke="#6c3483" stroke-width="2"/>
  <line x1="157" y1="51" x2="163" y2="47" stroke="#6c3483" stroke-width="2"/>
  <line x1="59" y1="129" x2="65" y2="125" stroke="#6c3483" stroke-width="2"/>
  <line x1="157" y1="129" x2="163" y2="133" stroke="#6c3483" stroke-width="2"/>
  <!-- Labels -->
  <text x="104" y="10" font-size="13" fill="#333" font-weight="bold">A</text>
  <text x="203" y="94" font-size="13" fill="#333" font-weight="bold">B</text>
  <text x="104" y="178" font-size="13" fill="#333" font-weight="bold">C</text>
  <text x="6" y="94" font-size="13" fill="#333" font-weight="bold">D</text>
  <text x="5" y="16" font-size="12" fill="#555" font-style="italic">Rhombus</text>
  <text x="85" y="85" font-size="10" fill="#c0392b">90°</text>
</svg>

Rectangle

<svg width="260" height="160" xmlns="http://www.w3.org/2000/svg">
  <rect x="40" y="30" width="180" height="100" fill="#e8f8ee" stroke="#1e8449" stroke-width="2"/>
  <!-- Diagonals -->
  <line x1="40" y1="30" x2="220" y2="130" stroke="#1e8449" stroke-width="1" stroke-dasharray="5,4"/>
  <line x1="220" y1="30" x2="40" y2="130" stroke="#1e8449" stroke-width="1" stroke-dasharray="5,4"/>
  <!-- Right angle markers -->
  <rect x="40" y="30" width="10" height="10" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <rect x="210" y="30" width="10" height="10" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <rect x="40" y="120" width="10" height="10" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <rect x="210" y="120" width="10" height="10" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <!-- Equal diagonal tick at centre -->
  <circle cx="130" cy="80" r="3" fill="#c0392b"/>
  <!-- Labels -->
  <text x="25" y="26" font-size="13" fill="#333" font-weight="bold">A</text>
  <text x="222" y="26" font-size="13" fill="#333" font-weight="bold">B</text>
  <text x="222" y="142" font-size="13" fill="#333" font-weight="bold">C</text>
  <text x="25" y="142" font-size="13" fill="#333" font-weight="bold">D</text>
  <text x="5" y="16" font-size="12" fill="#555" font-style="italic">Rectangle</text>
  <text x="118" y="72" font-size="9" fill="#1e8449">diag. equal</text>
</svg>

Square

<svg width="200" height="200" xmlns="http://www.w3.org/2000/svg">
  <rect x="40" y="30" width="130" height="130" fill="#fef9e7" stroke="#d4ac0d" stroke-width="2"/>
  <!-- Diagonals -->
  <line x1="40" y1="30" x2="170" y2="160" stroke="#d4ac0d" stroke-width="1" stroke-dasharray="5,4"/>
  <line x1="170" y1="30" x2="40" y2="160" stroke="#d4ac0d" stroke-width="1" stroke-dasharray="5,4"/>
  <!-- Right angle markers at corners -->
  <rect x="40" y="30" width="10" height="10" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <rect x="160" y="30" width="10" height="10" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <rect x="40" y="150" width="10" height="10" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <rect x="160" y="150" width="10" height="10" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <!-- Right angle at centre -->
  <rect x="105" y="95" width="8" height="8" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <!-- Tick marks on all sides -->
  <line x1="101" y1="30" x2="107" y2="36" stroke="#d4ac0d" stroke-width="2"/>
  <line x1="170" y1="91" x2="164" y2="97" stroke="#d4ac0d" stroke-width="2"/>
  <line x1="101" y1="160" x2="107" y2="154" stroke="#d4ac0d" stroke-width="2"/>
  <line x1="40" y1="91" x2="46" y2="97" stroke="#d4ac0d" stroke-width="2"/>
  <!-- Labels -->
  <text x="25" y="26" font-size="13" fill="#333" font-weight="bold">A</text>
  <text x="172" y="26" font-size="13" fill="#333" font-weight="bold">B</text>
  <text x="172" y="172" font-size="13" fill="#333" font-weight="bold">C</text>
  <text x="25" y="172" font-size="13" fill="#333" font-weight="bold">D</text>
  <text x="5" y="16" font-size="12" fill="#555" font-style="italic">Square</text>
</svg>

Trapezium

<svg width="260" height="160" xmlns="http://www.w3.org/2000/svg">
  <polygon points="70,130 200,130 170,30 100,30" fill="#fde8e8" stroke="#c0392b" stroke-width="2"/>
  <!-- Parallel arrows on top and bottom -->
  <line x1="100" y1="30" x2="170" y2="30" stroke="#c0392b" stroke-width="2"/>
  <line x1="70" y1="130" x2="200" y2="130" stroke="#c0392b" stroke-width="2"/>
  <text x="128" y="24" font-size="10" fill="#c0392b" text-anchor="middle">AB ∥ DC</text>
  <!-- Arrow heads to show parallel -->
  <polygon points="135,30 128,26 128,34" fill="#c0392b"/>
  <polygon points="130,130 123,126 123,134" fill="#c0392b"/>
  <!-- Labels -->
  <text x="93" y="24" font-size="13" fill="#333" font-weight="bold">A</text>
  <text x="171" y="24" font-size="13" fill="#333" font-weight="bold">B</text>
  <text x="202" y="143" font-size="13" fill="#333" font-weight="bold">C</text>
  <text x="55" y="143" font-size="13" fill="#333" font-weight="bold">D</text>
  <text x="5" y="16" font-size="12" fill="#555" font-style="italic">Trapezium</text>
  <text x="130" y="90" font-size="10" fill="#666" text-anchor="middle">one pair of ∥ sides</text>
</svg>

Kite

<svg width="200" height="220" xmlns="http://www.w3.org/2000/svg">
  <polygon points="100,15 170,90 100,200 30,90" fill="#e8f0fe" stroke="#2471a3" stroke-width="2"/>
  <!-- Axis of symmetry -->
  <line x1="100" y1="15" x2="100" y2="200" stroke="#2471a3" stroke-width="1" stroke-dasharray="5,4"/>
  <!-- Short diagonal -->
  <line x1="30" y1="90" x2="170" y2="90" stroke="#2471a3" stroke-width="1" stroke-dasharray="5,4"/>
  <!-- Right angle at intersection -->
  <rect x="100" y="90" width="8" height="8" fill="none" stroke="#c0392b" stroke-width="1.5"/>
  <!-- Tick marks: AB = AD, CB = CD -->
  <line x1="62" y1="50" x2="68" y2="55" stroke="#2471a3" stroke-width="2"/>
  <line x1="138" y1="50" x2="132" y2="55" stroke="#2471a3" stroke-width="2"/>
  <line x1="62" y1="147" x2="68" y2="142" stroke="#c0392b" stroke-width="2"/>
  <line x1="138" y1="147" x2="132" y2="142" stroke="#c0392b" stroke-width="2"/>
  <!-- Labels -->
  <text x="94" y="10" font-size="13" fill="#333" font-weight="bold">A</text>
  <text x="172" y="94" font-size="13" fill="#333" font-weight="bold">B</text>
  <text x="94" y="214" font-size="13" fill="#333" font-weight="bold">C</text>
  <text x="14" y="94" font-size="13" fill="#333" font-weight="bold">D</text>
  <text x="5" y="16" font-size="12" fill="#555" font-style="italic">Kite</text>
  <text x="108" y="86" font-size="9" fill="#c0392b">90°</text>
</svg>

3. Angle Sum of a Quadrilateral = 360°

Proof Using Triangles

Theorem: The sum of interior angles of any quadrilateral is 360°.

Proof:

Draw diagonal AC in quadrilateral ABCD. This divides it into two triangles: △ABC and △ACD.

   D ————————— C
   |          /|
   |         / |
   |        /  |
   |       /   |
   |      /    |
   A ————————— B
  • Sum of angles in △ABC = 180°
    • ∠BAC + ∠ABC + ∠BCA = 180°
  • Sum of angles in △ACD = 180°
    • ∠CAD + ∠ACD + ∠ADC = 180°

Adding both equations:

(∠BAC + ∠CAD) + ∠ABC + (∠BCA + ∠ACD) + ∠ADC = 360°

∠DAB + ∠ABC + ∠BCD + ∠CDA = 360°

Therefore, the sum of interior angles of a quadrilateral = 360°. ∎

4. Interior Angle Sum of Any Polygon

Formula

$$S = (n - 2) \times 180°$$

where n is the number of sides (and angles) of the polygon.

Why This Works

Any polygon with n sides can be divided into (n − 2) triangles by drawing all diagonals from one vertex.

  • Triangle (n = 3): S = (3 − 2) × 180° = 180°
  • Quadrilateral (n = 4): S = (4 − 2) × 180° = 360°
  • Pentagon (n = 5): S = (5 − 2) × 180° = 540°
  • Hexagon (n = 6): S = (6 − 2) × 180° = 720°

Derivation sketch:

From one vertex of an n-gon, draw (n − 3) diagonals. This creates (n − 2) non-overlapping triangles. Each triangle contributes 180°, so total = (n − 2) × 180°.

5. Exterior Angle Sum of Any Convex Polygon = 360°

The Rule

No matter how many sides a convex polygon has, its exterior angles always add up to 360°.

$$\text{Sum of exterior angles} = 360° \quad \text{(for any convex polygon)}$$

Intuitive Explanation

Imagine walking around the perimeter of any convex polygon. At each vertex you turn by one exterior angle. By the time you return to your starting point facing the original direction, you have turned a full 360°.

Relationship Between Interior and Exterior Angles

At each vertex of a polygon:

$$\text{Interior angle} + \text{Exterior angle} = 180°$$

(They form a straight line.)

So for a polygon with n sides:

Sum of (interior + exterior) at all vertices = n × 180°

(n − 2) × 180° + Sum of exterior angles = n × 180°

Sum of exterior angles = n × 180° − (n − 2) × 180° = 180°n − 180°n + 360° = 360° ✓

6. Regular Polygons — Angle Table

A regular polygon has all sides equal and all angles equal.

For a regular n-gon:

$$\text{Each interior angle} = \frac{(n-2) \times 180°}{n}$$

$$\text{Each exterior angle} = \frac{360°}{n}$$

n Name Each Interior Angle Each Exterior Angle
3 Equilateral Triangle 60° 120°
4 Square 90° 90°
5 Regular Pentagon 108° 72°
6 Regular Hexagon 120° 60°
7 Regular Heptagon ≈ 128.57° ≈ 51.43°
8 Regular Octagon 135° 45°
9 Regular Nonagon 140° 40°
10 Regular Decagon 144° 36°

Key check: Interior angle + Exterior angle must always equal 180°. Use this to verify your answers.

7. Finding Unknown Angles in Polygons — Worked Examples

Example 1: Unknown Angle in a Quadrilateral

Question: In quadrilateral ABCD, ∠A = 95°, ∠B = 110°, ∠C = 72°. Find ∠D.

Solution:

Sum of angles in a quadrilateral = 360°

∠A + ∠B + ∠C + ∠D = 360°

95° + 110° + 72° + ∠D = 360°

277° + ∠D = 360°

∠D = 360° − 277° = 83°

Example 2: Interior Angle of a Regular Polygon

Question: Find the interior angle of a regular 12-sided polygon (dodecagon).

Solution:

Each interior angle = (n − 2) × 180° ÷ n

= (12 − 2) × 180° ÷ 12

= 10 × 180° ÷ 12

= 1800° ÷ 12

= 150°

Check: Exterior angle = 360° ÷ 12 = 30°; 150° + 30° = 180° ✓

Example 3: Finding n from an Interior Angle

Question: The interior angle of a regular polygon is 156°. How many sides does it have?

Solution:

Each exterior angle = 180° − 156° = 24°

Number of sides = 360° ÷ exterior angle

= 360° ÷ 24°

= 15 sides

Example 4: Unknown Angles in a Polygon with Algebra

Question: A pentagon has angles x°, (x + 20)°, (2x − 10)°, (x + 30)°, and (x + 40)°. Find x and all the angles.

Solution:

Sum of interior angles of a pentagon = (5 − 2) × 180° = 540°

x + (x + 20) + (2x − 10) + (x + 30) + (x + 40) = 540

6x + 80 = 540

6x = 460

x = 460 ÷ 6 ≈ 76.67°

Wait — let us recount:

x + x + 20 + 2x − 10 + x + 30 + x + 40 = 540

(1 + 1 + 2 + 1 + 1)x + (20 − 10 + 30 + 40) = 540

6x + 80 = 540

6x = 460

x = 76⅔°

The five angles are:

  • x = 76.7°
  • x + 20 = 96.7°
  • 2x − 10 = 143.3°
  • x + 30 = 106.7°
  • x + 40 = 116.7°

Check: 76.7 + 96.7 + 143.3 + 106.7 + 116.7 = 540° ✓

8. Geometrical Constructions

Not in Sec 2 2026: Ruler-and-compass constructions (bisecting a line, bisecting an angle, perpendicular from a point, constructing triangles and quadrilaterals) are not part of the 2026 Sec 2 E-Mathematics syllabus. These constructions appear in some Sec 1 or enrichment programmes. The notes below are kept for reference only.

In all constructions below: use only a ruler (for straight lines, not measuring) and a pair of compasses. Show all arcs clearly — they earn marks.

Construction 1: Bisecting a Line Segment

Goal: Find the midpoint of line segment AB, and draw the perpendicular bisector.

Steps:

  1. Open compasses to more than half the length of AB.
  2. With centre A, draw an arc above and below AB.
  3. With the same radius and centre B, draw arcs above and below AB, cutting the previous arcs at points P and Q.
  4. Join P to Q with a straight line. PQ is the perpendicular bisector of AB, crossing AB at its midpoint M.
<svg width="300" height="200" xmlns="http://www.w3.org/2000/svg">
  <!-- Line AB -->
  <line x1="40" y1="100" x2="260" y2="100" stroke="#333" stroke-width="2"/>
  <!-- Points A and B -->
  <circle cx="40" cy="100" r="3" fill="#333"/>
  <circle cx="260" cy="100" r="3" fill="#333"/>
  <text x="28" y="115" font-size="13" fill="#333" font-weight="bold">A</text>
  <text x="262" y="115" font-size="13" fill="#333" font-weight="bold">B</text>
  <!-- Arcs from A -->
  <path d="M 40 100 m 120,-70 a 130,130 0 0,1 0,140" fill="none" stroke="#2c5f8a" stroke-width="1.5" stroke-dasharray="4,3"/>
  <!-- Arcs from B -->
  <path d="M 260 100 m -120,-70 a 130,130 0 0,0 0,140" fill="none" stroke="#c0392b" stroke-width="1.5" stroke-dasharray="4,3"/>
  <!-- Perpendicular bisector -->
  <line x1="150" y1="20" x2="150" y2="180" stroke="#1e8449" stroke-width="2"/>
  <!-- Points P and Q -->
  <circle cx="150" cy="30" r="3" fill="#1e8449"/>
  <circle cx="150" cy="170" r="3" fill="#1e8449"/>
  <text x="155" y="28" font-size="12" fill="#1e8449">P</text>
  <text x="155" y="178" font-size="12" fill="#1e8449">Q</text>
  <!-- Midpoint M -->
  <circle cx="150" cy="100" r="3" fill="#d4ac0d"/>
  <text x="154" y="95" font-size="12" fill="#d4ac0d" font-weight="bold">M</text>
  <!-- Right angle marker -->
  <rect x="150" y="100" width="8" height="8" fill="none" stroke="#333" stroke-width="1.5"/>
  <text x="5" y="16" font-size="11" fill="#555" font-style="italic">Perpendicular Bisector of AB</text>
</svg>

Construction 2: Bisecting an Angle

Goal: Bisect angle AOB.

Steps:

  1. With centre O, draw an arc cutting OA at P and OB at Q.
  2. With centre P, draw an arc inside the angle.
  3. With the same radius and centre Q, draw an arc to cut the previous arc at R.
  4. Join O to R. OR is the angle bisector.
<svg width="260" height="200" xmlns="http://www.w3.org/2000/svg">
  <!-- Angle arms -->
  <line x1="40" y1="170" x2="230" y2="170" stroke="#333" stroke-width="2"/>
  <line x1="40" y1="170" x2="150" y2="20" stroke="#333" stroke-width="2"/>
  <!-- Labels -->
  <text x="24" y="180" font-size="13" fill="#333" font-weight="bold">O</text>
  <text x="232" y="178" font-size="13" fill="#333" font-weight="bold">A</text>
  <text x="150" y="14" font-size="13" fill="#333" font-weight="bold">B</text>
  <!-- Arc from O -->
  <path d="M 40 170 m 80,0 a 80,80 0 0,0 -35,-70" fill="none" stroke="#2c5f8a" stroke-width="1.5" stroke-dasharray="4,3"/>
  <!-- Points P and Q -->
  <circle cx="120" cy="170" r="3" fill="#2c5f8a"/>
  <circle cx="84" cy="102" r="3" fill="#2c5f8a"/>
  <text x="120" y="184" font-size="12" fill="#2c5f8a">P</text>
  <text x="68" y="100" font-size="12" fill="#2c5f8a">Q</text>
  <!-- Arcs from P and Q to find R -->
  <path d="M 120 170 m -36,-68 a 80,80 0 0,1 0,0" fill="none" stroke="#c0392b" stroke-width="0"/>
  <circle cx="106" cy="118" r="3" fill="#c0392b"/>
  <text x="108" y="114" font-size="12" fill="#c0392b">R</text>
  <!-- Bisector OR -->
  <line x1="40" y1="170" x2="140" y2="80" stroke="#1e8449" stroke-width="2" stroke-dasharray="6,3"/>
  <text x="5" y="16" font-size="11" fill="#555" font-style="italic">Angle Bisector</text>
  <!-- Equal angle marks -->
  <path d="M 40 170 m 50,0 a 50,50 0 0,0 -13,-24" fill="none" stroke="#d4ac0d" stroke-width="2"/>
  <path d="M 40 170 m 37,-26 a 50,50 0 0,0 -14,-24" fill="none" stroke="#d4ac0d" stroke-width="2"/>
</svg>

Construction 3: Perpendicular from a Point to a Line

Goal: Drop a perpendicular from external point P to line AB.

Steps:

  1. With centre P, draw an arc that cuts line AB at two points, call them X and Y.
  2. With centre X, draw an arc below (or on the other side of) the line.
  3. With the same radius and centre Y, draw an arc to cut the previous one at Q.
  4. Join P to Q. PQ is perpendicular to AB, meeting it at foot F.
<svg width="300" height="200" xmlns="http://www.w3.org/2000/svg">
  <!-- Line AB -->
  <line x1="20" y1="150" x2="280" y2="150" stroke="#333" stroke-width="2"/>
  <text x="8" y="163" font-size="13" fill="#333" font-weight="bold">A</text>
  <text x="282" y="163" font-size="13" fill="#333" font-weight="bold">B</text>
  <!-- External point P -->
  <circle cx="150" cy="40" r="3" fill="#c0392b"/>
  <text x="154" y="36" font-size="13" fill="#c0392b" font-weight="bold">P</text>
  <!-- Arc from P cutting AB at X and Y -->
  <path d="M 90,150 a 70,70 0 0,1 120,0" fill="none" stroke="#2c5f8a" stroke-width="1.5" stroke-dasharray="4,3"/>
  <circle cx="90" cy="150" r="3" fill="#2c5f8a"/>
  <circle cx="210" cy="150" r="3" fill="#2c5f8a"/>
  <text x="84" y="168" font-size="12" fill="#2c5f8a">X</text>
  <text x="212" y="168" font-size="12" fill="#2c5f8a">Y</text>
  <!-- Arcs from X and Y below line -->
  <path d="M 90 150 m 60,60 a 80,80 0 0,0 0,0" fill="none" stroke="#6c3483" stroke-width="0"/>
  <circle cx="150" cy="190" r="3" fill="#6c3483"/>
  <text x="154" y="200" font-size="12" fill="#6c3483">Q</text>
  <!-- Perpendicular PQ -->
  <line x1="150" y1="40" x2="150" y2="195" stroke="#1e8449" stroke-width="2"/>
  <!-- Foot F -->
  <circle cx="150" cy="150" r="3" fill="#d4ac0d"/>
  <text x="154" y="145" font-size="12" fill="#d4ac0d" font-weight="bold">F</text>
  <!-- Right angle marker -->
  <rect x="150" y="150" width="8" height="8" fill="none" stroke="#333" stroke-width="1.5"/>
  <text x="5" y="16" font-size="11" fill="#555" font-style="italic">Perpendicular from P to AB</text>
</svg>

Construction 4: Constructing Triangles

SSS (Three sides given): e.g., AB = 6 cm, BC = 5 cm, AC = 4 cm

Steps:

  1. Draw base AB = 6 cm.
  2. With centre A, draw an arc of radius 4 cm.
  3. With centre B, draw an arc of radius 5 cm.
  4. The arcs intersect at C. Join AC and BC.

SAS (Two sides and included angle): e.g., AB = 6 cm, ∠A = 50°, AC = 4 cm

Steps:

  1. Draw base AB = 6 cm.
  2. At A, construct the angle 50° (use a protractor, or construct with compasses).
  3. On the angle ray from A, mark point C so that AC = 4 cm.
  4. Join BC.

ASA (Two angles and included side): e.g., BC = 6 cm, ∠B = 50°, ∠C = 60°

Steps:

  1. Draw base BC = 6 cm.
  2. At B, construct angle 50°.
  3. At C, construct angle 60°.
  4. The two rays from B and C meet at A.

Construction 5: Constructing a Quadrilateral

Example: Construct quadrilateral ABCD where AB = 6 cm, BC = 5 cm, CD = 4 cm, DA = 4.5 cm, and diagonal AC = 7 cm.

Strategy — divide into two triangles using a diagonal:

Steps:

  1. Construct △ABC using SSS: AB = 6, BC = 5, AC = 7.
    • Draw AB = 6 cm.
    • Arc from A (radius 7 cm) and arc from B (radius 5 cm) intersect at C.
    • Join AC and BC.
  2. Construct △ACD using SSS: AC = 7 (already drawn), CD = 4, DA = 4.5.
    • Arc from C (radius 4 cm) and arc from A (radius 4.5 cm) intersect at D (on the opposite side of AC from B).
    • Join CD and DA.

Key principle: Always split a quadrilateral into two triangles using a diagonal, then construct each triangle separately.

9. Common Exam Questions and Techniques

Type 1: Finding Interior/Exterior Angles

Approach:

  • Identify whether you are given the number of sides or an angle.
  • Use S = (n − 2) × 180° for total interior angle sum.
  • Use 360° ÷ n for each exterior angle of a regular polygon.
  • Remember: interior + exterior = 180° at any vertex.

Common trap: Students forget that the formula gives the total sum, not each angle. For each angle of a regular polygon, divide by n.

Type 2: Finding n from a Given Angle

If the interior angle of a regular polygon is given:

  1. Find the exterior angle: exterior = 180° − interior.
  2. Find n: n = 360° ÷ exterior angle.

If the exterior angle is given directly: n = 360° ÷ exterior angle.

Type 3: Proving Properties of Quadrilaterals

Example question: “ABCD is a rhombus. Prove that its diagonals bisect each other at right angles.”

Approach:

  • State the property you will use (e.g., all sides equal in a rhombus).
  • Use congruent triangles (SSS, SAS, AAS, RHS).
  • State the conclusion clearly.

Proof outline for rhombus diagonals:

In △AOB and △COB:

  • AB = CB (sides of rhombus)
  • OB = OB (common side)
  • AO = CO (diagonals of parallelogram bisect each other)

Therefore △AOB ≅ △COB (SSS)

So ∠AOB = ∠COB

But ∠AOB + ∠COB = 180° (straight line)

Therefore ∠AOB = 90° ✓

Type 4: Angles in Combined Figures

When a figure combines polygons and triangles:

  1. Break the figure into known shapes.
  2. Apply angle sum rules to each part.
  3. Use angle properties (vertically opposite, co-interior, alternate, corresponding) to link angles.

10. Practice Questions with Full Answers

Q1. The interior angles of a hexagon are 100°, 95°, 130°, 120°, 85°, and x°. Find x.

Answer:

Sum of interior angles of hexagon = (6 − 2) × 180° = 720°

100 + 95 + 130 + 120 + 85 + x = 720

530 + x = 720

x = 190°

Wait — 190° > 180°, which means this is a reflex angle and the hexagon is non-convex (concave). This is acceptable; the formula S = (n − 2) × 180° still holds for any simple polygon, including concave polygons.

Q2. The exterior angle of a regular polygon is 24°. How many sides does it have? What is each interior angle?

Answer:

Number of sides = 360° ÷ 24° = 15 sides

Each interior angle = 180° − 24° = 156°

Check: (15 − 2) × 180° ÷ 15 = 13 × 180° ÷ 15 = 2340° ÷ 15 = 156° ✓

Q3. In parallelogram ABCD, ∠ABC = 68°. Find all other angles.

Answer:

In a parallelogram, opposite angles are equal and co-interior angles are supplementary.

∠ADC = ∠ABC = 68° (opposite angles)

∠BAD = 180° − 68° = 112° (co-interior angles, AB ∥ DC)

∠BCD = ∠BAD = 112° (opposite angles)

Check: 68° + 112° + 68° + 112° = 360° ✓

Q4. ABCD is a kite where AB = AD = 5 cm and CB = CD = 8 cm. The diagonal AC = 10 cm. Find the length of diagonal BD.

Answer:

In a kite, the main diagonal AC is the axis of symmetry and bisects BD at right angles at point O.

In △ABO: AB = 5 cm, AO = part of AC.

First find AO using the axis of symmetry. In a kite, AO + OC = AC = 10 cm.

Use the Pythagorean theorem in △ABO and △CBO:

AO² + BO² = AB² = 25 → AO² + BO² = 25 … (1)

OC² + BO² = CB² = 64 → OC² + BO² = 64 … (2)

Also AO + OC = 10, so OC = 10 − AO.

From (2) − (1): OC² − AO² = 39

(10 − AO)² − AO² = 39

100 − 20·AO + AO² − AO² = 39

100 − 20·AO = 39

20·AO = 61

AO = 3.05 cm

From (1): BO² = 25 − (3.05)² = 25 − 9.3025 = 15.6975

BO = 3.962 cm

BD = 2 × BO = 7.92 cm (to 3 s.f.)

Q5. The sum of interior angles of a polygon is 1980°. How many sides does it have?

Answer:

(n − 2) × 180° = 1980°

n − 2 = 1980° ÷ 180° = 11

n = 13 sides

Q6. In the figure, ABCDE is a regular pentagon and ABFG is a square. Find ∠GAE.

Answer:

Each interior angle of a regular pentagon = (5−2) × 180° ÷ 5 = 108°

Each interior angle of a square = 90°

∠EAB = 108° (interior angle of pentagon)

∠GAB = 90° (interior angle of square)

∠GAE = ∠EAB − ∠GAB = 108° − 90° = 18°

Q7. In a rhombus ABCD, the diagonals meet at O. Given that ∠ABO = 35°, find ∠BAD.

Answer:

In a rhombus, the diagonals bisect the vertex angles and meet at 90°.

In △ABO: ∠AOB = 90°, ∠ABO = 35°

∠BAO = 180° − 90° − 35° = 55°

Since diagonal BD bisects ∠ABC: ∠ABC = 2 × 35° = 70°

Since AB ∥ DC: ∠BAD + ∠ABC = 180° (co-interior angles)

∠BAD = 180° − 70° = 110°

Q8. A polygon has n sides. Its interior angle sum exceeds its exterior angle sum by 720°. Find n.

Answer:

Interior angle sum = (n − 2) × 180°

Exterior angle sum = 360° (always)

(n − 2) × 180° − 360° = 720°

(n − 2) × 180° = 1080°

n − 2 = 6

n = 8

The polygon is an octagon.

Q9. In trapezium ABCD, AB ∥ DC. ∠ABC = 74° and ∠BCD = 106°. Find ∠CDA and ∠DAB.

Answer:

Since AB ∥ DC, the co-interior angles between the parallel sides are supplementary (sum to 180°).

On transversal BC (cutting the parallel lines AB and DC):

∠ABC + ∠BCD = 74° + 106° = 180° ✓ (this confirms AB ∥ DC)

On transversal AD (cutting the same parallel lines):

∠DAB + ∠CDA = 180°

We can also use the angle sum of a quadrilateral (360°):

∠DAB + ∠ABC + ∠BCD + ∠CDA = 360°

∠DAB + 74° + 106° + ∠CDA = 360°

∠DAB + ∠CDA = 360° − 180° = 180°

Both rules agree, as expected. The two given angles fix one pair of co-interior angles; the other pair sums to 180° but each individual angle is underdetermined unless more information is given (for example, that the trapezium is isosceles).

If ABCD is an isosceles trapezium (∠DAB = ∠ABC and ∠CDA = ∠BCD):

∠DAB = 74°, ∠CDA = 106°

Q10. Describe fully how to construct an equilateral triangle with side 5 cm using ruler and compasses only.

Answer:

Steps:

  1. Draw a line segment AB = 5 cm.
  2. Open compasses to 5 cm. With centre A, draw an arc above AB.
  3. With the same radius (5 cm) and centre B, draw an arc above AB, cutting the first arc at point C.
  4. Join AC and BC.

Triangle ABC is equilateral because AB = AC = BC = 5 cm (all radii are equal).

Why it works: All three sides are equal by construction (each is a radius of a circle of radius 5 cm), so all angles = 60° by the properties of an equilateral triangle.

Summary — Key Formulas to Remember

Formula What it gives
S = (n − 2) × 180° Total interior angle sum of any n-gon
Each interior angle = (n−2) × 180° ÷ n Interior angle of a regular n-gon
Each exterior angle = 360° ÷ n Exterior angle of a regular n-gon
Sum of exterior angles = 360° True for any convex polygon
Interior + Exterior = 180° At each vertex of any polygon
Sum of angles in quadrilateral = 360° Special case of S formula with n = 4

Quick Revision Checklist

  • [ ] I can state the properties (sides, angles, diagonals, symmetry) of all 6 special quadrilaterals.
  • [ ] I can prove the angle sum of a quadrilateral using two triangles.
  • [ ] I can apply S = (n − 2) × 180° to find interior angle sums.
  • [ ] I know that exterior angle sum = 360° for any convex polygon.
  • [ ] I can find each angle of a regular polygon.
  • [ ] I can find n given an interior or exterior angle.
  • [ ] (Not in Sec 2 2026) I can perform ruler-and-compass constructions (reference only).
  • [ ] I can solve algebraic angle problems in polygons.

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