Lesson 5: Polygons and Quadrilaterals

Polygons are everywhere: floor tiles, street signs, honeycombs, soccer balls. This lesson organises them. You will learn the angle rules that govern *any* polygon, then zoom into the quadrilateral family — parallelograms, rectangles, rhombuses, squares, trapezoids and kites — where a handful of properties (and one elegant coordinate proof) settle most questions.

Angle sums in any polygon

Triangulate an n-gon from one vertex and you get (n − 2) triangles, so the interior angle sum is (n − 2) · 180°. The exterior angles (one at each vertex, taken in the same direction around the shape) always sum to 360°, for every convex polygon. In a regular polygon (all sides and angles equal), each interior angle is (n − 2) · 180°/n and each exterior angle is 360°/n.

The quadrilateral family and its properties

A parallelogram has: both pairs of opposite sides parallel and congruent; opposite angles congruent; consecutive angles supplementary; diagonals that bisect each other. Special parallelograms add features: a rectangle adds congruent diagonals; a rhombus adds four congruent sides and perpendicular diagonals that bisect the vertex angles; a square has everything. A trapezoid has at least one pair of parallel sides (the bases); in an isosceles trapezoid the legs are congruent, base angles are congruent, and diagonals are congruent. A kite has two distinct pairs of adjacent congruent sides and one pair of opposite angles congruent, with perpendicular diagonals.

Worked example 1: Regular octagon angles

A STOP sign is a regular octagon (n = 8). Interior angle sum: (8 − 2) · 180° = 1080°. Each interior angle: 1080°/8 = 135°. Each exterior angle: 360°/8 = 45°. Notice 135° + 45° = 180°, as they must — interior and exterior angles at a vertex form a linear pair. That check catches most arithmetic slips in this topic.

Worked example 2: How many sides?

A polygon's interior angles sum to 1440°. Find n: (n − 2) · 180° = 1440°, so n − 2 = 8 and n = 10 — a decagon. Reverse problem: a regular polygon has each exterior angle 36°, so n = 360°/36° = 10 sides again. Exterior angles are usually the faster tool for regular polygons because 360 never changes.

Worked example 3: Coordinate proof — parallelogram diagonals bisect each other

Take a general parallelogram with vertices A(0, 0), B(a, 0), D(b, c) and C(a + b, c). (The fourth vertex must be (a + b, c): from A, going to C adds the run of AB and the run of AD.) The diagonals are AC and BD. Midpoint of AC: ((0 + a + b)/2, (0 + c)/2) = ((a + b)/2, c/2). Midpoint of BD: ((a + b)/2, (0 + c)/2) = ((a + b)/2, c/2). The midpoints are identical, so the diagonals share a midpoint — which is exactly what "bisect each other" means. Because a, b and c were arbitrary, this single computation proves the theorem for *every* parallelogram. That is the power of a coordinate proof.

Worked example 4: Algebra in a parallelogram

In parallelogram PQRS, side PQ measures (3x + 5) cm and the opposite side SR measures (5x − 9) cm. Opposite sides are congruent: 3x + 5 = 5x − 9, so 14 = 2x and x = 7. Each side measures 3(7) + 5 = 26 cm. Check: 5(7) − 9 = 26 cm. ✓

Geometry at work

How do you check that a wooden frame is truly a rectangle and not a leaning parallelogram? Carpenters measure the diagonals: a parallelogram is a rectangle exactly when its diagonals are congruent, so two equal diagonal measurements certify four right angles without ever measuring an angle. Tile setters use the exterior-angle rule in reverse — the angles of tiles meeting at a point must sum to 360°, which is why squares (90° × 4), hexagons (120° × 3) and triangles (60° × 6) tile a floor but regular pentagons (108°, which does not divide 360°) never can.

Common mistakes

• Using (n − 2) · 180° for one angle. That formula gives the *sum*; divide by n for one angle of a regular polygon.
• Mixing up which diagonals are special. Bisecting diagonals: every parallelogram. Congruent diagonals: rectangle (and isosceles trapezoid). Perpendicular diagonals: rhombus and kite. Keep the family tree straight.
• Assuming a quadrilateral is a parallelogram from one property. One pair of parallel sides makes a trapezoid; you need both pairs (or an equivalent property) for a parallelogram.
• Forgetting the exterior sum is always 360°. It does not grow with n.

Exam Experience

1. What is the sum of the interior angles of a pentagon?

• A) 360°
• B) 450°
• C) 540°
• D) 720°

Answer: C. (5 − 2) · 180° = 540°.

2. What is the measure of each interior angle of a regular hexagon?

• A) 108°
• B) 120°
• C) 135°
• D) 150°

Answer: B. (6 − 2) · 180°/6 = 720°/6 = 120°.

3. In which quadrilateral are the diagonals perpendicular but *not* necessarily congruent or bisecting both ways?

• A) Square
• B) Rectangle
• C) Kite
• D) Parallelogram

Answer: C. A kite's diagonals are perpendicular; only one diagonal is bisected.

Constructed response. A quadrilateral has vertices P(0, 0), Q(5, 0), R(7, 3) and S(2, 3). Prove, using slopes, that PQRS is a parallelogram, and then determine whether it is a rhombus by comparing side lengths.

Scoring guide (4 points):

• 1 point: Computes slope of PQ = 0 and slope of SR = (3 − 3)/(7 − 2) = 0, so PQ ∥ SR.
• 1 point: Computes slope of PS = (3 − 0)/(2 − 0) = 3/2 and slope of QR = (3 − 0)/(7 − 5) = 3/2, so PS ∥ QR; both pairs of opposite sides parallel, hence a parallelogram.
• 1 point: Computes PQ = 5 and PS = √(2² + 3²) = √13 ≈ 3.61.
• 1 point: Concludes adjacent sides are not congruent, so PQRS is a parallelogram but not a rhombus.

Going deeper: Rhombus diagonals, computed

A rhombus has side length 10 and one diagonal of length 12. Find the other diagonal and the area. The diagonals of a rhombus are perpendicular bisectors of each other, so they split the rhombus into four right triangles, each with hypotenuse 10 (the side) and one leg 6 (half of the known diagonal). The other leg is √(10² − 6²) = √(100 − 36) = √64 = 8, by the Pythagorean Theorem. So the second diagonal is 2 · 8 = 16. The area of any rhombus (in fact any kite-shaped figure with perpendicular diagonals) is half the product of the diagonals: A = (12 · 16)/2 = 96 square units. Verify another way: the four right triangles each have area (1/2) · 6 · 8 = 24, and 4 · 24 = 96. ✓ This example is a template: rhombus problems are Pythagoras problems wearing a diagonal disguise, because the perpendicular-bisector property manufactures right triangles for you.

The same trick runs in reverse for design: a kite maker who wants diagonals of 12 and 16 knows each side must be √(6² + 8²) = 10 before cutting a single spar — which is why the 6-8-10 triangle keeps reappearing throughout this course.

One more direction: from angle back to sides

Worked Example 2 solved for n from the angle sum; the reverse journey is just as common. A regular polygon has each exterior angle 24°. Then n = 360°/24° = 15 sides, and each interior angle is 180° − 24° = 156°. (That answers this lesson's Try It question style — always route through the exterior angle first, because 360 never changes while (n − 2) · 180° must be recomputed.) A harder flavour: a regular polygon's interior angle is five times its exterior angle. Let the exterior angle be e; the interior is 5e, and they sum to 180°, so 6e = 180° and e = 30°. Then n = 360°/30° = 12 — a dodecagon, with interior angle 150°. Notice the two-step rhythm of this whole topic: relate the two angles at one vertex (they are supplementary), then use the exterior angle to count sides. Polygon problems that look stuck usually surrender to exactly that rhythm.

Final drill: The trapezoid median

The segment joining the midpoints of the two legs of a trapezoid is called its median (or midsegment). It is parallel to both bases and its length is the *average* of the bases: m = (b₁ + b₂)/2. For bases 10 and 18, the median is 14 — and notice this is the same average that appeared in the trapezoid area formula of Lesson 9, A = h · m: area equals height times the median. One fact, two costumes. Companion computation for the isosceles trapezoid: its base angles are congruent and its same-side angles between the parallel bases are supplementary. If a lower base angle is 70°, the other lower base angle is 70°, and each upper base angle is 180° − 70° = 110° — the parallel bases turn the upper angles into same-side interior angles with the legs as transversals. Quadrilateral problems very often decompose into a Lesson 2 angle chase wearing a Lesson 5 costume; naming the costume is half the solve.