Grade 8

Quadrilaterals and polygons

Four-sided and many-sided figures, classified by their symmetry, such as parallelograms and rhombi.

IntuitionFrom triangles to four-sided and many-sided worlds

Cut any rectangular sheet of paper straight from one corner to the opposite corner and you hold two triangles in your hands. That simple diagonal cut is the master key to every quadrilateral $ABCD$ and every polygon: by drawing diagonals from a single vertex, any many-sided figure breaks cleanly into triangles whose angles and areas we already know how to compute! Within the family of four-sided shapes, adding symmetry step by step creates a rich hierarchy: a trapezoid has at least one pair of parallel sides, a parallelogram has both opposite pairs parallel so its diagonals $AC$ and $BD$ bisect each other, a rectangle makes the four corners right angles so the diagonals $AC$ and $BD$ become equal in length, a rhombus makes all four sides equal so the diagonals cross at right angles, and a square combines both to achieve maximum symmetry. The network diagram below shows the four vertices of $ABCD$ linked by its sides and diagonals, a preview of the hierarchy we will build next.

The four vertices of quadrilateral

ABCD

connected by its four sides and diagonals

AC

and

BD

; each diagonal splits the figure into two triangles.

SchoolInterior angle sums and the quadrilateral hierarchy

Definition: Quadrilateral and polygon interior angle sum

In any convex quadrilateral $ABCD$ , drawing one diagonal splits the figure into two triangles, so its four interior angles always add up to twice a straight angle:

\angle A + \angle B + \angle C + \angle D = 360^\circ

More generally, in any convex polygon with $n$ sides ( $n \ge 3$ ), the diagonals drawn from a single vertex divide the polygon into $n - 2$ non-overlapping triangles, so the sum of all $n$ interior angles is:

S_n = (n - 2) \cdot 180^\circ

Classification of special quadrilaterals by sides, diagonals, and area
Quadrilateral	Defining property	Diagonal characterization	Area formula
Trapezoid	At least one pair of parallel sides ( $AB \parallel CD$ )	Equal ( $AC = BD$ ) iff isosceles trapezoid	$S = \dfrac{(a + b) \cdot h}{2}$
Parallelogram	Both pairs of opposite sides parallel ( $AB \parallel CD$ , $AD \parallel BC$ )	Bisect each other ( $OA = OC$ , $OB = OD$ )	$S = a \cdot h$
Rectangle	Parallelogram with four right angles ( $90^\circ$ )	Bisect each other and equal ( $AC = BD$ )	$S = a \cdot b$
Rhombus	Parallelogram with four equal sides ( $AB = BC = CD = DA$ )	Perpendicular bisectors of each other ( $AC \perp BD$ )	$S = \dfrac{1}{2} d_1 d_2$
Square	Both a rectangle and a rhombus ( $90^\circ$ angles and equal sides)	Bisect each other, equal ( $AC = BD$ ), and perpendicular ( $AC \perp BD$ )	$S = a^2$

UndergraduateTwo key theorems and their proofs

Theorem: Interior angle sum of a quadrilateral and an n-gon

In any convex quadrilateral $ABCD$ , the four interior angles satisfy $\angle A + \angle B + \angle C + \angle D = 360^\circ$ . More generally, in any convex polygon with $n$ sides ( $n \ge 3$ ), the sum of the interior angles is $(n - 2) \cdot 180^\circ$ .

Why is it true?

You do not need a new geometric axiom to measure polygons with four or more sides: slicing the figure along diagonals from a single vertex reduces every polygon to a collection of triangles whose angle sums are already known to be $180^\circ$ each.

Proof

Step 1 (split the quadrilateral along a diagonal). In convex quadrilateral $ABCD$ , draw the diagonal $AC$ . Because the figure is convex, this segment lies entirely inside the quadrilateral and partitions it into two triangles $\triangle ABC$ and $\triangle ACD$ .

Step 2 (apply the triangle angle sum to each piece). In $\triangle ABC$ we have $\angle BAC + \angle B + \angle BCA = 180^\circ$ , and in $\triangle ACD$ we have $\angle CAD + \angle D + \angle DCA = 180^\circ$ .

Step 3 (add the two equations and generalize). Adding both equations together gives $(\angle BAC + \angle CAD) + \angle B + (\angle BCA + \angle DCA) + \angle D = 180^\circ + 180^\circ = 360^\circ$ . Since the adjacent angles at the two ends of the diagonal recombine into the full vertex angles $\angle BAC + \angle CAD = \angle A$ and $\angle BCA + \angle DCA = \angle C$ , this simplifies directly to $\angle A + \angle B + \angle C + \angle D = 360^\circ$ . For a convex polygon with $n$ vertices, drawing all $n - 3$ diagonals from one vertex cuts the interior into $n - 2$ triangles whose angles add up to $(n - 2) \cdot 180^\circ$ .

Theorem: Diagonal characterization of parallelograms

A convex quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ intersect at $O$ is a parallelogram ( $AB \parallel CD$ and $AD \parallel BC$ ) if and only if its diagonals bisect each other ( $OA = OC$ and $OB = OD$ ).

Why is it true?

This equivalence turns a statement about parallel directions (which requires measuring angles or slopes) into a statement about midpoints (which only requires checking equal lengths along the two diagonals).

Proof

Step 1 (parallelogram implies bisecting diagonals). Assume $ABCD$ is a parallelogram. Because opposite sides are parallel and equal, we have $AB = CD$ , and the alternate interior angles cut by the two diagonals satisfy $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$ . By the Angle-Side-Angle criterion, $\triangle AOB \cong \triangle COD$ , so corresponding sides give $OA = OC$ and $OB = OD$ .

Step 2 (bisecting diagonals imply congruent opposite triangles). Conversely, assume $OA = OC$ and $OB = OD$ . Because vertical angles at the intersection are equal ( $\angle AOB = \angle COD$ ), the Side-Angle-Side criterion yields $\triangle AOB \cong \triangle COD$ , which gives $\angle OAB = \angle OCD$ . By the exact same Side-Angle-Side argument on the other pair of vertical angles ( $\angle AOD = \angle COB$ ), we obtain $\triangle AOD \cong \triangle COB$ , which gives $\angle OAD = \angle OCB$ .

Step 3 (equal alternate interior angles force parallel sides). The equalities $\angle OAB = \angle OCD$ and $\angle OAD = \angle OCB$ state that the alternate interior angles formed by the diagonal transversal with both pairs of opposite sides are equal, forcing $AB \parallel CD$ and $AD \parallel BC$ . Thus $ABCD$ is a parallelogram.

UndergraduateReal-World Applications and Worked Examples

The quadrilateral hierarchy and polygon angle formulas are used daily in carpentry, engineering, and architecture. When carpenters build a rectangular window frame or pour a concrete slab with opposite sides equal (a parallelogram), they check whether the four corners are true $90^\circ$ right angles simply by pulling a tape measure across the two diagonals to verify $AC = BD$ — a parallelogram has equal diagonals if and only if it is a rectangle! Mechanical pantographs and scissor lifts use hinged parallelograms ( $AB = CD$ , $AD = BC$ ) so that a platform stays perfectly level as it rises. In architecture and materials science, the regular polygon interior angle formula $\dfrac{(n - 2) \cdot 180^\circ}{n}$ explains why only equilateral triangles ( $n = 3$ ), squares ( $n = 4$ ), and regular hexagons ( $n = 6$ ) can tile a flat floor edge-to-edge using a single regular shape, since their interior angles $60^\circ$ , $90^\circ$ , and $120^\circ$ are exact divisors of a full $360^\circ$ turn around each vertex.

Example: Squaring a rectangular frame with its diagonals

A carpenter builds a wooden door frame $ABCD$ with opposite sides $AB = CD = 80$ cm and $BC = AD = 150$ cm. What exact diagonal length $AC = BD$ must the tape measure show so that all four corners are true $90^\circ$ right angles?

Solution

Step 1: confirm the frame is a parallelogram. Because both pairs of opposite sides are equal ( $AB = CD = 80$ cm and $BC = AD = 150$ cm), $ABCD$ is already a parallelogram; it becomes a rectangle with $90^\circ$ corners if and only if its two diagonals have equal length $AC = BD$ .

Step 2: apply the Pythagorean theorem in right triangle $\triangle ABC$ . When $\angle B = 90^\circ$ , the diagonal satisfies $AC^2 = AB^2 + BC^2 = 80^2 + 150^2 = 6400 + 22500 = 28900$ .

Step 3: take the square root. Since $\sqrt{28900} = 170$ , the carpenter adjusts the frame until both diagonals measure $AC = BD = 170$ cm.

Example: Interior angles of a hexagonal honeycomb tile

A floor is tiled with regular hexagons ( $n = 6$ ). Find the sum of the interior angles of one hexagon, the measure of each interior angle, and how many hexagons meet at each corner vertex.

Solution

Step 1: compute the interior angle sum. Substituting $n = 6$ into $S_n = (n - 2) \cdot 180^\circ$ gives $S_6 = (6 - 2) \cdot 180^\circ = 4 \cdot 180^\circ = 720^\circ$ .

Step 2: find each interior angle of the regular hexagon. Because all $6$ interior angles of a regular hexagon are equal, each angle measures $720^\circ / 6 = 120^\circ$ .

Step 3: count how many tiles meet at one vertex. A full turn around a point is $360^\circ$ , so $360^\circ / 120^\circ = 3$ regular hexagons fit seamlessly around every corner without gaps or overlaps.

What is the sum of the interior angles of a convex octagon ( $n = 8$ sides)?

A rhombus-shaped kite has perpendicular diagonals of lengths $d_1 = 30$ cm and $d_2 = 40$ cm. What is the area $S$ of the kite?

In quadrilateral $ABCD$ , the diagonals $AC$ and $BD$ bisect each other ( $OA = OC$ , $OB = OD$ ), are equal in length ( $AC = BD$ ), and are perpendicular ( $AC \perp BD$ ). What is the most specific name for $ABCD$ ?

Which of the following properties holds for every parallelogram $ABCD$ without requiring it to be a rectangle or a rhombus?

References

Euclid (trans. Thomas L. Heath) (1956). Euclid's Elements (Books I–XIII)
H. S. M. Coxeter, S. L. Greitzer (1967). Geometry Revisited