Properties of Quadrilaterals
Each property of a special quadrilateral can be quoted as a reason in your working. In NSW Year 8 maths (NESA outcome MA4‑GEO‑C‑01) the focus is on the diagonals: they bisect each other in a parallelogram, are equal in a rectangle, and cross at right angles and cut the corner angles in half in a rhombus.
You will decide whether statements about quadrilaterals are true or false, halve or double diagonal lengths, and find angles inside the triangles the diagonals make. When two equal lengths are written as expressions, such as \(3y - 2\) and \(y + 10\), you will solve an equation. You will also name the most exact shape you can be sure of, and work with rhombus floor tiles and a garden trellis. It is part of Year 8 Geometry.
Theory
A property of a special quadrilateral is a reason. Naming it lets you write an equation: in a rectangle the diagonals are equal and bisect each other, so if \(KM = 34\) cm, then \(LN = 34\) cm and each half-diagonal is \(17\) cm. This Year 8 maths page shows how.
Kite: one axis of symmetry; the diagonals meet at right angles, and the axis cuts the other diagonal in half.
Parallelogram: no axis of symmetry; opposite angles are equal; the diagonals bisect each other, cutting it into four triangles.
Rectangle: two axes of symmetry; all the parallelogram properties; the diagonals are equal, so all four half-diagonals are equal and each triangle they make is isosceles.
Rhombus: two axes of symmetry; all the parallelogram properties; the diagonals meet at right angles and bisect the angles they pass through.
Square: four axes of symmetry; all the rhombus properties; the diagonals are equal. A diagonal splits each right angle into two \(45^\circ\) angles.
Two equal things make an equation, with pronumerals on either side. Name the triangle you are working in before using its angle sum.
Hedge the names. A square is both a rectangle and a rhombus, and every rhombus is a parallelogram, but a rhombus is not always a square, and a kite is not always a parallelogram.
Reasons to write (for ABCD with diagonals meeting at O):
| Reason | What it lets you write |
|---|---|
| the diagonals of a parallelogram bisect each other | \(AO = OC\) and \(BO = OD\) |
| the diagonals of a rectangle are equal | \(AC = BD\) |
| the diagonals of a rhombus meet at right angles | \(\angle AOB = 90^\circ\) |
| the diagonals of a rhombus bisect its angles | \(\angle DAC = \angle CAB\) |
| opposite angles of a parallelogram | \(\angle DAB = \angle BCD\) |
| co-interior angles, \(AB \parallel DC\) | \(\angle DAB + \angle ADC = 180^\circ\) |
Axes of symmetry:
| Shape | Axes of symmetry |
|---|---|
| kite | 1 |
| parallelogram | 0 |
| rectangle | 2 |
| rhombus | 2 |
| square | 4 |
How to use a property
- Name the shape and the property you need.
- Write the equation it gives, with the property as the reason.
- Solve, working in one named triangle at a time.
(i) \(LN\)
(ii) \(VL\)
| \(LN\) | \(=\) | \(KM = 34\text{ cm}\) |
| \(VL\) | \(=\) | \(34 \div 2 = 17\text{ cm}\) |
(i) \(34\) cm (the diagonals of a rectangle are equal)
(ii) \(17\) cm (the diagonals bisect each other)
(i) \(\angle XYZ\)
(ii) \(\angle WXV\)
(iii) \(\angle XWV\)
| \(\angle XYZ\) | \(=\) | \(180 - 76 = 104^\circ\) |
| \(\angle WXV\) | \(=\) | \(76 \div 2 = 38^\circ\) |
| \(\angle XWV\) | \(=\) | \(180 - 38 - 90 = 52^\circ\) |
(i) \(104^\circ\) (co-interior angles, \(WX \parallel ZY\))
(ii) \(38^\circ\) (the diagonals of a rhombus bisect its angles)
(iii) \(52^\circ\): \(\angle WVX = 90^\circ\) (the diagonals of a rhombus meet at right angles), then the angle sum of \(\triangle WXV\).
(i) \(p\)
(ii) the perimeter
| \(3p + 4\) | \(=\) | \(5p - 6\) |
| \(10\) | \(=\) | \(2p\) |
| \(p\) | \(=\) | \(5\) |
| \(2 \times 19 + 2 \times 9\) | \(=\) | \(56\) |
(i) \(p = 5\), so \(AB = AD = 19\) cm.
(ii) The perimeter is \(56\) cm.
| \(\angle QPR\) | \(=\) | \(90 \div 2 = 45^\circ\) |
| \(\angle PRQ\) | \(=\) | \(180 - 90 - 45 = 45^\circ\) |
Both are \(45^\circ\): the diagonal bisects the right angle at \(P\), and \(\triangle PQR\) is isosceles and right-angled.
(i) Name it as exactly as you can.
(ii) Is it a rhombus?
(iii) Is it a trapezium?
(i) Four equal sides make a rhombus, so \(EF \parallel HG\) and \(\angle EHG = 180 - 90 = 90^\circ\) (co-interior angles). The opposite angles are also \(90^\circ\), so it is a square.
(ii) Yes: a square has every rhombus property.
(iii) Yes: two pairs of parallel sides is at least one pair.
Common pitfalls
Frequently asked questions
Do the diagonals of a kite bisect each other?
Only one is cut in half: the diagonal along the axis of symmetry cuts the other diagonal into two equal parts.
Are the diagonals of a rhombus equal?
Not usually. They bisect each other at right angles; they are equal only in a square.
Which quadrilaterals have diagonals that meet at right angles?
Kites, rhombuses and squares.
How many axes of symmetry does a rhombus have?
Two: each diagonal is an axis of symmetry.
Is every rhombus a parallelogram?
Yes. Its opposite sides are parallel, so it has every parallelogram property.