Properties of Special Quadrilaterals

The angles of every quadrilateral add up to 360°. Learn the sides, angles, diagonals and symmetry of the parallelogram, rectangle, rhombus, square, trapezium and kite, and use them to find unknown angles and lengths.

Before you start

The angles of a triangle add up to 180° (∠ sum of △), and the base angles of an isosceles triangle are equal (base ∠s of isos. △). Conversely, two equal angles in a triangle have equal sides opposite them (isos. △) (the lessons Triangles and Their Angles, and Isosceles and Equilateral Triangles).
When a line crosses two parallel lines $AB$ and $DC$ , alternate angles are equal (alt. ∠s, AB // DC) and interior angles add up to 180° (int. ∠s, AB // DC). If $AB$ is parallel to $DC$ , we write $AB \parallel DC$ ; in a reason it is written AB // DC (Lesson 11.4).
A line of symmetry folds a figure exactly onto itself. The order of rotational symmetry is the number of times a figure fits onto itself in one complete turn; order 1 means it has no rotational symmetry (the lesson Line and Rotational Symmetry).

Quadrilaterals and Their Angle Sum

A quadrilateral is a closed plane figure with four straight sides. It has four vertices and four angles, and it is named by its vertices in order round it: the quadrilateral $ABCD$ has sides $AB$ , $BC$ , $CD$ and $DA$ , and angles $\angle DAB$ , $\angle ABC$ , $\angle BCD$ and $\angle CDA$ .

A diagonal joins two vertices that are not next to each other. $ABCD$ has two diagonals, $AC$ and $BD$ . Every quadrilateral in this lesson has each of its angles less than 180°.

The angle sum of a quadrilateral

The angles of a quadrilateral add up to 360°.

$\angle DAB + \angle ABC + \angle BCD + \angle CDA = 360^{\circ}$

The reason is written (∠ sum of quad.), short for "the angle sum of a quadrilateral is 360°".

Why the angles add up to 360°

The diagonal $AC$ cuts $ABCD$ into two triangles, $\triangle ABC$ and $\triangle ACD$ .

The angles of each triangle add up to 180° (∠ sum of △), so the six angles of the two triangles add up to $180^{\circ} + 180^{\circ} = 360^{\circ}$ .

Those six angles make up exactly the four angles of $ABCD$ : at $B$ and at $D$ the angle of one triangle is the whole angle of the quadrilateral, and at $A$ and at $C$ the diagonal splits the angle into two parts, one in each triangle.

So the four angles of a quadrilateral add up to 360°, whatever its shape.

Example 1: One unknown angle

In the quadrilateral $PQRS$ , $\angle SPQ = 85^{\circ}$ , $\angle PQR = 110^{\circ}$ and $\angle QRS = 72^{\circ}$ . Find $x$ .

Solution

$x + 85 + 110 + 72 = 360$ (∠ sum of quad.)

Add the numbers: $x + 267 = 360$

Subtract 267 from both sides: $x = 93$

Check: $93^{\circ} + 85^{\circ} + 110^{\circ} + 72^{\circ} = 360^{\circ}$ .

$x = 93$ , so $\angle RSP = 93^{\circ}$ .

Example 2: Angles in terms of x

The angles of the quadrilateral $ABCD$ are $x^{\circ}$ , $(x + 20)^{\circ}$ , $2x^{\circ}$ and $(x + 40)^{\circ}$ , as marked. Find $x$ and each angle of $ABCD$ .

Solution

$x + (x + 20) + 2x + (x + 40) = 360$ (∠ sum of quad.)

Collect like terms: $5x + 60 = 360$

Subtract 60 from both sides: $5x = 300$

Divide both sides by 5: $x = 60$

So $\angle DAB = 60^{\circ}$ , $\angle ABC = (60 + 20)^{\circ} = 80^{\circ}$ , $\angle BCD = 2 \times 60^{\circ} = 120^{\circ}$ and $\angle CDA = (60 + 40)^{\circ} = 100^{\circ}$ .

Check: $60^{\circ} + 80^{\circ} + 120^{\circ} + 100^{\circ} = 360^{\circ}$ .

$x = 60$ ; the angles are $\angle DAB = 60^{\circ}$ , $\angle ABC = 80^{\circ}$ , $\angle BCD = 120^{\circ}$ and $\angle CDA = 100^{\circ}$ .

The Parallelogram

Some quadrilaterals have parallel sides, equal sides or right angles, and have special names. In a drawing, sides marked with the same number of arrows are parallel, sides marked with the same number of ticks are equal, angles marked with the same arcs are equal, and a small square marks a right angle.

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. In the figure, $AB \parallel DC$ (one arrow) and $AD \parallel BC$ (two arrows).

Properties of a parallelogram

Both pairs of opposite sides are parallel: $AB \parallel DC$ and $AD \parallel BC$ .

Opposite sides are equal: $AB = DC$ and $AD = BC$ . The reason is written (opp. sides of //gram), short for "opposite sides of a parallelogram are equal"; //gram is short for parallelogram.

Opposite angles are equal: $\angle DAB = \angle BCD$ and $\angle ABC = \angle CDA$ , written (opp. ∠s of //gram), short for "opposite angles of a parallelogram are equal".

Angles next to each other add up to 180°: for example, $\angle DAB + \angle ABC = 180^{\circ}$ (int. ∠s, AD // BC).

The diagonals bisect each other, written (diags. of //gram bisect each other).

Why the opposite angles are equal

$\angle DAB + \angle ABC = 180^{\circ}$ (int. ∠s, AD // BC), because $AB$ crosses the parallel sides $AD$ and $BC$ .

$\angle BCD + \angle ABC = 180^{\circ}$ (int. ∠s, AB // DC), because $BC$ crosses the parallel sides $AB$ and $DC$ .

Both $\angle DAB$ and $\angle BCD$ are 180° take away $\angle ABC$ , so $\angle DAB = \angle BCD$ . In the same way, $\angle ABC = \angle CDA$ .

"The diagonals bisect each other" means each diagonal cuts the other into two equal parts. If the diagonals $AC$ and $BD$ meet at $O$ , then $AO = OC$ and $BO = OD$ . The two diagonals are not, in general, equal to each other.

Symmetry. A parallelogram that is not a rectangle or a rhombus has no line of symmetry: folded along a diagonal, or along a line through the midpoints of two opposite sides, its two halves do not match. A half turn about $O$ , the point where the diagonals meet, fits it onto itself, so it has rotational symmetry of order 2.

Example 3: The angles of a parallelogram

$ABCD$ is a parallelogram with $\angle DAB = 68^{\circ}$ . Find $\angle ABC$ , $\angle BCD$ and $\angle CDA$ .

Solution

$\angle ABC = 180^{\circ} - 68^{\circ} = 112^{\circ}$ (int. ∠s, AD // BC)

$\angle BCD = \angle DAB = 68^{\circ}$ (opp. ∠s of //gram)

$\angle CDA = \angle ABC = 112^{\circ}$ (opp. ∠s of //gram)

Check: $68^{\circ} + 112^{\circ} + 68^{\circ} + 112^{\circ} = 360^{\circ}$ (∠ sum of quad.).

$\angle ABC = 112^{\circ}$ , $\angle BCD = 68^{\circ}$ and $\angle CDA = 112^{\circ}$ .

Example 4: Sides and diagonals

$ABCD$ is a parallelogram with $AB = (2x + 3)$ cm, $DC = 15$ cm and $AD = (x + 4)$ cm. Its diagonals $AC$ and $BD$ meet at $O$ , $AO = 5.5$ cm and $BD = 23$ cm.

(a) Find $x$ . (b) Find $AD$ . (c) Find $AC$ and $OB$ .

Solution

(a) $AB = DC$ (opp. sides of //gram), so

$2x + 3 = 15$

Subtract 3 from both sides: $2x = 12$

Divide both sides by 2: $x = 6$

Check: $AB = 2 \times 6 + 3 = 15$ cm, the same as $DC$ .

(b) $AD = (6 + 4)$ cm $= 10$ cm

$AC = 2 \times AO = 2 \times 5.5$ cm $= 11$ cm

$OB = \frac{1}{2} \times BD = \frac{1}{2} \times 23$ cm $= 11.5$ cm

$x = 6$ , $AD = 10$ cm, $AC = 11$ cm and $OB = 11.5$ cm. (The diagonals, 11 cm and 23 cm, are not equal.)

The Rectangle

A rectangle is a parallelogram with four right angles. Because it is a parallelogram, it has every property of a parallelogram: its opposite sides are parallel and equal, and its diagonals bisect each other.

It has one more property: its diagonals are equal. Since they are equal and bisect each other, the four halves are all equal: $OA = OB = OC = OD$ , where $O$ is the point where they meet.

Properties of a rectangle

A rectangle is a parallelogram, so it has every property of a parallelogram. In addition:

All four angles are right angles.

The diagonals are equal and bisect each other: $AC = BD$ , and $OA = OB = OC = OD$ . The reason is written (diags. of rectangle are equal and bisect each other).

Symmetry. A rectangle that is not a square has 2 lines of symmetry: the lines through the midpoints of opposite sides. A half turn about $O$ fits it onto itself, so it has rotational symmetry of order 2.

Why a diagonal of a rectangle that is not a square is not a line of symmetry

Fold the rectangle $ABCD$ along its diagonal $AC$ . The triangle $ABC$ turns over onto the shaded triangle $AB'C$ . The corner $B$ lands at $B'$ , outside the rectangle, not on $D$ , because the side $AB$ is longer than $AD$ . The two halves do not match, so $AC$ is not a line of symmetry, and in the same way neither is $BD$ .

Example 5: The diagonals make isosceles triangles

The diagonals of the rectangle $ABCD$ meet at $O$ , and $\angle OAB = 32^{\circ}$ . Find $\angle OBA$ , $\angle AOB$ , $\angle BOC$ and $\angle OBC$ .

Solution

$OA = OB$ (diags. of rectangle are equal and bisect each other), so $\triangle OAB$ is isosceles.

$\angle OBA = \angle OAB = 32^{\circ}$ (base ∠s of isos. △)

$\angle AOB = 180^{\circ} - 32^{\circ} - 32^{\circ} = 116^{\circ}$ (∠ sum of △)

$AOC$ is a straight line, so $\angle BOC = 180^{\circ} - 116^{\circ} = 64^{\circ}$ (adj. ∠s on a str. line).

$\angle ABC = 90^{\circ}$ , so $\angle OBC = 90^{\circ} - 32^{\circ} = 58^{\circ}$ .

Check in $\triangle OBC$ , which is also isosceles ( $OB = OC$ ): $\angle OCB = 58^{\circ}$ , and $64^{\circ} + 58^{\circ} + 58^{\circ} = 180^{\circ}$ .

$\angle OBA = 32^{\circ}$ , $\angle AOB = 116^{\circ}$ , $\angle BOC = 64^{\circ}$ and $\angle OBC = 58^{\circ}$ .

The Rhombus

A rhombus is a parallelogram with four equal sides. Because it is a parallelogram, its opposite sides are parallel, its opposite angles are equal and its diagonals bisect each other. Its diagonals have two more properties.

Properties of a rhombus

A rhombus is a parallelogram, so it has every property of a parallelogram. In addition:

All four sides are equal.

The diagonals cut each other at right angles, written (diags. of rhombus are ⊥), short for "the diagonals of a rhombus are perpendicular".

Each diagonal bisects the angles at its two ends, written (diags. of rhombus bisect the ∠s). In the figure, $\angle BAC = \angle CAD$ and $\angle ABD = \angle DBC$ : each part marked $a$ is half of $\angle DAB$ , and each part marked $b$ is half of $\angle ABC$ .

The diagonals of a rhombus are not equal unless it is a square. Each diagonal cuts the rhombus into two isosceles triangles: $AC$ makes $\triangle ABC$ , with $AB = BC$ , and $\triangle ACD$ , with $AD = DC$ .

Symmetry. A rhombus that is not a square has 2 lines of symmetry: its two diagonals. Fold it along $AC$ and $B$ lands on $D$ , because $AB = AD$ and $AC$ bisects $\angle DAB$ . It has rotational symmetry of order 2.

Example 6: A diagonal of a rhombus

$ABCD$ is a rhombus with $\angle ABC = 124^{\circ}$ , and $AC$ is a diagonal. Find $\angle BAC$ , $\angle DAC$ and $\angle DAB$ .

Solution

$AB = BC$ (all sides of a rhombus are equal), so $\triangle ABC$ is isosceles with $\angle BAC = \angle BCA$ (base ∠s of isos. △).

$\angle BAC = \frac{1}{2} \times (180^{\circ} - 124^{\circ}) = \frac{1}{2} \times 56^{\circ} = 28^{\circ}$ (base ∠s of isos. △, ∠ sum of △)

$\angle DAC = \angle BAC = 28^{\circ}$ (diags. of rhombus bisect the ∠s)

$\angle DAB = 28^{\circ} + 28^{\circ} = 56^{\circ}$

Check: $\angle DAB + \angle ABC = 56^{\circ} + 124^{\circ} = 180^{\circ}$ (int. ∠s, AD // BC).

$\angle BAC = 28^{\circ}$ , $\angle DAC = 28^{\circ}$ and $\angle DAB = 56^{\circ}$ .

The Square

A square has four equal sides and four right angles. Each side meets the two sides beside it at 90°, and $90^{\circ} + 90^{\circ} = 180^{\circ}$ , so its opposite sides are parallel (int. ∠s supp.) (Lesson 11.4) and it is a parallelogram. With four right angles it is a rectangle, and with four equal sides it is a rhombus. It has every property of both.

Properties of a square

All four sides are equal, all four angles are right angles, and opposite sides are parallel.

The diagonals are equal, bisect each other and cut each other at right angles.

Each diagonal bisects the right angles at its two ends, making angles of 45°.

Symmetry. A square has 4 lines of symmetry: its two diagonals and the two lines through the midpoints of opposite sides. A quarter turn about $O$ fits it onto itself, so it fits 4 times in one complete turn: rotational symmetry of order 4.

The Trapezium and the Isosceles Trapezium

A trapezium is a quadrilateral with exactly one pair of parallel sides. In $ABCD$ below, $AB \parallel DC$ , and $AD$ and $BC$ are not parallel. So a parallelogram, which has two pairs of parallel sides, is not a trapezium.

Each non-parallel side crosses the two parallel sides, so the two angles on it are interior angles: they add up to 180°. In the figure, the two blue angles add up to $65^{\circ} + 115^{\circ} = 180^{\circ}$ , and the two green ones to $55^{\circ} + 125^{\circ} = 180^{\circ}$ .

Properties of a trapezium

Exactly one pair of opposite sides is parallel: here $AB \parallel DC$ .

The two angles on each non-parallel side add up to 180°: $\angle DAB + \angle ADC = 180^{\circ}$ and $\angle ABC + \angle BCD = 180^{\circ}$ (int. ∠s, AB // DC).

A trapezium need not have any equal sides or equal angles. A trapezium that is not isosceles has no line of symmetry and no rotational symmetry.

Example 7: The angles of a trapezium

$ABCD$ is a trapezium with $AB \parallel DC$ , $\angle DAB = 72^{\circ}$ and $\angle BCD = 115^{\circ}$ . Find $\angle ADC$ and $\angle ABC$ .

Solution

$AD$ crosses the parallel sides, so $\angle ADC = 180^{\circ} - 72^{\circ} = 108^{\circ}$ (int. ∠s, AB // DC).

$BC$ crosses them too, so $\angle ABC = 180^{\circ} - 115^{\circ} = 65^{\circ}$ (int. ∠s, AB // DC).

Check: $72^{\circ} + 65^{\circ} + 115^{\circ} + 108^{\circ} = 360^{\circ}$ (∠ sum of quad.).

$\angle ADC = 108^{\circ}$ and $\angle ABC = 65^{\circ}$ .

An isosceles trapezium is a trapezium whose two non-parallel sides are equal. Its two angles at the longer parallel side are called a pair of base angles, and so are its two angles at the shorter parallel side.

Properties of an isosceles trapezium

It is a trapezium, so it has every property of a trapezium. In addition:

The non-parallel sides are equal: $AD = BC$ .

Each pair of base angles is equal: $\angle DAB = \angle ABC$ at the longer parallel side, and $\angle ADC = \angle BCD$ at the shorter, written (base ∠s of isos. trapezium).

The diagonals are equal: $AC = BD$ , written (diags. of isos. trapezium are equal).

It has 1 line of symmetry, through the midpoints of the two parallel sides, and no rotational symmetry.

Example 8: An isosceles trapezium with x

$PQRS$ is an isosceles trapezium with $PQ \parallel SR$ and $PS = QR$ . $\angle SPQ = (3x + 5)^{\circ}$ and $\angle PQR = (x + 47)^{\circ}$ . Find $x$ , $\angle SPQ$ and $\angle QRS$ .

Solution

$\angle SPQ$ and $\angle PQR$ are the base angles at the parallel side $PQ$ , so they are equal (base ∠s of isos. trapezium):

$3x + 5 = x + 47$

Subtract $x$ from both sides: $2x + 5 = 47$

Subtract 5 from both sides: $2x = 42$

Divide both sides by 2: $x = 21$

$\angle SPQ = (3 \times 21 + 5)^{\circ} = 68^{\circ}$ ; check: $\angle PQR = (21 + 47)^{\circ} = 68^{\circ}$ .

$\angle QRS = 180^{\circ} - 68^{\circ} = 112^{\circ}$ (int. ∠s, PQ // SR)

Check: $\angle RSP = 112^{\circ}$ too (base ∠s of isos. trapezium), and $68^{\circ} + 68^{\circ} + 112^{\circ} + 112^{\circ} = 360^{\circ}$ .

$x = 21$ , $\angle SPQ = 68^{\circ}$ and $\angle QRS = 112^{\circ}$ .

The Kite

A kite is a quadrilateral with two pairs of equal adjacent sides. In the kite $ABCD$ below, $AB = AD$ and $CB = CD$ : one pair of equal sides meets at $A$ and the other at $C$ .

Properties of a kite

Two pairs of equal adjacent sides: $AB = AD$ and $CB = CD$ .

The diagonals are perpendicular: $AC$ and $BD$ meet at right angles.

The diagonal joining the vertices where the equal sides meet, $AC$ , is a line of symmetry. It bisects the other diagonal ( $BO = OD$ ) and bisects the angles at its two ends ( $\angle BAC = \angle CAD$ and $\angle BCA = \angle ACD$ ).

The two angles at the ends of the other diagonal are equal: $\angle ABC = \angle ADC$ .

In a solution, write the property you use in words, followed by (prop. of kite), short for "property of a kite".

In a kite that is not a rhombus, the other diagonal, $BD$ , is not a line of symmetry, and it does not bisect $AC$ or the angles at $B$ and $D$ .

Symmetry. A kite that is not a rhombus has 1 line of symmetry (the diagonal joining the vertices where the equal sides meet) and no rotational symmetry.

The line of symmetry is not always the longer diagonal. In the kite below, the equal sides again meet at $A$ and at $C$ , so $AC$ is the line of symmetry, although it is the shorter diagonal.

Example 9: The angles of a kite

$ABCD$ is a kite with $AB = AD$ and $CB = CD$ . $\angle DAB = 70^{\circ}$ and $\angle BCD = 46^{\circ}$ .

(a) Find $\angle ABC$ . (b) The diagonals $AC$ and $BD$ are now drawn (they are not shown above). Find $\angle BAC$ and $\angle ABD$ .

Solution

(a) $AC$ joins the vertices where the equal sides meet, so the angles at the ends of the other diagonal, $BD$ , are equal: $\angle ABC = \angle ADC$ (prop. of kite).

$2 \times \angle ABC = 360^{\circ} - 70^{\circ} - 46^{\circ} = 244^{\circ}$ (∠ sum of quad.)

$\angle ABC = 122^{\circ}$

(b) $AC$ joins the vertices where the equal sides meet, so it bisects $\angle DAB$ :

$\angle BAC = \frac{1}{2} \times 70^{\circ} = 35^{\circ}$ (prop. of kite)

$AB = AD$ , so $\triangle ABD$ is isosceles:

$\angle ABD = \frac{1}{2} \times (180^{\circ} - 70^{\circ}) = 55^{\circ}$ (base ∠s of isos. △, ∠ sum of △)

Check in $\triangle ABO$ : the diagonals are perpendicular (prop. of kite), and $35^{\circ} + 55^{\circ} + 90^{\circ} = 180^{\circ}$ .

(a) $\angle ABC = 122^{\circ}$ (b) $\angle BAC = 35^{\circ}$ and $\angle ABD = 55^{\circ}$ .

All the Properties in One Table

The table gathers the properties. It has two kinds of row. A row headed Every … gives the properties that every shape of that kind has, special cases included: every rectangle, squares among them, has equal diagonals. A row headed Symmetry gives the symmetry of the general shape only, because a special case can have more: a square has more lines of symmetry than a rectangle that is not a square.

Shape	Sides	Angles	Diagonals
Every parallelogram	Both pairs of opposite sides parallel; opposite sides equal	Opposite angles equal; angles next to each other add up to 180°	Bisect each other
Symmetry: A parallelogram that is not a rectangle or a rhombus	No line of symmetry; rotational symmetry of order 2
Every rectangle	Both pairs of opposite sides parallel; opposite sides equal	Four right angles	Equal; bisect each other
Symmetry: A rectangle that is not a square	2 lines of symmetry, through the midpoints of opposite sides (not the diagonals); order 2
Every rhombus	Four equal sides; both pairs of opposite sides parallel	Opposite angles equal; angles next to each other add up to 180°	Bisect each other at right angles; bisect the angles of the rhombus
Symmetry: A rhombus that is not a square	2 lines of symmetry (the diagonals); order 2
Every square	Four equal sides; both pairs of opposite sides parallel	Four right angles	Equal; bisect each other at right angles; bisect the angles (45°)
Symmetry: A square	4 lines of symmetry (the two diagonals and the two lines through the midpoints of opposite sides); order 4
Every trapezium	Exactly one pair of parallel sides	The two angles on each non-parallel side add up to 180°	No special property
Symmetry: A trapezium that is not isosceles	No line of symmetry; order 1
Every isosceles trapezium	Exactly one pair of parallel sides; the non-parallel sides equal	Each pair of base angles equal; the two angles on each non-parallel side add up to 180°	Equal
Symmetry: An isosceles trapezium	1 line of symmetry, through the midpoints of the parallel sides; order 1
Every kite	Two pairs of equal adjacent sides	A pair of opposite angles equal: the two that are not at the vertices where the equal sides meet	Perpendicular; the diagonal joining the vertices where the equal sides meet bisects the other diagonal and the angles at its ends
Symmetry: A kite that is not a rhombus	1 line of symmetry (the diagonal joining the vertices where the equal sides meet); order 1

Special cases

A special case has every property that every shape of the more general kind has (the rows headed Every …), and more. A square is a special rectangle and a special rhombus, so it has every property of both; a rectangle and a rhombus are special parallelograms; a rhombus is also a special kite (its four equal sides make two pairs of equal adjacent sides); an isosceles trapezium is a special trapezium.

The lesson Classifying Special Quadrilaterals arranges the shapes in a family tree and uses this table to classify them.

Example 10: The symmetry of floor tiles

Mdm Lim is choosing floor tiles. The tiles are (a) a rhombus that is not a square, (b) a rectangle that is not a square, (c) an isosceles trapezium and (d) a kite that is not a rhombus. For each tile, state the number of lines of symmetry, say where they are, and state the order of rotational symmetry.

Solution

(a) Fold the rhombus along either diagonal and the halves match: 2 lines of symmetry, the two diagonals. A half turn fits it onto itself: order 2.

(b) The lines through the midpoints of opposite sides: 2 lines of symmetry, not the diagonals. A half turn fits it onto itself: order 2.

(c) One line, through the midpoints of the two parallel sides: 1 line of symmetry. Only the complete turn fits it onto itself: order 1, no rotational symmetry.

(d) One line, the diagonal joining the vertices where the equal sides meet: 1 line of symmetry. Order 1, no rotational symmetry.

(a) 2 lines (the diagonals), order 2 (b) 2 lines (through the midpoints of opposite sides), order 2 (c) 1 line (through the midpoints of the parallel sides), order 1 (d) 1 line (the diagonal joining the vertices where the equal sides meet), order 1.

Common Mistakes

Taking the diagonals of a rectangle as lines of symmetry. Fold a rectangle that is not a square along a diagonal and the halves do not match. Its 2 lines of symmetry pass through the midpoints of opposite sides.
Making the diagonals of every parallelogram equal. The diagonals of a parallelogram bisect each other, but they are equal only in a rectangle (and so in a square). The diagonals of a rhombus that is not a square are not equal.
Calling a parallelogram a trapezium. A trapezium has exactly one pair of parallel sides; a parallelogram has two.
Adding the wrong angles of a trapezium. The angles that add up to 180° are the two on the same non-parallel side, between the parallel sides. Two opposite angles of a trapezium do not, in general, add up to 180°, and (int. ∠s) is never the reason for two opposite angles.
Using the wrong diagonal of a kite. In a kite that is not a rhombus, the line of symmetry joins the vertices where the equal sides meet, even when it is the shorter diagonal. In every kite, that diagonal bisects the other diagonal and the angles at its own ends; the equal angles are the two at the ends of the other diagonal.
Stopping at $x$ . When the angles are given in terms of $x$ , find $x$ , then work out each angle the question asks for, and check the angle sum.

Practice

The questions rise in difficulty. Give a reason for each step, then open the solution to check.

Practice 1: The angle sum

(a) Three angles of a quadrilateral are 96°, 124° and 58°. Find the fourth angle.

(b) In the quadrilateral $ABCD$ , $\angle DAB = 3x^{\circ}$ , $\angle ABC = (4x - 5)^{\circ}$ , $\angle BCD = (5x - 15)^{\circ}$ and $\angle CDA = (2x + 30)^{\circ}$ . Find $x$ and the largest angle of $ABCD$ .

ShowHide solution

(a) The fourth angle $= 360^{\circ} - 96^{\circ} - 124^{\circ} - 58^{\circ} = 82^{\circ}$ (∠ sum of quad.)

(b) $3x + (4x - 5) + (5x - 15) + (2x + 30) = 360$ (∠ sum of quad.)

Collect like terms: $14x + 10 = 360$

Subtract 10 from both sides: $14x = 350$

Divide both sides by 14: $x = 25$

The angles are $3 \times 25 = 75$ , $4 \times 25 - 5 = 95$ , $5 \times 25 - 15 = 110$ and $2 \times 25 + 30 = 80$ , in degrees.

Check: $75^{\circ} + 95^{\circ} + 110^{\circ} + 80^{\circ} = 360^{\circ}$ .

(a) 82° (b) $x = 25$ ; the largest angle is $\angle BCD = 110^{\circ}$ .

Practice 2: Angles of a parallelogram

$ABCD$ is a parallelogram with $\angle ABC = 57^{\circ}$ . Find $x$ and $y$ .

ShowHide solution

$\angle ABC$ and $\angle BCD$ are on the side $BC$ , which crosses the parallel sides $AB$ and $DC$ :

$x = 180 - 57 = 123$ (int. ∠s, AB // DC)

$\angle CDA$ is opposite $\angle ABC$ : $y = 57$ (opp. ∠s of //gram)

Check: $\angle DAB = 123^{\circ}$ (opp. ∠s of //gram), and $57^{\circ} + 123^{\circ} + 57^{\circ} + 123^{\circ} = 360^{\circ}$ .

$x = 123$ and $y = 57$ , so $\angle BCD = 123^{\circ}$ and $\angle CDA = 57^{\circ}$ .

Practice 3: True or false?

Say whether each statement is true or false, and give a reason.

(a) A rectangle that is not a square has 4 lines of symmetry: the two lines through the midpoints of opposite sides and its two diagonals.

(b) The diagonals of every parallelogram are equal.

(d) A kite that is not a rhombus has rotational symmetry of order 2.

(e) The diagonals of an isosceles trapezium are equal.

ShowHide solution

(a) False. Folded along a diagonal, the halves of a rectangle that is not a square do not match: as in the folded rectangle above, $B$ lands outside the rectangle. It has only 2 lines of symmetry, through the midpoints of opposite sides.

(b) False. The diagonals of a parallelogram bisect each other, but in Example 4 they were 11 cm and 23 cm. They are equal only in a rectangle.

(d) False. A half turn that fitted the kite onto itself would take each side onto the opposite side, $AB$ onto $CD$ . In a kite that is not a rhombus, $AB$ and $CD$ are not equal. Only the complete turn fits it onto itself: order 1, no rotational symmetry.

(e) True. Folding an isosceles trapezium along its line of symmetry carries the two ends of each parallel side onto each other, so each diagonal lands on the other: the diagonals of every isosceles trapezium are equal.

(a) False (b) False (c) True (d) False (e) True.

Practice 4: The diagonals of a rectangle

The diagonals of the rectangle $PQRS$ meet at $T$ . $\angle PTQ = 128^{\circ}$ and $PR = 13$ cm. Find $\angle TPQ$ , $\angle TQR$ and the length $QT$ .

ShowHide solution

$TP = TQ$ (diags. of rectangle are equal and bisect each other), so $\triangle TPQ$ is isosceles and $\angle TPQ = \angle TQP$ (base ∠s of isos. △).

$2 \times \angle TPQ = 180^{\circ} - 128^{\circ} = 52^{\circ}$ (∠ sum of △), so $\angle TPQ = 26^{\circ}$ .

$\angle TQP = 26^{\circ}$ and $\angle PQR = 90^{\circ}$ , so $\angle TQR = 90^{\circ} - 26^{\circ} = 64^{\circ}$ .

$QS = PR = 13$ cm, and $T$ is its midpoint, so $QT = 13 \div 2 = 6.5$ cm (diags. of rectangle are equal and bisect each other).

$\angle TPQ = 26^{\circ}$ , $\angle TQR = 64^{\circ}$ and $QT = 6.5$ cm.

Practice 5: A rhombus with x

$ABCD$ is a rhombus with $AB = (3x - 4)$ cm, $BC = (x + 10)$ cm and $\angle DAB = 50^{\circ}$ . The diagonal $BD$ is drawn.

(a) Find $x$ and the length of $AD$ . (b) Find $\angle ABD$ .

ShowHide solution

(a) $AB = BC$ (all sides of a rhombus are equal), so

$3x - 4 = x + 10$

Subtract $x$ from both sides: $2x - 4 = 10$

Add 4 to both sides: $2x = 14$

Divide both sides by 2: $x = 7$

$AB = 3 \times 7 - 4 = 17$ cm; check: $BC = 7 + 10 = 17$ cm. So $AD = 17$ cm.

(b) $AB = AD$ , so $\triangle ABD$ is isosceles:

$\angle ABD = \frac{1}{2} \times (180^{\circ} - 50^{\circ}) = 65^{\circ}$ (base ∠s of isos. △, ∠ sum of △)

Check: $\angle ABC = 180^{\circ} - 50^{\circ} = 130^{\circ}$ (int. ∠s, AD // BC), and $BD$ bisects it (diags. of rhombus bisect the ∠s): $130^{\circ} \div 2 = 65^{\circ}$ .

(a) $x = 7$ and $AD = 17$ cm (b) $\angle ABD = 65^{\circ}$ .

Practice 6: A trapezium with x

$ABCD$ is a trapezium with $AB \parallel DC$ . $\angle DAB = (2x + 10)^{\circ}$ , $\angle ADC = (x + 20)^{\circ}$ and $\angle ABC = 62^{\circ}$ . Find $x$ , $\angle ADC$ and $\angle BCD$ .

ShowHide solution

$\angle DAB$ and $\angle ADC$ are on the non-parallel side $AD$ :

$(2x + 10) + (x + 20) = 180$ (int. ∠s, AB // DC)

Collect like terms: $3x + 30 = 180$

Subtract 30 from both sides: $3x = 150$

Divide both sides by 3: $x = 50$

$\angle ADC = (50 + 20)^{\circ} = 70^{\circ}$ , and $\angle DAB = (2 \times 50 + 10)^{\circ} = 110^{\circ}$ .

$\angle BCD = 180^{\circ} - 62^{\circ} = 118^{\circ}$ (int. ∠s, AB // DC)

Check: $110^{\circ} + 62^{\circ} + 118^{\circ} + 70^{\circ} = 360^{\circ}$ (∠ sum of quad.).

$x = 50$ , $\angle ADC = 70^{\circ}$ and $\angle BCD = 118^{\circ}$ .

Practice 7: A kite with x

$PQRS$ is a kite with $PQ = PS$ and $RQ = RS$ . $\angle QPS = 2x^{\circ}$ , $\angle QRS = (x + 30)^{\circ}$ and $\angle PQR = 60^{\circ}$ . Find $\angle PSR$ , $x$ , $\angle QPS$ and $\angle QRS$ .

ShowHide solution

The equal sides meet at $P$ and at $R$ , so $PR$ joins the vertices where the equal sides meet, and $\angle PQR$ and $\angle PSR$ are the angles at the ends of the other diagonal, $QS$ :

$\angle PSR = \angle PQR = 60^{\circ}$ (prop. of kite)

$2x + 60 + (x + 30) + 60 = 360$ (∠ sum of quad.)

Collect like terms: $3x + 150 = 360$

Subtract 150 from both sides: $3x = 210$

Divide both sides by 3: $x = 70$

$\angle QPS = 2 \times 70^{\circ} = 140^{\circ}$ and $\angle QRS = (70 + 30)^{\circ} = 100^{\circ}$ .

Check: $140^{\circ} + 60^{\circ} + 100^{\circ} + 60^{\circ} = 360^{\circ}$ .

$\angle PSR = 60^{\circ}$ , $x = 70$ , $\angle QPS = 140^{\circ}$ and $\angle QRS = 100^{\circ}$ .

Practice 8: A garden trellis

The criss-cross wooden strips of a garden trellis form rhombuses. Ali looks at one rhombus $ABCD$ , with $\angle ABC = 136^{\circ}$ , and draws its diagonal $AC$ .

(a) Find $\angle BAD$ . (b) Find $\angle BAC$ . (c) Find $\angle ACD$ .

ShowHide solution

(a) $\angle BAD = 180^{\circ} - 136^{\circ} = 44^{\circ}$ (int. ∠s, AD // BC)

(b) $\angle BAC = \frac{1}{2} \times 44^{\circ} = 22^{\circ}$ (diags. of rhombus bisect the ∠s)

Check: $AB = BC$ , so in the isosceles $\triangle ABC$ , $\angle BAC = \frac{1}{2} \times (180^{\circ} - 136^{\circ}) = 22^{\circ}$ .

(a) $\angle BAD = 44^{\circ}$ (b) $\angle BAC = 22^{\circ}$ (c) $\angle ACD = 22^{\circ}$ .

Practice 9: A kite for flying

Wei Ming builds a kite $ABCD$ with $AB = AD$ and $CB = CD$ . Its two straight cross-pieces are the diagonals $AC$ and $BD$ , which meet at $O$ . $BD = 50$ cm, $\angle BAC = 40^{\circ}$ and $\angle BCA = 25^{\circ}$ .

(a) Find $OB$ . (b) Find $\angle AOB$ . (c) Find $\angle BAD$ . (d) Find $\angle ABC$ .

ShowHide solution

(a) $AC$ joins the vertices where the equal sides meet, so it bisects $BD$ (prop. of kite): $OB = 50 \div 2 = 25$ cm.

(b) The diagonals of a kite are perpendicular (prop. of kite): $\angle AOB = 90^{\circ}$ .

(d) In $\triangle ABO$ : $\angle ABO = 180^{\circ} - 90^{\circ} - 40^{\circ} = 50^{\circ}$ (∠ sum of △).

In $\triangle CBO$ : $\angle CBO = 180^{\circ} - 90^{\circ} - 25^{\circ} = 65^{\circ}$ (∠ sum of △).

$\angle ABC = 50^{\circ} + 65^{\circ} = 115^{\circ}$

Check: $\angle BCD = 2 \times 25^{\circ} = 50^{\circ}$ and $\angle ADC = \angle ABC = 115^{\circ}$ (prop. of kite); $80^{\circ} + 115^{\circ} + 50^{\circ} + 115^{\circ} = 360^{\circ}$ .

(a) $OB = 25$ cm (b) $\angle AOB = 90^{\circ}$ (c) $\angle BAD = 80^{\circ}$ (d) $\angle ABC = 115^{\circ}$ .

Practice 10: An isosceles trapezium and its diagonals

$ABCD$ is an isosceles trapezium with $AB \parallel DC$ and $AD = BC$ . $\angle ABC = 70^{\circ}$ , $\angle BAC = 40^{\circ}$ and the diagonal $BD$ (not drawn) is 9 cm long.

(a) Find $\angle ADC$ . (b) Find $\angle ACD$ and $\angle ACB$ . (c) Find the length of $AC$ . (d) Explain why $AB = AC$ , and write down the length of $AB$ .

ShowHide solution

(a) $\angle BCD = 180^{\circ} - 70^{\circ} = 110^{\circ}$ (int. ∠s, AB // DC)

$\angle ADC = \angle BCD = 110^{\circ}$ (base ∠s of isos. trapezium)

(b) $\angle ACD = \angle BAC = 40^{\circ}$ (alt. ∠s, AB // DC)

$\angle ACB = \angle BCD - \angle ACD = 110^{\circ} - 40^{\circ} = 70^{\circ}$

(d) In $\triangle ABC$ , $\angle ACB = \angle ABC = 70^{\circ}$ , so the sides opposite them are equal: $AB = AC$ (isos. △). So $AB = 9$ cm.

Check: in $\triangle ABC$ , $40^{\circ} + 70^{\circ} + 70^{\circ} = 180^{\circ}$ .

(a) $\angle ADC = 110^{\circ}$ (b) $\angle ACD = 40^{\circ}$ and $\angle ACB = 70^{\circ}$ (c) $AC = 9$ cm (d) $\angle ACB = \angle ABC$ , so $AB = AC = 9$ cm.

Practice 11: A stepladder

Mr Tan opens a stepladder on a level floor. Seen from the side, its two legs $AP$ and $AQ$ meet at the top $A$ and stand on the floor at $P$ and $Q$ , with $\angle APQ = \angle AQP = 72^{\circ}$ . A bar $BC$ joins the legs parallel to the floor, with $BP = CQ$ .

(a) Find $\angle ABC$ , giving a reason. (b) Find $\angle PAQ$ . (c) Name the special quadrilateral $PQCB$ , giving a reason. (d) Find $\angle PBC$ and $\angle QCB$ .

ShowHide solution

(a) $\angle ABC = \angle APQ = 72^{\circ}$ (corr. ∠s, BC // PQ)

(b) $\angle PAQ = 180^{\circ} - 72^{\circ} - 72^{\circ} = 36^{\circ}$ (∠ sum of △)

(c) $BC \parallel PQ$ , and $BP$ and $CQ$ are not parallel (they meet at $A$ when carried on), so $PQCB$ has exactly one pair of parallel sides: it is a trapezium. Its non-parallel sides are equal, $BP = CQ$ , so it is an isosceles trapezium.

(d) $\angle PBC = 180^{\circ} - 72^{\circ} = 108^{\circ}$ (int. ∠s, BC // PQ)

$\angle QCB = \angle PBC = 108^{\circ}$ (base ∠s of isos. trapezium)

Check: $\angle ABC + \angle PBC = 72^{\circ} + 108^{\circ} = 180^{\circ}$ (adj. ∠s on a str. line), and $72^{\circ} + 72^{\circ} + 108^{\circ} + 108^{\circ} = 360^{\circ}$ (∠ sum of quad.).

(a) $\angle ABC = 72^{\circ}$ (b) $\angle PAQ = 36^{\circ}$ (c) an isosceles trapezium (d) $\angle PBC = \angle QCB = 108^{\circ}$ .

Practice 12: A square and an equilateral triangle

$ABCD$ is a square. $E$ is a point inside it such that $\triangle ABE$ is equilateral. Find $\angle DAE$ , $\angle ADE$ , $\angle EDC$ and $\angle DEC$ .

ShowHide solution

$\angle BAE = 60^{\circ}$ (equilateral △), and $\angle DAB = 90^{\circ}$ , so $\angle DAE = 90^{\circ} - 60^{\circ} = 30^{\circ}$ .

$AD = AB$ (sides of the square) and $AB = AE$ (sides of the equilateral triangle), so $AD = AE$ and $\triangle ADE$ is isosceles.

$\angle ADE = \frac{1}{2} \times (180^{\circ} - 30^{\circ}) = 75^{\circ}$ (base ∠s of isos. △, ∠ sum of △)

$\angle EDC = 90^{\circ} - 75^{\circ} = 15^{\circ}$ , since $\angle ADC = 90^{\circ}$ .

In the same way at $B$ : $\angle CBE = 30^{\circ}$ , $BC = BE$ , $\angle BCE = 75^{\circ}$ and $\angle ECD = 15^{\circ}$ .

$\angle DEC = 180^{\circ} - 15^{\circ} - 15^{\circ} = 150^{\circ}$ (∠ sum of △)

Check at $E$ : $\angle AED = 75^{\circ}$ and $\angle BEC = 75^{\circ}$ (base ∠s of isos. △), and $60^{\circ} + 75^{\circ} + 150^{\circ} + 75^{\circ} = 360^{\circ}$ (∠s at a pt).

$\angle DAE = 30^{\circ}$ , $\angle ADE = 75^{\circ}$ , $\angle EDC = 15^{\circ}$ and $\angle DEC = 150^{\circ}$ .

Key Points

The angles of a quadrilateral add up to 360° (∠ sum of quad.): a diagonal cuts it into two triangles.
Parallelogram: opposite sides parallel and equal; opposite angles equal; angles next to each other add up to 180°; diagonals bisect each other. A parallelogram that is not a rectangle or a rhombus has no line of symmetry and rotational symmetry of order 2.
Rectangle: a parallelogram with four right angles; diagonals equal and bisecting each other. A rectangle that is not a square has 2 lines of symmetry, through the midpoints of opposite sides (not the diagonals), and order 2.
Rhombus: a parallelogram with four equal sides; diagonals bisect each other at right angles and bisect the angles. A rhombus that is not a square has 2 lines of symmetry (the diagonals) and order 2.
Square: four equal sides and four right angles; diagonals equal, perpendicular, bisecting each other and the angles (45°); 4 lines of symmetry and order 4.
Trapezium: exactly one pair of parallel sides; the two angles on each non-parallel side add up to 180°.
Isosceles trapezium: non-parallel sides equal; base angles equal in pairs; diagonals equal; 1 line of symmetry and no rotational symmetry.
Kite: two pairs of equal adjacent sides; diagonals perpendicular; the diagonal joining the vertices where the equal sides meet bisects the other diagonal and the angles at its ends; the two angles at the ends of the other diagonal are equal. A kite that is not a rhombus has 1 line of symmetry (that diagonal) and no rotational symmetry.
A special case has every property that every shape of the more general kind has, and more: a square is a special rectangle and a special rhombus; a rectangle and a rhombus are special parallelograms; a rhombus is a special kite; an isosceles trapezium is a special trapezium.