When a question says "$ABCD$ is a rhombus", it is handing you a whole list of facts for free — equal sides, parallel sides, bisecting diagonals — without stating any of them. Recognising the shape is often the entire question.
• Its sides — which are equal, which are parallel?
• Its angles — which are equal, which are right angles?
• Its diagonals — do they bisect each other, are they equal, do they meet at $90^\circ$, do they bisect the angles?
| Triangle | Sides | Angles | Symmetry |
|---|---|---|---|
| Equilateral | All three equal | All $60^\circ$ | 3 lines, order 3 |
| Isosceles | Two equal | Two equal base angles | 1 line, order 1 |
| Scalene | All different | All different | 0 lines, order 1 |
| Right-angled | Longest side is the hypotenuse | One angle is $90^\circ$ | Depends (isosceles version has 1 line) |
In triangle $PQR$, $\angle P = 50^\circ$ and $\angle Q = 65^\circ$. Show that the triangle is isosceles and state which two sides are equal.
$PR = PQ$
| Shape | Sides | Angles | Diagonals |
|---|---|---|---|
| Square | All $4$ equal; opposite sides parallel | All $90^\circ$ | Equal, bisect each other at $90^\circ$, bisect the angles |
| Rectangle | Opposite sides equal and parallel | All $90^\circ$ | Equal and bisect each other (not at $90^\circ$) |
| Parallelogram | Opposite sides equal and parallel | Opposite angles equal | Bisect each other (not equal, not at $90^\circ$) |
| Rhombus | All $4$ equal; opposite sides parallel | Opposite angles equal | Bisect each other at $90^\circ$, bisect the angles (not equal) |
| Trapezium | Exactly one pair of parallel sides | Co-interior pairs add to $180^\circ$ | No special property |
| Isosceles trapezium | One parallel pair; the other two equal | Two pairs of equal angles | Equal in length |
| Kite | Two pairs of adjacent equal sides | One pair of equal opposite angles | Cross at $90^\circ$; the long one bisects the short one |
Quadrilaterals form a hierarchy. A shape lower down inherits all the properties of everything above it.
• Every square is a rectangle, a rhombus, a parallelogram, a kite and a trapezium.
• Every rectangle and every rhombus is a parallelogram.
• Every parallelogram is a trapezium (it has a pair of parallel sides — indeed two pairs).
• The reverse is never automatically true: a rectangle need not be a square.
A quadrilateral has diagonals that bisect each other at right angles but are not equal in length. Name the shape.
It is a rhombus.
- Name the shape and write down which properties apply.
- Mark every equal side and equal angle on the diagram.
- Look for isosceles triangles created by the diagonals.
- Use angle sums ($180^\circ$ for a triangle, $360^\circ$ for a quadrilateral) to finish.
- Give a reason at each step.
In parallelogram $ABCD$, angle $A = 118^\circ$. Find angles $B$, $C$ and $D$.
Check: $118 + 62 + 118 + 62 = 360^\circ$ ✓
In rhombus $PQRS$ the diagonals meet at $M$. Angle $PQR = 76^\circ$. Find angle $QPM$.
Check: $\angle SPQ = 2 \times 52 = 104^\circ$, and $76 + 104 = 180^\circ$ as co-interior angles ✓
Kite $ABCD$ has $AB = AD$ and $CB = CD$. Angle $B = 105^\circ$ and angle $A = 62^\circ$. Find angles $C$ and $D$.
Isosceles rule
Equal sides sit opposite equal angles — and vice versa.
Quadrilateral sum
All four angles add to $360^\circ$.
Parallelogram
Opposite sides parallel and equal; opposite angles equal; diagonals bisect each other.
Rhombus
Parallelogram + all sides equal. Diagonals cross at $90^\circ$ and bisect the angles.
Rectangle
Parallelogram + all angles $90^\circ$. Diagonals are equal.
Square
Rectangle + rhombus. It has every property of both.
Kite
Two adjacent equal pairs; one equal angle pair; diagonals meet at $90^\circ$.
Trapezium
One pair of parallel sides, so co-interior angles add to $180^\circ$.
Family tree
Going down inherits properties; going up does not.
In parallelogram $WXYZ$, angle $W = 73^\circ$. Find the other three angles, giving reasons.
▶ Show solution
$\angle Y = 73^\circ$ — opposite angles of a parallelogram are equal.
$\angle X = 180 - 73 = 107^\circ$ — co-interior angles between parallel sides.
$\angle Z = 107^\circ$ — opposite angles are equal.
Check: $73 + 107 + 73 + 107 = 360^\circ$ ✓
A quadrilateral has four equal sides and four equal angles. Name it and give the size of each angle.
▶ Show solution
Four equal sides and four equal angles means it is a square.
$360 \div 4 = 90^\circ$ each.
Is the statement "every parallelogram is a rectangle" true or false? Explain.
▶ Show solution
False.
A rectangle needs all four angles to be $90^\circ$. A general parallelogram has two acute and two obtuse angles.
The correct statement is the reverse: every rectangle is a parallelogram.
In an isosceles triangle $ABC$, $AB = AC$ and angle $B = 4x$, angle $A = x + 20$. Find $x$.
▶ Show solution
$AB = AC$, so the base angles $B$ and $C$ are equal: $\angle C = 4x$.
$4x + 4x + (x + 20) = 180$
$9x + 20 = 180 \Rightarrow 9x = 160$
$x = \dfrac{160}{9} = 17.8$ (1 d.p.)
Angles: $B = C = 71.1^\circ$, $A = 37.8^\circ$. Check: $71.1 + 71.1 + 37.8 = 180$ ✓
Name every quadrilateral whose diagonals are equal in length.
▶ Show solution
Rectangle, square and isosceles trapezium.
(The square counts because it is a special rectangle.)
In trapezium $ABCD$, $AB \parallel DC$. Angle $A = 115^\circ$ and angle $C = 68^\circ$. Find angles $B$ and $D$.
▶ Show solution
$AB \parallel DC$, so $A$ and $D$ are co-interior: $\angle D = 180 - 115 = 65^\circ$.
Similarly $B$ and $C$ are co-interior: $\angle B = 180 - 68 = 112^\circ$.
Check: $115 + 112 + 68 + 65 = 360^\circ$ ✓
Kite $KLMN$ has $KL = KN$ and $ML = MN$. Angle $K = 40^\circ$ and angle $M = 100^\circ$. Find angles $L$ and $N$.
▶ Show solution
In a kite, the two angles not between a pair of equal sides are equal. Here $L$ and $N$ are that pair.
$L = N$, and all four angles add to $360^\circ$:
$40 + 100 + L + N = 360 \Rightarrow 2L = 220$
$L = N = 110^\circ$
The diagonals of quadrilateral $ABCD$ bisect each other but are not equal and do not meet at right angles. Name the shape and justify your answer.
▶ Show solution
Diagonals bisecting each other is exactly the defining property of a parallelogram.
Not equal rules out a rectangle (and a square).
Not at right angles rules out a rhombus (and a square).
So it is a general parallelogram.
Rhombus $ABCD$ has $\angle ABC = 128^\circ$. The diagonals meet at $O$.
(a) Find $\angle BAD$. (b) Find $\angle OBC$. (c) Find $\angle BOC$.
▶ Show solution
(a) $AB \parallel DC$, so $\angle ABC$ and $\angle BAD$ are co-interior: $\angle BAD = 180 - 128 = 52^\circ$.
(b) The diagonal $BD$ bisects $\angle ABC$: $\angle OBC = 128 \div 2 = 64^\circ$.
(c) The diagonals of a rhombus cross at right angles, so $\angle BOC = 90^\circ$.
$ABCD$ is a parallelogram. $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $DC$.
(a) Explain why $AP = QC$. (b) Explain why $APCQ$ is a parallelogram. (c) If $\angle DAB = 64^\circ$, find $\angle APC$.
▶ Show solution
(a) In a parallelogram opposite sides are equal, so $AB = DC$. Halving both gives $AP = \tfrac{1}{2}AB$ and $QC = \tfrac{1}{2}DC$, so $AP = QC$.
(b) $AP$ and $QC$ lie on $AB$ and $DC$, which are parallel — so $AP \parallel QC$. They are also equal in length by part (a). A quadrilateral with one pair of sides both equal and parallel is a parallelogram.
(c) $\angle APC$ and $\angle DAB$ are co-interior angles between the parallel lines $AD$ and $PC$… more directly, in parallelogram $APCQ$ the angle at $A$ is $\angle PAQ$. Using the straight line $AB$: $\angle APC$ is co-interior with $\angle PAQ = 64^\circ$.
$\angle APC = 180 - 64 = 116^\circ$