This note covers shapes that are the same size and shape (congruent), shapes that are enlargements of each other (similar), and the symmetry of shapes. Scale factors link similar shapes, and at Extended level they also link areas and volumes. These ideas appear in maps, models, scale drawings and design.
Key ideas
Congruence. Two shapes are congruent if one can be moved onto the other exactly, using a slide, a turn or a reflection. Matching sides are equal and matching angles are equal. For triangles, you only need to show one of these four sets of facts:
- SSS: all three sides are equal.
- SAS: two sides and the angle between them are equal.
- ASA (or AAS): two angles and one side are equal, where the side is in the same position in both triangles.
- RHS: both triangles have a right angle, equal hypotenuses and one other equal side.
Angle, angle, angle (AAA) is not enough for congruence. It only proves that the triangles are similar.
Similarity. Two shapes are similar if they have the same shape but possibly different sizes. All matching angles are equal, and all matching lengths are in the same ratio. This ratio is the scale factor \(k\).
- Scale factor \(k = \dfrac{\text{length on the new shape}}{\text{matching length on the original}}\).
- Two triangles are similar if their angles are equal in pairs. Two angles are enough, because the third is then fixed.
- Be careful to match the right sides. The longest side on one triangle matches the longest side on the other.
Areas and volumes of similar shapes (Extended). If the lengths of similar shapes are in the ratio \(1 : k\), then:
- the areas are in the ratio \(1 : k^2\)
- the volumes are in the ratio \(1 : k^3\)
To go backwards, take a square root of an area ratio to get a length ratio, and a cube root of a volume ratio to get a length ratio.
Symmetry. A shape has line symmetry if a mirror line splits it into two matching halves. A shape has rotational symmetry if it fits onto itself in more than one position during a full turn. The number of positions is the order of rotational symmetry. A shape with no rotational symmetry has order 1.
- A regular polygon with \(n\) sides has \(n\) lines of symmetry and rotational symmetry of order \(n\).
- A rectangle (not a square) has 2 lines of symmetry and order 2. A parallelogram (not a rectangle or rhombus) has no lines of symmetry and order 2. A kite has 1 line and order 1.
- In three dimensions, a plane of symmetry cuts a solid into two mirror-image halves. A cuboid with three different edge lengths has 3 planes of symmetry. A square-based pyramid has 4.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Congruent | Identical in shape and size. Tests for triangles: SSS, SAS, ASA, RHS. |
| Similar | Same shape, matching angles equal, matching lengths in a constant ratio. |
| Scale factor \(k\) | New length \(\div\) original length. |
| Area scale factor | \(k^2\) (Extended). |
| Volume scale factor | \(k^3\) (Extended). |
| Order of rotational symmetry | Number of positions in a full turn in which the shape looks unchanged. |
| Plane of symmetry | A flat surface that cuts a solid into two mirror-image halves. |
Worked examples
Example 1 Triangles \(ABC\) and \(DEF\) are similar, with \(A\) matching \(D\), \(B\) matching \(E\) and \(C\) matching \(F\). \(AB = 6\) cm, \(BC = 8\) cm, \(AC = 7\) cm and \(DE = 15\) cm. Find \(EF\) and \(DF\).
Step 1: \(AB\) matches \(DE\), so \(k = \dfrac{15}{6} = 2.5\).
Step 2: \(BC\) matches \(EF\), so \(EF = 8 \times 2.5 = 20\) cm.
Step 3: \(AC\) matches \(DF\), so \(DF = 7 \times 2.5 = 17.5\) cm.
Answer: \(EF = 20\) cm and \(DF = 17.5\) cm.
Example 2 In a quadrilateral \(ABCD\), \(AB\) is parallel to \(DC\) and \(AB = DC\). The diagonal \(AC\) is drawn. Show that triangles \(ABC\) and \(CDA\) are congruent, and state what this tells you about \(BC\) and \(AD\).
Step 1: \(AB = CD\) (given).
Step 2: \(AC\) is a side of both triangles, so \(AC = CA\) (common side).
Step 3: angle \(BAC\) = angle \(DCA\), because they are alternate angles between the parallel lines \(AB\) and \(DC\).
Step 4: two sides and the angle between them are equal, so the triangles are congruent (SAS).
Step 5: matching sides of congruent triangles are equal, so \(BC = AD\).
Answer: the triangles are congruent by SAS, and \(BC = AD\).
Example 3 (Extended) Two similar bottles have surface areas of 200 cm\(^2\) and 450 cm\(^2\). The smaller bottle holds 500 cm\(^3\). Find the volume of the larger bottle.
Step 1: the area ratio is \(450 : 200 = 2.25\), so \(k^2 = 2.25\).
Step 2: take the square root. \(k = 1.5\).
Step 3: the volume scale factor is \(k^3 = 1.5^3 = 3.375\).
Step 4: volume of the larger bottle \(= 500 \times 3.375 = 1687.5\) cm\(^3\).
Answer: about 1690 cm\(^3\) (3 s.f.).
Common errors
- Matching the wrong sides in similar triangles. Check which angles are equal and match the sides opposite them.
- Using the length scale factor for areas or volumes. Areas need \(k^2\) and volumes need \(k^3\).
- Writing "AAA" as a congruence test. Three equal angles only prove similarity.
- Giving SSA as a congruence test. The equal angle must be between the two equal sides (SAS), or the triangles must be right-angled (RHS).
- Saying a parallelogram has line symmetry. Its order of rotational symmetry is 2, but it has no mirror lines unless it is a rectangle or rhombus.
Exam tip: in a congruence proof, write each fact on its own line with its reason in brackets, then finish with the test name, for example "(SAS)". Always say which scale factor you are using before you apply it.
Check yourself
- How many lines of symmetry and what order of rotational symmetry does a regular hexagon have?
Answer
It has 6 lines of symmetry and rotational symmetry of order 6. - Rectangles \(4 \times 6\) and \(10 \times w\) are similar. Find \(w\).
Answer
\(k = 10 \div 4 = 2.5\), so \(w = 6 \times 2.5 = 15\). - (Extended) Two similar models have lengths in the ratio \(2 : 5\). The smaller has a surface area of 12 cm\(^2\). Find the surface area of the larger.
Answer
The area ratio is \(2^2 : 5^2 = 4 : 25\). The scale factor for area from small to large is \(25 \div 4 = 6.25\), so the area is \(12 \times 6.25 = 75\) cm\(^2\). - (Extended) Two similar solids have volumes in the ratio \(8 : 125\). What is the ratio of their surface areas?
Answer
The length ratio is \(\sqrt[3]{8} : \sqrt[3]{125} = 2 : 5\). The area ratio is \(4 : 25\).