Mathematics · AQA 8300 §G5
Congruence and Similarity
Learning Objectives
- State and apply the four congruence conditions (SSS, SAS, ASA/AAS, RHS)
- Write formal congruence proofs with a geometric reason for every step
- Identify similar shapes and calculate the linear scale factor
- Use area scale factor ($k^2$) and volume scale factor ($k^3$) to solve problems
- Solve multi-step problems using similarity to find unknown lengths in 2D and 3D
🔑 Core Concepts
What Is Congruence?
Two shapes are congruent if they are identical in shape and size — one can be mapped onto the other by a combination of rotations, reflections and translations (isometric transformations only; no enlargement). All corresponding sides are equal in length and all corresponding angles are equal.
DEFINITION — Congruent Shapes
Two geometric figures are congruent (symbol $\cong$) if every corresponding side and every corresponding angle is equal. The figures are the same shape and the same size.
COMMON MISTAKE
Congruence allows reflections. A shape and its mirror image are congruent — students often think reflections break congruence. Only enlargement (scale factor ≠ 1) destroys congruence.
The Four Congruence Conditions
To prove two triangles are congruent, you must demonstrate that one of the following four conditions holds. Each condition is a set of minimum information that guarantees the triangles are identical.
DEFINITION — SSS (Side-Side-Side)
If three sides of one triangle are equal to the three corresponding sides of another triangle, the triangles are congruent. No angle information is needed — the shape is fully determined by the three side lengths.
DEFINITION — SAS (Side-Angle-Side)
If two sides and the included angle (the angle between those two sides) of one triangle equal the corresponding two sides and included angle of another, the triangles are congruent. The angle must be between the two given sides.
DEFINITION — ASA / AAS (Angle-Side-Angle / Angle-Angle-Side)
If two angles and a side of one triangle equal the corresponding two angles and corresponding side of another, the triangles are congruent. In ASA the known side is between the two angles; in AAS it is not, but both conditions are equivalent and valid.
DEFINITION — RHS (Right angle-Hypotenuse-Side)
In a right-angled triangle, if the hypotenuse and one other side of one triangle equal the hypotenuse and corresponding side of another, the triangles are congruent. This condition applies only to right-angled triangles.
EXAM TIP — AAA is NOT a Congruence Condition
Three equal angles (AAA) proves similarity, not congruence. The triangles have the same shape but can be different sizes. Always check you have at least one side in your condition.
COMMON MISTAKE — SSA is not valid
Side-Side-Angle (SSA) — two sides and a non-included angle — is not a valid congruence condition. It can produce two different triangles (the ambiguous case). Do not use SSA in a proof.
Writing Formal Congruence Proofs
A formal congruence proof must list three pieces of matching information, give a reason for each piece, state the congruence condition used, and conclude with the congruence statement in correct notation. Every step must have a reason — losing even one reason loses marks at Grade 9.
Identify what information is given or follows from the diagram
→
State three matching facts with geometric reasons (e.g. "common side", "alternate angles", "vertically opposite angles")
→
Name the congruence condition (SSS / SAS / ASA / RHS)
→
Write the congruence statement: $\triangle ABC \cong \triangle DEF$, listing vertices in corresponding order
EXAM TIP — Vertex Order Matters
When writing $\triangle ABC \cong \triangle DEF$, vertices must be in matching order: $A$ corresponds to $D$, $B$ to $E$, $C$ to $F$. An incorrect vertex order will cost marks even if the rest of the proof is perfect.
Common Geometric Reasons Used in Proofs
| Reason | When to use it | Example statement |
|---|---|---|
| Common side | Same side shared by both triangles | "$BC$ is common to both triangles" |
| Alternate angles | Parallel lines cut by a transversal | "$\angle ABC = \angle DCB$ (alternate angles, $AB \parallel CD$)" |
| Corresponding angles | Parallel lines, same position at each crossing | "$\angle ABE = \angle DCE$ (corresponding angles, $AB \parallel DC$)" |
| Vertically opposite angles | Two lines crossing at a point | "$\angle AOB = \angle COD$ (vertically opposite)" |
| Co-interior angles | Parallel lines; angles sum to 180° | "$\angle ABC + \angle BCD = 180°$ (co-interior, $AB \parallel CD$)" |
| Base angles of isosceles triangle | Triangle with two equal sides | "$\angle ABD = \angle ACD$ (base angles, isosceles $\triangle$)" |
| Perpendicular bisector | Point on perpendicular bisector of a segment | "$OA = OB$ (radii of circle)" |
What Is Similarity?
Two shapes are similar if one is an enlargement of the other. They have the same shape (all corresponding angles are equal) but different sizes (all corresponding sides are in the same ratio). The ratio of corresponding sides is the linear scale factor, $k$.
DEFINITION — Similar Shapes
Two shapes are similar (symbol $\sim$) if all corresponding angles are equal and all corresponding sides are in the same ratio $k$ (the linear scale factor). For triangles: $k = \dfrac{\text{side in larger}}{\text{corresponding side in smaller}}$.
Conditions for Similar Triangles
Unlike congruence, similarity of triangles can be established in three ways:
DEFINITION — AA (Angle-Angle)
If two angles of one triangle equal two angles of another triangle, the triangles are similar. (The third angles are automatically equal since angles in a triangle sum to 180°.) AA is the most commonly used condition in exam questions.
DEFINITION — SAS Similarity
If two sides of one triangle are in the same ratio as two corresponding sides of another, and the included angles are equal, the triangles are similar.
DEFINITION — SSS Similarity
If all three pairs of corresponding sides are in the same ratio, the triangles are similar. Verify by checking $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$.
EXAM TIP — Always Prove Similarity First
In a Grade 9 question you must first prove the triangles are similar (state two equal angles with reasons), then set up the ratio. Do not simply state "the triangles are similar" without proof — that earns zero marks for that part.
Linear, Area and Volume Scale Factors
When the linear scale factor between two similar shapes is $k$, areas and volumes scale by fixed powers of $k$. This relationship is fundamental and must be memorised.
Linear Scale Factor
$$k = \frac{\text{length in image}}{\text{corresponding length in original}}$$
$k$ = linear scale factor (ratio of any pair of corresponding lengths)
Area Scale Factor
$$\text{Area SF} = k^2 \qquad \Rightarrow \qquad A_{\text{image}} = k^2 \times A_{\text{original}}$$
$k^2$ = area scale factor | If $k = 3$, areas are $9\times$ larger
Volume Scale Factor
$$\text{Volume SF} = k^3 \qquad \Rightarrow \qquad V_{\text{image}} = k^3 \times V_{\text{original}}$$
$k^3$ = volume scale factor | If $k = 2$, volumes are $8\times$ larger
Finding Unknown Lengths Using Similar Triangles
$$\frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'} = k$$
$a, b, c$ = sides of original | $a', b', c'$ = corresponding sides of image
Multiply original sides by $k$ to find image sides, or divide image by $k$ to find original
EXAM TIP — Reverse Scale Factor
If you are given an area ratio or volume ratio and need the length ratio, reverse the process: $k = \sqrt{\text{Area SF}}$ or $k = \sqrt[3]{\text{Volume SF}}$. Always establish $k$ before calculating lengths.
COMMON MISTAKE — Using $k$ instead of $k^2$ for Areas
A very common error is to multiply the area by $k$ rather than $k^2$. Remember: every dimension scales by $k$, and since area = length $\times$ length, you scale by $k \times k = k^2$.
Similarity in 3D Contexts
The same scale factor relationships apply to 3D similar solids. If two solids are similar with linear scale factor $k$, then their surface areas are in ratio $k^2$ and their volumes are in ratio $k^3$. Multi-step problems often require you to find $k$ from a volume ratio, then use it to find a surface area or a length.
IMPORTANT — 3D Strategy
Given volume ratio, find $k$: $k = \sqrt[3]{\dfrac{V_2}{V_1}}$ Then apply $k^2$ for surface areas and $k$ for any length. Never mix up which power to use.
MEMORY TRICK — "Lengths Lines, Areas Squares, Volumes Cubes"
Length $\to k^1$ Area $\to k^2$ Volume $\to k^3$. Count the dimensions: a length is 1D, an area is 2D (length×length), a volume is 3D (length×length×length). The power of $k$ equals the number of dimensions.
🗺 Visual Notes
Congruence
& Similarity
& Similarity
Congruence Conditions
- SSS — three equal sides
- SAS — two sides & included angle
- ASA/AAS — two angles & a side
- RHS — right angle, hypotenuse, side
Similarity Conditions
- AA — two equal angles
- SAS — ratio of sides + included ∠
- SSS — all three sides in ratio
- AAA proves similarity, not congruence
Scale Factors
- Linear: $k = $ length ratio
- Area: multiply by $k^2$
- Volume: multiply by $k^3$
- Reverse: $k = \sqrt{\text{Area SF}}$
Proof Structure
- State 3 matching facts with reasons
- Name the condition (SSS/SAS/…)
- Write $\triangle ABC \cong \triangle DEF$
- Vertex order = correspondence order
Key Reasons in Proofs
- Alternate angles (parallel lines)
- Vertically opposite angles
- Common side / radius
- Base angles of isosceles $\triangle$
3D Similarity
- Surface area ratio = $k^2$
- Volume ratio = $k^3$
- Find $k$ from $\sqrt[3]{\text{Vol ratio}}$
- Apply to any unknown length/area
Congruence vs Similarity — Key Differences
| Property | Congruent Shapes | Similar Shapes |
|---|---|---|
| Symbol | $\cong$ | $\sim$ |
| Corresponding angles | Equal | Equal |
| Corresponding sides | Equal (ratio = 1) | In constant ratio $k$ |
| Size | Identical | Proportional |
| Transformations | Rotation, reflection, translation | Plus enlargement (any $k$) |
| Scale factor $k$ | Always $k = 1$ | Any positive value |
| Area ratio | 1 : 1 | $k^2 : 1$ |
| Volume ratio | 1 : 1 : 1 | $k^3 : 1 : 1$ |
Deciding Which Condition to Use
What information do you have?
↓
Three sides given? → SSS
↓
Two sides + angle between them? → SAS
↓
Two angles + any side? → ASA or AAS
↓
Right angle + hypotenuse + another side? → RHS
↓
Two angles only (no side)? → AA (Similarity only)
Scale Factor Quick Reference
| Linear SF ($k$) | Area SF ($k^2$) | Volume SF ($k^3$) | Example context |
|---|---|---|---|
| $2$ | $4$ | $8$ | Doubled each dimension |
| $3$ | $9$ | $27$ | Tripled each dimension |
| $\frac{1}{2}$ | $\frac{1}{4}$ | $\frac{1}{8}$ | Halved each dimension |
| $\frac{3}{2}$ | $\frac{9}{4}$ | $\frac{27}{8}$ | 1.5× scale model |
| $\sqrt{2}$ | $2$ | $2\sqrt{2}$ | Area doubled |
| $\sqrt[3]{4}$ | $\sqrt[3]{16}$ | $4$ | Volume quadrupled |
✏ Worked Examples
Grade 4–5 — Simple
Question: Triangle $PQR$ has $PQ = 5$ cm, $QR = 7$ cm, $PR = 9$ cm. Triangle $STU$ has $ST = 5$ cm, $TU = 7$ cm, $SU = 9$ cm. Are the triangles congruent? Give a reason.
1
Compare corresponding sides
List the sides of each triangle in order: $PQ = ST = 5$ cm, $QR = TU = 7$ cm, $PR = SU = 9$ cm. All three pairs of corresponding sides are equal.
2
Identify the congruence condition
Three pairs of equal sides ⇒ this satisfies the SSS (Side-Side-Side) condition.
3
Write the conclusion
$\triangle PQR \cong \triangle STU$ (SSS)
Answer: Yes, $\triangle PQR \cong \triangle STU$ by SSS — all three pairs of corresponding sides are equal.
Grade 6–7 — Medium
Question: Triangles $ABC$ and $DEF$ are similar. $AB = 6$ cm, $BC = 8$ cm, $AC = 10$ cm, and $DE = 9$ cm. Find the lengths $EF$ and $DF$. Also find the ratio of the areas of the two triangles.
1
Find the linear scale factor
$AB$ corresponds to $DE$. Therefore: $$k = \frac{DE}{AB} = \frac{9}{6} = 1.5$$
2
Find EF (corresponding to BC)
$$EF = k \times BC = 1.5 \times 8 = 12 \text{ cm}$$
3
Find DF (corresponding to AC)
$$DF = k \times AC = 1.5 \times 10 = 15 \text{ cm}$$
4
Find the area scale factor
$$\text{Area SF} = k^2 = 1.5^2 = 2.25$$
So the area of $\triangle DEF$ is $2.25$ times the area of $\triangle ABC$.
$$\text{Area ratio: } \triangle ABC : \triangle DEF = 1 : 2.25 = 4 : 9$$
Answer: $EF = 12$ cm, $DF = 15$ cm. Area ratio $= 4 : 9$.
Grade 9 — Formal Proof + Multi-step
Question: In the diagram, $ABCD$ is a parallelogram. $E$ is the midpoint of $AB$ and $F$ is the midpoint of $CD$.
(a) Prove that $\triangle AEF \cong \triangle CFD$. [4 marks]
(b) Hence prove that $EF = FD$. [2 marks]
(a) Prove that $\triangle AEF \cong \triangle CFD$. [4 marks]
(b) Hence prove that $EF = FD$. [2 marks]
1
Part (a) — List given information and extract facts
$ABCD$ is a parallelogram so $AB \parallel DC$ and $AB = DC$. $E$ is the midpoint of $AB$ so $AE = \tfrac{1}{2}AB$. $F$ is the midpoint of $CD$ so $CF = \tfrac{1}{2}CD$.
2
State fact 1 with reason
$AE = CF$ because $E$ is the midpoint of $AB$, $F$ is the midpoint of $CD$, and $AB = CD$ (opposite sides of a parallelogram are equal), so $AE = \tfrac{1}{2}AB = \tfrac{1}{2}CD = CF$.
3
State fact 2 with reason
$\angle AEF = \angle CFD$ because $AB \parallel DC$ (opposite sides of a parallelogram) and $EF$ is a transversal, so these are alternate angles.
4
State fact 3 with reason
$AF = EF$... Actually, let us instead use the side $EF$ is common — wait. Let us re-approach: we need a second side or angle. $\angle EAF = \angle FCD$ (alternate angles, $AB \parallel DC$). Now we have two angles and the included side $AE = CF$. This satisfies ASA.
5
State the congruence condition and conclusion
We have: $AE = CF$, $\angle EAF = \angle FCD$ (alternate angles), $\angle AEF = \angle CFD$ (alternate angles). Two angles and the included side are equal $\Rightarrow$ ASA. Therefore $\triangle AEF \cong \triangle CFD$ (ASA).
6
Part (b) — Use congruence result
Since $\triangle AEF \cong \triangle CFD$, all corresponding sides are equal. $EF$ in $\triangle AEF$ corresponds to $FD$ in $\triangle CFD$. Therefore $EF = FD$.
Answer: (a) $\triangle AEF \cong \triangle CFD$ by ASA (two angles and included side equal; reasons: alternate angles and opposite sides of parallelogram). (b) $EF = FD$ as corresponding sides of congruent triangles.
❓ Exam Questions
Q1
1 mark
Which of the following is NOT a valid congruence condition for triangles?
A — SSS B — SAS C — AAA D — RHS
Answer: C — AAA
AAA (three equal angles) proves that triangles are similar, not necessarily congruent. The triangles have the same shape but may differ in size. All four congruence conditions require at least one side measurement. (1 mark for C.)
AAA (three equal angles) proves that triangles are similar, not necessarily congruent. The triangles have the same shape but may differ in size. All four congruence conditions require at least one side measurement. (1 mark for C.)
Q2
2 marks
Triangle $LMN$ is similar to triangle $PQR$. $LM = 4$ cm, $MN = 6$ cm and $PQ = 10$ cm. Find the length $QR$.
Mark Scheme:
M1: Correct scale factor $k = \dfrac{PQ}{LM} = \dfrac{10}{4} = 2.5$
A1: $QR = 2.5 \times 6 = 15$ cm
Full solution: Since the triangles are similar, $\dfrac{PQ}{LM} = \dfrac{QR}{MN}$. $k = \dfrac{10}{4} = 2.5$. $QR = 2.5 \times 6 = \mathbf{15}$ cm.
M1: Correct scale factor $k = \dfrac{PQ}{LM} = \dfrac{10}{4} = 2.5$
A1: $QR = 2.5 \times 6 = 15$ cm
Full solution: Since the triangles are similar, $\dfrac{PQ}{LM} = \dfrac{QR}{MN}$. $k = \dfrac{10}{4} = 2.5$. $QR = 2.5 \times 6 = \mathbf{15}$ cm.
Q3
3 marks
Two similar cylinders have heights $4$ cm and $10$ cm respectively. The surface area of the smaller cylinder is $96\pi$ cm². Find the surface area of the larger cylinder.
Mark Scheme:
M1: Correct linear scale factor $k = \dfrac{10}{4} = \dfrac{5}{2}$
M1: Area scale factor $= k^2 = \left(\dfrac{5}{2}\right)^2 = \dfrac{25}{4}$
A1: Surface area $= \dfrac{25}{4} \times 96\pi = 600\pi$ cm²
Full solution:
$k = \dfrac{10}{4} = 2.5$ $k^2 = 6.25$
Surface area of larger cylinder $= 6.25 \times 96\pi = \mathbf{600\pi}$ cm² $\approx 1885$ cm²
M1: Correct linear scale factor $k = \dfrac{10}{4} = \dfrac{5}{2}$
M1: Area scale factor $= k^2 = \left(\dfrac{5}{2}\right)^2 = \dfrac{25}{4}$
A1: Surface area $= \dfrac{25}{4} \times 96\pi = 600\pi$ cm²
Full solution:
$k = \dfrac{10}{4} = 2.5$ $k^2 = 6.25$
Surface area of larger cylinder $= 6.25 \times 96\pi = \mathbf{600\pi}$ cm² $\approx 1885$ cm²
Q4
4 marks
In triangle $ABC$, $D$ is a point on $AB$ and $E$ is a point on $AC$ such that $DE \parallel BC$. Given that $AD = 3$ cm, $DB = 5$ cm and $BC = 12$ cm, find the length $DE$. You must show that the triangles are similar before calculating.
Mark Scheme:
B1: $\angle ADE = \angle ABC$ (corresponding angles, $DE \parallel BC$)
B1: $\angle AED = \angle ACB$ (corresponding angles, $DE \parallel BC$) — OR angle $A$ is common
M1: $\triangle ADE \sim \triangle ABC$ (AA) — two angles proved equal
A1: $k = \dfrac{AD}{AB} = \dfrac{3}{3+5} = \dfrac{3}{8}$, so $DE = \dfrac{3}{8} \times 12 = 4.5$ cm
Full solution:
$\angle ADE = \angle ABC$ (corresponding angles, $DE \parallel BC$). $\angle DAE = \angle BAC$ (common angle). Therefore $\triangle ADE \sim \triangle ABC$ (AA). $k = \dfrac{3}{8}$. $DE = \dfrac{3}{8} \times 12 = \mathbf{4.5}$ cm.
B1: $\angle ADE = \angle ABC$ (corresponding angles, $DE \parallel BC$)
B1: $\angle AED = \angle ACB$ (corresponding angles, $DE \parallel BC$) — OR angle $A$ is common
M1: $\triangle ADE \sim \triangle ABC$ (AA) — two angles proved equal
A1: $k = \dfrac{AD}{AB} = \dfrac{3}{3+5} = \dfrac{3}{8}$, so $DE = \dfrac{3}{8} \times 12 = 4.5$ cm
Full solution:
$\angle ADE = \angle ABC$ (corresponding angles, $DE \parallel BC$). $\angle DAE = \angle BAC$ (common angle). Therefore $\triangle ADE \sim \triangle ABC$ (AA). $k = \dfrac{3}{8}$. $DE = \dfrac{3}{8} \times 12 = \mathbf{4.5}$ cm.
Q5
6 marks
Two similar cones $A$ and $B$ have volumes $54\pi$ cm³ and $128\pi$ cm³ respectively.
(a) Find the linear scale factor from cone $A$ to cone $B$. [2]
(b) The base radius of cone $A$ is $3$ cm. Find the base radius of cone $B$. [1]
(c) The curved surface area of cone $A$ is $15\pi$ cm². Find the curved surface area of cone $B$. [2]
(d) Explain why you cannot use the formula for the curved surface area of a cone directly without knowing the slant height of cone $B$, but the scale factor method is sufficient. [1]
Mark Scheme:
(a) M1: $k^3 = \dfrac{128\pi}{54\pi} = \dfrac{64}{27}$ A1: $k = \sqrt[3]{\dfrac{64}{27}} = \dfrac{4}{3}$ [2]
(b) A1: $r_B = \dfrac{4}{3} \times 3 = 4$ cm [1]
(c) M1: Area SF $= k^2 = \dfrac{16}{9}$ A1: $\text{CSA}_B = \dfrac{16}{9} \times 15\pi = \dfrac{80\pi}{3}$ cm² [2]
(d) B1: The scale factor method uses only the ratio — once we know $k^2$, we can find any area without needing to know the actual dimensions. The cones are similar, so all corresponding lengths scale by $k$, meaning any area scales by $k^2$ regardless of the specific formula used. [1]
Answers: (a) $k = \dfrac{4}{3}$ (b) $r_B = 4$ cm (c) $\dfrac{80\pi}{3}$ cm² $\approx 83.8$ cm²
(a) M1: $k^3 = \dfrac{128\pi}{54\pi} = \dfrac{64}{27}$ A1: $k = \sqrt[3]{\dfrac{64}{27}} = \dfrac{4}{3}$ [2]
(b) A1: $r_B = \dfrac{4}{3} \times 3 = 4$ cm [1]
(c) M1: Area SF $= k^2 = \dfrac{16}{9}$ A1: $\text{CSA}_B = \dfrac{16}{9} \times 15\pi = \dfrac{80\pi}{3}$ cm² [2]
(d) B1: The scale factor method uses only the ratio — once we know $k^2$, we can find any area without needing to know the actual dimensions. The cones are similar, so all corresponding lengths scale by $k$, meaning any area scales by $k^2$ regardless of the specific formula used. [1]
Answers: (a) $k = \dfrac{4}{3}$ (b) $r_B = 4$ cm (c) $\dfrac{80\pi}{3}$ cm² $\approx 83.8$ cm²
Q6
5 marks
$ABCD$ is a quadrilateral where diagonals $AC$ and $BD$ intersect at $E$. It is given that $AB \parallel DC$ and $AB = 2DC$.
Prove that $\triangle ABE \sim \triangle CDE$ and hence find the ratio $AE : EC$.
Mark Scheme:
B1: $\angle BAE = \angle DCE$ (alternate angles, $AB \parallel DC$) — must state reason
B1: $\angle ABE = \angle CDE$ (alternate angles, $AB \parallel DC$) — must state reason
B1: $\triangle ABE \sim \triangle CDE$ (AA) — must name condition
M1: Scale factor $k = \dfrac{AB}{DC} = \dfrac{2DC}{DC} = 2$
A1: $AE : EC = 2 : 1$ (corresponding sides; $AE$ in $\triangle ABE$ corresponds to $CE$ in $\triangle CDE$)
Key point: The ratio follows immediately from similarity — since $k=2$, the sides of $\triangle ABE$ are all twice those of $\triangle CDE$, so $AE = 2 \times CE$, giving $AE : EC = 2 : 1$.
B1: $\angle BAE = \angle DCE$ (alternate angles, $AB \parallel DC$) — must state reason
B1: $\angle ABE = \angle CDE$ (alternate angles, $AB \parallel DC$) — must state reason
B1: $\triangle ABE \sim \triangle CDE$ (AA) — must name condition
M1: Scale factor $k = \dfrac{AB}{DC} = \dfrac{2DC}{DC} = 2$
A1: $AE : EC = 2 : 1$ (corresponding sides; $AE$ in $\triangle ABE$ corresponds to $CE$ in $\triangle CDE$)
Key point: The ratio follows immediately from similarity — since $k=2$, the sides of $\triangle ABE$ are all twice those of $\triangle CDE$, so $AE = 2 \times CE$, giving $AE : EC = 2 : 1$.
⭐ Grade 9 Model Answers
Model Answer — Q5 (6 marks, Volume & Surface Area Scale Factor)
WORKED EXAMPLE — Full Grade 9 Response
Part (a): Find $k$The volumes are in the ratio $54\pi : 128\pi = 54 : 128 = 27 : 64$.
Volume scale factor $= k^3 = \dfrac{64}{27}$.
Therefore $k = \sqrt[3]{\dfrac{64}{27}} = \dfrac{\sqrt[3]{64}}{\sqrt[3]{27}} = \dfrac{4}{3}$.
Why this earns the method mark: The student correctly identifies that volume scales as $k^3$ and takes a cube root — not a square root — to recover $k$.
Part (b): Radius of cone B
$r_B = k \times r_A = \dfrac{4}{3} \times 3 = 4$ cm.
Why this earns the mark: The student applies the linear scale factor (not $k^2$ or $k^3$) to a length.
Part (c): Curved surface area of cone B
Area scale factor $= k^2 = \left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9}$.
$\text{CSA}_B = \dfrac{16}{9} \times 15\pi = \dfrac{240\pi}{9} = \dfrac{80\pi}{3}$ cm².
Why this earns both marks: The method mark requires $k^2$ (not $k$). The accuracy mark is for the correctly simplified exact answer $\dfrac{80\pi}{3}$. A decimal answer would also be accepted if exact form is not required.
Part (d): Conceptual explanation
The curved surface area formula is $\pi r l$ (radius $\times$ slant height). To use it directly, we would need $l_B$. However, since the cones are similar, every length — including slant height — scales by $k$, so the product $r_B \times l_B = k \cdot r_A \times k \cdot l_A = k^2(r_A l_A)$. This shows that any area, regardless of the specific formula, scales by $k^2$. The scale factor method therefore works without knowing individual dimensions.
Why this earns the quality of written communication mark: The student explains the underlying reason (all lengths scale by $k$) rather than simply asserting the result.
EXAM TIP — What Distinguishes a Grade 9 Answer
Grade 9 answers include: correct use of $k^3$ not $k^2$ for volumes; exact simplified answers (e.g. $\dfrac{80\pi}{3}$ not $83.78$); clear step-by-step reasoning; and for proofs, a reason on every single step. A student who writes the right answer without working can score 0 on method marks.
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Congruent | Same shape and size ($\cong$) |
| Similar | Same shape, different size ($\sim$) |
| Scale factor $k$ | Ratio of corresponding lengths |
| Included angle | Angle between two given sides |
| Corresponding | Matching sides/angles in same position |
| Isometric | Transformation preserving size (rot, refl, trans) |
Essential Formulae
Linear scale factor: $k = \dfrac{\text{image length}}{\text{original length}}$
Area scale factor: $k^2$
Volume scale factor: $k^3$
Finding $k$ from area: $k = \sqrt{\text{Area SF}}$
Finding $k$ from volume: $k = \sqrt[3]{\text{Vol SF}}$
Similar triangles: $\dfrac{a}{a'} = \dfrac{b}{b'} = \dfrac{c}{c'} = k$
Memory Hooks
- SSS, SAS, ASA, RHS — "Some Students Are Really Helpful" (initial letters)
- AAA = Similarity, never congruence
- SSA is invalid — the "ambiguous case"
- Powers match dimensions: 1D→$k$, 2D→$k^2$, 3D→$k^3$
- Vertex order matters in $\triangle ABC \cong \triangle DEF$
- Every step needs a reason in formal proofs
Exam Tips
- Always prove similarity before using ratios
- State the condition name (SSS/AA/etc.) explicitly
- Write congruence/similarity statement with vertices in matching order
- Use $k^2$ for areas — never $k$ — even if $k$ is a fraction
- Cube root (not square root) to find $k$ from volume ratio
- For parallel lines: name alternate/corresponding angles, then state the lines
- Common side = "shared side" = "same segment" — all valid reasons
- Give exact answers ($\pi$, fractions, surds) unless asked to round
🔄 Flashcards
Click a card to reveal the answer. Work through all 15 before your exam.
✗ Common Mistakes
MISTAKE 1 — Using AAA as a Congruence Condition
What students do: They establish three equal angles and conclude congruence.Why marks are lost: AAA only proves similarity. Two triangles can have the same angles but be completely different sizes. The examiner requires a side measurement in every congruence proof.
How to avoid it: After writing three equal angles, ask "do I have at least one side?" If not, you can only conclude similarity, not congruence.
MISTAKE 2 — Multiplying Area by $k$ Instead of $k^2$
What students do: They find the linear scale factor $k = 3$ and calculate the new area as $k \times \text{old area}$.Why marks are lost: Area involves two dimensions. If both length and width scale by $k$, area scales by $k \times k = k^2$. Using $k$ gives the wrong answer and loses the accuracy mark and likely the method mark.
How to avoid it: Write $k^2$ next to "area" every time. Underline it on your paper.
MISTAKE 3 — Wrong Vertex Order in Congruence Statement
What students do: They write $\triangle ABC \cong \triangle FDE$ when the correct correspondence is $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$.Why marks are lost: The vertex order encodes the correspondence. Writing the wrong order means your statement describes a different (incorrect) congruence, even if the triangles are genuinely congruent. This typically loses 1 mark at Grade 9.
How to avoid it: Before writing the statement, list corresponding vertices explicitly: $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$, then read off the order.
MISTAKE 4 — Omitting Reasons in Congruence Proofs
What students do: They write "angle $A$ = angle $D$" without explaining why (e.g., without saying "alternate angles, $AB \parallel CD$").Why marks are lost: In a formal proof, each statement must be justified. A statement without a reason earns zero marks for that step at higher grades, even if the statement is factually correct.
How to avoid it: For every angle or side equality, ask "why?" and write the answer in brackets: (alternate angles, $PQ \parallel RS$), (common side), (radii of same circle), etc.
MISTAKE 5 — Using Square Root Instead of Cube Root for Volume
What students do: Given a volume ratio, they find $k$ by taking $\sqrt{\text{Vol ratio}}$ instead of $\sqrt[3]{\text{Vol ratio}}$.Why marks are lost: Volume scales by $k^3$, so you must undo a cube, not a square. Using $\sqrt{}$ gives a completely wrong scale factor and all subsequent calculations will be wrong.
How to avoid it: Write "Vol SF = $k^3$, so $k = \sqrt[3]{\text{Vol SF}}$" at the top of every volume problem to remind yourself.
MISTAKE 6 — Failing to Prove Similarity Before Using Ratios
What students do: They spot two "obviously similar" triangles in a diagram and immediately set up a ratio without proving similarity first.Why marks are lost: Most multi-step similarity questions (especially proof questions) allocate 2–3 marks specifically for proving similarity. Simply stating "the triangles are similar" without reasoning loses all those marks, even if the final length calculation is correct.
How to avoid it: Always write two angle equalities with reasons, name the condition (AA/SAS/SSS), state the similarity, then calculate.
✅ Final Checklist
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- I can state all four congruence conditions: SSS, SAS, ASA/AAS, RHS
- I know that AAA is NOT a valid congruence condition
- I can write a formal congruence proof with a reason on every step
- I write congruence statements with vertices in corresponding order
- I can identify similar triangles using AA, SAS or SSS similarity
- I always prove similarity before applying scale factors in a proof question
- I can calculate the linear scale factor from two corresponding sides
- I use $k^2$ (not $k$) as the area scale factor
- I use $k^3$ (not $k^2$) as the volume scale factor
- I can find $k$ from an area ratio using $k = \sqrt{\text{Area SF}}$
- I can find $k$ from a volume ratio using $k = \sqrt[3]{\text{Vol SF}}$
- I can solve problems combining similarity with algebra (e.g. forming equations)
- I can apply similarity to 3D similar solids (surface area and volume)
- I know common geometric reasons: alternate angles, corresponding angles, vertically opposite, common side
- I give exact answers ($\pi$, surds, fractions) in similarity problems unless told to round