Congruence, Similarity and Geometrical Proof · Study Notes · Edexcel IGCSE 4MA1 Higher

Congruent (same size + shape) vs similar (same shape, scaled). Four congruence tests, scale factors for area and volume, proof technique.

At a glance

Congruent: same size + shape.
Similar: same shape, scaled.
Congruence tests: SSS, SAS, ASA, RHS (NOT SSA).
Linear scale: $k$ . Area: $k^{2}$ . Volume: $k^{3}$ .
Similar triangles: AA criterion (or SSS for similar).
Proof: state reasons explicitly.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.3 — Recognise and prove congruent triangles using SSS, SAS, ASA, RHS.
4.3 — Recognise similar shapes; apply scale factors to lengths, areas, volumes.
4.3 — Use geometrical reasoning to prove results.

Congruent triangles

▼

Same size + shape. Four valid tests.

Congruent = identical (one fits exactly on the other).

Four valid congruence tests:

Test	What's needed
SSS	Three corresponding sides equal
SAS	Two sides and the INCLUDED angle equal
ASA	Two angles and one corresponding side equal
RHS	Right-angle, Hypotenuse, one Side equal (right-angled only)

SAS works only when the equal angle lies between the two equal sides.

NOT a valid test: SSA (ambiguous — could give two different triangles).

Important. For SAS, the angle must be BETWEEN the two sides. SSA (where angle is NOT between) doesn't always work.

Worked qualitative. Why isn't SSA a valid test?

Two triangles can share two sides and a non-included angle, but be different triangles.
Example: with two sides $5, 7$ and angle $30°$ NOT between them, you can construct two distinct triangles.
This is the 'ambiguous case' from the sine rule.

Edexcel tip. State the test name (SSS, SAS, etc.) explicitly when proving congruence. Mark schemes give M1 for naming, A1 for justification.

SSS, SAS, ASA, RHS.
SAS: angle BETWEEN sides.
RHS for right-angled triangles.
SSA NOT valid.

Similar shapes

▼

Same shape, different size. Scale factor connects them.

Similar = same SHAPE but possibly different SIZE.

Conditions for similar triangles:

All corresponding ANGLES equal (AA — third is automatic).
OR: all corresponding SIDES in proportion (SSS for similar).

Scale factor $k$ = ratio of corresponding lengths.

Example. Two similar triangles. Smaller has sides $4, 5, 6$ . Larger has $\_, 10, \_$ .

Scale factor: $10/5 = 2$ .
Other sides: $4 \times 2 = 8$ , $6 \times 2 = 12$ .

Similar triangles share all angles; every length multiplies by the same scale factor.

Scale factors for area and volume:

If linear scale factor is $k$ :

AREA scale factor: $k^{2}$ .
VOLUME scale factor: $k^{3}$ .

Memorise:

Linear $k$	Area $k^{2}$	Volume $k^{3}$
$2$	$4$	$8$
$3$	$9$	$27$
$1.5$	$2.25$	$3.375$
$\dfrac{1}{2}$	$\dfrac{1}{4}$	$\dfrac{1}{8}$

Worked example. Two similar shapes. Linear ratio $1:3$ . Larger has area $90$ cm². Smaller area?

Area scale factor: $1/9$ (or $9:1$ from larger to smaller).
Smaller: $90 / 9 = 10$ cm².

Why $k^{2}$ for area? Area depends on TWO linear dimensions. Both scale by $k$ , so area scales by $k \times k = k^{2}$ .

Why $k^{3}$ for volume? Volume depends on THREE linear dimensions. $k \times k \times k = k^{3}$ .

Edexcel tip. Always identify whether the question is asking about LENGTH, AREA, or VOLUME. Use the right power of $k$ .

Similar = same shape, different size.
AA suffices for similar triangles.
Scale factor: $k, k^{2}, k^{3}$ .
Linear, area, volume.

Geometrical proof

▼

Step-by-step justification with reasons.

Goal: Prove a geometric result rigorously.

Method:

Identify what's given.
Identify what to prove.
State reasons for each deduction (e.g. 'isosceles, so angles equal', 'parallel lines, alternate angles').
Conclude clearly.

Common reasons:

'Angles in a triangle sum to $180°$ .'
'Vertically opposite angles are equal.'
'Corresponding angles (parallel lines).'
'Alternate angles (parallel lines).'
'Co-interior angles sum to $180°$ .'
'Angles in a quadrilateral sum to $360°$ .'
'Isosceles triangle: base angles equal.'
'Sum of exterior angles of polygon = $360°$ .'

Worked example. Prove the angles in a quadrilateral sum to $360°$ .

Draw a diagonal, splitting the quadrilateral into two triangles.
Each triangle has angle sum $180°$ .
Total: $180° + 180° = 360°$ .
QED.

Worked qualitative. Why is geometrical proof useful?

Establishes results CONFIRMED for ALL cases, not just one example.
Develops logical reasoning.
Foundation for higher mathematics.

Edexcel tip. Edexcel mark schemes for proofs award marks for SPECIFIC REASONS, not generic 'because of geometry'. Use the standard reason language.

Reason for each step.
Standard reason language.
Show full chain of logic.
Conclude with QED or ' $\therefore$ '.

Quick recap

Congruent: same size + shape.
Similar: same shape, different size.
Tests: SSS, SAS, ASA, RHS (NOT SSA).
Scale: $k, k^{2}, k^{3}$ .
Proof: state reasons.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Congruent: SSS, SAS, ASA, RHS.
Similar: AA (or SSS for similar).
Linear: $k$ .
Area: $k^{2}$ .
Volume: $k^{3}$ .

How it’s examined

Congruent/similar appears Higher Tier (5-7 marks). Scale factors for area/volume regularly tested. Examiner reports flag (1) wrong scale factor for area/volume, (2) using SSA, (3) not naming the test.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024, 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on congruence, similarity and geometrical proof, written exactly the way a tutor would explain them at the board.

1Identify a congruence test
Higher• congruent
▼
Question
Two triangles ABC and DEF have AB = DE, BC = EF, AC = DF. Which congruence test applies?
Step-by-step solution
1. Step 1
  Three pairs of corresponding SIDES equal → SSS test.
Answer
SSS (Side-Side-Side)
Examiner tip
Memorise the four tests: SSS, SAS, ASA, RHS. Identify which applies to each given diagram.
2Find a length using similarity
Higher• Adapted from 4MA1/1H May/Jun 2024 Q11• similar
▼
Question
Two triangles are similar. Smaller has sides $4, 5, 6$ . Larger has corresponding sides $\_, 10, \_$ . Find the missing sides.
Step-by-step solution
1. Step 1
  Find scale factor: $10 / 5 = 2$ .
2. Step 2
  Other sides scale by same factor:
  $4 \times 2 = 8,\ 6 \times 2 = 12$
Answer
$8$ and $12$ .
Examiner tip
Always identify the SCALE FACTOR first. Mark schemes credit explicit calculation.
3Area scale factor
Higher• area scale factor
▼
Question
Two similar shapes have linear scale factor $3$ . The smaller has area $20$ cm². Find the area of the larger.
Step-by-step solution
1. Step 1
  Area scale factor = (linear scale factor)² = $3^{2} = 9$ .
2. Step 2
  Larger area = $20 \times 9 = 180$ cm².
Answer
$180$ cm²
4Volume scale factor
Higher• Adapted from 4MA1/2H Jan 2024 Q16• volume scale factor
▼
Question
Two similar containers have heights $20$ cm and $30$ cm. The smaller holds $4000$ ml. How much does the larger hold?
Step-by-step solution
1. Step 1
  Linear scale factor: $30/20 = 1.5$ .
2. Step 2
  Volume scale factor: $1.5^{3} = 3.375$ .
3. Step 3
  Larger: $4000 \times 3.375 = 13{,}500$ ml.
Answer
$13{,}500$ ml
Examiner tip
VOLUME scales as CUBE of linear factor. Don't confuse with area (square).
5Prove triangles are similar
Higher• similar, proof
▼
Question
Two triangles share a vertex and have sides in the ratio $2:3$ . Prove similarity by showing all angles are equal.
Step-by-step solution
1. Step 1
  Show ANGLES are equal — three pairs of corresponding angles equal makes triangles similar (AA criterion).
2. Step 2
  Or show all three pairs of corresponding SIDES are in proportion (SSS for similar).
Answer
AA (or SSS for similar) → similar.
Examiner tip
For SIMILAR (not congruent), AA suffices because the third angle is automatic. Mark schemes credit naming the criterion.
6Ratio of areas from a ratio of sides [1 mark]
Getting started• Paper 1H/2H style — 1 mark• similarity, area scale factor
▼
Question
Two triangles are mathematically similar. The lengths of their corresponding sides are in the ratio $2 : 5$

Write down the ratio of their areas. [1]
Step-by-step solution
1. Step 1
  B1 Areas are in the ratio of the squares of corresponding lengths.
  $2^{2} : 5^{2} = 4 : 25$
Answer
$4 : 25$
Examiner tip
Examiner tip: Square each side of the ratio separately. A quick check: two squares of side $2$ cm and $5$ cm have areas $4 \text{ cm}^2$ and $25 \text{ cm}^2$ .

Common mistake: Leaving the ratio as $2 : 5$ , or doubling it to $4 : 10$ . Doubling is not squaring.
7Congruent by ASA, and which sides match [2 marks]
Getting started• Paper 1H/2H style — 2 marks• congruence, asa
▼
Question
In triangle $PQR$ , $PQ = 8$ cm, angle $QPR = 40^{\circ}$ and angle $PQR = 75^{\circ}$ .

In triangle $XYZ$ , $YZ = 8$ cm, angle $XYZ = 40^{\circ}$ and angle $YZX = 75^{\circ}$ .

(a) Give a reason why triangles $PQR$ and $XYZ$ are congruent. [1]

(b) Write down the side of triangle $XYZ$ that is equal in length to $PR$ . [1]
Step-by-step solution
1. Step 1
  B1 (a) In each triangle the $8$ cm side lies between the $40^{\circ}$ angle and the $75^{\circ}$ angle. Two angles and the side between them are equal, so the triangles are congruent (ASA).
2. Step 2
  B1 (b) Match the vertices by their angles: $P$ (the $40^{\circ}$ angle) matches $Y$ , $Q$ (the $75^{\circ}$ angle) matches $Z$ , so $R$ matches $X$ . Then $PR$ matches $YX$ .
  $PR = XY$
Answer
(a) Two angles and the included side are equal (ASA). (b) $XY$
Examiner tip
Examiner tip: Congruent triangles are often labelled in a different order. Match each vertex by its angle, never by its letter's position in the alphabet.

Common mistake: Answering $XZ$ in (b) because $P$ and $X$ are both first letters. $X$ is where the third angle, $65^{\circ}$ , is, which matches $R$ , not $P$ .
8Height of a lamp post from a shadow [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• similarity, similar triangles, problem solving
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Question
A vertical lamp post stands on horizontal ground.

Maya is $1.8$ m tall. She stands upright $3$ m from the foot of the lamp post. The light from the top of the lamp post makes her shadow on the ground $2$ m long, pointing directly away from the lamp post.

The top of Maya's head, the end of her shadow and the top of the lamp post lie on a straight line.

Work out the height of the lamp post. [3]
Step-by-step solution
1. Step 1
  Two similar right-angled triangles share the angle at the end of the shadow. The small one has Maya as its height and base $2$ m. The large one has the lamp post as its height and a base that runs all the way from the shadow's end to the lamp post.
2. Step 2
  M1 The base of the large triangle is the whole distance.
  $3 + 2 = 5 \text{ m}$
3. Step 3
  M1 Scale factor $\dfrac{5}{2}$ , applied to Maya's height.
  $1.8 \times \dfrac{5}{2}$
4. Step 4
  A1 Evaluate.
  $= 4.5$
Answer
$4.5$ m
Examiner tip
Examiner tip: Sketch both triangles separately with the shared angle in the same corner. The large triangle's base is $5$ m, not $3$ m.

Common mistake: Using $\dfrac{3}{2}$ as the scale factor, which gives $2.7$ m. The $3$ m is only the gap between Maya and the lamp post, not a side of either triangle.
9Area on a model boat and on the real boat [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• similarity, area scale factor, units
▼
Question
A model of a boat is mathematically similar to the real boat.

Every length on the real boat is $25$ times the corresponding length on the model.

The deck of the model has an area of $180 \text{ cm}^2$ .

Work out the area of the deck of the real boat. Give your answer in $\text{m}^2$ . [3]
Step-by-step solution
1. Step 1
  M1 Areas scale by the square of the length scale factor.
  $180 \times 25^{2} = 180 \times 625 = 112\,500 \text{ cm}^2$
2. Step 2
  M1 Convert to square metres. $1 \text{ m}^2 = 100 \times 100 = 10\,000 \text{ cm}^2$ .
  $112\,500 \div 10\,000$
3. Step 3
  A1 Evaluate.
  $= 11.25 \text{ m}^2$
Answer
$11.25 \text{ m}^2$
Examiner tip
Examiner tip: Two separate squarings happen here: the scale factor ( $25^2$ ) and the unit conversion ( $100^2$ ). Do them one at a time.

Common mistake: Dividing by $100$ instead of $10\,000$ when converting, giving $1125 \text{ m}^2$ , or multiplying by $25$ only, giving $0.45 \text{ m}^2$ after conversion.
10Diagonals of a parallelogram bisect each other [4 marks]
Stretch• Paper 1H/2H style — 4 marks• congruence, geometrical proof, parallelogram
▼
Question
$ABCD$ is a parallelogram. The diagonals $AC$ and $BD$ meet at $M$ .

Prove that $AM = CM$ . [4]
Step-by-step solution
1. Step 1
  Plan: prove triangles $ABM$ and $CDM$ congruent, then use the fact that corresponding sides of congruent triangles are equal.
2. Step 2
  M1 A pair of equal sides with a reason: $AB = CD$ because opposite sides of a parallelogram are equal.
  $AB = CD$
3. Step 3
  M1 Two pairs of equal angles with reasons. $AB$ is parallel to $DC$ , so angle $BAM$ = angle $DCM$ and angle $ABM$ = angle $CDM$ because alternate angles are equal.
  $\angle BAM = \angle DCM, \qquad \angle ABM = \angle CDM$
4. Step 4
  A1 Two angles and the side between them are equal, so triangle $ABM$ is congruent to triangle $CDM$ (ASA).
  $\triangle ABM \equiv \triangle CDM$
5. Step 5
  A1 $AM$ and $CM$ are corresponding sides of the congruent triangles, so they are equal.
  $AM = CM$
Answer
Triangles $ABM$ and $CDM$ are congruent (ASA), so $AM = CM$ .
Examiner tip
Examiner tip: To prove two lengths are equal, look for two congruent triangles that contain them as corresponding sides. The proof then ends with "corresponding sides of congruent triangles are equal".

Common mistake: Giving "Z angles" as the reason. Pearson accepts only "alternate angles are equal". Another loss is using angle $AMB$ = angle $CMD$ (vertically opposite) together with the two alternate pairs and no side, which proves only similarity.

Question 1
Paper 1H/2H style — 1 mark1 mark
Two cones are mathematically similar. Their volumes are in the ratio $8 : 125$

Write down the ratio of their heights. [1]
Model answer
$\sqrt[3]{8} = 2$ and $\sqrt[3]{125} = 5$

Answer: $2 : 5$
Why this scores
Mark scheme:
- B1 for $2 : 5$ oe
Why this scores full marks: Heights are lengths, and volumes are in the ratio of the cubes of lengths, so the script takes the cube root of each part of the ratio.
Question 2
Paper 1H/2H style — 2 marks2 marks
The sides of triangle $T$ are $5$ cm, $6$ cm and $7$ cm.

Triangle $U$ is mathematically similar to triangle $T$ . The longest side of triangle $U$ is $17.5$ cm.

Work out the perimeter of triangle $U$ . [2]
Model answer
Scale factor $= 17.5 \div 7 = 2.5$

$(5 + 6 + 7) \times 2.5 = 18 \times 2.5 = 45$

Answer: $45$ cm
Why this scores
Mark scheme:
- M1 for $17.5 \div 7$ (= $2.5$ ) or for finding both missing sides, $12.5$ and $15$
- A1 for $45$ cao
Why this scores full marks: Perimeter is a length, so it scales by the same factor as the sides. The script multiplies the whole perimeter once rather than working out each side, which is quicker and less error-prone.
Question 3
Paper 1H/2H style — 2 marks2 marks
In triangle $ABC$ , $AB = 9$ cm, $BC = 7$ cm and angle $BAC = 48^{\circ}$ .

In triangle $DEF$ , $DE = 9$ cm, $EF = 7$ cm and angle $EDF = 48^{\circ}$ .

Sam says, "The triangles are congruent because of SAS."

(a) Explain why Sam's reason is wrong. [1]

(b) Write down one more fact about the triangles that would prove they are congruent. [1]
Model answer
(a) The $48^{\circ}$ angle is not between the two given sides. It is at $A$ , but the sides meet at $B$ . So this is not SAS, and two sides with a non-included angle do not prove congruence.

(b) $AC = DF$ (then the triangles are congruent by SAS: $AB$ , $AC$ and the angle between them)
Why this scores
Mark scheme:
- B1 for (a) stating that the angle is not between the two sides (not the included angle) oe
- B1 for (b) $AC = DF$ , or angle $ABC$ = angle $DEF$ , or angle $ACB$ = angle $DFE$ oe
Why this scores full marks: Part (a) names the precise fault: SAS needs the angle to be between the two sides. With a $48^{\circ}$ angle at $A$ and a $7$ cm side opposite it, two different triangles can actually be drawn, which is why the information is not enough. Part (b) gives a fact that turns it into a genuine congruence condition.
Question 4
Paper 1H/2H style — 3 marks3 marks
Two regular hexagons are mathematically similar.

The area of the smaller hexagon is $50 \text{ cm}^2$ . The area of the larger hexagon is $72 \text{ cm}^2$ .

The perimeter of the larger hexagon is $42$ cm.

Work out the perimeter of the smaller hexagon. [3]
Model answer
$k^{2} = \dfrac{72}{50} = 1.44 \quad\Rightarrow\quad k = 1.2$

$42 \div 1.2 = 35$

Answer: $35$ cm
Why this scores
Mark scheme:
- M1 for $\dfrac{72}{50}$ or $\dfrac{50}{72}$ oe
- M1 for $\sqrt{1.44}$ (= $1.2$ ) or $\sqrt{\dfrac{50}{72}}$ oe
- A1 for $35$ cao
Why this scores full marks: The script turns the area ratio into a length ratio with a square root, then divides because it is going from the larger shape to the smaller one. A perimeter of $35$ cm, less than $42$ cm, is the right way round.
Question 5
Paper 1H/2H style — 3 marks3 marks
A company makes a larger version of its tin. The larger tin is mathematically similar to the original tin.

The volume of the larger tin is $72.8\%$ greater than the volume of the original tin.

Work out the percentage increase in the surface area. [3]
Model answer
$k^{3} = 1.728 \quad\Rightarrow\quad k = \sqrt[3]{1.728} = 1.2$

$k^{2} = 1.2^{2} = 1.44$

The surface area is multiplied by $1.44$

Answer: $44\%$
Why this scores
Mark scheme:
- M1 for $\sqrt[3]{1.728}$ (= $1.2$ )
- M1 for $1.2^2$ (= $1.44$ )
- A1 for $44$ cao
Why this scores full marks: A $72.8\%$ increase is a multiplier of $1.728$ , and that multiplier is the volume scale factor $k^3$ . The script cube-roots it to get the length scale factor, squares that for area, and converts the multiplier $1.44$ back into a percentage increase. Weaker scripts take two-thirds of the percentage, $\dfrac{2}{3} \times 72.8 = 48.5\%$ , or cube-root $72.8$ itself; percentages cannot be scaled like that, only multipliers.
Question 6
Paper 1H/2H style — 3 marks3 marks
$ABCD$ is a kite with $AB = AD$ and $CB = CD$ .

The diagonal $AC$ is drawn.

Prove that triangle $ABC$ is congruent to triangle $ADC$ . [3]
Model answer
$AB = AD$ (given)

$CB = CD$ (given)

$AC$ is common to both triangles

All three pairs of sides are equal, so triangle $ABC$ is congruent to triangle $ADC$ (SSS).
Why this scores
Mark scheme:
- M1 for $AB = AD$ and $CB = CD$ (given)
- M1 for $AC$ is common (a shared side)
- A1 for a complete proof with the condition SSS
Why this scores full marks: Every statement has a reason, and the shared side is stated explicitly. Many scripts use $AC$ without saying it belongs to both triangles, which leaves only two sides and cannot earn the final mark.
Question 7
Paper 1H/2H style — 4 marks4 marks
$ABC$ is an equilateral triangle.

$D$ is a point on $BC$ , $E$ is a point on $CA$ and $F$ is a point on $AB$ such that $BD = CE = AF$ .

Prove that triangle $DEF$ is equilateral. [4]
Model answer
$AF = BD = CE$ (given)

$AB = BC = CA$ (sides of an equilateral triangle), so $AE = CA - CE$ , $BF = AB - AF$ and $CD = BC - BD$ are also equal

Angle $FAE$ = angle $DBF$ = angle $ECD = 60^{\circ}$ (angles in an equilateral triangle are $60^{\circ}$ )

In triangles $AFE$ , $BDF$ and $CED$ , two sides and the angle between them are equal, so the three triangles are congruent (SAS).

So $FE = DF = ED$ (corresponding sides of congruent triangles are equal), and triangle $DEF$ is equilateral.
Why this scores
Mark scheme:
- M1 for $AF = BD = CE$ and $AE = BF = CD$ with reasons oe
- M1 for angle $A$ = angle $B$ = angle $C = 60^{\circ}$ (angles in an equilateral triangle)
- A1 for triangles $AFE$ , $BDF$ and $CED$ congruent (SAS), with all reasons
- A1 for $FE = DF = ED$ (corresponding sides of congruent triangles) and a conclusion
Why this scores full marks: The script finds the second pair of equal sides by subtraction, each side of $ABC$ minus an equal length, and gives that reason. The $60^{\circ}$ angle is the angle between the two known sides in every triangle, so the condition is genuinely SAS. It then uses the congruence to reach the sides of $DEF$ , which is what the question asks. Weaker scripts say $DEF$ "looks equilateral", or prove only two of the corner triangles congruent and so show only two sides of $DEF$ are equal.
Question 8
Paper 1H/2H style — 4 marks4 marks
$ABC$ is a triangle. $D$ is a point on $AB$ and $E$ is a point on $AC$ such that $DE$ is parallel to $BC$ .

The ratio of the area of triangle $ADE$ to the area of the quadrilateral $DBCE$ is $4 : 5$

Work out the ratio $AD : DB$ . [4]
Model answer
Triangles $ADE$ and $ABC$ are similar (corresponding angles, $DE$ parallel to $BC$ ).

Area $ADE$ : area $ABC = 4 : (4 + 5) = 4 : 9$

$AD : AB = \sqrt{4} : \sqrt{9} = 2 : 3$

$DB = AB - AD \quad\Rightarrow\quad AD : DB = 2 : (3 - 2)$

Answer: $2 : 1$
Why this scores
Mark scheme:
- M1 for area $ADE$ : area $ABC = 4 : 9$ oe
- M1 for $AD : AB = 2 : 3$ oe (square root of the area ratio)
- M1 for $AD : DB = 2 : (3 - 2)$ oe
- A1 for $2 : 1$ oe
Why this scores full marks: The area ratio compares similar shapes only when both are triangles, so the script first converts $4 : 5$ into triangle to triangle, $4 : 9$ . Square-rooting $4 : 5$ directly is the most common error; it gives $2 : \sqrt{5}$ , which has no meaning here.
Question 9
Paper 1H/2H style — 5 marks5 marks
$ABC$ is a triangle. $D$ is a point on $AB$ and $E$ is a point on $AC$ such that $DE$ is parallel to $BC$ .

$AD = x$ cm, $DB = 4$ cm, $DE = 6$ cm and $BC = (x + 1)$ cm.

(a) Show that $x^2 - 5x - 24 = 0$ [3]

(b) Work out the length of $BC$ . Show clear algebraic working. [2]
Model answer
(a) Triangles $ADE$ and $ABC$ are similar, and $AB = x + 4$

$\dfrac{x}{x + 4} = \dfrac{6}{x + 1}$

$x(x + 1) = 6(x + 4)$

$x^{2} + x = 6x + 24$

$x^{2} - 5x - 24 = 0$

(b)

$(x - 8)(x + 3) = 0$

$x = 8$ or $x = -3$ , but $x$ is a length so $x = 8$

$BC = 8 + 1 = 9$

Answer: $9$ cm
Why this scores
Mark scheme:
- M1 for a correct equation using corresponding sides, e.g. $\dfrac{x}{x + 4} = \dfrac{6}{x + 1}$ oe
- M1 for $x(x + 1) = 6(x + 4)$ oe, with brackets
- A1 for reaching $x^2 - 5x - 24 = 0$ with no errors seen
- M1 for $(x - 8)(x + 3)$ or a correct use of the quadratic formula
- A1 for $9$ cao
Why this scores full marks: The large triangle's side is written as $x + 4$ before the ratio is formed, and the cross-multiplication keeps the brackets. In (b) the negative root is rejected with a reason, and the script answers the question asked (the length of $BC$ ), not just $x = 8$ . Check: $\dfrac{8}{12} = \dfrac{6}{9}$ .
Question 10
Paper 1H/2H style — 5 marks5 marks
A glass is in the shape of a cone with its vertex at the bottom. The height of the glass is $12$ cm, and it holds $270$ ml when full.

The liquid in the glass always forms a cone that is mathematically similar to the glass.

(a) The glass is filled with juice to a depth of $8$ cm. Work out the volume of juice in the glass. [2]

(b) More juice is poured in until the depth is $10$ cm. Work out the volume of juice that was added. Give your answer correct to $3$ significant figures. [3]
Model answer
(a)

$270 \times \left(\dfrac{8}{12}\right)^{3} = 270 \times \dfrac{8}{27} = 80$

Answer: $80$ ml

(b)

$270 \times \left(\dfrac{10}{12}\right)^{3} = 270 \times \dfrac{125}{216} = 156.25$

$156.25 - 80 = 76.25$

Answer: $76.3$ ml
Why this scores
Mark scheme:
- M1 for (a) $270 \times \left(\dfrac{8}{12}\right)^3$ oe
- A1 for (a) $80$ cao
- M1 for (b) $270 \times \left(\dfrac{10}{12}\right)^3$ oe (= $156.25$ )
- M1 for (b) their $156.25$ − their $80$
- A1 for (b) awrt $76.3$ (accept $76.25$ )
Why this scores full marks: The script uses the full glass as the reference cone both times and cubes the ratio of depths. It never needs the radius or the cone-volume formula. Scaling the liquid from $8$ cm to $10$ cm by $\left(\dfrac{10}{8}\right)^3$ from the $80$ ml is also valid and gives the same $156.25$ ml.

4MA1 Congruence, Similarity and Geometrical Proof

At a glance

What you’ll learn

Congruent triangles

Similar shapes

Geometrical proof

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Identify a congruence test

2Find a length using similarity

3Area scale factor

4Volume scale factor

5Prove triangles are similar

6Ratio of areas from a ratio of sides [1 mark]

7Congruent by ASA, and which sides match [2 marks]

8Height of a lamp post from a shadow [3 marks]

9Area on a model boat and on the real boat [3 marks]

10Diagonals of a parallelogram bisect each other [4 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10