What is similarity in geometry according to the Cambridge IGCSE Mathematics curriculum?

In geometry, similarity refers to the relationship between two shapes that have the same form but may differ in size. According to the Cambridge IGCSE Mathematics curriculum, two figures are similar if their corresponding angles are equal and their corresponding sides are in proportion.

How do you determine if two triangles are similar in IGCSE Mathematics?

To determine if two triangles are similar in IGCSE Mathematics, you can use criteria such as Angle-Angle (AA), Side-Angle-Side (SAS), or Side-Side-Side (SSS). If two angles of one triangle are equal to two angles of another triangle, or if the sides of one triangle are proportional to the sides of another, the triangles are similar.

What is the significance of scale factor in similarity?

The scale factor is a crucial concept in similarity as it describes the ratio of the lengths of corresponding sides of similar figures. In the context of Cambridge IGCSE Mathematics, the scale factor helps in calculating unknown side lengths and understanding the proportionality between similar shapes.

Can similarity be applied to real-world problems in IGCSE Mathematics?

Yes, similarity can be applied to solve real-world problems in IGCSE Mathematics. For example, it is used in map reading, creating models, and understanding proportions in architectural designs. By applying the principles of similarity, students can solve problems involving scale drawings and models.

How does similarity differ from congruence in geometry?

Similarity and congruence are both concepts in geometry, but they differ in terms of size. Similar figures have the same shape but different sizes, with corresponding angles equal and sides proportional. In contrast, congruent figures are identical in both shape and size, with all corresponding sides and angles equal. This distinction is important in the Cambridge IGCSE Mathematics curriculum.

Why is "Multiplying area by the linear scale factor (instead of its square)" a common mistake in Similarity?

Why it happens: Forgetting to square. How to avoid it: Linear k → Area uses k^2, Volume uses k^3. ALWAYS. Source: 0580/42 — every series

Why is "Multiplying volume by the linear scale factor" a common mistake in Similarity?

Why it happens: Same root cause as above. How to avoid it: Volume scale factor is k^3.

Why is "Pairing the wrong sides between two similar figures" a common mistake in Similarity?

Why it happens: Not aligning by equal angles first. How to avoid it: Identify equal angles first; sides BETWEEN those equal angles are corresponding.

Why is "Using $k = \frac{\text{old}}{\text{new}}$ instead of $\frac{\text{new}}{\text{old}}$" a common mistake in Similarity?

Why it happens: Forgetting the convention. How to avoid it: Decide first: are you scaling UP (factor > 1) or DOWN (< 1)? Use the value that matches.

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Short Notes - Similarity

Detailed Study Notes

Detailed notes on Geometry for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Similarity — Cambridge IGCSE 0580 Maths Extended (2026)

Same shape, different size. Match angles, scale corresponding sides, and use the squared and cubed scale factors for area and volume. The cleanest way to find unknown lengths in a diagram with parallel lines.

At a glance

Similar = same shape (equal angles, sides in proportion).
Linear scale factor $k$ : ratio of corresponding sides.
Area scale factor $= k^{2}$ .
Volume scale factor $= k^{3}$ .
Conditions for similarity: AA, SSS proportional, SAS proportional.
Parallel lines inside a triangle create similar sub-triangles.
Always match corresponding vertices when writing ratios.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E4.4 — Use the relationship between similar figures: ratios of sides; areas; volumes.

What 'similar' means

Same angles, sides in the same ratio. Often denoted with $\sim$ .

Similar figures have:

Equal corresponding angles.
Corresponding sides in proportion — there's a single scale factor $k$ that converts one figure's sides into the other's.

Notation: $\triangle ABC \sim \triangle DEF$ means triangle $ABC$ is similar to triangle $DEF$ , with $A \leftrightarrow D$ , $B \leftrightarrow E$ , $C \leftrightarrow F$ .

Conditions for similarity (triangles).

AA (angle-angle): two pairs of equal angles. The third pair is automatic ( $180\degree$ sum).
SSS proportional: all three pairs of sides in the same ratio.
SAS proportional: two pairs of sides in the same ratio AND the included angle equal.

Test	What you need	Example
AA	two pairs of equal angles	$60\degree$ and $80\degree$ in both
SSS	all three side ratios equal	$\frac{6}{4} = \frac{9}{6} = \frac{12}{8}$
SAS	two side ratios equal + included angle equal	ratio $1.5$ with a shared $50\degree$

Worked. Two triangles share two equal angles ( $60\degree$ and $80\degree$ ). Are they similar?

AA satisfied → similar.

Worked. Triangle $A$ has sides $4, 6, 8$ . Triangle $B$ has sides $6, 9, 12$ . Are they similar?

Ratios: $\dfrac{6}{4} = 1.5$ , $\dfrac{9}{6} = 1.5$ , $\dfrac{12}{8} = 1.5$ . All equal.
Yes, similar (SSS proportional). Scale factor $k = 1.5$ from $A$ to $B$ .

Similar triangles have equal angles; every side of the large triangle is k times the matching side of the small one.

Similar = same shape, different size.
Equal corresponding angles + proportional corresponding sides.
AA, SSS proportional, SAS proportional are the three triangle tests.
Match vertices in the correct order: $A \leftrightarrow D$ , $B \leftrightarrow E$ , $C \leftrightarrow F$ .

Finding unknown sides in similar figures

Match corresponding sides, set up a proportion, cross-multiply.

Method.

Identify which figures are similar (often given).
Match corresponding vertices.
Write equal ratios of corresponding sides.
Cross-multiply to find the unknown.

Worked. Triangles $ABC$ and $DEF$ are similar with $A \leftrightarrow D$ , etc. $AB = 6$ , $DE = 9$ , $BC = 8$ . Find $EF$ .

$\dfrac{AB}{DE} = \dfrac{BC}{EF}$ .
$\dfrac{6}{9} = \dfrac{8}{EF}$ .
$EF = \dfrac{8 \times 9}{6} = 12$ .

Tip. Sketch the two triangles with corresponding sides aligned the same way. Then it's clear which side maps to which.

Match corresponding sides.
Equal ratios.
Cross-multiply for the unknown.
Always re-sketch if the original diagram is rotated.

Area and volume scale factors

Linear scale $k$ → area $k^{2}$ , volume $k^{3}$ . The trap that catches everyone.

If two figures are similar with linear scale factor $k$ :

Area scale factor $= k^{2}$ .
Volume scale factor $= k^{3}$ .

Doubling the lengths quadruples the area and multiplies the volume by eight.

Linear scale $k$	Length	Area ( $k^{2}$ )	Volume ( $k^{3}$ )
$2$	$\times 2$	$\times 4$	$\times 8$
$3$	$\times 3$	$\times 9$	$\times 27$
$\tfrac{1}{2}$	$\times 0.5$	$\times 0.25$	$\times 0.125$

Worked. Two similar triangles. The smaller has area $20\,\text{cm}^{2}$ . Lengths in the larger are $3$ times those in the smaller. Find the area of the larger.

$k = 3$ . Area scale: $k^{2} = 9$ .
Larger area: $20 \times 9 = 180\,\text{cm}^{2}$ .

Worked. Two similar cylinders. Volume of smaller is $50\,\text{cm}^{3}$ . The larger has lengths $2$ times bigger. Volume of larger?

$k = 2$ . Volume scale: $k^{3} = 8$ .
Larger volume: $50 \times 8 = 400\,\text{cm}^{3}$ .

Reverse direction. "Two similar triangles have areas $25$ and $100$ . Find the linear scale factor."

Area ratio: $4$ . Linear ratio: $\sqrt{4} = 2$ .

Reverse for volume. Volume ratio $27$ → linear ratio $\sqrt[3]{27} = 3$ .

Linear $k$ → area $k^{2}$ → volume $k^{3}$ .
Reverse: $k_{\text{linear}} = \sqrt{k_{\text{area}}} = \sqrt[3]{k_{\text{volume}}}$ .
Common Paper 4 trap: confusing $k^{2}$ with $k$ .

Similar triangles inside parallel lines

Parallel lines crossing two transversals create similar triangles. Use to find unknown lengths.

When parallel lines cut across two converging transversals (like rays from a single point), they create similar triangles by AA.

A line parallel to the base splits off a smaller triangle similar to the whole.

Worked example. A small triangle is nested inside a larger one, sharing the same apex, with a base parallel to the larger triangle's base. The two triangles are similar.

If the smaller triangle has base $4$ and height $3$ , and the larger has height $9$ :

Linear scale $k = 9/3 = 3$ .
Larger base = $3 \times 4 = 12$ .

Common scenario. Cambridge gives you a triangle with a line drawn parallel to one side, dividing the triangle into a smaller triangle plus a trapezium. The smaller triangle is similar to the whole.

Parallel-line setups → similar triangles by AA.
Same apex, parallel base → similar.
Use linear scale to find unknown lengths.

Quick recap

Similar = same angles + sides in proportion.
Triangle similarity tests: AA, SSS proportional, SAS proportional.
Match corresponding vertices when writing ratios.
Linear scale $k$ → area $k^{2}$ , volume $k^{3}$ .
Reverse: take the right root.
Parallel lines often create similar triangles.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Similar figures — same shape, sides in proportion, equal corresponding angles.
Linear scale factor — ratio of corresponding sides between two similar figures.
Area scale factor — $k^{2}$ , where $k$ is the linear scale factor.
Volume scale factor — $k^{3}$ .
AA similarity — two pairs of equal corresponding angles imply similar triangles.

How it’s examined

Similarity appears most years on Paper 4 as a 4-6 mark question — often combining length finding with area/volume scaling. Paper 2 has simpler matching-sides questions (2-3 marks). Examiner reports flag confusing $k^{2}$ and $k^{3}$ with the linear scale.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E4.4); 0580/42 Oct/Nov 2024 — Q11 (similar volumes); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Similarity

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

Question type:

Question patterns to master — Similarity

Almost every similarity exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the missing side / area / ratio — on a given pair of similar figures, with the scale factor obtainable in one step.

How to approach it

Form the linear scale factor $k = \frac{\text{new}}{\text{old}}$ from a matched pair, then multiply by $k$ for lengths, $k^{2}$ for areas, $k^{3}$ for volumes; square- or cube-root to reverse.

Common trap

Scaling an area by $k$ instead of $k^{2}$ (or a volume by $k$ ). Examiner reports flag this every series — lengths, areas and volumes use three different multipliers.

Multi-step problem

Recognise it by

Two conversions chained — area ratio to linear to volume ratio, or one figure enlarged and both its area and volume required.

How to approach it

Recover the linear factor first (square-root an area ratio), then apply $k^{2}$ and $k^{3}$ separately to reach the area and volume answers.

Common trap

Applying a single factor to both area and volume. Examiner reports stress the area and volume multipliers ( $k^{2}$ and $k^{3}$ ) are distinct.

Word problem

Recognise it by

A real-world scale-model context — model car, scale drawing — usually with a unit conversion buried in it.

How to approach it

Convert both quantities to the same unit before forming the scale factor, then apply $k^{3}$ for a capacity or volume.

Common trap

Forgetting the unit conversion (metres vs centimetres), or scaling a volume by the linear factor. Examiner reports flag both — convert units first.

Show that / prove

Recognise it by

The words show that, prove or justify — typically that two triangles are similar.

How to approach it

Find two pairs of equal angles, naming the theorem for each (vertically opposite, alternate angles), then state that the triangles are similar by AA.

Common trap

Claiming similarity from a single pair of equal angles, or from a vague "they look similar". Examiner reports stress AA needs two named, justified angle pairs.

1Find a missing side from similarity
CoreDirect calculation• scale factor
Question
Two similar triangles. The first has sides $4, 6, 8$ . The second has the corresponding side $4$ matched to $10$ . Find the other two sides.
Step-by-step solution
1. Step 1
  Linear scale factor.
  $k = \dfrac{10}{4} = 2.5$
2. Step 2
  Apply $k$ to each side.
  $6 \times 2.5 = 15,\quad 8 \times 2.5 = 20$
Answer
Sides are $15$ and $20$
2Use the area scale factor
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q10• area scale factor
Question
Two similar shapes have lengths in ratio $3:5$ . The smaller shape has area $36\,\text{cm}^{2}$ . Find the area of the larger.
Step-by-step solution
1. Step 1
  Area scale factor is the SQUARE of the linear scale factor.
  $\left(\dfrac{5}{3}\right)^{2} = \dfrac{25}{9}$
2. Step 2
  Multiply.
  $36 \times \dfrac{25}{9} = 100$
Answer
$100\,\text{cm}^{2}$
3Use the volume scale factor
ExtendedDirect calculation• volume scale factor
Question
Two similar bottles. The smaller holds $200\,\text{ml}$ . The bottles have a height ratio $1:2$ . Find the volume of the larger.
Step-by-step solution
1. Step 1
  Volume scale factor is the CUBE of the linear ratio.
  $2^{3} = 8$
2. Step 2
  Multiply.
  $200 \times 8 = 1600$
Answer
$1600\,\text{ml}\ (= 1.6\,\text{litres})$
4Reverse: find linear ratio from area ratio
ExtendedDirect calculation• area to linear
Question
Two similar shapes have areas in ratio $9 : 25$ . Find the ratio of their corresponding lengths.
Step-by-step solution
1. Step 1
  Take the square root of each part.
  $\sqrt{9} : \sqrt{25} = 3 : 5$
Answer
$3 : 5$
5Identify similar triangles from equal angles
CoreShow that / prove• similar triangles, identify
Question
Triangles $ABC$ and $DEF$ have $\angle A = \angle D = 62\degree$ and $\angle B = \angle E = 47\degree$ . Are the triangles similar? Justify your answer.
Step-by-step solution
1. Step 1
  Two angles in $\triangle ABC$ match two angles in $\triangle DEF$ .
2. Step 2
  Since the angles of a triangle sum to $180\degree$ , the third angles must also match: $\angle C = \angle F = 180 - 62 - 47 = 71\degree$ .
3. Step 3
  All three pairs of corresponding angles are equal — the AAA (or AA) test confirms similarity.
Answer
Yes — by AA, $\triangle ABC \sim \triangle DEF$ (all three corresponding angles are equal).
Examiner tip
The examiner report flags candidates who claim similarity from only one pair of equal angles. AA needs TWO pairs of equal angles (the third follows automatically).
6Find a missing length in similar triangles
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2023 Q9• similar triangles, ratio
Question
$\triangle ABC \sim \triangle DEF$ with $AB$ corresponding to $DE$ . $AB = 8\,\text{cm}$ , $BC = 12\,\text{cm}$ , $DE = 14\,\text{cm}$ . Find $EF$ .
Step-by-step solution
1. Step 1
  Linear scale factor from $\triangle ABC$ to $\triangle DEF$ .
  $k = \dfrac{DE}{AB} = \dfrac{14}{8} = \dfrac{7}{4}$
2. Step 2
  Apply the scale factor to the corresponding side $BC$ .
  $EF = 12 \times \dfrac{7}{4} = 21\,\text{cm}$
Answer
$EF = 21\,\text{cm}$
Examiner tip
The examiner report flags candidates who match sides by length instead of by position. Always identify CORRESPONDING sides via the equal angles before forming the ratio.
7Scaling word problem with units conversion
ExtendedWord problem• word problem, scale model
Question
A model car is a scale model of a real car. The real car is $4.5$ m long and the model is $15$ cm long. The real car has a fuel tank of capacity $54$ litres. Find the capacity of the model's fuel tank in millilitres.
Step-by-step solution
1. Step 1
  Convert both lengths to the same unit. Real car $4.5$ m $= 450$ cm.
2. Step 2
  Linear scale factor from real to model.
  $k = \dfrac{15}{450} = \dfrac{1}{30}$
3. Step 3
  Volumes scale by $k^{3}$ .
  $k^{3} = \left(\dfrac{1}{30}\right)^{3} = \dfrac{1}{27{,}000}$
4. Step 4
  Real capacity in millilitres: $54$ litres $= 54{,}000$ ml.
  $V_{\text{model}} = 54{,}000 \times \dfrac{1}{27{,}000} = 2\,\text{ml}$
Answer
$2\,\text{ml}$
Examiner tip
The examiner report flags candidates who scale the volume by the linear factor or forget to convert metres to centimetres. Always convert units BEFORE forming the scale factor.
8Volume ratio from surface-area ratio
ExtendedMulti-step problem• volume, area, ratio
Question
Two similar solids have surface areas in ratio $16 : 49$ . Find the ratio of their volumes.
Step-by-step solution
1. Step 1
  Take square roots to recover the linear ratio.
  $k = \sqrt{16} : \sqrt{49} = 4 : 7$
2. Step 2
  Cube the linear ratio for volumes.
  $k^{3} = 4^{3} : 7^{3} = 64 : 343$
Answer
$64 : 343$
9Prove two triangles are similar (proof-style)
ChallengeShow that / prove• Adapted from 0580/42 May/Jun 2024 Q14• proof, similar triangles
Question
$ABCD$ is a trapezium with $AB \parallel DC$ . The diagonals $AC$ and $BD$ intersect at $X$ . Show that $\triangle ABX$ is similar to $\triangle CDX$ .
Step-by-step solution
1. Step 1
  $\angle AXB = \angle CXD$ (vertically opposite angles).
2. Step 2
  $\angle BAX = \angle DCX$ (alternate angles between parallels $AB$ and $DC$ , transversal $AC$ ).
3. Step 3
  Two pairs of corresponding angles are equal, so by AA the triangles are similar.
Answer
$\triangle ABX \sim \triangle CDX$ by AA (vertically opposite at $X$ ; alternate angles from $AB \parallel DC$ ).
Examiner tip
The examiner report flags candidates who claim similarity from a vague "they look similar". You must NAME each equal-angle theorem and give a reason; the AA argument is the only path that scores full marks.
10Enlargement of a solid by area and volume
ChallengeMulti-step problem• enlargement, volume, area
Question
A cylindrical tin has surface area $150\,\text{cm}^{2}$ and volume $200\,\text{cm}^{3}$ . A larger similar tin is made by enlarging the original by linear scale factor $1.5$ . Find the surface area and volume of the larger tin.
Step-by-step solution
1. Step 1
  Area scales by $k^{2}$ .
  $k^{2} = 1.5^{2} = 2.25$
2. Step 2
  New surface area.
  $A_{\text{new}} = 150 \times 2.25 = 337.5\,\text{cm}^{2}$
3. Step 3
  Volume scales by $k^{3}$ .
  $k^{3} = 1.5^{3} = 3.375$
4. Step 4
  New volume.
  $V_{\text{new}} = 200 \times 3.375 = 675\,\text{cm}^{3}$
Answer
Surface area $= 337.5\,\text{cm}^{2}$ ; volume $= 675\,\text{cm}^{3}$ .
Examiner tip
The examiner report flags candidates who apply the same factor $1.5$ to both area and volume. Linear $k$ scales lengths; $k^{2}$ scales areas; $k^{3}$ scales volumes — three different multipliers.

Key Formulae — Similarity

The formulae you need to memorise for similarity on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Linear scale factor
$k = \dfrac{\text{new length}}{\text{old length}}$
When to use
Use to convert between corresponding lengths in similar figures.
Area scale factor
$k_{\text{area}} = k^{2}$
When to use
When converting areas of similar figures.
Volume scale factor
$k_{\text{volume}} = k^{3}$
When to use
When converting volumes (or capacities) of similar solids.

Key Definitions and Keywords — Similarity

Definitions to memorise and the exact keywords mark schemes credit for similarity answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Similar figures
Examiner keyword
Two figures with corresponding angles equal and corresponding sides in the same ratio.
Corresponding sides
Examiner keyword
Sides in matching positions in two similar figures — found by aligning equal angles.
Scale factor
Examiner keyword
The constant ratio between corresponding lengths of two similar figures.
Congruent figures
Similar figures with scale factor $1$ — identical in size and shape.

Common Mistakes and Misconceptions — Similarity

The traps other students keep falling into on similarity questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Multiplying area by the linear scale factor (instead of its square)
0580/42 — every series
Why it happens
Forgetting to square.
How to avoid it
Linear $k$ → Area uses $k^{2}$ , Volume uses $k^{3}$ . ALWAYS.
✕Multiplying volume by the linear scale factor
Why it happens
Same root cause as above.
How to avoid it
Volume scale factor is $k^{3}$ .
✕Pairing the wrong sides between two similar figures
Why it happens
Not aligning by equal angles first.
How to avoid it
Identify equal angles first; sides BETWEEN those equal angles are corresponding.
✕Using $k = \frac{\text{old}}{\text{new}}$ instead of $\frac{\text{new}}{\text{old}}$
Why it happens
Forgetting the convention.
How to avoid it
Decide first: are you scaling UP (factor $> 1$ ) or DOWN ( $< 1$ )? Use the value that matches.

Similarity — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Test

What you need

Example

two pairs of equal angles

60\degree

and

80\degree

in both

SSS

all three side ratios equal

\frac{6}{4} = \frac{9}{6} = \frac{12}{8}

SAS

two side ratios equal + included angle equal

ratio

1.5

with a shared

50\degree

Linear scale

k

Length

Area (

k^{2}

)

Volume (

k^{3}

)

2

\times 2

\times 4

\times 8

3

\times 3

\times 9

\times 27

\tfrac{1}{2}

\times 0.5

\times 0.25

\times 0.125

When parallel lines cut across two converging transversals (like rays from a single point), they create similar triangles by AA.

A line parallel to the base splits off a smaller triangle similar to the whole.

Worked example. A small triangle is nested inside a larger one, sharing the same apex, with a base parallel to the larger triangle's base. The two triangles are similar.

If the smaller triangle has base $4$ and height $3$ , and the larger has height $9$ :

Linear scale $k = 9/3 = 3$ .
Larger base = $3 \times 4 = 12$ .

Similarity

Short Notes - Similarity

Detailed Study Notes

At a glance

What you’ll learn

What 'similar' means

Finding unknown sides in similar figures

Area and volume scale factors

Similar triangles inside parallel lines

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Similarity

1Find a missing side from similarity

2Use the area scale factor

3Use the volume scale factor

4Reverse: find linear ratio from area ratio

5Identify similar triangles from equal angles

6Find a missing length in similar triangles

7Scaling word problem with units conversion

8Volume ratio from surface-area ratio

9Prove two triangles are similar (proof-style)

10Enlargement of a solid by area and volume

Linear scale factor

Area scale factor

Volume scale factor

Similar figures

Corresponding sides

Scale factor

Congruent figures

✕Multiplying area by the linear scale factor (instead of its square)

✕Multiplying volume by the linear scale factor

✕Pairing the wrong sides between two similar figures

✕Using k=oldnewk = \frac{\text{old}}{\text{new}}k=newold​ instead of newold\frac{\text{new}}{\text{old}}oldnew​

Similarity — frequently asked questions

Similarity

Short Notes - Similarity

Detailed Study Notes

At a glance

What you’ll learn

What 'similar' means

Finding unknown sides in similar figures

Area and volume scale factors

Similar triangles inside parallel lines

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Similarity

1Find a missing side from similarity

2Use the area scale factor

3Use the volume scale factor

4Reverse: find linear ratio from area ratio

5Identify similar triangles from equal angles

6Find a missing length in similar triangles

7Scaling word problem with units conversion

8Volume ratio from surface-area ratio

9Prove two triangles are similar (proof-style)

10Enlargement of a solid by area and volume

Linear scale factor

Area scale factor

Volume scale factor

Similar figures

Corresponding sides

Scale factor

Congruent figures

✕Multiplying area by the linear scale factor (instead of its square)

✕Multiplying volume by the linear scale factor

✕Pairing the wrong sides between two similar figures

✕Using k=oldnewk = \frac{\text{old}}{\text{new}}k=newold​ instead of newold\frac{\text{new}}{\text{old}}oldnew​

Similarity — frequently asked questions

✕Using $k = \frac{\text{old}}{\text{new}}$ instead of $\frac{\text{new}}{\text{old}}$

✕Using $k = \frac{\text{old}}{\text{new}}$ instead of $\frac{\text{new}}{\text{old}}$