Similarity notes: ICSE Class 10 Mathematics

1. Similar figures and scale factor

Two figures are similar when they have the same shape: corresponding angles are equal and corresponding sides are in the same ratio $k$ (the scale factor).
An enlargement ( $k > 1$ ) or a reduction ( $k < 1$ ) keeps all angles.
Congruent figures are similar with $k = 1$ .
For triangles, one test is enough (see section 3). For polygons with four or more sides, both equal angles and proportional sides are needed: a square and a rhombus of equal sides are not similar.

Example: triangles with sides $4, 5, 6$ and $8, 10, 12$ cm are similar with $k = 2$ .

Scale factor $k = 2$: $\dfrac{8}{4} = \dfrac{10}{5} = \dfrac{12}{6} = 2$. Lengths in cm.

Equal sides are not enough: the angles differ ($90^\circ$ against $60^\circ$).

Formulas

Corresponding sides
$\dfrac{PQ}{AB} = \dfrac{QR}{BC} = \dfrac{RP}{CA} = k$

Key terms

Similar figures: Figures of the same shape: equal corresponding angles, corresponding sides in a constant ratio.
Scale factor: The constant ratio $k$ of corresponding sides.

2. Similar and congruent; naming in order

Similar: same shape, size may differ. Congruent: same shape and size.
Every congruent pair is similar ( $k = 1$ ); not every similar pair is congruent.
Write $\triangle ABC \sim \triangle PQR$ with vertices in corresponding order: $A \to P$ , $B \to Q$ , $C \to R$ . Then $\angle A = \angle P$ and so on, and $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP}$ .

Example: $\triangle ABC \sim \triangle PQR$ , $AB = 5$ , $BC = 6$ , $CA = 7$ cm, $PQ = 7.5$ cm: $k = \dfrac{7.5}{5} = \dfrac{3}{2}$ , $QR = 9$ cm, $RP = 10.5$ cm.

Congruent means similar with scale factor $1$.
	Similar figures	Congruent figures
Shape	same	same
Size	may differ	same
Corresponding angles	equal	equal
Corresponding sides	in the ratio $k$	equal ( $k = 1$ )
Relation	congruent figures are similar with $k = 1$	not every similar pair is congruent

$\triangle ABC \sim \triangle PQR$ with $A \to P$, $B \to Q$, $C \to R$ and $k = \dfrac{3}{2}$. Lengths in cm.

Given: $AB = 5$, $BC = 6$, $CA = 7$ and $PQ = 7.5$ (cm).

Formulas

Similarity, in order
$\triangle ABC \sim \triangle PQR$

Key terms

Corresponding order: Writing the vertices of similar triangles so that the first letters match, the second letters match and the third letters match.

3. Conditions for similar triangles: AA, SSS, SAS

AA: two pairs of equal angles. The third angles are then equal.
SSS: three pairs of sides in the same ratio (compare shortest with shortest).
SAS: one equal angle, and the two sides including it in the same ratio.
These conditions are used, not proved, in the ICSE 2027 syllabus.

Example: $6, 8, 12$ and $9, 12, 18$ cm: ratios $\dfrac{2}{3}, \dfrac{2}{3}, \dfrac{2}{3}$ , so similar (SSS). $4, 5, 7$ and $8, 10, 15$ : $\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{7}{15}$ , so not similar.

Part 1 on one card.
Condition	What to check	Notation
AA	two pairs of equal angles	$\angle A = \angle P$ , $\angle B = \angle Q$
SSS	three pairs of sides in one ratio	$\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP}$
SAS	an equal angle between two proportional pairs of sides	$\angle B = \angle Q$ , $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$
Result	matching sides in the ratio $k$ ; matching angles equal	$\triangle ABC \sim \triangle PQR$

AA: two equal angles. The third angles are $180^\circ - 50^\circ - 60^\circ = 70^\circ$ in both.

SSS: $\dfrac{5}{10} = \dfrac{6}{12} = \dfrac{7}{14} = \dfrac{1}{2}$. Lengths in cm.

Compare shortest with shortest, longest with longest.
Triangles (cm)	Ratios, shortest to longest	Result
$6, 8, 12$ and $9, 12, 18$	$\dfrac{6}{9} = \dfrac{2}{3}$ , $\dfrac{8}{12} = \dfrac{2}{3}$ , $\dfrac{12}{18} = \dfrac{2}{3}$	equal: similar (SSS)
$4, 5, 7$ and $8, 10, 15$	$\dfrac{4}{8} = \dfrac{1}{2}$ , $\dfrac{5}{10} = \dfrac{1}{2}$ , $\dfrac{7}{15} \ne \dfrac{1}{2}$	not equal: not similar

SAS: $\angle B = \angle Q$, $\dfrac{3}{6} = \dfrac{5}{10}$. Lengths in cm.

Match the facts to the condition. No need to prove the conditions: just use them.

Formulas

SSS
$\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP}$
SAS
$\angle B = \angle Q, \; \dfrac{AB}{PQ} = \dfrac{BC}{QR}$

Key terms

AA condition: Two pairs of equal angles make two triangles similar.
SSS condition: Three pairs of sides in the same ratio make two triangles similar.
SAS condition: An equal angle between two proportional pairs of sides makes two triangles similar.

4. Using similar triangles

Name the similar triangles, state the condition, write the proportion in matching order, then solve.
Shadows: the sun's rays are parallel and objects are vertical, so the triangles are similar (AA): $\dfrac{\text{height}}{\text{shadow}}$ is the same for both.
Common angle: if $\angle A$ is shared and $\angle ADE = \angle ACB$ , then $\triangle ADE \sim \triangle ACB$ with $D \to C$ and $E \to B$ .
Right angle: a perpendicular from the right angle to the hypotenuse splits the triangle into two triangles similar to it (AA).

Example: a $1.2$ m stick has a $1.6$ m shadow; a tree's shadow is $12$ m, so the tree is $\dfrac{12 \times 1.2}{1.6} = 9$ m high.

Stick $AB$ and tree $PQ$; shadows $AC$ and $PR$. Lengths in m.

Two equal pairs of angles: $\triangle ABC \sim \triangle PQR$ (AA). Lengths in cm.

Equal included angles and $\dfrac{5}{7.5} = \dfrac{8}{12}$: SAS. Lengths in cm.

Angle $A$ is common and $\angle ADE = \angle ACB$: $\triangle ADE \sim \triangle ACB$. Lengths in cm.

Key terms

Angle of elevation of the sun: The same for all objects at one moment: the reason shadow triangles are similar.

5. Basic Proportionality Theorem and its converse

Theorem: a line parallel to one side of a triangle divides the other two sides in the same ratio: $DE \parallel BC \Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC}$ . Equivalent forms: $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}$ .
Converse: if $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ (same order), then $DE \parallel BC$ .
Trapezium: in $ABCD$ with $AB \parallel DC$ , diagonals meet at $O$ : $\triangle AOB \sim \triangle COD$ and $\dfrac{AO}{OC} = \dfrac{BO}{OD} = \dfrac{AB}{DC}$ .

Example: $AD = 3$ , $DB = 4.5$ , $AE = 2.4$ cm: $EC = \dfrac{2.4 \times 4.5}{3} = 3.6$ cm. With $AD = 5$ , $DB = 7.5$ , $AE = 4$ , $EC = 6$ both ratios are $\dfrac{2}{3}$ , so $DE \parallel BC$ .

Basic Proportionality Theorem: $DE \parallel BC \Rightarrow \dfrac{AD}{DB} = \dfrac{AE}{EC}$.

Four ways to write the same fact.
Form	Statement	Why
Parts of the sides	$\dfrac{AD}{DB} = \dfrac{AE}{EC}$	the theorem
Part to whole	$\dfrac{AD}{AB} = \dfrac{AE}{AC}$	add $1$ to both sides, or use $\triangle ADE \sim \triangle ABC$
With the parallel line	$\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}$	$\triangle ADE \sim \triangle ABC$ (AA)
Other parts	$\dfrac{DB}{AB} = \dfrac{EC}{AC}$	subtract from $1$

Given $AD = 3$, $DB = 4.5$, $AE = 2.4$ (cm).

Unknown $x$ in the proportion. Lengths in cm.

Converse: equal ratios on the two sides $\Rightarrow$ $DE \parallel BC$.

If $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ then $DE \parallel BC$. Lengths in cm.

Triangles $AOB$ and $COD$ are similar (AA): $\dfrac{AO}{OC} = \dfrac{BO}{OD} = \dfrac{AB}{DC} = \dfrac{3}{2}$. Lengths in cm.

Given $AB = 15$, $DC = 9$, $AO = 10$ (cm).

Formulas

Basic Proportionality Theorem
$\dfrac{AD}{DB} = \dfrac{AE}{EC}$
Diagonals of a trapezium
$\dfrac{AO}{OC} = \dfrac{BO}{OD} = \dfrac{AB}{DC}$

Key terms

Basic Proportionality Theorem: A line parallel to one side of a triangle cuts the other two sides in the same ratio.
Converse: If a line cuts two sides in the same ratio, it is parallel to the third side.

6. Areas of similar triangles

The areas of similar triangles are in the ratio of the squares of corresponding sides: $\dfrac{\text{area}_1}{\text{area}_2} = \left(\dfrac{a_1}{a_2}\right)^2$ .
The same ratio holds for the squares of corresponding altitudes and medians; perimeters follow the sides (not squared).
To go from areas to sides, take the square root.
For $DE \parallel BC$ : $\triangle ADE \sim \triangle ABC$ , so $\dfrac{\text{area } ADE}{\text{area } ABC} = \left(\dfrac{AD}{AB}\right)^2$ , and the trapezium $DBCE$ is the difference.

Example: sides $6 : 10$ , larger area $75$ : smaller $= \dfrac{9}{25} \times 75 = 27$ . Areas $20 : 45$ , side $8$ : larger side $= 8 \times \dfrac{3}{2} = 12$ .

Sides $\times 2$, area $\times 4$: $6 \to 24$ cm$^2$.

Square the ratio of sides to get the ratio of areas.
Step	Working
Ratio of sides	$6 : 10 = 3 : 5$
Ratio of areas	$\left(\dfrac{6}{10}\right)^2 = \dfrac{9}{25}$
Area of the smaller triangle	$\dfrac{9}{25} \times 75 = 27$ cm $^2$

Square root of the area ratio gives the side ratio.
Step	Working
Ratio of areas	$\dfrac{20}{45} = \dfrac{4}{9}$
Ratio of sides: square root	$\sqrt{\dfrac{4}{9}} = \dfrac{2}{3}$
Corresponding side of the larger triangle	$8 \times \dfrac{3}{2} = 12$ cm

The shaded triangle $ADE$ is similar to $ABC$ with ratio $\dfrac{3}{5}$.

Formulas

Areas of similar triangles
$\dfrac{\text{area } \triangle ABC}{\text{area } \triangle PQR} = \left(\dfrac{AB}{PQ}\right)^2$

Key terms

Ratio of areas: For similar triangles, the square of the ratio of corresponding sides.

7. Maps and models

A scale $1 : n$ means lengths on the map or model are $\dfrac{1}{n}$ of the real lengths; areas are $\dfrac{1}{n^2}$ ; volumes are $\dfrac{1}{n^3}$ .
With scale factor $k$ (model : real): lengths $k$ , areas $k^2$ , volumes $k^3$ .
Convert units at the end: $1$ km $= 100\,000$ cm, $1$ m $^2 = 10\,000$ cm $^2$ , $1$ m $^3 = 1\,000\,000$ cm $^3$ .
Going from real to map divides by $n$ (lengths), $n^2$ (areas), $n^3$ (volumes); going from map to real multiplies.

Example: on a $1 : 30\,000$ map, $6.5$ cm is $1.95$ km and $8$ cm $^2$ is $0.72$ km $^2$ . A $1 : 50$ model of a $125$ m $^3$ tank holds $1000$ cm $^3$ .

Map length : real length $= 1 : n$; map area : real area $= 1 : n^2$.
Scale (representative fraction)	Meaning of 1 cm on the map	1 cm $^2$ on the map
$1 : 20\,000$	0.2 km	$0.04$ km $^2$
$1 : 25\,000$	0.25 km	$0.0625$ km $^2$
$1 : 50\,000$	0.5 km	$0.25$ km $^2$
$1 : 100\,000$	1 km	$1$ km $^2$

Lengths use $n$, areas use $n^2$.
Part	Working	Answer
(i) Real length of a 6.5 cm path	$6.5 \times 30000 = 195000$ cm; $\dfrac{195000}{100000} = 1.95$	1.95 km
(ii) Real area of an 8 cm $^2$ lake	$1$ cm $= 0.3$ km, so $1$ cm $^2$ $= 0.3 \times 0.3 = 0.09$ km $^2$ ; $8 \times 0.09 = 0.72$	0.72 km $^2$
(iii) Map length of a 4.5 km road	$\dfrac{450000}{30000} = 15$	15 cm

Lengths $k$, areas $k^2$, volumes $k^3$.
Quantity	Ratio, model : real	For $k = \dfrac{1}{20}$
Length	$k$	$1 : 20$
Area	$k^2$	$1 : 400$
Volume	$k^3$	$1 : 8000$

Divide lengths by $50$, areas by $50^2$, volumes by $50^3$.
Quantity	Real	Factor	Model
Height	45 m $= 4500$ cm	$\div 50$	$\dfrac{4500}{50} = 90$ cm
Roof area	$180$ m $^2$ $= 1\,800\,000$ cm $^2$	$\div 2500$	$\dfrac{1800000}{2500} = 720$ cm $^2$
Tank volume	$125$ m $^3$ $= 125\,000\,000$ cm $^3$	$\div 125\,000$	$\dfrac{125000000}{125000} = 1000$ cm $^3$

Formulas

Lengths, areas, volumes
$k, \; k^2, \; k^3$

Key terms

Representative fraction: The scale $1 : n$ of a map: map length to real length.
Scale factor of a model: The ratio $k$ of a model length to the matching real length.

Exam tips

How marks are awarded

Similar triangles (2–3 marks): state the condition with the equal angles or ratios 1; correspondence in order 1; proportion and answer 1.
Basic Proportionality Theorem (2–3 marks): quote the theorem or the similar triangles 1; proportion 1; answer 1.
Ratio of areas (2–3 marks): ratio of sides 1; squared 1; answer 1.
Maps and models (2–4 marks): scale factor for length 1; squared or cubed for area or volume 1–2; units 1.
10-mark structured question: parts (a), (b), (c), usually $3 + 3 + 4$ ; show every ratio and substitution.

Common mistakes

Matching sides in the wrong order because the vertices are not written in corresponding order.
Using SAS when the equal angle is not between the two proportional sides.
Using the ratio of sides as the ratio of areas (areas need the square) or as the ratio of volumes (volumes need the cube).
Writing the ratios in a different order on the two sides when using the converse of the theorem.
Dividing a map area by $n$ instead of $n^2$ , or forgetting to change cm to km.
Taking the ratio of areas instead of its square root when finding a side.

1. Similar figures and scale factor

Formulas

Key terms

2. Similar and congruent; naming in order

Formulas

Key terms

3. Conditions for similar triangles: AA, SSS, SAS

Formulas

Key terms

4. Using similar triangles

Key terms

5. Basic Proportionality Theorem and its converse

Formulas

Key terms

6. Areas of similar triangles

Formulas

Key terms

7. Maps and models

Formulas

Key terms

Exam tips

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