Similarity ICSE Class 10: Similar Triangles, Area Ratios, Maps and Models

What Similarity is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Similarity is in the 3. Geometry (a) Similarity part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Similarity as a size transformation

If every length of figure $F$ is multiplied by $k$ to give figure $F^{\prime }$, then $F^{\prime }$ is similar to $F$ with scale factor $k$.

• $k>1$: enlargement; $0<k<1$: reduction; $k=1$: congruent (same size).
• Angles do not change.
• Lengths (sides, perimeters, heights, medians) are multiplied by $k$.
• Areas are multiplied by $k^2$ (and volumes of models by $k^3$).

Congruent triangles: equal angles and equal sides. Similar triangles: equal angles and proportional sides.

Worked example

A triangle with sides 3 cm, 4 cm and 5 cm is enlarged with scale factor 2.5. Find the new sides and the ratio of the perimeters.

Try it first, then show the solutionSolution

1. New sides: $3\times 2.5=7.5$, $4\times 2.5=10$, $5\times 2.5=12.5$ cm.
2. Perimeters: 12 cm and 30 cm, ratio $12:30=2:5$, which is $1:k$.

Answer: 7.5 cm, 10 cm, 12.5 cm; perimeters in the ratio 2 : 5

The conditions AA, SAS and SSS

Two triangles are similar if any one of these holds:

• AA: two angles of one equal two angles of the other (the third angles are then equal too).
• SAS: one angle equal and the sides including it proportional.
• SSS: all three pairs of corresponding sides proportional.

Write the similarity statement with vertices in matching order: $△ABC\sim △PQR$ means $\angle A=\angle P$, $\angle B=\angle Q$, $\angle C=\angle R$ and $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$.

Worked example

In $△ABC$, $AB=6$, $BC=8$, $CA=10$; in $△PQR$, $PQ=9$, $QR=12$, $RP=15$ (all in cm). Are they similar?

Try it first, then show the solutionSolution

1. $\frac{AB}{PQ}=\frac{6}{9}=\frac{2}{3}$, $\frac{BC}{QR}=\frac{8}{12}=\frac{2}{3}$, $\frac{CA}{RP}=\frac{10}{15}=\frac{2}{3}$.
2. All three ratios are equal, so $△ABC\sim △PQR$ (SSS).

Answer: Yes: △ABC ∼ △PQR by SSS, scale factor 3/2 from ABC to PQR

Finding equal angles: the usual reasons

Most proofs use AA. The equal angles usually come from:

• a common angle (shared by both triangles);
• vertically opposite angles (where two lines cross);
• alternate or corresponding angles (parallel lines);
• two right angles (given perpendiculars).

State the correct reason for each pair. Name angles with three letters (such as $\angle AOB$), not one, when several angles meet at a point.

Worked example

In trapezium $ABCD$, $AB\parallel DC$ and the diagonals meet at $O$. $AO=6$ cm, $OC=4$ cm and $DC=5$ cm. Find $AB$.

Try it first, then show the solutionSolution

1. $\angle OAB=\angle OCD$ (alternate angles, $AB\parallel DC$); $\angle AOB=\angle COD$ (vertically opposite).
2. So $△AOB\sim △COD$ (AA).
3. $\frac{AB}{CD}=\frac{AO}{CO}=\frac{6}{4}$, so $AB=5\times 1.5=7.5$ cm.

Answer: AB = 7.5 cm

Basic proportionality theorem (BPT)

If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio. In $△ABC$ with $DE\parallel BC$ ($D$ on $AB$, $E$ on $AC$):
$\frac{AD}{DB}=\frac{AE}{EC}.$
Also $△ADE\sim △ABC$, so $\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$. Take care: $\frac{DE}{BC}$ equals $\frac{AD}{AB}$, not $\frac{AD}{DB}$.

The converse also holds: if a line divides two sides in the same ratio, it is parallel to the third side.

Worked example

In $△ABC$, $DE\parallel BC$, $AD=3$ cm, $DB=5$ cm and $AE=4.5$ cm. Find $EC$ and the ratio $DE:BC$.

Try it first, then show the solutionSolution

1. BPT: $\frac{3}{5}=\frac{4.5}{EC}$, so $EC=7.5$ cm.
2. $\frac{DE}{BC}=\frac{AD}{AB}=\frac{3}{8}$.

Answer: EC = 7.5 cm; DE : BC = 3 : 8

Areas of similar triangles

The ratio of the areas of two similar triangles equals the square of the ratio of corresponding sides:
$\frac{\text{area}(△ABC)}{\text{area}(△PQR)}={\left(\frac{AB}{PQ}\right)}^2.$
The same holds for corresponding altitudes, medians and perimeters in place of sides. To go from areas back to sides, take the square root.

Worked example

$△ABC\sim △PQR$, $AB=4$ cm, $PQ=6$ cm and the area of $△ABC$ is 32 cm². Find the area of $△PQR$.

Try it first, then show the solutionSolution

1. $\frac{32}{\text{area }PQR}={\left(\frac{4}{6}\right)}^2=\frac{4}{9}$.
2. Area of $△PQR=\frac{32\times 9}{4}=72$ cm².

Answer: 72 cm²

Maps and models

A map or model with scale $1:n$ is a similar figure with scale factor $k=\frac{1}{n}$.

• Length on map : actual length $=1:n$.
• Area on map : actual area $=1:n^2$.
• Volume of model : actual volume $=1:n^3$.

Convert units carefully: $1\text{ m}=100\text{ cm}$, $1{\text{ m}}^2=10^4{\text{ cm}}^2$, $1{\text{ km}}^2=10^{10}{\text{ cm}}^2$, $1{\text{ m}}^3=10^6{\text{ cm}}^3$.

Worked example

A map is drawn to the scale $1:20000$. Find (a) the actual distance for 6 cm on the map, (b) the actual area, in m², of a park shown as 2 cm².

Try it first, then show the solutionSolution

1. (a) $6\times 20000=120000$ cm $=1.2$ km.
2. (b) Area factor $=20000^2=4\times 10^8$. Actual area $=2\times 4\times 10^8=8\times 10^8$ cm².
3. $8\times 10^8÷10^4=80000$ m².

Answer: (a) 1.2 km (b) 80 000 m²

Right triangle with an altitude to the hypotenuse

In $△ABC$ with $\angle BAC=90^{\circ }$ and $AD\perp BC$, the three triangles $ABC$, $DBA$ and $DAC$ are similar (each pair shares an angle and has a right angle: AA). This gives useful results:
$A{D}^2=BD\cdot DC,A{B}^2=BD\cdot BC,A{C}^2=DC\cdot BC.$
Derive the one you need from the similar triangles in your answer, rather than quoting it.

Worked example

In $△ABC$, $\angle BAC=90^{\circ }$ and $AD\perp BC$. $BD=4$ cm and $DC=9$ cm. Find $AD$.

Try it first, then show the solutionSolution

1. $△DBA\sim △DAC$ (AA: right angles at $D$, and $\angle DBA=\angle DAC$ since each is $90^{\circ }-\angle C$).
2. $\frac{BD}{AD}=\frac{AD}{DC}$, so $A{D}^2=4\times 9=36$.
3. $AD=6$ cm.

Answer: AD = 6 cm

Maths in real life: Similarity

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Similar Triangles: AA, SAS and SSS
• Areas of Similar Triangles
• Maps and Models (Scale Factor)

Similarity formula sheet

Similar triangles (△ABC ∼ △PQR)

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}=k$

Write the vertices in matching order.

Basic proportionality theorem

$DE\parallel BC\Rightarrow \frac{AD}{DB}=\frac{AE}{EC}$

And DE/BC = AD/AB (whole side, not DB).

Areas of similar triangles

$\frac{A_1}{A_2}={\left(\frac{s_1}{s_2}\right)}^2$

s = any pair of corresponding sides, altitudes, medians or perimeters.

Map scale 1 : n

$\text{length}\times n,\text{area}\times n^2$

Convert cm² to m² by dividing by 10 000.

Model scale 1 : n

$\text{volume}\times n^3$

1 m³ = 1 000 000 cm³.

Altitude on the hypotenuse

$A{D}^2=BD\cdot DC$

From △DBA ∼ △DAC.

How the board asks Similarity

The question patterns to expect, and the method that works for each.

Tricks to remember Similarity

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Order is everything

  In △ABC ∼ △PQR, read pairs straight down: A–P, B–Q, C–R; AB–PQ, BC–QR, CA–RP.

  Use it when: Writing proportional sides correctly.

• 1, 2, 3 for L, A, V

  Lengths k¹, areas k², volumes k³.

  Use it when: Maps, models and area ratios.

• Part to part, part to whole

  BPT: part : part on both sides (AD/DB = AE/EC). For DE : BC use part : whole (AD/AB).

  Use it when: Avoiding the most common BPT slip.

• CAVR for angles

  Common, Alternate (or corresponding), Vertically opposite, Right angle: the four usual reasons in an AA proof.

  Use it when: Writing reasons in proofs.

Similarity: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Similarity

The slips examiners see most often, and what to write instead.

• ✗Mistake: Wrong reason for an equal angle

  CISCE examiners' comments, 2024: in proving △APD ∼ △BPC, candidates gave '∠P = ∠P (common)' when the angles were vertically opposite.

  Wrong working

  ∠APD = ∠BPC (common)

  Right working

  ∠APD = ∠BPC (vertically opposite angles)

  ✓Do this: Look at the figure: a common angle is shared by both triangles; vertically opposite angles are where two lines cross. Name angles with three letters.

  Usually costs: 1 mark in the proof.

• ✗Mistake: Matching the wrong sides

  CISCE examiners' comments, 2024 and 2025: proportional sides were written incorrectly, e.g. AD/BC = AE/AC = DE/AB, and in a 2025 MCQ candidates paired AB with FG instead of FE when the triangles were drawn in different orientations.

  ✓Do this: Write the similarity statement in matching order first, then read the side pairs straight from it.

  Usually costs: Every length that follows.

• ✗Mistake: Not squaring for areas

  CISCE examiners' comments, 2024: candidates did not use 'ratio of areas = ratio of squares of corresponding sides'.

  Wrong working

  Sides 3 : 4, so areas 3 : 4

  Right working

  Areas $3^2:4^2=9:16$

  ✓Do this: Lengths k, areas k², volumes k³.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Unit slips on map scales

  CISCE examiners' comments, 2025: in a map question, candidates found the area scale factor k² incorrectly and made errors converting 576 m² to cm².

  ✓Do this: Convert to the same unit first (1 m² = 10 000 cm²), then use n² for areas.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Missing a Pythagoras step or ignoring 'Hence'

  CISCE examiners' comments, 2024: candidates did not use Pythagoras in the right triangle to find a needed side, and ignored 'Hence' in a later part, using a different method.

  ✓Do this: Check for right angles in the figure; when the question says 'Hence', use the result just found.

  Usually costs: 1 to 2 marks.

Practice questions on Similarity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   If $△ABC\sim △PQR$ and $\frac{AB}{PQ}=\frac{2}{3}$, then area $△ABC$ : area $△PQR$ is:
   (a) $2:3$
   (b) $4:9$
   (c) $8:27$
   (d) $3:2$

   Hint

   Square the ratio.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   On a map with scale $1:10000$, two points are 5 cm apart. The actual distance is:
   (a) 50 m
   (b) 0.5 km
   (c) 5 km
   (d) 50 km

   Hint

   50 000 cm.

   Answer

   (b)

3. Q3 · 1 mark · MCQHarder board level

   In $△ABC$, $DE\parallel BC$, $AD=x$, $DB=x-2$, $AE=x+2$ and $EC=x-1$ (all in cm). Then $x$ is:
   (a) 2
   (b) 3
   (c) 4
   (d) 5

   Hint

   Cross-multiply AD/DB = AE/EC.

   Answer

   (c): $x(x-1)=(x-2)(x+2)$ gives $-x=-4$.

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Two triangles whose corresponding angles are equal are always congruent.
   Reason (R): If two angles of one triangle are equal to two angles of another, the triangles are similar.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Equal angles fix the shape, not the size.

   Answer

   (b)

5. Q5 · 4 marks · Short answerHarder board level

   In $△ABC$, $D$ is a point on $BC$ such that $\angle ADC=\angle BAC$. $BC=9$ cm, $CD=4$ cm and $AB=7.5$ cm. (a) Prove $△ADC\sim △BAC$. (b) Find $AC$ and $AD$.

   Hint

   ∠C is common.

   Answer

   (a) $\angle ADC=\angle BAC$ (given) and $\angle ACD=\angle BCA$ (common): AA. (b) $\frac{AC}{BC}=\frac{DC}{AC}$ gives $A{C}^2=36$, $AC=6$ cm. $\frac{AD}{BA}=\frac{AC}{BC}=\frac{2}{3}$, so $AD=5$ cm.

6. Q6 · 4 marks · Short answerHarder board level

   A model of a car is made to the scale $1:25$. (a) The model is 18 cm long; find the length of the car in m. (b) The model windscreen has area 48 cm²; find the actual area in m². (c) The model boot holds 32 cm³; find the actual capacity in m³.

   Hint

   25, 625, 15 625.

   Answer

   (a) $450$ cm $=4.5$ m (b) $48\times 625=30000$ cm² $=3$ m² (c) $32\times 15625=500000$ cm³ $=0.5$ m³

7. Q7 · 3 marks · Long answerBoard level

   Two similar triangles have corresponding sides 4 cm and 7 cm. The area of the smaller is 48 cm². Find the area of the larger.

   Hint

   Multiply by 49/16.

   Answer

   $48\times \frac{49}{16}=147$ cm²

8. Q8 · 4 marks · Long answerHarder board level

   In $△ABC$, $\angle ACB=90^{\circ }$ and $CD\perp AB$. $AD=9$ cm and $DB=16$ cm. (a) Prove $△ACD\sim △CBD$. (b) Find $CD$. (c) Find area $△ACD$ : area $△CBD$.

   Hint

   ∠CAD = ∠BCD (each is 90° − ∠B).

   Answer

   (a) Right angles at $D$, and $\angle CAD=\angle BCD$ (each is $90^{\circ }-\angle B$): AA. (b) $\frac{AD}{CD}=\frac{CD}{BD}$, $C{D}^2=9\times 16=144$, $CD=12$ cm. (c) Ratio of corresponding sides $\frac{AD}{CD}=\frac{9}{12}=\frac{3}{4}$, so areas $9:16$.

9. Q9 · 4 marks · Long answerHOTS / competency

   Two vertical poles $AB=6$ m and $CD=9$ m stand on level ground. The line from the top of each pole to the foot of the other cross at $P$. Using similar triangles, find the height of $P$ above the ground. Does the answer depend on the distance between the poles?

   Hint

   Let the poles be d apart and the foot of the perpendicular from P split d into p and q.

   Answer

   Let $PM=h$ be the perpendicular from $P$ to the ground, with $BM=p$, $MD=q$, $p+q=d$. From similar triangles, $\frac{h}{9}=\frac{p}{d}$ and $\frac{h}{6}=\frac{q}{d}$. Adding: $h\left(\frac{1}{9}+\frac{1}{6}\right)=1$, so $h=\frac{54}{15}=3.6$ m. It does not depend on $d$.

10. Q10 · 4 marks · Long answerHarder board level

    In trapezium $PQRS$, $PQ\parallel SR$ and $PQ=3SR$. The diagonals meet at $O$. If area $△POQ=54$ cm², find area $△ROS$ and the ratio $PO:OR$.

    Hint

    △POQ ∼ △ROS by AA.

    Answer

    Scale factor 3, area factor 9: area $△ROS=54÷9=6$ cm². $PO:OR=3:1$.

11. Q11 · 1 mark · MCQHOTS / competency

    A field of 0.36 km² is shown as 4 cm² on a map. The scale of the map is:
    (a) $1:3000$
    (b) $1:30000$
    (c) $1:300000$
    (d) $1:9\times 10^8$

    Hint

    Convert km² to cm², divide, take the square root.

    Answer

    (b): $0.36$ km² $=3.6\times 10^9$ cm²; $÷4=9\times 10^8$; $\sqrt{9\times 10^8}=30000$.

12. Q12 · 5 marks · Long answerHOTS / competency

    In $△ABC$, $P$ is on $AB$ and $Q$ on $AC$ with $AP=4$ cm, $PB=6$ cm, $AQ=5$ cm and $QC=7.5$ cm. (a) Show that $PQ\parallel BC$. (b) If $BC=12$ cm, find $PQ$. (c) Find area $△APQ$ : area of quadrilateral $PBCQ$.

    Hint

    Converse of BPT.

    Answer

    (a) $\frac{AP}{PB}=\frac{4}{6}=\frac{2}{3}$ and $\frac{AQ}{QC}=\frac{5}{7.5}=\frac{2}{3}$: equal ratios, so $PQ\parallel BC$ (converse of BPT). (b) $\frac{PQ}{BC}=\frac{AP}{AB}=\frac{4}{10}$, $PQ=4.8$ cm. (c) Areas $△APQ:△ABC=4:25$, so $△APQ:PBCQ=4:21$.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Three conditions for similar triangles?Show answer

   AA, SAS, SSS.

2. 2.Key word distinguishing similarity from congruency?Show answer

   Proportionality: sides in the same ratio (not equal).

3. 3.State the basic proportionality theorem.Show answer

   A line parallel to one side of a triangle divides the other two sides in the same ratio.

4. 4.Ratio of areas of similar triangles?Show answer

   The square of the ratio of corresponding sides.

5. 5.Map scale 1 : n — area factor?Show answer

   n²

6. 6.Model scale 1 : n — volume factor?Show answer

   n³

7. 7.1 m² in cm²?Show answer

   10 000 cm²

8. 8.Are proofs of SSS, SAS, AA in the syllabus?Show answer

   No: only applications.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write △ABC ∼ △PQR with matching vertices?
• ☐Do you give the right reason for each equal angle?
• ☐Do you use AD/AB (not AD/DB) for DE : BC?
• ☐Do you square the scale factor for areas?
• ☐Can you convert km², m² and cm² confidently?
• ☐Can you find the area of a trapezium cut off by a parallel line?

Similarity: frequently asked questions