Similarity

Q: Define similar triangles. How does the statement △ ABC sim △ PQR fix which sides correspond?

Similar triangles are triangles whose corresponding angles are equal and whose corresponding sides are proportional. The order of letters pairs A with P, B with Q and C with R, so AB leftrightarrow PQ, BC leftrightarrow QR and CA leftrightarrow RP.

Q: Are all congruent triangles similar? Are all similar triangles congruent? Give a reason for each answer.

All congruent triangles are similar, with scale factor 1. Similar triangles are congruent only when the scale factor is 1; for example, triangles with sides 3, 4, 5 cm and 6, 8, 10 cm are similar but not congruent.

Q: Why are a square and a rhombus of the same side length not similar?

Their sides are in the same ratio, but their angles are not equal (the square has 90° angles and the rhombus does not), and similar polygons must have equal corresponding angles.

Q: A photograph 8 cm wide is reduced so that its width becomes 6 cm. Find the scale factor, and say whether the transformation is an enlargement or a reduction.

k = (6)/(8) = 0.75. Since 0 < k < 1, it is a reduction.

Q: Name the three conditions of similarity of triangles and state the AA condition.

SSS, SAS and AA. AA: if two angles of one triangle are equal to two angles of another triangle, the triangles are similar.

Q: △ XYZ has sides 5 cm, 6 cm, 8 cm and △ LMN has sides 7.5 cm, 9 cm, 12 cm. Are they similar?

Yes. (7.5)/(5) = (9)/(6) = (12)/(8) = 1.5, so they are similar by SSS.

Q: A vertical stick 2 m long casts a shadow 1.6 m long, while a tower casts a shadow 40 m long at the same time. Find the height of the tower.

By AA similarity, (h)/(2) = (40)/(1.6) = 25, so h = 50 m.

Q: In △ ABC, DE ∥ BC with D on AB and E on AC. AD = 3 cm, AB = 7.5 cm and AC = 10 cm. Find AE and EC.

(AE)/(AC) = (AD)/(AB) = (3)/(7.5) = 0.4, so AE = 4 cm and EC = 6 cm.

ICSE Class 10 · Mathematics · Chapter 15

What this chapter covers

Similarity as a size transformation
Similar triangles and corresponding parts
Similarity compared with congruency
Conditions of similarity: the AA test
Conditions of similarity: the SAS and SSS tests
Simple applications of similar triangles
Applications of the Basic Proportionality Theorem
Areas of similar triangles
Applications to maps
Applications to models

Key points

Similarity is a size transformation: every length is multiplied by the same scale factor k and every angle is kept; k > 1 is an enlargement and 0 < k < 1 a reduction.
Similar figures have the same shape but not necessarily the same size; polygons need equal angles and proportional sides.
Congruency is similarity with k = 1; the keyword for similarity is proportionality, not equality.
In △ ABC sim △ PQR the order of letters fixes the corresponding vertices, sides and angles.
Conditions of similarity of triangles: AA, SAS (equal included angle, proportional including sides) and SSS (all three sides proportional). Proofs are not required.
Corresponding sides, altitudes, medians and perimeters of similar triangles are all in the same ratio k.
In a right-angled triangle, the perpendicular from the right angle to the hypotenuse makes two triangles similar to the whole and to each other, giving AB² = BD × BC and AD² = BD × DC.
Basic Proportionality Theorem: if DE ∥ BC, then (AD)/(DB) = (AE)/(EC); for the parallel side use (DE)/(BC) = (AD)/(AB).
Converse: a line dividing two sides of a triangle in the same ratio is parallel to the third side.
Areas of similar triangles are in the ratio of the squares of corresponding sides, so area changes by k².
Maps and models: lengths change by k, areas by k², and volumes or capacities of models by k³; angles do not change.

Questions answered

Define similar triangles. How does the statement △ ABC sim △ PQR fix which sides correspond?

Similar triangles are triangles whose corresponding angles are equal and whose corresponding sides are proportional. The order of letters pairs A with P, B with Q and C with R, so AB leftrightarrow PQ, BC leftrightarrow QR and CA leftrightarrow RP.

Are all congruent triangles similar? Are all similar triangles congruent? Give a reason for each answer.

All congruent triangles are similar, with scale factor 1. Similar triangles are congruent only when the scale factor is 1; for example, triangles with sides 3, 4, 5 cm and 6, 8, 10 cm are similar but not congruent.

Why are a square and a rhombus of the same side length not similar?

Their sides are in the same ratio, but their angles are not equal (the square has 90^° angles and the rhombus does not), and similar polygons must have equal corresponding angles.

A photograph 8 cm wide is reduced so that its width becomes 6 cm. Find the scale factor, and say whether the transformation is an enlargement or a reduction.

k = (6)/(8) = 0.75. Since 0 < k < 1, it is a reduction.

Name the three conditions of similarity of triangles and state the AA condition.

SSS, SAS and AA. AA: if two angles of one triangle are equal to two angles of another triangle, the triangles are similar.

△ XYZ has sides 5 cm, 6 cm, 8 cm and △ LMN has sides 7.5 cm, 9 cm, 12 cm. Are they similar?

Yes. (7.5)/(5) = (9)/(6) = (12)/(8) = 1.5, so they are similar by SSS.

A vertical stick 2 m long casts a shadow 1.6 m long, while a tower casts a shadow 40 m long at the same time. Find the height of the tower.

By AA similarity, (h)/(2) = (40)/(1.6) = 25, so h = 50 m.

In △ ABC, DE ∥ BC with D on AB and E on AC. AD = 3 cm, AB = 7.5 cm and AC = 10 cm. Find AE and EC.

(AE)/(AC) = (AD)/(AB) = (3)/(7.5) = 0.4, so AE = 4 cm and EC = 6 cm.

Past board exam questions

ICSE 2027 sample paper · 1 mark

On a map of scale 1: 50,000, a rectangular plot has sides 4 cm by 3 cm. The actual area of the plot on ground is:

3 km²
9 km²
12 km²
16 km²

Show answer

3 km²

ICSE 2026 board paper · 1 mark

If the areas of two similar triangles are in the ratio 9 : 64, then the ratio of their corresponding altitudes is:

3 : 8
2 : 1
9 : 64
8 : 3

Show answer

3 : 8

ICSE 2026 board paper · 4 marks

In ΔABC, D is a point on AB and E is a point on AC such that DE ∥ BC and AD : DB = 2 : 3. BE and CD intersect at F. (a) Prove that: ΔADE ~ ΔABC and hence find DE : BC (b) Prove: ΔDFE ~ ΔCFB (c) Given, area of ΔDFE = 16 square units, find the area of ΔCFB.

Show answer

(a) In ΔADE and ΔABC: ∠ A is common; ∠ ADE = ∠ ABC (corresponding angles, DE ∥ BC). So ΔADE ~ ΔABC (AA). AD : AB = 2 : (2 + 3) = 2 : 5, so DE : BC = AD : AB = 2 : 5. (b) In ΔDFE and ΔCFB: ∠ DFE = ∠ CFB (vertically opposite); ∠ EDF = ∠ BCF (alternate angles, DE ∥ BC, transversal DC). So ΔDFE ~ ΔCFB (AA). (c) (ar(Δ DFE))/(ar(Δ CFB)) = ((DE)/(BC))² = (4)/(25) ⇒ ar(Δ CFB) = (16 × 25)/(4) = 100 square units.

ICSE 2025 compartment paper · 1 mark

Assertion (A): In ΔABC and ΔPQR, if ∠BAC = ∠QPR and ∠ABC = ∠PQR, then ΔABC ~ ΔPQR Reason (R): ΔABC ~ ΔPQR by SSS axiom

(A) is true, (R) is false.
(A) is false, (R) is true.
Both (A) and (R) are true, and (R) is the correct reason for (A).
Both (A) and (R) are true, and (R) is the incorrect reason for (A).

Show answer

(A) is true, (R) is false.

ICSE 2025 sample paper · 1 mark

AC is a diameter of a circle and AB and BC are chords. P is a point on AB and Q is a point on AC. AP = 3 cm and PB = 4 cm and QP ⊥ AB. If the area of ΔAPQ is 18 cm², then the area of the shaded portion QPBC (the quadrilateral bounded by QP, PB, BC and CQ) is:

32 cm²
49 cm²
80 cm²
98 cm²

Show answer

80 cm²

ICSE 2025 sample paper · 3 marks

A, B, C and D are points on a circle. Chords AD and BC intersect at P, and AB and CD are joined (the figure is not drawn to scale), where AB = 9 cm, PB = 3 cm and PD = 2 cm. (a) Prove that ΔAPB ~ ΔCPD. (b) Find the length of CD. (c) Find area ΔAPB : area ΔCPD.

Show answer

(a) In ΔAPB and ΔCPD: ∠BAP = ∠DCP (angles in the same segment, on chord BD) ∠ABP = ∠CDP (angles in the same segment, on chord AC) ∴ ΔAPB ~ ΔCPD (AA similarity). (b) Corresponding sides are proportional: (AB)/(CD) = (PB)/(PD) = (3)/(2), so CD = (2 × 9)/(3) = 6 cm. (c) (ar(Δ APB))/(ar(Δ CPD)) = (PB²)/(PD²) = (9)/(4), i.e. 9 : 4.

ICSE 2024 sample paper · 1 mark

∆ ABC ~ ∆ PQR, AD ⊥ BC (D on BC), PS ⊥ QR (S on QR) and (AD)/(PS) = (3)/(8). The value of AB : PQ is:

8 : 3
3 : 5
3 : 8
5 : 8

Show answer

3 : 8

ICSE 2024 board paper · 1 mark

Δ ABC sim Δ PQR. If AD and PS are bisectors of ∠ BAC and ∠ QPR respectively then:

Δ ABC sim Δ PQS
Δ ABD sim Δ PQS
Δ ABD sim Δ PSR
Δ ABC sim Δ PSR

Show answer

Δ ABD sim Δ PQS