Chapter 6 of 14

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Rajasthan (RBSE)Class 10 Mathematics★ 6–8 marks · RBSE Class 10 Maths board (Unit — Geometry)

Triangles

RBSE Class 10 Maths on similar triangles — the Basic Proportionality Theorem (Thales), the AA/SSS/SAS similarity criteria, and the ratio of areas of similar triangles. Learn to prove similarity, set up proportions and solve height-and-shadow problems, with fully solved board-pattern questions and proofs.

⏱ 33 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Define similar triangles and write correct correspondence
2State and apply the Basic Proportionality Theorem (Thales) and its converse
3Prove triangles similar using the AA, SSS and SAS criteria
4Use the area-ratio = (side-ratio)² theorem
5Apply the Pythagoras theorem and its converse; solve height-and-shadow problems

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Why this chapter matters

One of the most heavily weighted geometry chapters. It typically carries a theorem proof (BPT or area-ratio), a similarity numerical, and MCQs — a large, reliable block of marks that rewards clear proof-writing.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Lines and Angles (Class 9)

Angle relationships used to establish equal angles for AA similarity.

Triangles — RBSE Class 10 (Mathematics)

Two triangles can be different sizes yet exactly the same shape — a photograph and its enlargement, a tree and its scale drawing. Such triangles are similar: equal angles, proportional sides. This chapter turns that idea into precise theorems you can prove and use, and it is one of the most proof-heavy — and rewarding — chapters of the year.

1. Similar figures and similar triangles

Two polygons are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (proportional). For triangles, either condition forces the other.

Notation: $△ A B C \sim △ D E F$ means $∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F$ and $\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D}$ .

Congruent ⊂ similar: congruent triangles are similar with ratio 1.

2. Basic Proportionality Theorem (Thales)

If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, it divides those two sides in the same ratio.

If $D E ∥ B C$ in $△ A B C$ (D on AB, E on AC), then $\frac{A D}{D B} = \frac{A E}{E C}$

The converse is equally examinable: if a line divides two sides in the same ratio, it is parallel to the third side. BPT is the engine behind most proofs in this chapter.

3. Criteria for similarity of triangles

You do not need all six equalities — any one criterion suffices:

AA — two angles of one equal two angles of the other. (Most used.)
SSS — all three pairs of sides in the same ratio.
SAS — one pair of angles equal and the two including sides proportional.

Once similarity is established, all corresponding sides are proportional — the source of most numerical answers.

4. Area of similar triangles

The ratio of the areas of two similar triangles equals the ratio of the squares of their corresponding sides.

$\frac{ar ( △ A B C )}{ar ( △ D E F )} = (\frac{A B}{D E})^{2} = (\frac{B C}{E F})^{2} = (\frac{C A}{F D})^{2}$

It also equals the square of the ratio of corresponding medians, altitudes or angle bisectors. A common trap: students use the side ratio for areas — remember to square it.

RBSE note (2026-27). In the rationalised NCERT, the Pythagoras theorem and its converse are stated and used (e.g. to test right angles), though the formal similarity proof of Pythagoras is no longer part of the core exercises. Know the statement and applications.

5. Using it all — a worked idea

To prove two triangles similar, hunt for equal angles first (common angle, vertically opposite angles, alternate angles from parallels) — AA is usually the shortest path. Then write the proportion with corresponding vertices in the same order, and cross-multiply for the unknown side.

Example (height by shadows): a 6 m pole casts a 4 m shadow while a tower casts a 28 m shadow at the same time. The sun's rays make equal angles, so the triangles are similar: $\frac{height of tower}{28} = \frac{6}{4} \Rightarrow height = 42 m .$

6. Closing thought

Similarity rests on a short toolkit: BPT and its converse, the AA/SSS/SAS criteria, and the area-ratio = (side-ratio)² rule. The chapter rewards clean, well-justified proofs — state the criterion by name, keep vertices in corresponding order, and quote the theorem you use. In the RBSE board it is a heavyweight, often carrying a full theorem-proof plus a numerical similarity problem.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Similarity

△ABC ~ △DEF ⇒ equal angles, AB/DE = BC/EF = CA/FD

Same shape, proportional sides.

Basic Proportionality (Thales)

DE ∥ BC ⇒ AD/DB = AE/EC

A parallel line splits two sides in equal ratio.

Converse of BPT

AD/DB = AE/EC ⇒ DE ∥ BC

Equal ratios force parallelism.

Similarity criteria

AA, SSS, SAS

Any one proves similarity.

Area ratio

ar(△ABC)/ar(△DEF) = (AB/DE)²

Ratio of areas = square of ratio of corresponding sides/medians/altitudes.

Pythagoras

hyp² = base² + perpendicular²

Converse tests for a right angle.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Using the side ratio (not its square) for areas

✓ ar ratio = (side ratio)². If sides are in ratio 2:3, areas are in ratio 4:9.

WATCH OUT

✗ Writing similarity with vertices out of order

✓ △ABC ~ △DEF means A↔D, B↔E, C↔F. Getting the order wrong scrambles which sides are proportional.

WATCH OUT

✗ Quoting SAS similarity like SAS congruence

✓ For similarity, SAS needs the included angles equal AND the two including sides PROPORTIONAL — not equal.

WATCH OUT

✗ Applying BPT when the line is not parallel

✓ BPT requires the dividing line parallel to the third side. To conclude parallelism, use the CONVERSE from equal ratios.

WATCH OUT

✗ Forgetting to name the criterion in a proof

✓ Always state 'by AA similarity' (or SSS/SAS) — examiners award marks for the justification, not just the conclusion.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 6.1

Exercise 6.1

Similar figures and basic definitions

Questions

Ex 6.2

Exercise 6.2

Basic Proportionality Theorem and its converse

Questions

Ex 6.3

Exercise 6.3

Similarity criteria and proofs

Questions

Examples (in-text)

Worked proofs and numerical similarity problems

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Definition

If △ABC ~ △PQR and AB/PQ = 3/5, what is the ratio of their perimeters?

▶ Show solution

Step 1 — In similar triangles all sides are in the same ratio, so perimeters share that ratio.

✦ Answer: 3 : 5.

Q2EASY· BPT

In △ABC, DE ∥ BC with AD = 2 cm, DB = 3 cm, AE = 4 cm. Find EC.

▶ Show solution

Step 1 — By BPT, AD/DB = AE/EC ⇒ 2/3 = 4/EC.

Step 2 — EC = 6 cm.

✦ Answer: EC = 6 cm.

Q3EASY· Area ratio

Two similar triangles have corresponding sides in ratio 4:9. Find the ratio of their areas.

▶ Show solution

Step 1 — ar ratio = (4/9)² = 16/81.

✦ Answer: 16 : 81.

Q4MEDIUM· Height by shadow

A vertical pole 6 m high casts a shadow 4 m long; at the same time a tower casts a 28 m shadow. Find the tower's height.

▶ Show solution

Step 1 — Equal sun angles ⇒ triangles similar: height/28 = 6/4.

Step 2 — height = 28 × 6/4 = 42 m.

✦ Answer: 42 m.

Q5MEDIUM· Similarity numerical

In △ABC, DE ∥ BC. If AD = x, DB = x−2, AE = x+2, EC = x−1, find x.

▶ Show solution

Step 1 — By BPT, x/(x−2) = (x+2)/(x−1).

Step 2 — x(x−1) = (x+2)(x−2) ⇒ x² − x = x² − 4 ⇒ −x = −4 ⇒ x = 4.

✦ Answer: x = 4.

Q6MEDIUM· Pythagoras converse

Sides of a triangle are 7 cm, 24 cm, 25 cm. Is it right-angled?

▶ Show solution

Step 1 — Check largest side: 25² = 625.

Step 2 — 7² + 24² = 49 + 576 = 625 = 25².

Step 3 — By the converse of Pythagoras, the triangle is right-angled (right angle opposite the 25 cm side).

✦ Answer: Yes, right-angled.

Q7MEDIUM· Area problem

The areas of two similar triangles are 81 cm² and 49 cm². If a side of the larger is 6.3 cm, find the corresponding side of the smaller.

▶ Show solution

Step 1 — (side ratio)² = 81/49 ⇒ side ratio = 9/7.

Step 2 — 6.3 / x = 9/7 ⇒ x = 6.3 × 7/9 = 4.9 cm.

✦ Answer: 4.9 cm.

Q8HARD· Proof (BPT)

State and prove the Basic Proportionality Theorem.

▶ Show solution

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

Step 1 — In △ABC, let DE ∥ BC (D on AB, E on AC). Join BE and CD; draw EM ⊥ AB and DN ⊥ AC.

Step 2 — ar(ADE)/ar(BDE) = AD/DB (same height EM, bases AD, DB). Similarly ar(ADE)/ar(CDE) = AE/EC.

Step 3 — △BDE and △CDE are on the same base DE and between the same parallels DE, BC, so ar(BDE) = ar(CDE).

Step 4 — Hence AD/DB = AE/EC. ∎

✦ Answer: proof by equal-area triangles as above.

Q9HARD· Proof (area ratio)

Prove that the ratio of areas of two similar triangles equals the ratio of the squares of their corresponding sides.

▶ Show solution

Step 1 — Let △ABC ~ △DEF; draw altitudes AP ⊥ BC and DQ ⊥ EF.

Step 2 — ar(ABC)/ar(DEF) = (½·BC·AP)/(½·EF·DQ) = (BC/EF)(AP/DQ).

Step 3 — In △ABP and △DEQ: ∠B = ∠E (similar) and ∠P = ∠Q = 90° ⇒ △ABP ~ △DEQ (AA) ⇒ AP/DQ = AB/DE = BC/EF.

Step 4 — Substitute: ar(ABC)/ar(DEF) = (BC/EF)(BC/EF) = (BC/EF)². ∎

✦ Answer: area ratio = (corresponding side ratio)².

Q10HARD· Combined

In △ABC, D and E are points on AB and AC such that DE ∥ BC and AD:DB = 3:2. Find ar(△ADE) : ar(△ABC).

▶ Show solution

Step 1 — DE ∥ BC ⇒ △ADE ~ △ABC (AA: common angle A and equal corresponding angles).

Step 2 — AD/AB = 3/(3+2) = 3/5.

Step 3 — ar(ADE)/ar(ABC) = (3/5)² = 9/25.

✦ Answer: 9 : 25.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Similar triangles: equal angles and proportional sides.
•BPT: DE ∥ BC ⇒ AD/DB = AE/EC; converse gives parallelism.
•Similarity criteria: AA, SSS, SAS (SAS needs proportional including sides).
•Area ratio = (side ratio)² = (median/altitude/bisector ratio)².
•Pythagoras and its converse identify and use right angles.
•Congruent triangles are similar with ratio 1.
•State the criterion/theorem by name in every proof.

Rajasthan (RBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6–8 marks

Question type	Marks each	Typical count	What it tests
MCQ / very short	1	1–2	Similarity ratio, area ratio, BPT
Short answer	2–3	1	Similarity numerical or Pythagoras converse
Long answer / proof	3–4	1	BPT or area-ratio theorem proof

Prep strategy

Memorise the BPT proof and the area-ratio proof — one usually appears
For similarity, look for equal angles first and use AA
Always square the side ratio for areas
Write correspondence in the correct vertex order every time

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Measuring inaccessible heights

Shadow-and-similar-triangle methods find the height of trees, poles and towers.

Scale models and maps

Architectural models and map distances rely on similar figures and proportional scaling.

Photography and screens

Enlargements and aspect ratios keep shapes similar while changing size.

Engineering drawings

Blueprints scale real objects up or down while preserving shape.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Name the similarity criterion (AA/SSS/SAS) explicitly in proofs.
Keep corresponding vertices in the same order when writing ratios.
For area questions, write the (side ratio)² step clearly.
In BPT problems, state whether you are using the theorem or its converse.
Draw a clear, labelled figure — many marks hinge on correct construction lines.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

The full similarity proof of the Pythagoras theorem.
Ceva's and Menelaus' theorems on ratios along a triangle's sides.
The angle-bisector length and the internal/external division ratios.
Similar triangles in circle geometry (power of a point).

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

RBSE Class 10 Board (BSER Ajmer)High — a theorem proof plus a similarity numerical almost every year

NTSE / state scholarshipMedium — geometry reasoning MCQs

JEE FoundationMedium — similarity underpins coordinate and trigonometric geometry

Maths Olympiad (IMO/NMTC)High — triangle geometry is a staple

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Yes — RBSE (BSER, Ajmer) prescribes the NCERT Mathematics textbook; chapters and exercises match the national syllabus while RBSE sets its own exam pattern.

AA — it needs only two pairs of equal angles, which are often easy to spot (common angles, parallel-line angles, vertically opposite angles).

In the rationalised NCERT the theorem and its converse are used and stated, but the formal similarity-based proof is no longer in the core exercises. Know the statement and its applications well.

Area depends on two length dimensions. If every length scales by k, area scales by k², so the area ratio is the square of the side ratio.

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