CBSE • Class 10 • Mathematics • Chapter 6 (Triangles) • Exercise 6.1

Exercise 6.1: Triangles — NCERT Solutions

Definitions of similar figures and the basic proportionality (Thales) theorem.

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What this exercise covers

Similar figuresBasic Proportionality TheoremConverse of BPT

Step-by-step solutions — Exercise 6.1

3 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 6.1 Q1 • 3 marks

Fill in the blanks: (i) All circles are ______. (ii) All squares are ______. (iii) Two polygons of the same number of sides are similar if their corresponding angles are ______ and their corresponding sides are ______.
Hint (Socratic — try this first)▾
What must be true about both the shape (angles) and the size relation (sides) for figures to be called similar?
Step-by-step solution▾

(i) All circles are similar — any two circles have the same shape and differ only in radius.

(ii) All squares are similar — all angles are 90∘90^\circ and sides are always in equal ratio.

(iii) Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are in the same ratio (proportional).

Both conditions must hold together for polygons.

Common mistake:
Writing that all triangles or all rectangles are similar — they are not, because their angle/side ratios can differ.
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Exercise 6.1 Q2 • 2 marks

In triangle ABCABC, DE∥BCDE \parallel BC. If AD=3AD = 3 cm, DB=4DB = 4 cm and AE=4.5AE = 4.5 cm, find ECEC.
Hint (Socratic — try this first)▾
Which theorem tells you that a line parallel to one side divides the other two sides in the same ratio?
Step-by-step solution▾

By the Basic Proportionality Theorem (Thales' Theorem), since DE∥BCDE \parallel BC:

ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}

Substitute the values:

34=4.5EC\frac{3}{4} = \frac{4.5}{EC}

Cross-multiply:

3×EC=4×4.5=183 \times EC = 4 \times 4.5 = 18

EC=183=6 cmEC = \frac{18}{3} = 6 \text{ cm}

Common mistake:
Setting up the ratio as ADAB=AEEC\frac{AD}{AB} = \frac{AE}{EC} by mixing a full side with a segment instead of keeping ADDB=AEEC\frac{AD}{DB}=\frac{AE}{EC}.
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Exercise 6.1 Q3 • 3 marks

In triangle ABCABC, points DD and EE lie on ABAB and ACAC. Given AD=2AD = 2 cm, AB=6AB = 6 cm, AE=3AE = 3 cm and AC=9AC = 9 cm, show that DE∥BCDE \parallel BC.
Hint (Socratic — try this first)▾
If a line divides two sides of a triangle in the same ratio, what can you conclude about that line and the third side?
Step-by-step solution▾

First find the segment ratios.

DB=AB−AD=6−2=4 cmDB = AB - AD = 6 - 2 = 4 \text{ cm} EC=AC−AE=9−3=6 cmEC = AC - AE = 9 - 3 = 6 \text{ cm}

Now compare:

ADDB=24=12,AEEC=36=12\frac{AD}{DB} = \frac{2}{4} = \frac{1}{2}, \qquad \frac{AE}{EC} = \frac{3}{6} = \frac{1}{2}

Since ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}, by the converse of the Basic Proportionality Theorem, DE∥BCDE \parallel BC.

Common mistake:
Forgetting to subtract to get DBDB and ECEC, and instead comparing ADAB\frac{AD}{AB} with AEEC\frac{AE}{EC}, which mixes different types of segments.
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How to approach Exercise 6.1

  1. Re-read the chapter summary first. Open Triangles and refresh the key concepts: Similar triangles, AA criterion, SSS criterion, SAS criterion.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Triangles

  1. Exercise 6.1 — Definitions of similar figures and the basic proportionality (Thales) theorem.
  2. Exercise 6.2 — Criteria for similarity of triangles — AA, SSS, SAS.
  3. Exercise 6.3 — Application problems on similar triangles, including height and ratio problems.

Exercise 6.2 →

Criteria for similarity of triangles — AA, SSS, SAS.

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