Class 10 · Chapter 6 · Geometry unit (15 of 80 marks)

Triangles Class 10: notes and important questions

Revision notes, 10 free important questions with step-marked answers, and a four-step route from the basics to 95+ through 46 board-style questions.

Practise this chapter Join freeFree account, no card: +5 practice questions on every chapter, with instant feedback.

46 questions
13 multiple choice, 4 assertion–reason, 11 very short answer, 9 short answer, 6 long answer, 3 case study
About 14 hours to master

Geometry unit: 15 of 80 theory marks (Triangles and Circles).

NCERT textbook: Chapter 6 on ncert.nic.in (PDF)

Revision notes

Triangles — revision notes

1. Similar figures

Two polygons with the same number of sides are similar if (i) corresponding angles are equal and (ii) corresponding sides are in the same ratio. Both conditions are needed for polygons (a square and a rhombus fail (i)).
All circles, all squares and all equilateral triangles are similar. Congruent shapes are similar (ratio $1 : 1$ ), but not conversely.

2. Basic Proportionality Theorem (BPT / Thales)

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

Proof idea (examinable): compare areas. $\frac{ar (A D E)}{ar (B D E)} = \frac{A D}{D B}$ and $\frac{ar (A D E)}{ar (D E C)} = \frac{A E}{E C}$ ; and $ar (B D E) = ar (D E C)$ (same base $D E$ , same parallels).

Converse: if $\frac{A D}{D B} = \frac{A E}{E C}$ , then $D E ∥ B C$ .

Note: $D E$ itself is not in the ratio $A D : D B$ ; use $\frac{D E}{B C} = \frac{A D}{A B}$ (from similarity).

3. Criteria for similarity of triangles

AAA / AA: two pairs of equal angles (the third then matches).
SSS: all three pairs of sides in the same ratio.
SAS: two pairs of sides in the same ratio and the included angles equal.

Write the correspondence in order: $△ A B C \sim △ P Q R$ means $A \leftrightarrow P$ , $B \leftrightarrow Q$ , $C \leftrightarrow R$ .

4. Useful consequences

Ratio of perimeters = ratio of corresponding sides = ratio of corresponding altitudes/medians.
Trapezium $A B C D$ ( $A B ∥ D C$ ) with diagonals meeting at $O$ : $△ A O B \sim △ C O D$ , so $\frac{O A}{O C} = \frac{O B}{O D} = \frac{A B}{C D}$ .
Shadow problems: vertical objects and parallel sun rays give similar triangles (AA).

Worked example 1

$D E ∥ B C$ , $A D = 5$ , $D B = 3$ , $A C = 12$ . Then $\frac{A E}{A C} = \frac{A D}{A B} = \frac{5}{8} \Rightarrow A E = 7.5$ .

Worked example 2

Sides $5, 12, 13$ and $7.5, 18, 19.5$ : ratios all $\frac{2}{3}$ , so similar by SSS.

Worked example 3

A $1.8$ m man casts a $2.4$ m shadow; a pole casts a $10$ m shadow at the same time: height $= \frac{1.8}{2.4} \times 10 = 7.5$ m.

Common errors

Writing $\frac{A D}{D B} = \frac{D E}{B C}$ — wrong; use $\frac{A D}{A B} = \frac{D E}{B C}$ .
Matching vertices in the wrong order, leading to wrong side ratios.
Using SAS similarity with an angle that is not between the two sides.
Claiming two rhombuses/rectangles are similar from sides alone.

Board-exam tips

In proofs, state the reason for every equality (given, common, alternate angles, corresponding angles, BPT, criterion).
Draw and label your own rough sketch for every triangle question, even when none is printed.
The BPT proof is a frequently asked 3-mark item: learn it with the construction.

Topics in this chapter: Similar figures · Basic Proportionality Theorem and its converse · Criteria for similarity of triangles · Applications of similarity.

Formulas for this chapter: Chapter 6 on the Class 10 formula sheet (free PDF and A3 poster).

Triangles Class 10: important questions

10 original questions in the board paper's question types, from 1-mark MCQs to a case study, each with a step-marked worked answer. Try each one first. 5 worked answers open here; the other 5 open with a free account (no card).

Download these questions (PDF) 10 questions, 5 worked answers · 6 pages · 467 KB

Section A style · Multiple choice and assertion–reason (1 mark each)

Q1·1 mark·Multiple choiceBasic Proportionality Theorem and its converse

In $△ A B C$ , $D$ lies on $A B$ and $E$ on $A C$ with $D E ∥ B C$ . If $A D = 4$ cm, $D B = 6$ cm and $A E = 6$ cm, then $E C$ is

Triangle ABC with D on AB and E on AC, and DE parallel to BC. AD is marked 4 cm, DB 6 cm and AE 6 cm; EC is marked with a question mark. — DE ∥ BC: AD = 4 cm, DB = 6 cm, AE = 6 cm

Show worked answer

By BPT, $\frac{A D}{D B} = \frac{A E}{E C}$ : $\begin{aligned} \frac{4}{6} = \frac{6}{E C} \\ \Rightarrow & E C = 9 \end{aligned}$ cm A1
Correct option: (c) $9$ cm

Q2·1 mark·Multiple choiceSimilar figures

$△ A B C \sim △ P Q R$ with $\frac{A B}{P Q} = \frac{2}{3}$ . If the perimeter of $△ P Q R$ is $36$ cm, the perimeter of $△ A B C$ is

Q3·1 mark·Assertion–reasonCriteria for similarity of triangles

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Assertion (A): A triangle with sides $3$ cm, $4$ cm, $5$ cm is similar to a triangle with sides $6$ cm, $8$ cm, $10$ cm.

Reason (R): All equilateral triangles are similar.

Section B style · Short answer (2 marks each)

Q4·2 marks·Short answerBasic Proportionality Theorem and its converse

In $△ P Q R$ , $S$ and $T$ are points on $P Q$ and $P R$ with $S T ∥ Q R$ . If $P S = (x + 1)$ cm, $S Q = (3 x - 1)$ cm, $P T = (x + 3)$ cm and $T R = (3 x + 4)$ cm, find $x$ .

Triangle PQR with S on PQ and T on PR, and ST parallel to QR. PS is marked (x + 1) cm, SQ (3x − 1) cm, PT (x + 3) cm and TR (3x + 4) cm. — ST ∥ QR with sides in terms of x

Show worked answer

Answer:

x = 7

By BPT, $\frac{x + 1}{3 x - 1} = \frac{x + 3}{3 x + 4}$ , so $(x + 1) (3 x + 4) = (x + 3) (3 x - 1)$ M1
$\begin{aligned} 3 x^{2} + 7 x + 4 = 3 x^{2} + 8 x - 3 \\ \Rightarrow & x = 7 \end{aligned}$ A1

Q5·2 marks·Short answerApplications of similarity

A girl $160$ cm tall casts a shadow $120$ cm long. At the same time a building casts a shadow $18$ m long. Find the height of the building.

Section C style · Short answer (3 marks each)

Q6·3 marks·Short answerBasic Proportionality Theorem and its converse

$D$ and $E$ are points on the sides $A B$ and $A C$ of $△ A B C$ with $A D = 2$ cm, $D B = 3$ cm, $A E = 3$ cm and $E C = 4.5$ cm. Show that $D E ∥ B C$ .

Triangle ABC with D on AB and E on AC, joined by the line DE. AD is marked 2 cm, DB 3 cm, AE 3 cm and EC 4.5 cm. — AD = 2 cm, DB = 3 cm, AE = 3 cm, EC = 4.5 cm

Show worked answer

Answer:

D E ∥ B C

$\frac{A D}{D B} = \frac{2}{3}$ and $\begin{aligned} \frac{A E}{E C} & = \frac{3}{4.5} \\ = \frac{2}{3} \end{aligned}$ M1
So $\frac{A D}{D B} = \frac{A E}{E C}$ A1
By the converse of the Basic Proportionality Theorem, $D E ∥ B C$ R1

Q7·3 marks·Short answerSimilar figures

$△ A B C \sim △ D E F$ , with $A B = 6$ cm, $B C = 9$ cm, $C A = 12$ cm and $D E = 4$ cm. Find $E F$ , $F D$ and the perimeter of $△ D E F$ .

Section D style · Long answer (5 marks each)

Q8·5 marks·Long answerBasic Proportionality Theorem and its converse

State and prove the Basic Proportionality Theorem.

Show worked answer

Answer: Proved

Statement: a line drawn parallel to one side of a triangle, cutting the other two sides at distinct points, divides those sides in the same ratio. Given $△ A B C$ with $D E ∥ B C$ , prove $\frac{A D}{D B} = \frac{A E}{E C}$ M1
Construction: join $B E$ and $C D$ ; draw $E N ⊥ A B$ and $D M ⊥ A C$ M1
$\begin{aligned} \frac{ar (A D E)}{ar (B D E)} & = \frac{\frac{1}{2} A D \cdot E N}{\frac{1}{2} D B \cdot E N} \\ = \frac{A D}{D B} \end{aligned}$ and similarly $\frac{ar (A D E)}{ar (D E C)} = \frac{A E}{E C}$ A1
$△ B D E$ and $△ D E C$ are on the same base $D E$ and between the same parallels $D E$ and $B C$ , so $ar (B D E) = ar (D E C)$ R1
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ A1

Q9·5 marks·Long answerCriteria for similarity of triangles

$A B C D$ is a trapezium with $A B ∥ D C$ , and its diagonals meet at $O$ . (a) Prove that $△ A O B \sim △ C O D$ and hence that $\frac{A O}{O C} = \frac{B O}{O D}$ . (b) If $A B = 12$ cm, $D C = 8$ cm and $A C = 15$ cm, find $A O$ and $O C$ .

Trapezium ABCD with AB parallel to DC, AB = 12 cm on top and DC = 8 cm at the bottom. The diagonals AC and BD cross at O. — Trapezium ABCD: AB ∥ DC, AB = 12 cm, DC = 8 cm

Section E style · Case study (4 marks)

Q10·4 marks·Case studyApplications of similarity

Wheelchair ramp. A straight ramp starts at point $A$ on level ground and rises to point $C$ , which is $75$ cm above the point $B$ on the ground directly below it, with $A B = 9$ m. A vertical support $P Q$ stands at $P$ on $A B$ , with $Q$ on the ramp and $A P = 3$ m.

Name two similar triangles and the similarity criterion. [1 mark]
Find the height of the support $P Q$ . [1 mark]
A second support $R S$ is $50$ cm high. How far from $A$ must it stand? [2 marks]

Show worked answer

$△ A P Q \sim △ A B C$ (AA: common $∠ A$ , and $∠ A P Q = ∠ A B C = 90^{\circ}$ ) A1
$\begin{aligned} \frac{P Q}{75} = \frac{A P}{A B} = \frac{3}{9} \\ \Rightarrow & P Q = 25 \end{aligned}$ cm A1
$\frac{A R}{9} = \frac{50}{75}$ M1
$A R = 6$ m A1

Next: practise by Route to 95 step · the revision notes · free MCQs and case studies · all Class 10 assertion–reason questions · the Class 10 syllabus checklist.

Route to 95: four steps

Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.

Step 1

Secure the basics

You know when figures are similar and can use the Basic Proportionality Theorem to find missing lengths.

Read first: 1. Similar figures; 2. Basic Proportionality Theorem (BPT / Thales) 15 practice questions · checkpoint: 4 questions, 8 marks, pass 80%

Practise step 1

Step 2

Board standard

You can prove triangles similar using AA, SAS or SSS with a reason for every step, and use the converse of BPT.

Read first: 2. Basic Proportionality Theorem; 3. Criteria for similarity of triangles; Worked examples 1-2 14 practice questions · checkpoint: 4 questions, 9 marks, pass 80%

Practise step 2

Step 3

Full marks on long answers

You can write the BPT proof and the long 'prove that' answers the board asks, with figure, given, to prove, construction and proof.

Read first: 2. Basic Proportionality Theorem (proof); 4. Useful consequences; Worked example 3; Board-exam tips 6 practice questions · checkpoint: 3 questions, 14 marks, pass 80%

Practise step 3

Step 4

95+ stretch (HOTS)

You can handle the hard proofs: altitudes, parallelograms with extended sides, and algebraic BPT problems.

Read first: 3. Criteria for similarity of triangles; 4. Useful consequences; Common errors 5 practice questions · checkpoint: 3 questions, 11 marks, pass 80%

Practise step 4

Practice questions

Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 9 of the 46 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.

Q1·1 mark·Multiple choiceSimilar figures

All ________ triangles are similar.

(a)isosceles
(b)equilateral
(c)right-angled
(d)scalene

Q2·1 mark·Multiple choiceSimilar figures

Which of the following pairs of shapes are always similar?

(a)Two rectangles
(b)Two rhombuses
(c)Two isosceles triangles
(d)Two squares

Q3·1 mark·Multiple choiceSimilar figures

Which of the following statements is true?

(a)Any two similar triangles are congruent.
(b)Any two right triangles are similar.
(c)Any two congruent triangles are similar.
(d)Any two isosceles triangles are similar.

Worked solution·2 marks·Very short answerSimilar figures

Quadrilateral $A B C D$ is similar to quadrilateral $P Q R S$ , with $A B = 6$ cm, $B C = 8$ cm and $P Q = 9$ cm. Find $Q R$ and the ratio of the perimeters of $A B C D$ and $P Q R S$ .

Full worked solution Free, no sign-in

Scale factor $\frac{P Q}{A B} = \frac{3}{2}$ , so $Q R = \frac{3}{2} \times 8 = 12$ cm A1
Perimeters are in the ratio of the sides: $2 : 3$ A1

Every question here has a full worked solution. See them all →

Stretch yourself: original geometry problems at Intermediate level (ages 13 to 16), with hints and full solutions: Problem I18 · Problem I86 · Problem I93 · Problem I100. All 28 →

Themed lesson: National Science Day (India) maths lesson · Eclipse geometry maths lesson (starter, activity and worked answers). All themed maths

Knowledge organiser: Triangles on one page (key facts, formulas and mistakes to avoid; printable PDF)

Where marks are lost in Triangles

Writing AD/DB = DE/BC. Fix: DE/BC matches AD/AB (whole side), not AD/DB; write the similarity statement first and read ratios from it.
Similarity statement with vertices in the wrong order (△ABC ~ △QPR when angle A matches angle P). Fix: match equal angles first, then write the letters in that order.
Steps in a proof with no reasons. Fix: every equality needs a bracket: (given), (common), (corresponding angles), (BPT), (AA criterion). R marks depend on it.
Using SAS with an angle that is not between the two sides. Fix: check that the angle is the included angle before quoting SAS.
BPT proof without the construction (join BE, CD; draw EN ⟂ AB, DM ⟂ AC). Fix: learn the proof with the construction written as a separate line.
No figure in a geometry proof. Fix: draw and label a rough figure even when none is printed; it usually carries a mark in 3- and 5-mark proofs.

Exam strategy: how answers are step-marked · the proofs, written out · competency-based questions

Examiner Insights: common mistakes in CBSE Class 10 Maths, with fixes (our analysis of public sources) →

Triangles in our sample papers

Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).

Sample paper 1 (Board standard): questions 8, 23, 28, 33
Sample paper 2 (Board standard): questions 9, 23, 29, 33
Sample paper 3 (Above board): questions 10, 12, 19, 28, 33
Sample paper 4 (Above board): questions 9, 29, 33
Sample paper 5 (95+ challenge): questions 9, 10, 28, 33

Want it against the clock? Take a timed 30-mark chapter test on Triangles, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Pro or the free trial).

Triangles: common questions

What is in CBSE Class 10 Maths Chapter 6, Triangles?

The topics are: Similar figures, Basic Proportionality Theorem and its converse, Criteria for similarity of triangles, Applications of similarity. This page has revision notes for each, and 46 board-style practice questions with step-marked solutions.

How many marks is Triangles worth?

It is in the Geometry unit, which carries 15 of the 80 theory marks. CBSE gives marks by unit, not by chapter.

Where do students lose marks in Triangles?

Writing AD/DB = DE/BC. Fix: DE/BC matches AD/AB (whole side), not AD/DB; write the similarity statement first and read ratios from it. Similarity statement with vertices in the wrong order (△ABC ~ △QPR when angle A matches angle P). Fix: match equal angles first, then write the letters in that order.

Are the Triangles important questions free?

Yes. The 10 important questions on this page are free to try, from 1-mark MCQs to a case study. 5 worked answers are open on the page; the others open with a free account (no card).

Are these NCERT questions?

No. Every practice question here is original, written in the board's question styles with step marks. For the textbook itself, use the NCERT link on this page.

Work through the Route to 95 with your progress saved

Save your answers and track your progressA free account remembers the questions you have done and brings you back to this page.

Join free

Already have an account? Sign in

Try it first

Revision notes

Triangles — revision notes

1. Similar figures

2. Basic Proportionality Theorem (BPT / Thales)

3. Criteria for similarity of triangles

4. Useful consequences

Worked example 1

Worked example 2

Worked example 3

Common errors

Board-exam tips

Triangles Class 10: important questions

Section A style · Multiple choice and assertion–reason (1 mark each)

Section B style · Short answer (2 marks each)

Section C style · Short answer (3 marks each)

Section D style · Long answer (5 marks each)

Section E style · Case study (4 marks)

Route to 95: four steps

Secure the basics

Board standard

Full marks on long answers

95+ stretch (HOTS)

Practice questions

Full worked solution Free, no sign-in

Where marks are lost in Triangles

Triangles in our sample papers

Triangles: common questions

Work through the Route to 95 with your progress saved