CBSE Class 10 · Chapter 6 · Geometry · 2026-27

Triangles Class 10: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceSimilar figures

Two polygons with the same number of sides are similar when

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Answer: (c) their corresponding angles are equal and their corresponding sides are proportional

Why: Polygons in general need both conditions; only for triangles does one imply the other.

A square and a non-square rectangle have equal angles but are not similar; a square and a non-square rhombus have proportional sides but are not similar. Both conditions are needed.

Q2

·1 mark·Multiple choiceSimilar figures

Which of these pairs of figures need not be similar?

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Answer: (d) Two rhombuses

Why: Two rhombuses can have different angles (e.g. a square and a 60°–120° rhombus).

All circles, all squares and all equilateral triangles are similar. Two rhombuses with equal sides can have different angles, so they need not be similar.

Q3

·1 mark·Multiple choiceBasic Proportionality Theorem and its converse

In $△ A B C$ , $D E ∥ B C$ with $D$ on $A B$ and $E$ on $A C$ . If $A D = 5$ cm, $D B = 8$ cm and $A E = 10$ cm, then $E C$ is

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Answer: (b)

16

Why: BPT: AD/DB = AE/EC.

$\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow \frac{5}{8} = \frac{10}{E C} \Rightarrow E C = 16$ cm.

Q4

·1 mark·Multiple choiceBasic Proportionality Theorem and its converse

In $△ A B C$ , $D E ∥ B C$ , $D$ on $A B$ , $E$ on $A C$ . If $A D = 3$ cm, $A B = 9$ cm and $A C = 12$ cm, then $A E$ is

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Answer: (a)

4

Why: By BPT, AD/AB = AE/AC.

$\frac{A D}{A B} = \frac{A E}{A C} \Rightarrow \frac{3}{9} = \frac{A E}{12} \Rightarrow A E = 4$ cm.

Q5

·1 mark·Multiple choiceSimilar figures

$△ A B C \sim △ D E F$ with $A B = 8$ cm, $D E = 12$ cm and $B C = 10$ cm. Then $E F$ is

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Answer: (c)

15

Why: Corresponding sides are in the ratio DE/AB = 3/2.

$\frac{E F}{B C} = \frac{D E}{A B} = \frac{12}{8} = \frac{3}{2}$ , so $E F = \frac{3}{2} \times 10 = 15$ cm.

Q6

·1 mark·Multiple choiceCriteria for similarity (AA, SSS, SAS)

A triangle with sides $6$ cm, $8$ cm and $10$ cm is similar to a triangle whose shortest side is $9$ cm. The longest side of the second triangle is

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Answer: (d)

15

Why: Scale factor 9/6 = 3/2 applies to every side.

Scale factor $= \frac{9}{6} = \frac{3}{2}$ . Longest side $= \frac{3}{2} \times 10 = 15$ cm.

Q7

·1 mark·Multiple choiceCriteria for similarity (AA, SSS, SAS)

At the same moment, a girl $1.5$ m tall casts a shadow $2$ m long and a water tank casts a shadow $24$ m long. The height of the tank is

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Answer: (a)

18

Why: The sun's rays make equal angles, so the two triangles are similar (AA).

$\frac{h}{24} = \frac{1.5}{2} \Rightarrow h = 24 \times \frac{3}{4} = 18$ m.

Q8

·1 mark·Multiple choiceBasic Proportionality Theorem and its converse

In $△ A B C$ , $D E ∥ B C$ with $D$ on $A B$ , $E$ on $A C$ and $\frac{A D}{D B} = \frac{2}{3}$ . Then $\frac{A E}{A C}$ equals

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Answer: (c)

\frac{2}{5}

Why: AE/EC = 2/3, so AE is 2 parts out of 2 + 3 = 5.

By BPT $\frac{A E}{E C} = \frac{2}{3}$ . So $\frac{A E}{A C} = \frac{2}{2 + 3} = \frac{2}{5}$ .

Q9

·1 mark·Multiple choiceBasic Proportionality Theorem and its converse

In $△ P Q R$ , $D$ is on $P Q$ and $E$ is on $P R$ with $P D = 3$ cm, $D Q = 4.5$ cm, $P E = 4$ cm and $E R = 6$ cm. Then

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Answer: (a)

D E ∥ Q R

Why: PD/DQ = 3/4.5 = 2/3 = PE/ER, so the converse of BPT applies.

$\frac{P D}{D Q} = \frac{3}{4.5} = \frac{2}{3}$ and $\frac{P E}{E R} = \frac{4}{6} = \frac{2}{3}$ . A line dividing two sides in the same ratio is parallel to the third side: $D E ∥ Q R$ .

Q10

·1 mark·Multiple choiceCriteria for similarity (AA, SSS, SAS)

Which of the following is not a criterion for the similarity of two triangles?

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Answer: (d) SSA

Why: Two sides in proportion with a non-included equal angle does not force similarity.

AA, SSS and SAS are similarity criteria. SSA is not: two proportional sides and an equal angle that is not between them can give triangles of different shapes.

Q11

·1 mark·Multiple choiceSimilar figures

If $△ A B C \sim △ Q R P$ , which of the following is correct?

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Answer: (b)

\frac{A B}{Q R} = \frac{B C}{R P}

Why: The order of letters gives A↔Q, B↔R, C↔P, so AB↔QR and BC↔RP.

$A \leftrightarrow Q$ , $B \leftrightarrow R$ , $C \leftrightarrow P$ . So $\frac{A B}{Q R} = \frac{B C}{R P} = \frac{C A}{P Q}$ .

Q12

·1 mark·Multiple choiceBasic Proportionality Theorem and its converse

In $△ A B C$ , $D E ∥ B C$ with $A D = x$ , $D B = x - 2$ , $A E = x + 2$ and $E C = x - 1$ (all in cm). The value of $x$ is

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Answer: (a)

4

Why: BPT gives x(x − 1) = (x − 2)(x + 2), so −x = −4.

$\frac{x}{x - 2} = \frac{x + 2}{x - 1} \Rightarrow x^{2} - x = x^{2} - 4 \Rightarrow x = 4$ . (Then $D B = 2$ , $E C = 3$ , both positive.)

Q13

·1 mark·Multiple choiceCriteria for similarity (AA, SSS, SAS)

In $△ A B C$ , $A B = 3$ cm, $A C = 5$ cm and $∠ A = 70^{\circ}$ . In $△ P Q R$ , $P Q = 6$ cm, $P R = 10$ cm and $∠ P = 70^{\circ}$ . Which statement is true?

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Answer: (b)

△ A B C \sim △ P Q R

by SAS

Why: AB/PQ = AC/PR = 1/2 and the included angles are equal.

$\frac{A B}{P Q} = \frac{3}{6} = \frac{1}{2}$ , $\frac{A C}{P R} = \frac{5}{10} = \frac{1}{2}$ and the included angles $∠ A = ∠ P$ . So $△ A B C \sim △ P Q R$ (SAS), with $A \leftrightarrow P$ , $B \leftrightarrow Q$ , $C \leftrightarrow R$ . They are not congruent (ratio $\frac{1}{2}$ ).

Q14

·1 mark·Multiple choiceBasic Proportionality Theorem and its converse

In $△ P Q R$ , $S$ and $T$ lie on $P Q$ and $P R$ with $S T ∥ Q R$ . If $P S = 2$ cm, $S Q = 3$ cm and $S T = 4$ cm, then $Q R$ is

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Answer: (b)

10

Why: △PST ∼ △PQR (AA), so ST/QR = PS/PQ = 2/5.

$△ P S T \sim △ P Q R$ , so $\frac{S T}{Q R} = \frac{P S}{P Q} = \frac{2}{5}$ . $Q R = 4 \times \frac{5}{2} = 10$ cm. (Not $\frac{P S}{S Q}$ : that is the BPT ratio of the side pieces.)

Q15

·1 mark·Multiple choiceSimilar figures

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

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Answer: (c)

18

Why: Scale factor AB/DE = 2/3, so BC = 6 and CA = 8.

Scale factor $\frac{4}{6} = \frac{2}{3}$ : $B C = \frac{2}{3} \times 9 = 6$ , $C A = \frac{2}{3} \times 12 = 8$ . Perimeter $= 4 + 6 + 8 = 18$ cm.

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 10 case study questions by chapter.

Case study 1: Pinhole camera (4 marks)

In a science project, Kabir builds a pinhole camera. Light from the top $A$ and bottom $B$ of a tree passes through the pinhole $O$ and forms an inverted image $A^{'} B^{'}$ on a screen $20$ cm behind the pinhole. The tree is $9$ m tall and stands $30$ m in front of the pinhole, and the tree and the screen are both vertical, so $A B ∥ A^{'} B^{'}$ .

(i) Which similarity criterion shows that

△ O A B \sim △ O A^{'} B^{'}

? [1 mark]

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Answer: AA

∠ A O B = ∠ A^{'} O B^{'}

(vertically opposite) and

∠ O A B = ∠ O A^{'} B^{'}

(alternate angles,

A B ∥ A^{'} B^{'}

), so AA. A1

(ii) Find the ratio of the height of the image to the height of the tree. [1 mark]

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Answer:

1 : 150

△ O A B \sim △ O A^{'} B^{'}

, so their corresponding altitudes (the perpendicular distances of

O

from

A B

and from

A^{'} B^{'}

) are in the same ratio as corresponding sides:

\frac{0.2}{30} = \frac{1}{150}

. A1

(iii) Find the height of the image in centimetres. [2 marks]

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Answer:

6

\frac{A^{'} B^{'}}{9} = \frac{1}{150}

M1, so

A^{'} B^{'} = \frac{9}{150}

= 0.06

= 6

cm. A1

OR Kabir moves back so that the tree is

45

m from the pinhole (the screen stays

20

cm behind it). Find the new height of the image. [2 marks]

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Answer:

4

\frac{A^{'} B^{'}}{900 cm} = \frac{20}{4500}

M1, so

A^{'} B^{'} = 4

cm. A1

Case study 2: Width of a river (4 marks)

To find the width of a river without crossing it, Ravi stands at $Q$ on one bank, exactly opposite a tree $P$ on the far bank. He walks $12$ m along the bank to a point $R$ , puts a stick there, walks a further $4$ m to $S$ , and then walks $6$ m away from the river at right angles to the bank to a point $T$ from which the stick $R$ and the tree $P$ are in a straight line. $P Q$ and $T S$ are both perpendicular to the bank $Q S$ .

(i) Name the criterion by which

△ P Q R \sim △ T S R

. [1 mark]

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Answer: AA

∠ P Q R = ∠ T S R = 90^{\circ}

and

∠ P R Q = ∠ T R S

(vertically opposite), so AA. A1

(ii) Write the equal ratios of corresponding sides. [1 mark]

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Answer:

\frac{P Q}{T S} = \frac{Q R}{S R} = \frac{P R}{T R}

\frac{P Q}{T S} = \frac{Q R}{S R} = \frac{P R}{T R}

. A1

(iii) Find the width

P Q

of the river. [2 marks]

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Answer:

18

\frac{P Q}{6} = \frac{12}{4}

M1, so

P Q = 18

m. A1

OR If Ravi had walked

5

m (instead of

4

m) from

R

S

, how far from the bank would he have had to walk for

T

R

P

to be in line? [2 marks]

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Answer:

7.5

\frac{P Q}{T S} = \frac{Q R}{S R} \Rightarrow \frac{18}{T S} = \frac{12}{5}

M1, so

T S = \frac{90}{12} = 7.5

m. A1

Next steps for Triangles

This free set is separate from the chapter's question bank. On the Triangles chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 46 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Triangles: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.