Triangles: CBSE Class 10 Previous Year Questions

210 different questions from Triangles (NCERT Chapter 6) asked in CBSE Class 10 Maths board exams 2022–2026. 8 of them came up in more than one year. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions70 questions 2 marks questions56 questions 3 marks questions21 questions 4 marks questions4 questions 5 marks questions59 questions

Most asked Triangles questions

Q35 (OR) (OR)5 marksLong AnswerTriangles CBSE 2023 · Standard 30/4/2

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

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Answer: Proved.

Given: In $△ A B C$ , a line DE $∥$ BC meets AB at D and AC at E. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: join BE and CD; draw EM $⊥$ AB and DN $⊥$ AC.
ar(ADE) $= \frac{1}{2} A D \cdot E M$ , ar(BDE) $= \frac{1}{2} D B \cdot E M$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ .
ar(ADE) $= \frac{1}{2} A E \cdot D N$ , ar(DEC) $= \frac{1}{2} E C \cdot D N$ , so $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ .
$△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
Therefore $\frac{A D}{D B} = \frac{A E}{E C}$ .

Q335 marksLong AnswerTriangles CBSE 2024 · Standard 30/3/1

State and prove Basic Proportionality theorem.

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Answer: Proved.

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In $△ A B C$ , DE $∥$ BC, with D on AB and E on AC. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ .
Construction: Join BE and CD; draw EM $⊥$ AB and DN $⊥$ AC.
ar(ADE) = $\frac{1}{2} \times A D \times E M$ and ar(BDE) = $\frac{1}{2} \times D B \times E M$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ .
ar(ADE) = $\frac{1}{2} \times A E \times D N$ and ar(DEC) = $\frac{1}{2} \times E C \times D N$ , so $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ .
Triangles BDE and DEC are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC).
Hence $\frac{A D}{D B} = \frac{A E}{E C}$ .

Also asked in: 2026 Standard 30/1/1, 2026 Standard 30/1/2, 2026 Standard 30/1/3, 2025 Basic 430/2/3, 2025 Basic 430/3/1, 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q283 marksShort AnswerTriangles CBSE 2023 · Basic 430/6/1

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $△ A B E \sim △ C F B$ .

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Answer: Proved.

In $△ A B E$ and $△ C F B$ :
$∠ A = ∠ C$ (opposite angles of parallelogram ABCD).
AD $∥$ BC, so AE $∥$ BC and BE is a transversal; hence $∠ A E B = ∠ C B F$ (alternate angles).
By AA similarity, $△ A B E \sim △ C F B$ .

Also asked in: 2024 Standard 30/4/1, 2024 Standard 30/4/2, 2024 Standard 30/4/3, 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q35 (OR) (OR)5 marksLong AnswerTriangles CBSE 2023 · Basic 430/6/1

In the given figure, $\frac{QR}{QS} = \frac{QT}{P R}$ and $∠1 = ∠2$ . Prove that $△ P QS \sim △ T QR$ .

Diagram for CBSE 2023 Class 10 Maths question 35 (OR)

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Answer: Proved.

In $△ P QR$ , $∠1 = ∠2$ i.e. $∠ P QR = ∠ P R Q$ , so $P Q = P R$ (sides opposite equal angles).
Given $\frac{QR}{QS} = \frac{QT}{P R} = \frac{QT}{P Q}$ , so $\frac{QS}{QR} = \frac{P Q}{QT}$ .
In $△ P QS$ and $△ T QR$ : $\frac{P Q}{T Q} = \frac{QS}{QR}$ and $∠ P QS = ∠ T QR$ (common angle $∠1$ ).
By SAS similarity, $△ P QS \sim △ T QR$ .

Also asked in: 2025 Basic 430/3/1, 2025 Basic 430/3/2, 2025 Basic 430/3/3, 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q355 marksLong AnswerTriangles CBSE 2024 · Standard 30/3/3

In $△ A B C$ , if AD $⊥$ BC and $A D^{2} = B D \times D C$ , then prove that $∠ B A C = 9 0^{\circ}$ .

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Answer: Proved.

$A D^{2} = B D \times D C \Rightarrow \frac{A D}{B D} = \frac{D C}{A D}$ .
In $△ A D B$ and $△ C D A$ : $∠ A D B = ∠ C D A = 9 0^{\circ}$ and $\frac{A D}{C D} = \frac{B D}{A D}$ .
So $△ A D B \sim △ C D A$ (SAS similarity).
Hence $∠ B A D = ∠ A C D$ and $∠ A B D = ∠ C A D$ .
$∠ B A C = ∠ B A D + ∠ C A D = ∠ C + ∠ B$ .
In $△ A B C$ : $∠ A + ∠ B + ∠ C = 18 0^{\circ} \Rightarrow 2∠ B A C = 18 0^{\circ}$ .
So $∠ B A C = 9 0^{\circ}$ .

Also asked in: 2025 Standard 30/1/1, 2025 Standard 30/1/2, 2025 Standard 30/1/3

Q242 marksVery Short AnswerTriangles CBSE 2023 · Basic 430/2/2

In the given figure, $L M ∥ C B$ and $L N ∥ C D$ . Prove that $\frac{A M}{A N} = \frac{A B}{A D}$ .

Diagram for CBSE 2023 Class 10 Maths question 24

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Answer: Proved.

In $△ A B C$ , $L M ∥ C B$ , so by BPT, $\frac{A M}{A B} = \frac{A L}{A C}$ ...(1)
In $△ A C D$ , $L N ∥ C D$ , so by BPT, $\frac{A N}{A D} = \frac{A L}{A C}$ ...(2)
From (1) and (2), $\frac{A M}{A B} = \frac{A N}{A D}$
Hence $\frac{A M}{A N} = \frac{A B}{A D}$ .

Also asked in: 2024 Basic 430/4/1, 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q35 (OR) (OR)5 marksLong AnswerTriangles CBSE 2023 · Standard 30/2/1

Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD (produced) in E. Prove that EL = 2BL.

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Answer: Proved.

In $△ B M C$ and $△ E M D$ : MC = MD, $∠ B M C = ∠ E M D$ (vertically opposite), $∠ B C M = ∠ E D M$ (alternate angles, $B C ∥ A E$ ).
So $△ B M C ≅ △ E M D$ (ASA), giving DE = BC.
AD = BC, so AE = AD + DE = 2BC.
In $△ A L E$ and $△ C L B$ : $∠ A L E = ∠ C L B$ (vertically opposite), $∠ L A E = ∠ L C B$ (alternate angles).
So $△ A L E \sim △ C L B$ (AA) and $\frac{E L}{B L} = \frac{A E}{C B} = 2$ .
Hence EL = 2BL.

Also asked in: 2024 Standard 30/3/2, 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q313 marksShort AnswerTriangles CBSE 2023 · Basic 430/1/1

In the given figure, DE $∥$ AC and DF $∥$ AE
Prove that
$\frac{B F}{F E} = \frac{B E}{E C}$

Diagram for CBSE 2023 Class 10 Maths question 31

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Answer: Proved.

In $△ A B C$ , DE $∥$ AC, so by BPT $\frac{B D}{D A} = \frac{B E}{E C}$ ... (1)
In $△ A B E$ , DF $∥$ AE, so by BPT $\frac{B D}{D A} = \frac{B F}{F E}$ ... (2)
From (1) and (2), $\frac{B F}{F E} = \frac{B E}{E C}$ .

Also asked in: 2026 Standard 30/4/2, 2023 Basic 430/1/3

Q151 markMCQTriangles CBSE 2026 · Basic 430/4/1

In the given figure, PQ $∥$ BC. If AP : AB = 3 : 7 then, AQ : QC equals :

Diagram for CBSE 2026 Class 10 Maths question 15

(A)3 : 7
(B)3 : 10
(C)7 : 3
(D)3 : 4

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Answer: (D) 3 : 4

AP : AB = 3 : 7, so AP : PB = 3 : 4.
By BPT, $\frac{A Q}{QC} = \frac{A P}{P B} = \frac{3}{4}$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q51 markMCQTriangles CBSE 2026 · Basic 430/5/1

Which of the following statements is not always true ?

(A)Two circles are similar.
(B)Two isosceles right triangles are similar.
(C)Two rectangles are similar.
(D)Two equilateral triangles are similar.

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Answer: (C) Two rectangles are similar.

All circles, all isosceles right triangles (angles 45°, 45°, 90°) and all equilateral triangles have the same shape, so they are always similar.
Two rectangles have equal angles, but their sides need not be in the same ratio (e.g. 1 × 2 and 1 × 3).
So 'Two rectangles are similar' is not always true.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

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