Triangles: 3 marks Questions (CBSE Class 10)

21 different 3 marks questions on Triangles from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (70)2 marks (56)3 marks (21)4 marks (4)5 marks (59)

Q313 marksShort AnswerTriangles CBSE 2025 · Basic 430/5/1

Point E lies on the extended side AD of parallelogram ABCD. BE intersects CD at F. Show that (i) $△ D F E \sim △ C F B$ (ii) $△ A E B \sim △ C B F$ .

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Answer: Proved.

ABCD is a parallelogram, so AD $∥$ BC (hence AE $∥$ BC) and $∠ A = ∠ C$ .
(i) In $△ D F E$ and $△ C F B$ : $∠ D F E = ∠ C F B$ (vertically opposite angles).
$∠ F D E = ∠ F C B$ (alternate angles, DE $∥$ BC with transversal DC).
So $△ D F E \sim △ C F B$ (AA).
(ii) In $△ A E B$ and $△ C B F$ : $∠ E A B = ∠ B C F$ (opposite angles of a parallelogram).
$∠ A E B = ∠ C B F$ (alternate angles, AE $∥$ BC with transversal BE).
So $△ A E B \sim △ C B F$ (AA).

Also asked in: 2025 Basic 430/5/2

Q293 marksShort AnswerTriangles CBSE 2025 · Basic 430/5/3

It is given that $△ A C D ≅ △ A B E$ . Prove that $△ A D E \sim △ A B C$ .

Diagram for CBSE 2025 Class 10 Maths question 29

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Answer: Proved.

Since $△ A C D ≅ △ A B E$ , corresponding parts are equal: AD = AE and AC = AB.
So $\frac{A D}{A B} = \frac{A E}{A C}$ .
In $△ A D E$ and $△ A B C$ , $∠ D A E = ∠ B A C$ (common angle).
Hence $△ A D E \sim △ A B C$ (SAS similarity criterion).

Q303 marksShort AnswerTriangles CBSE 2024 · Basic 430/3/1

In the given figure, $∠ A B C = ∠ A C B$ and $\frac{B C}{B E} = \frac{B D}{A C}$ .
Show that $△ A B E \sim △ D B C$ and $A E ∥ D C$ .

Diagram for CBSE 2024 Class 10 Maths question 30

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Answer: Proved.

$∠ A B C = ∠ A C B$ , so $A B = A C$ (sides opposite equal angles).
Given $\frac{B C}{B E} = \frac{B D}{A C}$ ; replacing AC by AB gives $\frac{B C}{B E} = \frac{B D}{A B}$ , i.e. $\frac{A B}{B D} = \frac{B E}{B C}$ .
In $△ A B E$ and $△ D B C$ , $∠ B$ is common and $\frac{A B}{D B} = \frac{B E}{B C}$ .
So $△ A B E \sim △ D B C$ (SAS similarity).
Hence $∠ B A E = ∠ B D C$ ; these are corresponding angles for lines AE and DC with transversal BD.
Therefore $A E ∥ D C$ .

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

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

Q31 (OR) (OR)3 marksShort AnswerTriangles CBSE 2023 · Basic 430/1/1

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{A O}{B O} = \frac{C O}{O D}$
Show that quadrilateral ABCD is a trapezium.

Diagram for CBSE 2023 Class 10 Maths question 31 (OR)

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Answer: Proved.

Given $\frac{A O}{B O} = \frac{C O}{O D}$ , so $\frac{A O}{C O} = \frac{B O}{O D}$ ... (1)
Through O draw OE $∥$ AB, meeting AD at E.
In $△ A B D$ , EO $∥$ AB, so by BPT $\frac{A E}{E D} = \frac{B O}{O D}$ ... (2)
From (1) and (2), $\frac{A E}{E D} = \frac{A O}{O C}$ .
So in $△ A D C$ , EO divides AD and AC in the same ratio; by the converse of BPT, EO $∥$ DC.
Thus AB $∥$ EO $∥$ DC, so ABCD is a trapezium.

Also asked in: 2023 Basic 430/1/3

Q273 marksShort AnswerTriangles CBSE 2023 · Basic 430/1/2

S and T are points on sides PR and QR of $△ P QR$ such that $∠ P = ∠ R T S$ . Show that $△ R P Q \sim △ R T S$ .

Diagram for CBSE 2023 Class 10 Maths question 27

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Answer: Proved.

In $△ R P Q$ and $△ R T S$ :
$∠ R P Q = ∠ R T S$ (given, $∠ P = ∠ R T S$ )
$∠ P R Q = ∠ T R S$ (common angle R)
So $△ R P Q \sim △ R T S$ (AA similarity).

Q27 (OR) (OR)3 marksShort AnswerTriangles CBSE 2023 · Basic 430/1/2

The diagonal BD of a quadrilateral ABCD bisects both $∠ B$ and $∠ D$ . Show that $\frac{A B}{B C} = \frac{A D}{C D}$ .

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Answer: Proved.

In $△ A B D$ and $△ C B D$ :
$∠ A B D = ∠ C B D$ (BD bisects $∠ B$ )
$∠ A D B = ∠ C D B$ (BD bisects $∠ D$ )
BD = BD (common)
So $△ A B D ≅ △ C B D$ (ASA), giving AB = BC and AD = CD (CPCT).
Hence $\frac{A B}{B C} = 1 = \frac{A D}{C D}$ .

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

Q28 (OR) (OR)3 marksShort AnswerTriangles CBSE 2023 · Basic 430/6/1

In the given figure, CM and RN are respectively the medians of $△ A B C$ and $△ P QR$ . If $△ A B C \sim △ P QR$ , then prove that $△ A M C \sim △ P N R$ .

Diagram for CBSE 2023 Class 10 Maths question 28 (OR)

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Answer: Proved.

Since $△ A B C \sim △ P QR$ : $\frac{A B}{P Q} = \frac{A C}{P R}$ and $∠ A = ∠ P$ .
M and N are mid-points of AB and PQ, so $A M = \frac{1}{2} A B$ and $P N = \frac{1}{2} P Q$ .
Hence $\frac{A M}{P N} = \frac{A B}{P Q} = \frac{A C}{P R}$ .
In $△ A M C$ and $△ P N R$ : $\frac{A M}{P N} = \frac{A C}{P R}$ and $∠ M A C = ∠ N P R$ .
By SAS similarity, $△ A M C \sim △ P N R$ .

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

Q283 marksShort AnswerTriangles CBSE 2023 · Standard 30/5/1

In the given figure, CD is the perpendicular bisector of AB. EF is perpendicular to CD. AE intersects CD at G. Prove that $\frac{C F}{C D} = \frac{F G}{D G}$ .

Diagram for CBSE 2023 Class 10 Maths question 28

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Answer: Proved.

EF $⊥$ CD and BD $⊥$ CD, so EF $∥$ BD and EF $∥$ AD.
In $△ C F E$ and $△ C D B$ : $∠ C$ common, $∠ C F E = ∠ C D B = 9 0^{\circ}$ , so $△ C F E \sim △ C D B$ (AA). Hence $\frac{C F}{C D} = \frac{E F}{B D}$ ... (1)
In $△ GF E$ and $△ G D A$ : $∠ F GE = ∠ D G A$ (vertically opposite), $∠ GF E = ∠ G D A = 9 0^{\circ}$ , so $△ GF E \sim △ G D A$ (AA). Hence $\frac{F G}{D G} = \frac{E F}{A D}$ ... (2)
D is the mid-point of AB, so AD = BD.
From (1) and (2), $\frac{C F}{C D} = \frac{F G}{D G}$ .

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q28 (OR) (OR)3 marksShort AnswerTriangles CBSE 2023 · Standard 30/5/1

In the given figure, ABCD is a parallelogram. BE bisects CD at M and intersects AC at L. Prove that EL = 2BL.

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Answer: Proved.

E lies on AD produced. In $△ E D M$ and $△ B C M$ : DM = CM (M is the mid-point of CD), $∠ D M E = ∠ C M B$ (vertically opposite), $∠ E D M = ∠ B C M$ (alternate angles, ED $∥$ BC).
So $△ E D M ≅ △ B C M$ (ASA), giving ED = BC.
BC = AD (opposite sides of parallelogram), 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, AE $∥$ BC). So $△ A L E \sim △ C L B$ (AA).
$\frac{E L}{B L} = \frac{A E}{C B} = \frac{2 B C}{B C} = 2$ .
Hence EL = 2BL.

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q283 marksShort AnswerTriangles CBSE 2023 · Standard 30/6/1

In the given figure, E is a point on the side CB produced of an isosceles triangle ABC with AB = AC. If AD $⊥$ BC and EF $⊥$ AC, then prove that $△ A B D \sim △ E C F$ .

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Answer: Proved.

AB = AC, so $∠ A B C = ∠ A C B$ (angles opposite equal sides).
In $△ A B D$ and $△ E C F$ :
$∠ A B D = ∠ A B C = ∠ A C B = ∠ E C F$
$∠ A D B = ∠ E F C = 9 0^{\circ}$
So $△ A B D \sim △ E C F$ (AA similarity).

Also asked in: 2023 Standard 30/6/2, 2023 Standard 30/6/3

Q8 (OR) (OR)3 marksShort AnswerTriangles CBSE 2022 · Basic 430/1/1Not in current syllabus

Draw a line segment PQ = 7.5 cm. Divide it in the ratio 3 : 1.

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Answer: Construction (see steps).

Draw PQ = 7.5 cm and a ray PX making an acute angle with PQ
Mark 4 equal points $A_{1}, A_{2}, A_{3}, A_{4}$ on PX
Join $A_{4} Q$ ; through $A_{3}$ draw a line parallel to $A_{4} Q$ meeting PQ at R
Then PR : RQ = 3 : 1 (by BPT), i.e. PR = 5.625 cm, RQ = 1.875 cm

Also asked in: 2022 Basic 430/1/2

Q9 (OR) (OR)3 marksShort AnswerTriangles CBSE 2022 · Basic 430/1/3Not in current syllabus

Draw a line segment AB = 8.5 cm. Divide it in the ratio 1 : 3.

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Answer: Construction (see steps).

Draw AB = 8.5 cm and a ray AX making an acute angle with AB
Mark 4 equal points $A_{1}, A_{2}, A_{3}, A_{4}$ on AX
Join $A_{4} B$ ; through $A_{1}$ draw a line parallel to $A_{4} B$ meeting AB at C
Then AC : CB = 1 : 3 (by BPT), i.e. AC = 2.125 cm, CB = 6.375 cm

Q83 marksShort AnswerTriangles CBSE 2022 · Basic 430/2/2Not in current syllabus

Draw a line segment of length 7.5 cm and divide it in the ratio 1 : 3.

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Answer: Construction (the dividing point is 1.875 cm from one end).

Draw AB = 7.5 cm and a ray AX making an acute angle with AB.
Mark 4 (= 1 + 3) equal arcs $A_{1}, A_{2}, A_{3}, A_{4}$ along AX.
Join $A_{4} B$ .
Through $A_{1}$ draw a line parallel to $A_{4} B$ meeting AB at P.
Then AP : PB = 1 : 3 (by the Basic Proportionality Theorem); AP = 1.875 cm.

Q7 (OR) (OR)3 marksShort AnswerTriangles CBSE 2022 · Basic 430/3/1Not in current syllabus

Draw a line segment of length 9.5 cm and divide it in the ratio 2 : 3.

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Answer: Construction (parts 3.8 cm and 5.7 cm)

Draw AB = 9.5 cm and a ray AX making an acute angle with AB.
Mark 5 equal arcs $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ on AX.
Join $A_{5} B$ ; through $A_{2}$ draw a line parallel to $A_{5} B$ meeting AB at C.
Then AC : CB = 2 : 3 (by BPT), i.e. AC = 3.8 cm and CB = 5.7 cm.

Also asked in: 2022 Basic 430/3/3

Q9 (OR) (OR)3 marksShort AnswerTriangles CBSE 2022 · Basic 430/3/2Not in current syllabus

Draw a line segment AB = 9 cm. Divide AB in the ratio 2 : 3.

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Answer: Construction (parts 3.6 cm and 5.4 cm)

Draw AB = 9 cm and a ray AX making an acute angle with AB.
Mark 5 equal arcs $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ on AX.
Join $A_{5} B$ ; through $A_{2}$ draw a line parallel to $A_{5} B$ meeting AB at C.
Then AC : CB = 2 : 3 (by BPT), i.e. AC = 3.6 cm and CB = 5.4 cm.

Q83 marksShort AnswerTriangles CBSE 2022 · Basic 430/4/1Not in current syllabus

Draw a line segment of length 7 cm and divide it in the ratio 5 : 3.

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Answer: Point P on AB with AP : PB = 5 : 3 (AP = 4.375 cm, PB = 2.625 cm)

Draw AB = 7 cm.
Draw a ray AX making an acute angle with AB.
Mark 8 (= 5 + 3) equal arcs $A_{1}, A_{2}, \dots, A_{8}$ along AX.
Join $A_{8} B$ .
Through $A_{5}$ draw a line parallel to $A_{8} B$ meeting AB at P.
By the Basic Proportionality Theorem, AP : PB = 5 : 3.

Also asked in: 2022 Basic 430/4/3

Q73 marksShort AnswerTriangles CBSE 2022 · Basic 430/4/2Not in current syllabus

Draw a line segment AB of length 6 cm and mark a point X on it such that AX $= \frac{4}{5}$ AB. [Using a scale & compass]

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Answer: Point X on AB with AX : XB = 4 : 1 (AX = 4.8 cm)

AX $= \frac{4}{5}$ AB means AX : XB = 4 : 1.
Draw AB = 6 cm and a ray AY making an acute angle with AB.
Mark 5 equal arcs $A_{1}, \dots, A_{5}$ along AY and join $A_{5} B$ .
Through $A_{4}$ draw a line parallel to $A_{5} B$ meeting AB at X.
By the Basic Proportionality Theorem, AX : XB = 4 : 1, so AX $= \frac{4}{5}$ AB = 4.8 cm.

Q83 marksShort AnswerTriangles CBSE 2022 · Standard 30/1/3Not in current syllabus

Draw a line segment AB = 7 cm. Divide it in the ratio 3 : 2.

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Answer: Construction: point P on AB with AP : PB = 3 : 2 (AP = 4.2 cm, PB = 2.8 cm).

Draw AB = 7 cm. Draw a ray AX making an acute angle with AB.
Mark 5 (= 3 + 2) equal arcs on AX: $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ .
Join $A_{5} B$ . Through $A_{3}$ draw a line parallel to $A_{5} B$ , meeting AB at P.
By BPT, AP : PB = $A A_{3} : A_{3} A_{5}$ = 3 : 2.
Check: AP $= \frac{3}{5} \times 7 = 4.2$ cm, PB = 2.8 cm.

Q73 marksShort AnswerTriangles CBSE 2022 · Standard 30/3/1Not in current syllabus

Draw a line segment AB of length 8 cm and locate a point P on AB such that AP : PB = 1 : 5.

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Answer: P is located on AB with AP : PB = 1 : 5 (AP =

\frac{4}{3}

cm ≈ 1.33 cm, PB =

\frac{20}{3}

cm ≈ 6.67 cm).

Draw AB = 8 cm.
Draw a ray AX making an acute angle with AB.
Mark 6 (= 1 + 5) equal arcs on AX at points $A_{1}, A_{2}, \dots, A_{6}$ .
Join $A_{6} B$ .
Through $A_{1}$ draw a line parallel to $A_{6} B$ , meeting AB at P.
By BPT, $A P : P B = A A_{1} : A_{1} A_{6} = 1 : 5$ . So $A P = \frac{8}{6} = \frac{4}{3}$ cm.

Also asked in: 2022 Standard 30/3/2, 2022 Standard 30/3/3

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