Triangles: Important Questions and Answers for Class 10

Triangles is one of the most important chapters in Class 10 Mathematics, as it introduces students to the concepts of similarity, proportionality, and geometric proofs. The chapter includes important topics such as similarity criteria, Basic Proportionality Theorem (BPT), and area relationships between similar triangles. In this guide, you will find carefully selected important questions covering all major concepts from the Class 10 chapter, Triangles, designed according to the latest syllabus and exam pattern to support effective preparation and revision.

Our subject experts have provided detailed solutions for these problems based on the latest CBSE syllabus and the NCERT textbook. This material helps students revise the chapter easily and perform well in the final examination.

Table of Contents

• Key Concepts on Triangles for Class 10
• Section A: 1-Mark Questions (MCQ & Very Short Answer)
• Section B: 2-Mark Questions (Short Answer)
• Section C: 3-Mark Questions
• Section D: 5-Mark Questions (Long Answer & Word Problems)
• Section E: Previous Year Questions (PYQs) with Solutions
• Section F: HOTS & Case-Study Based Questions
• Exam Tips & Common Mistakes to Avoid

Key Concepts on Triangles for Class 10

• Similar Figures vs Congruent Figures

Two figures are congruent if they are identical in shape and size. Two figures are similar if they have the same shape but not necessarily the same size. All congruent figures are similar, but the reverse is not always true. For example, all circles are similar. All squares are similar. But two rectangles with different aspect ratios are not similar even though they are both rectangles.

• Basic Proportionality Theorem (BPT) or Thales' Theorem

Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio.

If DE ∥ BC in △ABC, then   AD/DB = AE/EC

The converse is equally important: if a line divides two sides of a triangle in the same ratio, it is parallel to the third side. 

Both the theorem and its converse are tested as 3–5 mark questions.

• Criteria for Similarity of Triangles

Angle-Angle (AA): Two angles of one triangle equal two angles of another, then the triangles are similar

Side-Side-Side (SSS): All three pairs of corresponding sides are proportional, then the triangles are similar

Side-Angle-Side (SAS): One pair of corresponding angles is equal and the sides including those angles are proportional, then the triangles are similar

• Areas of Similar Triangles

Area Ratio Theorem: 

If △ABC ~ △PQR, then

ar(△ABC) / ar(△PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)²

The ratio of areas = the square of the ratio of corresponding sides

Section A: 1-Mark Questions (MCQ & Very Short Answer)

Q1: If △ABC ~ △PQR with BC/QR = 1/4, find ar(△PRQ)/ar(△ABC).

Solution: Given: △ABC ~ △PRQ and BC/QR = 1/4

Ratio of areas = (ratio of corresponding sides)²

ar(△ABC)/ar(△PRQ) = (BC/QR)² = (1/4)² = 1/16

∴ ar(△PRQ)/ar(△ABC) = 16/1 = 16

Q2: The sides of two similar triangles are in the ratio 7 : 10. Find the ratio of their areas.

Solution: Ratio of areas = (Ratio of sides)²

 = (7/10)²

 = 49 : 100        

Q3: In △ABC, ∠BAC = 90° and AD ⊥ BC. Then which of the following is true? (a) BD · CD = BC² (b) AB · AC = BC² (c) BD · CD = AD² (d) AB · AC = AD²

Solution: When the altitude is drawn from the right angle to the hypotenuse in a right-angled triangle, the square of the altitude = product of the two segments of the hypotenuse.

Answer: (c) BD · CD = AD²

Q4: If areas of two similar triangles are in the ratio 25 : 64, write the ratio of their corresponding sides.

Solution: Ratio of areas = (Ratio of sides)²

25/64 = (side ratio)²

Side ratio = √(25/64) = 5/8

Ratio of corresponding sides = 5 : 8

Q5: In △ABC, DE ∥ BC. If AD = 3 cm, DB = 9 cm and AC = 12 cm, find AE.

Solution: By BPT: AD/DB = AE/EC

3/9   = AE/EC  

⇒ AE/EC = 1/3

Let AE = x, then EC = 12 − x

x/(12 − x) = 1/3

⇒ 3x = 12 − x

⇒ 4x = 12

⇒ AE = 3 cm

Section B: 2-Mark Questions (Short Answer)

Q7: In △ABC, PQ ∥ BC and AP : PB = 1 : 2. Find the ratio of ar(△APQ) to ar(△ABC).

Solution: Since PQ ∥ BC, by AA similarity:

△APQ ~ △ABC

AP/AB = AP/(AP + PB) = 1/(1 + 2) = 1/3

⇒ Ratio of areas = (AP/AB)² = (1/3)² = 1/9

⇒ ar(△APQ) : ar(△ABC) = 1 : 9    

Q8: R and S are points on sides DE and EF respectively of △DEF such that ER = 5 cm, RD = 2.5 cm, SE = 1.5 cm and FS = 3.5 cm. Check whether RS ∥ DF or not.

Solution: For RS ∥ DF, by converse of BPT, we need: ER/RD = ES/SF

⇒ ER/RD = 5/2.5 = 2

⇒ ES/SF = 1.5/3.5 = 3/7

⇒ Since 2 ≠ 3/7, the ratios are NOT equal.

RS is NOT parallel to DF.

Q9: The perimeters of two similar triangles ABC and LMN are 60 cm and 48 cm respectively. If LM = 8 cm, find AB.

Solution: For similar triangles, the ratio of perimeters = ratio of corresponding sides

⇒ AB/LM = Perimeter of △ABC / Perimeter of △LMN

⇒ AB/8  = 60/48 = 5/4

⇒ AB    = 8 × 5/4

⇒ AB = 10 cm

Q10: The lengths of the diagonals of a rhombus are 30 cm and 40 cm. Find the side of the rhombus.

Solution: The diagonals of a rhombus bisect each other at right angles. 

So, half-diagonals are 15 cm and 20 cm, forming the legs of a right triangle. The side of the rhombus is the hypotenuse.

Side² = 15² + 20² = 225 + 400 = 625

⇒ Side  = √625

⇒ Side of rhombus = 25 cm

Q11: △ABC ~ △DEF. If BC = 3 cm, EF = 4 cm and area of △ABC = 54 cm², find the area of △DEF.

Solution:  ar(△ABC)/ar(△DEF) = (BC/EF)²

⇒ 54/ar(△DEF) = (3/4)² = 9/16

⇒ ar(△DEF) = 54 × 16/9

⇒ ar(△DEF) = 96 cm²

Section C: 3-Mark Questions

Q12: In △ABC, if ∠ADE = ∠B, prove that △ADE ~ △ABC.

Solution: Given: In △ABC, ∠ADE = ∠B

To prove: △ADE ~ △ABC

In △ADE and △ABC:

∠A = ∠A          (common angle)

∠ADE = ∠ABC      (given)

By AA criterion:

△ADE ~ △ABC

Q13: In △ABC, from altitudes AD and BE, prove that △ADC ~ △BEC.

Solution: Given: In △ABC, AD ⊥ BC and BE ⊥ AC

To prove: △ADC ~ △BEC

In △ADC and △BEC:

∠ADC = ∠BEC = 90°   (altitudes)

∠ACD = ∠BCE          (same angle C, common to both)

By AA criterion:

△ADC ~ △BEC 

Q14: In an equilateral triangle ABC of side 8 cm, find the altitude.

Solution: Given: AB = BC = CA = 8 cm

Draw altitude AD ⊥ BC. D is the midpoint of BC.

BD = BC/2 = 4 cm (altitude bisects base in equilateral △)

In right △ADB:

AB² = AD² + BD²    (Pythagoras)

⇒ 8²  = AD² + 4²

⇒ 64  = AD² + 16

⇒ AD² = 48

⇒ AD  = √48 = 4√3

Altitude = 4√3 cm ≈ 6.93 cm

Q15: Diagonals of a trapezium PQRS intersect each other at O. PQ ∥ RS and PQ = 3RS. Find the ratio of ar(△POQ) to ar(△ROS).

Solution: In △POQ and △ROS:

∠OPQ = ∠ORS    (alternate interior angles, PQ ∥ RS)

∠OQP = ∠OSR    (alternate interior angles, PQ ∥ RS)

By AA criterion: △POQ ~ △ROS

PQ/RS = 3/1 (given PQ = 3RS)

⇒ ar(△POQ)/ar(△ROS) = (PQ/RS)² = (3/1)² = 9/1

⇒ ar(△POQ) : ar(△ROS) = 9 : 1     

Q16: A pole 4 m high casts a shadow 2.5 m long. At the same time, a tower casts a shadow 15 m long. Find the height of the tower.

Solution: At the same time of day, the sun's rays strike the ground at the same angle. This means the pole and tower, along with their shadows, form similar right triangles.

Height of pole / Shadow of pole = Height of tower / Shadow of tower

(Same sun angle ⇒ similar triangles)

⇒ 4 / 2.5 = Height of tower / 15

⇒ Height of tower = 4 × 15 / 2.5

= 60 / 2.5

Height of tower = 24 m

Section D: 5-Mark Questions (Long Answer & Word Problems)

Q17: State and prove the Basic Proportionality Theorem (Thales' Theorem).

Solution: 

Statement: If a line is drawn parallel to one side of a triangle to intersect

the other two sides in distinct points, then the other two sides are divided

in the same ratio.

Given: In △ABC, DE ∥ BC, where D is on AB and E is on AC.

To prove: AD/DB = AE/EC

Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.

Proof:

ar(△ADE) = ½ × AD × EN   ...(base AD, height EN)

ar(△BDE) = ½ × DB × EN   ...(base DB, height EN)

∴ ar(△ADE)/ar(△BDE) = AD/DB   ...(1)

ar(△ADE) = ½ × AE × DM   ...(base AE, height DM)

ar(△CDE) = ½ × EC × DM   ...(base EC, height DM)

∴ ar(△ADE)/ar(△CDE) = AE/EC   ...(2)

Now, △BDE and △CDE are on the same base DE and between the same

parallel lines DE and BC (since DE ∥ BC).

∴ ar(△BDE) = ar(△CDE)   ...(3)

From (1), (2), and (3):

AD/DB = AE/EC

Hence proved.

Q18: Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. (Frequently Asked in Board Exams)

Solution: 

Given: △ABC ~ △PQR

To prove: ar(△ABC)/ar(△PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)²

Construction: Draw altitudes AM ⊥ BC and PN ⊥ QR.

Step 1: Express areas in terms of base and height.

ar(△ABC) = ½ × BC × AM

ar(△PQR) = ½ × QR × PN

ar(△ABC)/ar(△PQR) = (BC × AM)/(QR × PN)   ...(1)

Step 2: Show AM/PN = AB/PQ using similarity.

In △ABM and △PQN:

∠AMB = ∠PNQ = 90°

∠B = ∠Q  (△ABC ~ △PQR corresponding angles equal)

By AA: △ABM ~ △PQN

∴ AM/PN = AB/PQ   ...(2)

Step 3: Since △ABC ~ △PQR:

AB/PQ = BC/QR = AC/PR   ...(3)

Step 4: Substitute (2) and (3) in (1):

ar(△ABC)/ar(△PQR) = (BC/QR) × (AM/PN)

= (BC/QR) × (AB/PQ)

= (BC/QR) × (BC/QR)    [from (3): AB/PQ = BC/QR]

= (BC/QR)²

∴ ar(△ABC)/ar(△PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)² 

Q19: O is a point inside rectangle ABCD. Prove that OB² + OD² = OA² + OC².

Solution: Given: ABCD is a rectangle. O is any point inside it.

To prove: OB² + OD² = OA² + OC²

Construction: Through O, draw PQ ∥ BC such that P lies on AB and Q lies on DC.

Since PQ ∥ BC and BC ∥ AD:

BPQC is a rectangle ⇒ BP ⊥ PQ and BQ = PC

APQD is a rectangle  ⇒  AP ⊥ PQ and AQ = PD

In right △OPB:  OB² = OP² + BP²   ...(1)

In right △OQD:  OD² = OQ² + DQ²   ...(2)

Adding: OB² + OD² = OP² + BP² + OQ² + DQ²   ...(3)

In right △OPA:  OA² = OP² + AP²   ...(4)

In right △OQC:  OC² = OQ² + QC²   ...(5)

Adding: OA² + OC² = OP² + AP² + OQ² + QC²   ...(6)

Since BPQC is a rectangle: BP = QC

Since APQD is a rectangle: AP = DQ

Substituting in (3) and (6):

OB² + OD² = OP² + BP² + OQ² + AP²   [since DQ = AP]

OA² + OC² = OP² + AP² + OQ² + BP²   [since QC = BP]

Both expressions are equal.

∴ OB² + OD² = OA² + OC² 

Section E: Previous Year Questions (PYQs) with Solutions

Q 20: In △ABC, D and E are points on sides AB and AC respectively. If AD = 6 cm, DB = 9 cm, AE = 8 cm and EC = 12 cm, show that DE ∥ BC. (CBSE 2020)

Solution: Converse of BPT

AD/DB = 6/9 = 2/3

AE/EC = 8/12 = 2/3

Since AD/DB = AE/EC, by the Converse of Basic Proportionality Theorem,

DE ∥ BC

Q 21: △ABC ~ △PQR. If AB/PQ = 1/3, find ar(△PQR)/ar(△ABC). (CBSE 2019)

Solution: ar(△ABC)/ar(△PQR) = (AB/PQ)² = (1/3)² = 1/9

∴ ar(△PQR)/ar(△ABC) = 9/1 = 9

Q 22: In △ABC ~ △XYZ and AD, XE are angle bisectors of ∠A and ∠X respectively such that AD = 4 cm and XE = 3 cm. Find the ratio of ar(△ABD) to ar(△XYE). (CBSE 2019)

Solution: Since △ABC ~ △XYZ, the angle bisectors are also in proportion with the sides. 

For similar triangles, the ratio of angle bisectors = ratio of corresponding sides.

△ABC ~ △XYZ 

⇒ △ABD ~ △XYE (AA same angles, bisected halves)

⇒ ar(△ABD)/ar(△XYE) = (AD/XE)²

= (4/3)²

⇒ ar(△ABD) : ar(△XYE) = 16 : 9

Q 23: In equilateral △ABC, D is a point on BC such that BD = (1/3)BC. Prove that 9AD² = 7AB². (CBSE 2018)

Solution: Let each side of equilateral △ABC = a.

BD = a/3, so DC = 2a/3.

Draw AE ⊥ BC. Since △ABC is equilateral, E is the midpoint of BC.

So BE = a/2.

⇒ DE = BE − BD = a/2 − a/3 = a/6

In right △ADE (AE ⊥ BC):

AD² = AE² + DE²   ...(Pythagoras)

⇒ AE² = AB² − BE² = a² − (a/2)² = a² − a²/4 = 3a²/4

⇒ AD² = 3a²/4 + (a/6)²

    = 3a²/4 + a²/36

    = 27a²/36 + a²/36

    = 28a²/36

    = 7a²/9

∴ 9AD² = 9 × (7a²/9) = 7a² = 7AB²

Hence, 9AD² = 7AB² 

Q 24: If PQ || RS, prove that △POQ ~ △SOR (where O is the intersection of lines PR and QS). (CBSE 2018)

Solution: Given: PQ ∥ RS. O is the point where PR and QS intersect.

To prove: △POQ ~ △SOR

In △POQ and △SOR:

∠OPQ = ∠OSR  (alternate interior angles, PQ ∥ RS)

∠OQP = ∠ORS  (alternate interior angles, PQ ∥ RS)

∠POQ = ∠SOR  (vertically opposite angles)

By AA criterion:

△POQ ~ △SOR

Section F: HOTS & Case-Study Based Questions 

Q 25: In △ABC, D is a point on side BC such that ∠ADC = ∠BAC. Prove that CA² = CB × CD.

Solution: Given: In △ABC, ∠ADC = ∠BAC

To prove: CA² = CB × CD

In △BAC and △ADC:

∠BAC = ∠ADC   (given)

∠ACB = ∠DCA   (common angle C)

By AA criterion: △BAC ~ △ADC

∴ CA/CD = CB/CA   (corresponding sides of similar triangles)

Cross-multiplying:

CA² = CB × CD

Q 26: ABC is a triangle. PQ is a line segment intersecting AB at P and AC at Q such that PQ ∥ BC and divides △ABC into two parts equal in area. Find BP/AB.

Solution: Since PQ ∥ BC: △APQ ~ △ABC

Let AP/AB = k

ar(△APQ)/ar(△ABC) = k²   ...(area ratio = square of side ratio)

Given: ar(△APQ) = ar(trapezium BPQC) = ½ × ar(△ABC)

∴ ar(△APQ)/ar(△ABC) = 1/2

So: k² = 1/2  

⇒  k = 1/√2

AP/AB = 1/√2

∴ AP = AB/√2

∴ BP = AB − AP = AB − AB/√2 = AB(1 − 1/√2)

⇒ BP/AB = 1 − 1/√2 = (√2 − 1)/√2

⇒ BP/AB = (√2 − 1)/√2

Q 27: Case Study

A civil engineer is designing two triangular plots (Plot A and Plot B) that are similar in shape. The sides of Plot A are 30 m, 40 m, and 50 m. The longest side of Plot B is 75 m. (i) Are the sides of Plot A a right triangle? (ii) Find the other two sides of Plot B. (iii) Find the ratio of areas of Plot A to Plot B.

Solution: 

(i) Check if Plot A is a right triangle:

30² + 40² = 900 + 1600 = 2500 = 50²

Yes. Plot A is a right-angled triangle (right angle opposite 50 m side).

(ii) Ratio of corresponding sides (Plot A : Plot B):

50/75 = 2/3

Other sides of Plot B:

30 × (3/2) = 45 m

40 × (3/2) = 60 m

Plot B sides: 45 m, 60 m, 75 m

(iii) Ratio of areas:

ar(Plot A)/ar(Plot B) = (2/3)² = 4/9

⇒ Plot A : Plot B = 4 : 9

Q 28: Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Solution: Let ABCD be a rhombus with side a. 

Let diagonals AC = 2p and BD = 2q.

The diagonals bisect each other at right angles at point O.

So AO = p, BO = q, and ∠AOB = 90°.

In right △AOB (by Pythagoras):

⇒ AB² = AO² + BO² = p² + q²

Since all sides of a rhombus are equal:

AB = BC = CD = DA = a

Sum of squares of sides = 4a² = 4(p² + q²)   ...(1)

Sum of squares of diagonals = AC² + BD²

= (2p)² + (2q)²

= 4p² + 4q²

= 4(p² + q²)       ...(2)

From (1) and (2):

Sum of squares of sides = Sum of squares of diagonals

Exam Tips & Common Mistakes to Avoid

• Given below are exam tips for Triangles help students approach problems in a structured way, avoid common mistakes, and improve accuracy:

• State What's Given and to Prove: Begin every proof with ‘Given:’ and ‘To prove:’. Then ‘Construction:’ if applicable. Then ‘Proof:’. This structure signals to the examiner that your reasoning is systematic.

• Match Vertices Correctly: The single most common mistake is wrong vertex correspondence. If ∠A = ∠P and ∠B = ∠Q, write △ABC ~ △PQR, not △ABC ~ △QPR. Wrong correspondence loses all CPST marks.

• Cite the Criterion by Name: Always write ‘By AA criterion’ or ‘By SSS criterion’ or ‘By SAS criterion’. Declaring similarity without citing the criterion costs 1 mark in every proof.

• Areas = Square of Side Ratios: This is tested in every board exam. If side ratio is m:n, area ratio is m²:n². Conversely, if area ratio is m²:n², side ratio is m:n. Memorise this.

• BPT vs Converse: BPT: line parallel: divides sides proportionally. Converse: sides divided proportionally: the line is parallel. Know which direction of reasoning the question requires before writing.

Common Mistakes Students Make: