Important Questions Class 10 Maths Chapter 6 Triangles

Similar triangles are triangles with the same shape, equal corresponding angles, and proportional corresponding sides. The ratio of corresponding sides helps find unknown lengths, prove parallel lines, and measure inaccessible heights.

A triangle can reveal a height or distance that cannot be measured directly. Important Questions Class 10 Maths Chapter 6 help students practise similar figures, similar triangles, Basic Proportionality Theorem, converse of BPT, AA similarity, SSS similarity, SAS similarity, medians, trapezium proofs, and shadow-based problems. The 2026 NCERT Chapter 6, Triangles, explains similarity through same-shape figures, proportional sides, equal angles, and indirect measurement.

Key Takeaways

• Similar Figures: Similar figures have the same shape but may have different sizes.
• Basic Proportionality Theorem: If a line is parallel to one side of a triangle, it divides the other two sides in the same ratio.
• Similarity Criteria: AA, AAA, SSS, SAS, and RHS help prove triangle similarity.
• Indirect Measurement: Similar triangles help find heights, shadows, and distances without direct measurement.

Important Questions Class 10 Maths Chapter 6 Structure 2026

Triangles Class 10 Chapter Overview

Triangles Class 10 focuses on figures that have the same shape but different sizes. The chapter begins with similar figures and then moves into triangle similarity.

Students use ratios, parallel lines, equal angles, and side correspondence. NCERT also connects triangle similarity with indirect measurement, such as heights and shadows.

Class 10 Maths Chapter 6 Important Questions On Similar Figures

Similar figures are not judged by size. They are judged by shape, equal corresponding angles, and proportional corresponding sides.

This idea appears in circles, squares, equilateral triangles, photographs, maps, and blueprints.

Q1. What Are Similar Figures In Class 10 Maths?

Similar figures are figures that have the same shape but not necessarily the same size.

Congruent figures have the same shape and same size. Similar figures may have different sizes.

Example: All circles are similar, but circles of different radii are not congruent.

Q2. Are All Congruent Figures Similar?

Yes, all congruent figures are similar.

Congruent figures have the same shape and same size. Similarity needs only the same shape.

Example: Two squares of side 6 cm are congruent and similar.

Q3. Are All Similar Figures Congruent?

No, all similar figures are not congruent.

Similar figures can have different sizes. Congruent figures must have equal size.

Example: Circles of radii 3 cm and 7 cm are similar but not congruent.

Q4. What Is The Condition For Two Polygons To Be Similar?

Two polygons are similar when their corresponding angles are equal and corresponding sides are proportional.

Both conditions are needed for polygons. Equal angles alone do not prove polygon similarity.

Example: A square and a rectangle have equal angles, but their sides may not be proportional.

Q5. Why Are Enlarged Photographs Similar To The Original Photograph?

Enlarged photographs are similar because every corresponding length changes by the same scale factor.

The angles between corresponding parts remain equal. The shape stays unchanged.

Example: A 35 mm photograph enlarged to 45 mm keeps all lengths in the ratio 35 : 45.

Similar Triangles Class 10 Important Questions

Similar triangles follow a stricter pattern than general polygons. In triangles, angle equality can prove side proportionality.

Correct vertex order matters because it fixes corresponding sides and angles.

Q6. What Are Similar Triangles Class 10?

Similar triangles are triangles whose corresponding angles are equal and corresponding sides are proportional.

If ∆ABC ~ ∆DEF, then A corresponds to D, B corresponds to E, and C corresponds to F.

Corresponding side ratio:
AB/DE = BC/EF = CA/FD

Q7. Why Is Vertex Order Important In Similar Triangles?

Vertex order is important because it shows the correct correspondence.

If ∆ABC ~ ∆DEF, then A ↔ D, B ↔ E, and C ↔ F. Wrong order gives wrong side ratios.

Correct ratio:
AB/DE = BC/EF = AC/DF

Q8. If ∆ABC ~ ∆PQR, Write The Corresponding Side Ratios.

The corresponding side ratios are AB/PQ = BC/QR = CA/RP.

The order ∆ABC ~ ∆PQR fixes the matching vertices. A matches P, B matches Q, and C matches R.

Final Answer: AB/PQ = BC/QR = CA/RP

Q9. If ∆ABC ~ ∆RQP, Which Angle Equals ∠C?

∠C equals ∠P.

The order ∆ABC ~ ∆RQP gives A ↔ R, B ↔ Q, and C ↔ P.

Final Answer: ∠C = ∠P

Basic Proportionality Theorem Class 10 Questions

Basic Proportionality Theorem is also called Thales theorem. It links parallel lines with equal ratios on two sides of a triangle.

Most questions ask students to find a missing side or prove a ratio.

Q10. State The Basic Proportionality Theorem Class 10.

Basic Proportionality Theorem states that a line parallel to one side of a triangle divides the other two sides in the same ratio.

In ∆ABC, if DE ∥ BC, then:

AD/DB = AE/EC

This theorem is also known as Thales theorem.

Q11. In ∆ABC, DE ∥ BC, AD = 3 cm, DB = 2 cm, And AE = 6 cm. Find EC.

The value of EC is 4 cm.

1. Given Data:
   AD = 3 cm, DB = 2 cm, AE = 6 cm, DE ∥ BC
2. Formula Used:
   AD/DB = AE/EC
3. Substitute values:
   3/2 = 6/EC
4. Cross multiply:
   3 × EC = 2 × 6
   3EC = 12
5. Divide by 3:
   EC = 12 ÷ 3 = 4

Final Result: EC = 4 cm

Q12. In ∆ABC, DE ∥ BC, AD = 4 cm, AE = 5 cm, And EC = 10 cm. Find DB.

The value of DB is 8 cm.

1. Given Data:
   AD = 4 cm, AE = 5 cm, EC = 10 cm, DE ∥ BC
2. Formula Used:
   AD/DB = AE/EC
3. Substitute values:
   4/DB = 5/10
4. Cross multiply:
   5 × DB = 4 × 10
   5DB = 40
5. Divide by 5:
   DB = 40 ÷ 5 = 8

Final Result: DB = 8 cm

Q13. If DE ∥ BC In ∆ABC, Prove That AD/AB = AE/AC.

The result follows from BPT by converting part ratios into whole-side ratios.

1. Given Data:
   DE ∥ BC
2. By BPT:
   AD/DB = AE/EC
3. Invert both sides:
   DB/AD = EC/AE
4. Add 1 to both sides:
   DB/AD + 1 = EC/AE + 1
5. Combine terms:
   (DB + AD)/AD = (EC + AE)/AE
6. Use whole sides:
   AB/AD = AC/AE
7. Invert both sides:
   AD/AB = AE/AC

Final Result: AD/AB = AE/AC

Q14. Why Does The Proof Of BPT Use Areas?

The proof uses areas because triangles on the same base and between the same parallels have equal areas.

The proof compares ∆ADE, ∆BDE, and ∆DEC. Their area ratios become side ratios.

Example:
ar(ADE)/ar(BDE) = AD/DB
ar(ADE)/ar(DEC) = AE/EC

Converse Of BPT Class 10 Questions

The converse of BPT works from ratios to parallel lines. If two sides are divided in the same ratio, the joining line is parallel to the third side.

This theorem is useful in proof-based NCERT questions.

Q15. State The Converse Of BPT Class 10.

The converse of BPT states that a line dividing two sides of a triangle in the same ratio is parallel to the third side.

In ∆ABC, if:

AD/DB = AE/EC

then:

DE ∥ BC

This proves parallelism from side ratios.

Q16. In ∆PQR, E Lies On PQ And F Lies On PR. PE = 4 cm, EQ = 5 cm, PF = 8 cm, FR = 10 cm. Is EF ∥ QR?

Yes, EF ∥ QR.

1. Given Data:
   PE = 4 cm, EQ = 5 cm, PF = 8 cm, FR = 10 cm
2. Check first ratio:
   PE/EQ = 4/5
3. Check second ratio:
   PF/FR = 8/10 = 4/5
4. Compare ratios:
   PE/EQ = PF/FR
5. Apply converse of BPT:
   EF ∥ QR

Final Result: EF ∥ QR

Q17. In ∆PQR, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm. Is EF ∥ QR?

No, EF is not parallel to QR.

1. Given Data:
   PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
2. Check first ratio:
   PE/EQ = 3.9/3 = 1.3
3. Check second ratio:
   PF/FR = 3.6/2.4 = 1.5
4. Compare ratios:
   PE/EQ ≠ PF/FR

Final Result: EF is not parallel to QR

Q18. Prove That The Line Joining The Mid-Points Of Two Sides Of A Triangle Is Parallel To The Third Side.

The line joining the mid-points is parallel to the third side by the converse of BPT.

1. In ∆ABC, let D be the mid-point of AB.
   So, AD = DB.
2. Let E be the mid-point of AC.
   So, AE = EC.
3. Write the ratios:
   AD/DB = 1
   AE/EC = 1
4. Therefore:
   AD/DB = AE/EC
5. By converse of BPT:
   DE ∥ BC

Final Result: DE ∥ BC

Q19. Prove That A Line Through The Mid-Point Of One Side Parallel To Another Side Bisects The Third Side.

The line bisects the third side by Basic Proportionality Theorem.

1. In ∆ABC, D is the mid-point of AB.
   So, AD = DB.
2. Draw DE ∥ BC, where E lies on AC.
3. By BPT:
   AD/DB = AE/EC
4. Since AD = DB:
   AD/DB = 1
5. Therefore:
   AE/EC = 1
6. So:
   AE = EC

Final Result: E is the mid-point of AC

AA Similarity Criterion Class 10 Questions

AA similarity is the most used similarity test in diagram-based questions. Two equal angles are enough because the third angle becomes equal automatically.

Students should look for parallel-line angles, vertically opposite angles, and right angles.

Q20. State The AA Similarity Criterion Class 10.

AA similarity criterion states that two triangles are similar if two angles of one triangle equal two angles of another triangle.

The third angle also becomes equal because each triangle has angle sum 180°. This makes the triangles equiangular.

Example: If ∠A = ∠D and ∠B = ∠E, then ∆ABC ~ ∆DEF.

Q21. If PQ ∥ RS, Prove That ∆POQ ~ ∆SOR.

∆POQ ~ ∆SOR by AA similarity.

1. Given Data:
   PQ ∥ RS
2. Since PQ ∥ RS:
   ∠P = ∠S
   ∠Q = ∠R
3. Vertically opposite angles:
   ∠POQ = ∠SOR
4. Two corresponding angles are equal.
5. Therefore:
   ∆POQ ~ ∆SOR

Final Result: ∆POQ ~ ∆SOR

Q22. In Two Right Triangles, One Acute Angle Is Equal. Are They Similar?

Yes, the two right triangles are similar by AA similarity.

Each triangle has one angle equal to 90°. If one acute angle is equal, the third angle also becomes equal.

Example: Two triangles with angles 90°, 40°, and 50° are similar.

Q23. Why Is AAA Similarity Also Written As AA Similarity In Triangles?

AAA similarity is written as AA similarity because two equal angles force the third angle to be equal.

A triangle has total angle measure 180°. Once two angles match, the remaining angle must match.

Example: If two angles are 60° and 40°, the third angle is 80°.

SSS Similarity Criterion Class 10 Questions

SSS similarity compares three pairs of corresponding sides. The side order must match the triangle order.

If all three side ratios are equal, the corresponding angles also become equal.

Q24. State The SSS Similarity Criterion Class 10.

SSS similarity criterion states that two triangles are similar if their corresponding sides are proportional.

If:

AB/DE = BC/EF = CA/FD

then:

∆ABC ~ ∆DEF

This criterion proves equality of corresponding angles.

Q25. Are Triangles With Sides 3 cm, 6 cm, 8 cm And 4.5 cm, 9 cm, 12 cm Similar?

Yes, the triangles are similar by SSS similarity.

1. Given Data:
   First triangle sides = 3 cm, 6 cm, 8 cm
   Second triangle sides = 4.5 cm, 9 cm, 12 cm
2. Compare side ratios:
   3/4.5 = 2/3
   6/9 = 2/3
   8/12 = 2/3
3. All three ratios are equal.

Final Result: The triangles are similar by SSS similarity

Q26. In ∆ABC And ∆PQR, AB/RQ = BC/QP = CA/PR. If ∠C = 40°, Find ∠P.

∠P = 40°.

1. Given Data:
   AB/RQ = BC/QP = CA/PR
2. Correct correspondence:
   AB ↔ RQ
   BC ↔ QP
   CA ↔ PR
3. Therefore:
   ∆ABC ~ ∆RQP
4. Matching vertices:
   A ↔ R, B ↔ Q, C ↔ P
5. Since ∠C = 40°:
   ∠P = 40°

Final Result: ∠P = 40°

Q27. Why Is Checking Only Two Side Ratios Not Enough For SSS Similarity?

Checking only two side ratios is not enough because SSS similarity needs all three side ratios.

Two side pairs may be proportional, but the included or opposite angle can change. The triangle shape may differ.

Example: Two sides fixed in ratio can still form different triangles.

SAS Similarity Criterion Class 10 Questions

SAS similarity uses two proportional side pairs and one equal included angle. The included angle must lie between the two compared sides.

This criterion appears in product-relation and vertically opposite angle questions.

Q28. State The SAS Similarity Criterion Class 10.

SAS similarity criterion states that two triangles are similar when one included angle is equal and the including sides are proportional.

The equal angle must lie between the two proportional sides. Otherwise, the criterion does not apply.

Example: If AB/DE = AC/DF and ∠A = ∠D, then ∆ABC ~ ∆DEF.

Q29. If OA × OB = OC × OD, Prove That ∠A = ∠C And ∠B = ∠D.

The angles are equal because ∆AOD ~ ∆COB by SAS similarity.

1. Given Data:
   OA × OB = OC × OD
2. Rearrange the equation:
   OA/OC = OD/OB
3. Vertically opposite angles:
   ∠AOD = ∠COB
4. Apply SAS similarity:
   ∆AOD ~ ∆COB
5. Corresponding angles are equal:
   ∠A = ∠C
   ∠D = ∠B

Final Result: ∠A = ∠C and ∠B = ∠D

Q30. Why Must The Equal Angle Be Included In SAS Similarity?

The equal angle must be included because it fixes the shape between the two proportional sides.

If the equal angle lies elsewhere, the third side may change. The triangles may fail to be similar.

Example: ∠A must lie between AB and AC in ∆ABC.

Q31. Are Triangles With AB = 2 cm, AC = 4 cm, ∠A = 50°, DE = 3 cm, DF = 6 cm, ∠D = 50° Similar?

Yes, the triangles are similar by SAS similarity.

1. Given Data:
   AB = 2 cm, AC = 4 cm, ∠A = 50°
   DE = 3 cm, DF = 6 cm, ∠D = 50°
2. Compare side ratios:
   AB/DE = 2/3
   AC/DF = 4/6 = 2/3
3. Included angles:
   ∠A = ∠D = 50°
4. Apply SAS similarity.

Final Result: ∆ABC ~ ∆DEF

Triangles Class 10 Important Questions From NCERT Exercises

NCERT exercise questions often combine BPT, similarity criteria, and one construction. Many proofs need a diagonal, a median, or a parallel-line argument.

Students should write the theorem name after each ratio or similarity step.

Q32. In A Trapezium ABCD, AB ∥ DC, Diagonals Meet At O. Prove AO/BO = CO/DO.

The result follows by proving ∆AOB ~ ∆COD.

1. Given Data:
   AB ∥ DC
2. Alternate interior angles:
   ∠ABO = ∠CDO
   ∠BAO = ∠DCO
3. Vertically opposite angles:
   ∠AOB = ∠COD
4. Therefore:
   ∆AOB ~ ∆COD
5. Corresponding sides give:
   AO/CO = BO/DO
6. Rearrange:
   AO/BO = CO/DO

Final Result: AO/BO = CO/DO

Q33. In A Trapezium ABCD, AB ∥ DC, E And F Lie On AD And BC. If EF ∥ AB, Prove AE/ED = BF/FC.

The result follows by drawing diagonal AC and applying BPT twice.

1. Draw diagonal AC.
   Let AC meet EF at G.
2. Since EF ∥ DC in ∆ADC:
   AE/ED = AG/GC
3. Since EF ∥ AB in ∆CAB:
   CG/GA = CF/FB
4. Invert the second ratio:
   AG/GC = BF/FC
5. Therefore:
   AE/ED = BF/FC

Final Result: AE/ED = BF/FC

Q34. If PS/SQ = PT/TR And ∠PST = ∠PRQ, Prove That ∆PQR Is Isosceles.

∆PQR is isosceles because ∠PQR = ∠PRQ.

1. Given Data:
   PS/SQ = PT/TR
2. By converse of BPT:
   ST ∥ QR
3. Corresponding angles:
   ∠PST = ∠PQR
4. Given:
   ∠PST = ∠PRQ
5. Therefore:
   ∠PQR = ∠PRQ
6. Sides opposite equal angles are equal:
   PQ = PR

Final Result: ∆PQR is isosceles

Q35. D Lies On BC Of ∆ABC Such That ∠ADC = ∠BAC. Prove CA² = CB × CD.

The result follows by proving ∆ADC ~ ∆BAC.

1. Given Data:
   ∠ADC = ∠BAC
2. Since D lies on BC:
   ∠ACD = ∠BCA
3. By AA similarity:
   ∆ADC ~ ∆BAC
4. Corresponding sides:
   AC/BC = CD/AC
5. Cross multiply:
   AC × AC = BC × CD

Final Result: CA² = CB × CD

Class 10 Maths Triangles Questions With Answers On Shadows

Shadow questions use similar right triangles. Vertical objects make right angles with the ground.

NCERT connects this idea with indirect measurement of heights and distances.

Q36. A 6 m Pole Casts A 4 m Shadow. A Tower Casts A 28 m Shadow At The Same Time. Find The Tower Height.

The height of the tower is 42 m.

1. Given Data:
   Pole height = 6 m
   Pole shadow = 4 m
   Tower shadow = 28 m
   Tower height = x m
2. Similar triangle ratio:
   Height/Shadow is same.
3. Form the equation:
   6/4 = x/28
4. Cross multiply:
   4x = 6 × 28
   4x = 168
5. Solve:
   x = 168 ÷ 4 = 42

Final Result: Tower height = 42 m

Q37. A Girl Of Height 90 cm Walks Away From A 3.6 m Lamp-Post At 1.2 m/s. Find Her Shadow After 4 Seconds.

The length of her shadow is 1.6 m.

1. Convert height:
   90 cm = 0.9 m
2. Distance walked in 4 seconds:
   1.2 × 4 = 4.8 m
3. Let shadow length be x m.
4. Total distance from lamp base to shadow tip:
   4.8 + x
5. Use similar triangles:
   (4.8 + x)/x = 3.6/0.9
6. Simplify:
   (4.8 + x)/x = 4
7. Solve:
   4.8 + x = 4x
   3x = 4.8
   x = 1.6

Final Result: Shadow length = 1.6 m

Q38. Why Are Shadow Triangles Similar?

Shadow triangles are similar because they have one right angle and one equal light-ray angle.

The object and ground form 90°. The same sunlight or lamp ray gives another equal angle.

Example: A pole and a tower at the same time form similar right triangles.

NCERT Class 10 Maths Chapter 6 Questions On Medians

Median questions test whether similarity transfers to smaller triangles. A median divides the opposite side into two equal parts.

NCERT uses medians to prove ratios between corresponding medians and sides.

Q39. If ∆ABC ~ ∆PQR And CM, RN Are Medians, Prove ∆AMC ~ ∆PNR.

∆AMC ~ ∆PNR by SAS similarity.

1. Given Data:
   ∆ABC ~ ∆PQR
   CM and RN are medians
2. From similarity:
   AB/PQ = AC/PR
3. Since CM and RN are medians:
   AB = 2AM
   PQ = 2PN
4. Substitute in side ratio:
   2AM/2PN = AC/PR
5. Simplify:
   AM/PN = AC/PR
6. Corresponding included angles:
   ∠MAC = ∠NPR
7. Apply SAS similarity:
   ∆AMC ~ ∆PNR

Final Result: ∆AMC ~ ∆PNR

Q40. If ∆ABC ~ ∆PQR, Prove CM/RN = AB/PQ, Where CM And RN Are Medians.

The ratio of corresponding medians equals the ratio of corresponding sides.

1. From Q39:
   ∆AMC ~ ∆PNR
2. Corresponding sides give:
   CM/RN = AC/PR
3. Since ∆ABC ~ ∆PQR:
   AC/PR = AB/PQ
4. Therefore:
   CM/RN = AB/PQ

Final Result: CM/RN = AB/PQ

Important Questions Class 10 Maths All Chapters