Chapter 6 · Written-answer practice
Practise 52 selected Triangles questions worth 2 marks each, from CBSE Class 10 Basic and Standard board papers, and official sample papers. Practise BPT, AA and SAS similarity, corresponding-side ratios, and shadow problems. Try each question before opening its solution.
52 numbered questions · 74 board-paper appearances + 7 sample pairings
71 worked solutions, including both alternatives in 19 OR pairs
By Siddharth Agrawal, Lead Maths Tutor, TutorVertex · Solutions prepared with Team TutorVertex
Updated
| Use | Result |
|---|---|
| BPT | If DE ∥ BC in △ABC, then AD/DB = AE/EC. |
| Converse of BPT | If AD/DB = AE/EC, then DE ∥ BC. |
| AA similarity | Two pairs of equal corresponding angles establish similarity. |
| SAS similarity | Two pairs of corresponding sides are proportional and their included angles are equal. |
| Corresponding sides | If △ABC ∼ △PQR, then AB/PQ = BC/QR = AC/PR. |
| Perimeter ratio | For similar triangles, the perimeter ratio equals the corresponding-side ratio. |
| Given | Starting point |
|---|---|
| A parallel line in a triangle | Use BPT for divided sides or AA similarity for whole-side ratios. |
| Equal division ratios | Use the converse of BPT to prove the line parallel. |
| Two equal angle pairs | Write the triangles in matching order and use AA. |
| Two proportional side pairs | Check their included angles before using SAS. |
| A length or a shadow | Establish similarity, then write one proportion with matching sides. |
Showing all 52 questions
No questions match these filters. Try another combination or reset the filters.
Q1. In the given figure, DE ∥ AC and = . Prove that DC ∥ AP. [2 marks]
Appeared in: 2023 · Basic · Main
In △BAC, DE ∥ AC, so BD/DA = BE/EC by BPT.
Given BE/EC = BC/CP, hence BD/DA = BC/CP.
In △BAP, D lies on BA and C lies on BP.
By the converse of BPT, DC ∥ AP.
DC ∥ AP
Q2. In the given figure, LM ∥ CB and LN ∥ CD. Prove that = . [2 marks]
Appeared in: 2023 · Basic · Main
In △ABC, ML ∥ BC, so AM/AB = AL/AC by similarity.
In △ACD, LN ∥ CD, so AN/AD = AL/AC.
Therefore AM/AB = AN/AD.
Rearranging gives AM/AN = AB/AD.
AM/AN = AB/AD
Q3. (a) In the given figure, PQ ∥ RS. Prove that OP × OR = OQ × OS. [2 marks]
∠OPQ = ∠OSR (alternate interior angles).
∠POQ = ∠SOR (vertically opposite angles).
Therefore △OPQ ∼ △OSR by AA.
OP/OS = OQ/OR.
OP × OR = OQ × OS.
OP × OR = OQ × OS
OR
(b) In the given figure, LM ∥ CB and LN ∥ CD. Prove that = . [2 marks]
Appeared in: 2024 · Basic · Main
In △ABC, ML ∥ BC, so AM/AB = AL/AC.
In △ACD, LN ∥ CD, so AN/AD = AL/AC.
Equating the two ratios gives AM/AB = AN/AD.
Rearranging gives AM/AN = AB/AD.
AM/AN = AB/AD
Q4. E and F are points on sides PQ and PR respectively of △PQR. If PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and PR = 6 cm, find whether EF ∥ QR. [2 marks]
Appeared in: 2024 · Standard · Supplementary
FR = PR − PF = 6 − 3.6 = 2.4 cm.
PE/EQ = 3.9/3 = 1.3.
PF/FR = 3.6/2.4 = 1.5.
The ratios are unequal, so the sides are not divided proportionally.
Therefore EF is not parallel to QR.
EF is not parallel to QR
Q5. In the given figure, ∠ADE = ∠ACB and = . Prove that △ABC is an isosceles triangle. [2 marks]
Appeared in: 2025 · Basic · Main
AD/DB = AE/EC, so DE ∥ BC by the converse of BPT.
Thus ∠ADE = ∠ABC (corresponding angles).
Given ∠ADE = ∠ACB, therefore ∠ABC = ∠ACB.
Sides opposite equal angles are equal, so AB = AC.
AB = AC; △ABC is isosceles
Q6. In the given figure, AB ∥ DE and BD ∥ EF. Prove that DC² = CF × AC. [2 marks]
Appeared in: 2025 · Basic · Main
In △CAB, DE ∥ AB, so CD/CA = CE/CB.
In △CBD, EF ∥ BD, so CF/CD = CE/CB.
Therefore CD/CA = CF/CD.
Cross-multiplying gives DC² = CF × AC.
DC² = CF × AC
Q7. (a) In the given figure, = and ∠PST = ∠PRQ. Prove that △PQR is isosceles. [2 marks]
PS/SQ = PT/TR, so ST ∥ QR by the converse of BPT.
Hence ∠PST = ∠PQR (corresponding angles).
Given ∠PST = ∠PRQ, so ∠PQR = ∠PRQ.
Therefore PQ = PR.
PQ = PR; △PQR is isosceles
OR
(b) In the given figure, △ABE ≅ △ACD. Show that △ADE ∼ △ABC. [2 marks]
Appeared in: 2025 · Standard · Main
Since △ABE ≅ △ACD, AB = AC and AE = AD (CPCT).
Thus AD/AB = AE/AC.
∠DAE = ∠BAC, as D lies on AB and E lies on AC.
Therefore △ADE ∼ △ABC by SAS similarity.
△ADE ∼ △ABC
Q8. (a) P and Q are points on sides DE and DF respectively of △DEF such that DP = 5 cm, DQ = 6 cm, DE = 15 cm and QF = 18 cm. Find whether PQ ∥ EF or not. [2 marks]
PE = DE − DP = 15 − 5 = 10 cm.
DP/PE = 5/10 = 1/2.
DQ/QF = 6/18 = 1/3.
The ratios are unequal, so PQ is not parallel to EF.
PQ is not parallel to EF
OR
(b) Prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. [2 marks]
Appeared in: 2025 · Basic · Supplementary
Let D and E be midpoints of AB and AC in △ABC.
AD = DB and AE = EC.
Thus AD/DB = AE/EC = 1.
By the converse of BPT, DE ∥ BC.
The line joining the midpoints is parallel to the third side
Q9. (a) S and T are points on sides PR and QR of △PQR such that ∠P = ∠RTS. Show that △RPQ ∼ △RTS. [2 marks]
∠RPQ = ∠RTS (given).
∠PRQ = ∠TRS, since R, S, P and R, T, Q are collinear.
Therefore △RPQ ∼ △RTS by AA.
△RPQ ∼ △RTS
OR
(b) If a line intersects sides AB and AC of △ABC at D and E respectively and is parallel to BC, prove that = . [2 marks]
Appeared in: 2025 · Standard · Supplementary
By BPT, AD/DB = AE/EC, so DB/AD = EC/AE.
Adding 1 gives (AD + DB)/AD = (AE + EC)/AE.
Hence AB/AD = AC/AE.
Taking reciprocals gives AD/AB = AE/AC.
AD/AB = AE/AC
Q10. In the given figure, DE ∥ AC and DF ∥ AE. Prove that = . [2 marks]
Appeared in: 2026 · Standard · Main
In △BAE, DF ∥ AE, so BF/FE = BD/DA by BPT.
In △BAC, DE ∥ AC, so BE/EC = BD/DA by BPT.
Equating gives BF/FE = BE/EC.
BF/FE = BE/EC
Q11. (a) In the given figure, EF ∥ CA and DF ∥ EA. Prove that = . [2 marks]
In △BEA, DF ∥ EA, so BD/DE = BF/FA by BPT.
In △BCA, EF ∥ CA, so BE/EC = BF/FA by BPT.
Thus BD/DE = BE/EC.
BD/DE = BE/EC
OR
(b) In the given figure, △ABE ≅ △ACD. Show that △ADE ∼ △ABC. [2 marks]
Appeared in: 2026 · Basic · Second board examination
Since △ABE ≅ △ACD, AB = AC and AE = AD (CPCT).
Thus AD/AB = AE/AC.
∠DAE = ∠BAC, as D lies on AB and E lies on AC.
Therefore △ADE ∼ △ABC by SAS similarity.
△ADE ∼ △ABC
Q12. (a) X and Y are points on sides PQ and PR respectively of △PQR. If PX, QX, PY and YR are 4, 4.5, 8 and 9 cm respectively, show that XY ∥ QR. [2 marks]
PX/XQ = 4/4.5 = 8/9.
PY/YR = 8/9.
Thus PX/XQ = PY/YR.
By the converse of BPT, XY ∥ QR.
XY ∥ QR
OR
(b) ABCD is a trapezium in which AB ∥ CD and P and Q are points on AD and BC respectively such that PQ ∥ DC. If PD = 18 cm, BQ = 3.5 cm and QC = 15 cm, find AD. [2 marks]
Appeared in: 2026 · Standard · Second board examination
Join BD and let it meet PQ at R.
In △DAB, PR ∥ AB, so AP/PD = BR/RD by BPT.
In △BCD, RQ ∥ CD, so BQ/QC = BR/RD by BPT.
Thus AP/18 = 3.5/15, giving AP = 4.2 cm.
AD = AP + PD = 4.2 + 18 = 22.2 cm.
AD = 22.2 cm
Q13. (a) In the given figure, ABCD is a trapezium in which AB ∥ CD ∥ EF. Prove that = . [2 marks]
Draw AC and let it meet EF at G.
In △ADC, EG ∥ DC, so AE/ED = AG/GC by BPT.
In △ABC, GF ∥ AB, so CG/GA = CF/FB.
Taking reciprocals gives AG/GC = BF/FC.
Therefore AE/ED = BF/FC.
AE/ED = BF/FC
OR
(b) In the given figure, AD = 6 cm, DB = 9 cm, AE = 8 cm, EC = 12 cm and ∠ADE = 48°. Find ∠ABC. [2 marks]
Appeared in: 2023–24 · Basic · Official sample
AD/DB = 6/9 = 2/3 and AE/EC = 8/12 = 2/3.
By the converse of BPT, DE ∥ BC.
Therefore ∠ABC = ∠ADE = 48° (corresponding angles).
∠ABC = 48°
Q14. D is a point on the side BC of △ABC such that ∠ADC = ∠BAC. Show that AC² = BC × DC. [2 marks]
Appeared in: 2025 · Basic · Main; 2026 · Standard · Main
In △ADC and △BAC, ∠ADC = ∠BAC (given).
∠ACD = ∠BCA, since B, D and C are collinear.
Therefore △ADC ∼ △BAC by AA similarity.
AC/BC = DC/AC (corresponding sides).
AC² = BC × DC.
AC² = BC × DC
Q15. In the given figure, ABCD is a quadrilateral. Diagonal BD bisects ∠B and ∠D both. Prove that (i) △ABD ∼ △CBD and (ii) AB = BC. [2 marks]
Appeared in: 2024 · Standard · Main
∠ABD = ∠CBD and ∠ADB = ∠CDB, since BD bisects the two angles.
Therefore △ABD ∼ △CBD by AA.
AB/CB = BD/BD = 1 (corresponding sides).
Hence AB = BC.
△ABD ∼ △CBD; AB = BC
Q16. (a) In △ABC, D and E are points on the sides AB and AC respectively such that BD = CE. If ∠B = ∠C, show that DE ∥ BC. [2 marks]
Since ∠B = ∠C, AB = AC (sides opposite equal angles).
AD = AB − BD and AE = AC − CE.
Thus AD = AE, and AD/DB = AE/EC.
By the converse of BPT, DE ∥ BC.
DE ∥ BC
OR
(b) If △ABC ∼ △DEF and AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of △ABC. [2 marks]
Appeared in: 2024 · Standard · Main
The ratio of corresponding sides is AB/DE = 4/6 = 2/3.
BC = (2/3) × 9 = 6 cm.
CA = (2/3) × 12 = 8 cm.
Perimeter = 4 + 6 + 8 = 18 cm.
Perimeter of △ABC = 18 cm
Q17. (a) In the given figure, if PQ ∥ RS, prove that △POQ ∼ △SOR. [2 marks]
∠OPQ = ∠OSR (alternate interior angles, PQ ∥ RS).
∠POQ = ∠SOR (vertically opposite angles).
Therefore △POQ ∼ △SOR by AA similarity.
△POQ ∼ △SOR
OR
(b) In the given figure, △OSR ∼ △OQP, ∠ROQ = 125° and ∠ORS = 70°. Find the measures of ∠OSR and ∠OQP. [2 marks]
Appeared in: 2025 · Basic · Main
∠SOR = 180° − ∠ROQ = 55° (linear pair).
In △OSR, ∠OSR = 180° − 55° − 70° = 55°.
Since △OSR ∼ △OQP, ∠OQP = ∠OSR = 55°.
∠OSR = 55°; ∠OQP = 55°
Q18. Two chords BA and CD intersect at a point P outside the circle. Prove that △PDA ∼ △PBC. [2 marks]
Appeared in: 2025 · Basic · Main
This proof also uses the cyclic-quadrilateral angle property from earlier geometry.
∠DAB + ∠DCB = 180° (opposite angles of cyclic quadrilateral).
∠PAD + ∠DAB = 180° (linear pair), so ∠PAD = ∠PCB.
∠DPA = ∠BPC, since P, A, B and P, D, C are collinear.
Therefore △PDA ∼ △PBC by AA.
△PDA ∼ △PBC
Q19. In a trapezium ABCD, AB ∥ DC and its diagonals intersect at O. Prove that = . [2 marks]
Appeared in: 2025 · Basic · Main
Since AB ∥ CD, ∠ABO = ∠CDO and ∠BAO = ∠DCO (alternate interior angles).
Therefore △AOB ∼ △COD by AA.
Corresponding sides give OA/OC = OB/OD.
OA/OC = OB/OD
Q20. (a) Prove that diagonals of a trapezium divide each other proportionally. [2 marks]
Let diagonals AC and BD of trapezium ABCD meet at O, with AB ∥ CD.
Since AB ∥ CD, ∠ABO = ∠CDO and ∠BAO = ∠DCO (alternate interior angles).
Therefore △AOB ∼ △COD by AA.
Corresponding sides give OA/OC = OB/OD.
OA/OC = OB/OD
OR
(b) S is a point on the side QR of △PQR such that ∠PSR = ∠QPR. Prove that PR² = QR × SR. [2 marks]
Appeared in: 2025 · Basic · Main
In △PSR and △QPR, ∠PSR = ∠QPR (given).
∠PRS = ∠QRP, since Q, S and R are collinear.
Therefore △PSR ∼ △QPR by AA.
PR/QR = SR/PR.
Hence PR² = QR × SR.
PR² = QR × SR
Q21. (a) In the given figure, D is a point on side BC of △ABC such that ∠ADC = ∠BAC. Show that CA² = CD × CB. [2 marks]
In △ADC and △BAC, ∠ADC = ∠BAC (given).
∠ACD = ∠BCA, since B, D and C are collinear.
Therefore △ADC ∼ △BAC by AA similarity.
AC/BC = DC/AC (corresponding sides).
AC² = BC × DC.
CA² = CD × CB
OR
(b) In the given figure, OA × OB = OC × OD. Show that ∠A = ∠C and ∠B = ∠D. [2 marks]
Appeared in: 2025 · Standard · Main
Given OA × OB = OC × OD, so OA/OC = OD/OB.
∠AOD = ∠COB (vertically opposite angles).
Therefore △AOD ∼ △COB by SAS similarity.
Corresponding angles give ∠A = ∠C and ∠D = ∠B.
∠A = ∠C; ∠B = ∠D
Q22. AD and PS are medians of triangles ABC and PQR respectively such that △ABD ∼ △PQS. Prove that △ABC ∼ △PQR. [2 marks]
Appeared in: 2025 · Standard · Main
Since AD and PS are medians, BC = 2BD and QR = 2QS.
Given △ABD ∼ △PQS, so AB/PQ = BD/QS and ∠ABD = ∠PQS.
Thus AB/PQ = BC/QR and ∠ABC = ∠PQR.
Therefore △ABC ∼ △PQR by SAS similarity.
△ABC ∼ △PQR
Q23. AD and PS are angle bisectors of ∠A and ∠P of triangles ABC and PQR. If △ABC ∼ △PQR, prove that △ACD ∼ △PRS. [2 marks]
Appeared in: 2025 · Standard · Main
Since △ABC ∼ △PQR, ∠A = ∠P and ∠C = ∠R.
AD and PS bisect ∠A and ∠P, so ∠CAD = ∠RPS.
Also ∠ACD = ∠PRS, since D lies on BC and S lies on QR.
Therefore △ACD ∼ △PRS by AA.
△ACD ∼ △PRS
Q24. In △ABC and △PQR, AD and PS are altitudes such that △ABD ∼ △PQS and △ACD ∼ △PRS. Prove that △ABC ∼ △PQR. [2 marks]
Appeared in: 2025 · Standard · Main
From △ABD ∼ △PQS, ∠ABD = ∠PQS.
From △ACD ∼ △PRS, ∠ACD = ∠PRS.
As D and S are the feet on BC and QR, ∠ABC = ∠PQR and ∠ACB = ∠PRQ.
Therefore △ABC ∼ △PQR by AA.
△ABC ∼ △PQR
Q25. In the given figure, two triangles ABC and PQR are such that ∠A = ∠P and ∠C = ∠R. If AD ⊥ BC and PS ⊥ QR, prove that (i) △ADB ∼ △PSQ and (ii) AD × QS = BD × PS. [2 marks]
Appeared in: 2026 · Basic · Main
In △ABC and △PQR, ∠B = ∠Q by the angle sum property.
∠ADB = ∠PSQ = 90° and ∠ABD = ∠PQS.
Therefore △ADB ∼ △PSQ by AA.
AD/PS = BD/QS (corresponding sides).
Cross-multiplying gives AD × QS = BD × PS.
△ADB ∼ △PSQ; AD × QS = BD × PS
Q26. In the given figure, DEFG is a square. ABC is a right triangle with ∠A = 90°. Prove that AG × DG = AF × DB. [2 marks]
Appeared in: 2026 · Standard · Main
∠GAF = ∠BDG = 90°, since ∠A = 90° and DEFG is a square.
∠AGF = ∠DBG, since A, G, B are collinear and GF ∥ BD.
Therefore △AGF ∼ △DBG by AA similarity.
AG/DB = AF/DG (corresponding sides).
Hence AG × DG = AF × DB.
AG × DG = AF × DB
Q27. Two right triangles PRQ and PSQ are drawn on the same hypotenuse PQ. If PR and QS intersect at T, prove that ST × TQ = PT × TR. [2 marks]
Appeared in: 2026 · Standard · Main
∠PST = ∠QRT = 90°.
∠PTS = ∠QTR (vertically opposite angles).
Therefore △PST ∼ △QRT by AA.
ST/TR = PT/TQ.
Hence ST × TQ = PT × TR.
ST × TQ = PT × TR
Q28. In the given figure, △ABE ≅ △ACD. Show that △ADE ∼ △ABC. [2 marks]
Appeared in: 2026 · Standard · Main
Since △ABE ≅ △ACD, AB = AC and AE = AD (CPCT).
Thus AD/AB = AE/AC.
∠DAE = ∠BAC, as D lies on AB and E lies on AC.
Therefore △ADE ∼ △ABC by SAS similarity.
△ADE ∼ △ABC
Q29. (a) In the figure, altitudes AD and CE of △ABC intersect at P. Prove that (i) △ABD ∼ △CBE and (ii) △PDC ∼ △BEC. [2 marks]
∠ADB = ∠CEB = 90° and ∠ABD = ∠CBE, so △ABD ∼ △CBE by AA.
Also ∠PDC = ∠BEC = 90°.
∠PCD = ∠BCE, since C, P, E and C, D, B are collinear.
Therefore △PDC ∼ △BEC by AA.
△ABD ∼ △CBE; △PDC ∼ △BEC
OR
(b) In the given figure, DE ∥ AC and DF ∥ AE. Prove that = . [2 marks]
Appeared in: 2022–23 · Basic · Official sample
In △BAE, DF ∥ AE, so BF/FE = BD/DA by BPT.
In △BAC, DE ∥ AC, so BE/EC = BD/DA by BPT.
Equating gives BF/FE = BE/EC.
BF/FE = BE/EC
Q30. In the given figure, = and ∠1 = ∠2. Show that △BAE ∼ △CAD. [2 marks]
Appeared in: 2022–23 · Standard · Official sample
In △ABD, ∠BAD = ∠ADB (given ∠1 = ∠2), so BD = AB.
Given AD/AE = AC/BD, hence AB/AC = AE/AD.
∠BAE = ∠CAD, since A, B, C are collinear and A, E, D are collinear.
Therefore △BAE ∼ △CAD by SAS similarity.
△BAE ∼ △CAD
Q31. ABCD is a parallelogram. Point P divides AB in the ratio 2 : 3 and point Q divides DC in the ratio 4 : 1. Prove that OC is half of OA. [2 marks]
Appeared in: 2023–24 · Standard · Official sample
AB = DC (opposite sides of a parallelogram).
AP = 2AB/5 and CQ = DC/5, so AP/CQ = 2.
AP ∥ CQ, so ∠APO = ∠CQO and ∠PAO = ∠QCO.
Therefore △APO ∼ △CQO by AA.
AO/CO = AP/CQ = 2.
OC = OA/2
Q32. If AP and DQ are medians of triangles ABC and DEF, where △ABC ∼ △DEF, prove that = . [2 marks]
Appeared in: 2025–26 · Standard · Official sample
From △ABC ∼ △DEF, AB/DE = BC/EF and ∠ABC = ∠DEF.
Since AP and DQ are medians, BP = BC/2 and EQ = EF/2.
Therefore AB/DE = BP/EQ and ∠ABP = ∠DEQ.
Thus △ABP ∼ △DEQ by SAS.
Hence AB/DE = AP/DQ.
AB/DE = AP/DQ
Q33. (a) ABCD is a trapezium in which AB is parallel to DC and its diagonals intersect at O. Show that = . [2 marks]
Since AB ∥ CD, ∠ABO = ∠CDO and ∠BAO = ∠DCO (alternate interior angles).
Therefore △AOB ∼ △COD by AA.
Corresponding sides give OA/OC = OB/OD.
Rearranging OA/OC = OB/OD gives AO/BO = CO/DO.
AO/BO = CO/DO
OR
(b) In the given figure, if ∠PQR = ∠QSP, PQ = 6 cm and PS = 3 cm, find PR. [2 marks]
Appeared in: 2026–27 · Standard · Official sample
∠PQR = ∠QSP (given) and ∠QPR = ∠SPQ (common angle).
Therefore △PQR ∼ △PSQ by AA.
PQ/PS = PR/PQ.
6/3 = PR/6.
PR = 12 cm.
PR = 12 cm
Q34. (a) ABCD is a trapezium in which AB is parallel to DC and its diagonals intersect at O. Show that = . [2 marks]
Since AB ∥ CD, ∠ABO = ∠CDO and ∠BAO = ∠DCO (alternate interior angles).
Therefore △AOB ∼ △COD by AA.
Corresponding sides give OA/OC = OB/OD.
Rearranging OA/OC = OB/OD gives AO/BO = CO/DO.
AO/BO = CO/DO
OR
(b) In a right-angled triangle ABC, right-angled at B, a perpendicular BD is drawn to hypotenuse AC. Prove that △BCD ∼ △ACB. [2 marks]
Appeared in: 2026–27 · Standard · Official sample
Accessible alternative to the preceding sample question. Part (a) is shared; the full OR pairing is retained.
∠BDC = ∠ABC = 90°.
∠BCD = ∠ACB, since A, D and C are collinear.
Therefore △BCD ∼ △ACB by AA similarity.
△BCD ∼ △ACB
Q35. P and Q are respectively the points on the sides AB and AC of a triangle ABC such that AB = 12.5 cm, AP = 5 cm and CQ = 6 cm. If PQ ∥ BC, find the length of AQ. [2 marks]
Appeared in: 2023 · Standard · Main
PB = AB − AP = 12.5 − 5 = 7.5 cm.
By BPT, AP/PB = AQ/QC.
5/7.5 = AQ/6.
AQ = 6 × 5/7.5 = 4 cm.
AQ = 4 cm
Q36. In the given figure, AP and BQ are perpendicular to AB. If OA = 15 cm, BO = 12 cm and AP = 10 cm, find BQ. [2 marks]
Appeared in: 2024 · Basic · Main
∠OAP = ∠OBQ = 90° and ∠AOP = ∠BOQ (vertically opposite).
Therefore △OAP ∼ △OBQ by AA.
AP/BQ = OA/OB.
10/BQ = 15/12.
BQ = 8 cm.
BQ = 8 cm
Q37. A line l intersects sides PQ and PR of △PQR at L and M respectively such that LM ∥ QR. If PQ = 12 cm, PR = 10 cm and PL = 7.2 cm, find the length of PM. [2 marks]
Appeared in: 2024 · Basic · Main
Since LM ∥ QR, △PLM ∼ △PQR.
PL/PQ = PM/PR.
7.2/12 = PM/10.
PM = 6 cm.
PM = 6 cm
Q38. Diagonals AC and BD of a trapezium ABCD intersect at O, where AB ∥ DC. If = , show that AB = 2CD. [2 marks]
Appeared in: 2024 · Standard · Main
∠ABO = ∠CDO (alternate interior angles).
∠AOB = ∠COD (vertically opposite angles).
Therefore △AOB ∼ △COD.
AB/CD = OB/OD = 2.
Thus AB = 2CD.
AB = 2CD
Q39. In △ABC, altitudes AD and BE are drawn. If AD = 7 cm, BE = 9 cm and EC = 12 cm, find the length of CD. [2 marks]
Appeared in: 2024 · Standard · Main
∠ADC = ∠BEC = 90° and ∠ACD = ∠BCE.
Therefore △ADC ∼ △BEC by AA.
AD/BE = CD/CE.
7/9 = CD/12.
CD = 28/3 cm.
CD = 28/3 cm
Q40. In the given figure, △AHK ∼ △ABC. If AK = 8 cm, BC = 3.2 cm and HK = 6.4 cm, find the length of AC. [2 marks]
Appeared in: 2024 · Standard · Main
Corresponding sides of similar triangles are proportional.
AK/AC = HK/BC.
8/AC = 6.4/3.2 = 2.
AC = 4 cm.
AC = 4 cm
Q41. E and F are points on the sides PQ and PR respectively of △PQR and EF ∥ QR. If PE = 2.5 cm, PQ = 6 cm and PF = 4.5 cm, find the lengths of RF and PR. [2 marks]
Appeared in: 2024 · Basic · Supplementary
Since EF ∥ QR, PE/PQ = PF/PR.
2.5/6 = 4.5/PR.
PR = 10.8 cm.
RF = PR − PF = 10.8 − 4.5 = 6.3 cm.
RF = 6.3 cm; PR = 10.8 cm
Q42. The diagonal BD of parallelogram ABCD is divided by segment AE in the ratio 1 : 2. If BE = 1.8 cm, find the length of AD. [2 marks]
Appeared in: 2025 · Basic · Main
Let AE meet BD at O. From the figure, BO/OD = 1/2.
Since BE ∥ AD, ∠OBE = ∠ODA and ∠BOE = ∠DOA.
Therefore △OBE ∼ △ODA.
BE/AD = BO/OD = 1/2.
AD = 2 × 1.8 = 3.6 cm.
AD = 3.6 cm
Q43. The perimeters of two similar triangles are 22 cm and 33 cm respectively. If one side of the first triangle is 9 cm, find the length of the corresponding side of the second triangle. [2 marks]
Appeared in: 2025 · Basic · Main
For similar triangles, the ratio of corresponding sides equals the ratio of perimeters.
9/x = 22/33.
x = 9 × 33/22 = 13.5 cm.
Corresponding side = 13.5 cm
Q44. (a) If △ABC ∼ △PQR in which AB = 6 cm, BC = 4 cm, AC = 8 cm and PR = 6 cm, find the length of (PQ + QR). [2 marks]
PQ/AB = QR/BC = PR/AC = 6/8 = 3/4.
PQ = (3/4) × 6 = 4.5 cm.
QR = (3/4) × 4 = 3 cm.
PQ + QR = 4.5 + 3 = 7.5 cm.
PQ + QR = 7.5 cm
OR
(b) In the given figure, = and ∠1 = ∠2. Prove that △PQS ∼ △TQR. [2 marks]
Appeared in: 2025 · Standard · Main
In △PQR, ∠PRQ = ∠PQR (given ∠1 = ∠2), so PR = PQ.
Given QR/QS = QT/PR, hence QS/QR = PR/QT = PQ/QT.
∠PQS = ∠TQR, since Q, S, R and Q, P, T are collinear.
Therefore △PQS ∼ △TQR by SAS similarity.
△PQS ∼ △TQR
Q45. In the given figure, QR ∥ CB and RP ∥ AC. If BR = 10 cm, QA = 12 cm, BP = 12 cm and PC = 18 cm, find the lengths of AR and QC. [2 marks]
Appeared in: 2026 · Basic · Main
BC = BP + PC = 12 + 18 = 30 cm.
Since RP ∥ AC, △BRP ∼ △BAC, so BR/BA = BP/BC = 2/5.
BA = 25 cm, hence AR = 25 − 10 = 15 cm.
Since RQ ∥ BC, AQ/AC = AR/AB = 15/25 = 3/5.
AC = 20 cm, hence QC = 20 − 12 = 8 cm.
AR = 15 cm; QC = 8 cm
Q46. In the given figure, AB ∥ DC. If OB = 3OD and CD = 1.8 cm, find the length AB. [2 marks]
Appeared in: 2026 · Basic · Main
∠AOB = ∠COD and ∠ABO = ∠CDO.
Therefore △AOB ∼ △COD by AA.
AB/CD = OB/OD = 3.
AB = 3 × 1.8 = 5.4 cm.
AB = 5.4 cm
Q47. In the given figure, △ODC ∼ △OBA. If ∠BOC = 110°, ∠ODC = 45° and AB = 2CD, find (i) ∠OAB and (ii) OB : OD. [2 marks]
Appeared in: 2026 · Basic · Main
∠DOC = 180° − 110° = 70° (linear pair).
∠OCD = 180° − 70° − 45° = 65°.
By similarity, ∠OAB = ∠OCD = 65°.
OB/OD = AB/CD = 2.
∠OAB = 65°; OB : OD = 2 : 1
Q48. In the given figure, AB ∥ DE and AC ∥ DF. Show that △ABC ∼ △DEF. If BC = 10 cm, EB = CF = 5 cm and AB = 7 cm, find the length DE. [2 marks]
Appeared in: 2026 · Standard · Main
∠ABC = ∠DEF and ∠ACB = ∠DFE (corresponding angles).
Thus △ABC ∼ △DEF by AA.
EF = EB + BC + CF = 5 + 10 + 5 = 20 cm.
AB/DE = BC/EF = 10/20 = 1/2.
DE = 14 cm.
△ABC ∼ △DEF; DE = 14 cm
Q49. (a) In △ABC, DE ∥ BC. If AD = x, DB = x − 2, AE = x + 2 and EC = x − 1, find x. [2 marks]
By BPT, AD/DB = AE/EC.
x/(x − 2) = (x + 2)/(x − 1).
x(x − 1) = (x + 2)(x − 2).
x² − x = x² − 4, so x = 4.
All four segment lengths are positive at x = 4.
x = 4
OR
(b) In the figure, △ABC ∼ △XYZ. Find x and y. [2 marks]
Appeared in: 2026 · Standard · Main
From the diagram, AB = 4 cm, BC = 6 cm, AC = y, XY = x, YZ = 7.2 cm and XZ = 6 cm.
AB/XY = BC/YZ = AC/XZ (corresponding sides).
4/x = 6/7.2, so x = 4.8 cm.
y/6 = 6/7.2, so y = 5 cm.
x = 4.8 cm; y = 5 cm
Q50. (a) In the given figure, △AHK ∼ △ABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find AC. [2 marks]
By similarity, AK/AC = HK/BC.
10/AC = 7/3.5 = 2.
AC = 5 cm.
AC = 5 cm
OR
(b) In the given figure, XY ∥ QR, = and PR = 6.3 cm. Find YR. [2 marks]
Appeared in: 2026 · Standard · Main
By similarity, PX/PQ = PY/PR.
Subtracting from 1 gives XQ/PQ = YR/PR.
3/7 = YR/6.3.
YR = 2.7 cm.
YR = 2.7 cm
Q51. (a) A vertical pole of height 10 m casts a shadow of 15 m on the ground and at the same time a tower casts a shadow of 45 m. Find the height of the tower. [2 marks]
The pole and tower are vertical, and the Sun’s elevation is the same.
The two right triangles are similar by AA.
Height of tower/45 = 10/15.
Height of tower = 30 m.
Height of tower = 30 m
OR
(b) In the given figure, = and ∠1 = ∠2. Prove that △PQS ∼ △TQR. [2 marks]
Appeared in: 2025 · Basic · Main
In △PQR, ∠PRQ = ∠PQR (given ∠1 = ∠2), so PR = PQ.
Given QR/QS = QT/PR, hence QS/QR = PR/QT = PQ/QT.
∠PQS = ∠TQR, since Q, S, R and Q, P, T are collinear.
Therefore △PQS ∼ △TQR by SAS similarity.
△PQS ∼ △TQR
Q52. (a) A 1.5 m tall boy is walking away from the base of a lamp post which is 12 m high, at a speed of 2.5 m/s. Find the length of his shadow after 3 seconds. [2 marks]
Distance walked = speed × time = 2.5 × 3 = 7.5 m.
Let the shadow length be x m.
By similarity, 1.5/x = 12/(7.5 + x).
1.5(7.5 + x) = 12x, so 10.5x = 11.25.
x = 15/14 m.
Shadow length = 15/14 m
OR
(b) In parallelogram ABCD, side AD is produced to a point E and BE intersects CD at F. Prove that △ABE ∼ △CFB. [2 marks]
Appeared in: 2025 · Standard · Main
AB ∥ CD and AD ∥ BC (opposite sides of a parallelogram).
∠BAE = ∠FCB, since AE lies on AD and CF lies on CD.
∠ABE = ∠CFB, since B, F and E are collinear and AB ∥ CF.
Therefore △ABE ∼ △CFB by AA.
△ABE ∼ △CFB
Use BPT when a line is already known to be parallel to one side of a triangle and you need a side ratio. Use its converse when the side-division ratios are equal and you must prove the joining line parallel to the third side.
Match equal angles first. If angle A equals angle P, B equals Q and C equals R, write △ABC ∼ △PQR. The corresponding sides are AB with PQ, BC with QR, and AC with PR. Keep that order when forming ratios.
Yes, after establishing the correct relationship. If DE is parallel to BC, BPT gives AD/DB = AE/EC, while similarity gives AD/AB = AE/AC. Do not pair a divided segment on one side with an unrelated whole side on the other.
Use SAS similarity when two pairs of corresponding sides have the same ratio and the angles included between those side pairs are equal. A side-side ratio alone does not establish similarity. Name the equal included angles before writing the similarity conclusion.
No. Similar triangles have equal corresponding angles and proportional corresponding sides, but their sizes may differ. Congruent triangles are similar with a scale factor of 1. Two equilateral triangles of sides 2 cm and 4 cm are similar but not congruent.
For vertical objects on level ground at the same time, the Sun’s elevation is the same. The two right triangles are therefore similar by AA, so the ratio of heights equals the ratio of shadow lengths. Use the same units throughout.
Yes. Corresponding medians of similar triangles are in the same ratio as corresponding sides. To prove it, use the midpoint property to halve the corresponding bases, establish similarity of the smaller triangles by SAS, and then compare the medians.
State the angle equalities or side proportions, give their reasons, name the similarity criterion or BPT result, and write the required conclusion. A short proof still needs the logical link; a diagram that merely looks parallel or similar is insufficient.
Answer one alternative where an OR choice is printed. Both alternatives are solved here for practice. A full OR pair is one numbered question, while each alternative has its own worked solution and answer-summary entry.
| Question | Answer |
|---|---|
| Q1 | DC ∥ AP |
| Q2 | AM/AN = AB/AD |
| Q3(a) | OP × OR = OQ × OS |
| Q3(b) OR | AM/AN = AB/AD |
| Q4 | EF is not parallel to QR |
| Q5 | AB = AC; △ABC is isosceles |
| Q6 | DC² = CF × AC |
| Q7(a) | PQ = PR; △PQR is isosceles |
| Q7(b) OR | △ADE ∼ △ABC |
| Q8(a) | PQ is not parallel to EF |
| Q8(b) OR | The line joining the midpoints is parallel to the third side |
| Q9(a) | △RPQ ∼ △RTS |
| Q9(b) OR | AD/AB = AE/AC |
| Q10 | BF/FE = BE/EC |
| Q11(a) | BD/DE = BE/EC |
| Q11(b) OR | △ADE ∼ △ABC |
| Q12(a) | XY ∥ QR |
| Q12(b) OR | AD = 22.2 cm |
| Q13(a) | AE/ED = BF/FC |
| Q13(b) OR | ∠ABC = 48° |
| Q14 | AC² = BC × DC |
| Q15 | △ABD ∼ △CBD; AB = BC |
| Q16(a) | DE ∥ BC |
| Q16(b) OR | Perimeter of △ABC = 18 cm |
| Q17(a) | △POQ ∼ △SOR |
| Q17(b) OR | ∠OSR = 55°; ∠OQP = 55° |
| Q18 | △PDA ∼ △PBC |
| Q19 | OA/OC = OB/OD |
| Q20(a) | OA/OC = OB/OD |
| Q20(b) OR | PR² = QR × SR |
| Q21(a) | CA² = CD × CB |
| Q21(b) OR | ∠A = ∠C; ∠B = ∠D |
| Q22 | △ABC ∼ △PQR |
| Q23 | △ACD ∼ △PRS |
| Q24 | △ABC ∼ △PQR |
| Q25 | △ADB ∼ △PSQ; AD × QS = BD × PS |
| Q26 | AG × DG = AF × DB |
| Q27 | ST × TQ = PT × TR |
| Q28 | △ADE ∼ △ABC |
| Q29(a) | △ABD ∼ △CBE; △PDC ∼ △BEC |
| Q29(b) OR | BF/FE = BE/EC |
| Q30 | △BAE ∼ △CAD |
| Q31 | OC = OA/2 |
| Q32 | AB/DE = AP/DQ |
| Q33(a) | AO/BO = CO/DO |
| Q33(b) OR | PR = 12 cm |
| Q34(a) | AO/BO = CO/DO |
| Q34(b) OR | △BCD ∼ △ACB |
| Q35 | AQ = 4 cm |
| Q36 | BQ = 8 cm |
| Q37 | PM = 6 cm |
| Q38 | AB = 2CD |
| Q39 | CD = 28/3 cm |
| Q40 | AC = 4 cm |
| Q41 | RF = 6.3 cm; PR = 10.8 cm |
| Q42 | AD = 3.6 cm |
| Q43 | Corresponding side = 13.5 cm |
| Q44(a) | PQ + QR = 7.5 cm |
| Q44(b) OR | △PQS ∼ △TQR |
| Q45 | AR = 15 cm; QC = 8 cm |
| Q46 | AB = 5.4 cm |
| Q47 | ∠OAB = 65°; OB : OD = 2 : 1 |
| Q48 | △ABC ∼ △DEF; DE = 14 cm |
| Q49(a) | x = 4 |
| Q49(b) OR | x = 4.8 cm; y = 5 cm |
| Q50(a) | AC = 5 cm |
| Q50(b) OR | YR = 2.7 cm |
| Q51(a) | Height of tower = 30 m |
| Q51(b) OR | △PQS ∼ △TQR |
| Q52(a) | Shadow length = 15/14 m |
| Q52(b) OR | △ABE ∼ △CFB |
Both OR alternatives are listed for practice. The answer summary follows your selected filters.
Selected official CBSE 2-mark questions; exact repeated questions are merged, and complete OR pairings are retained.
Solutions prepared by Team TutorVertex in NCERT style, with the displayed answers compared with official marking-scheme working. How we check solutions.
PDF pages are given for the displayed version. Repeat appearances are identified by paper code and question number. Archive member names identify the exact files inside official ZIPs. Question wording follows the first listed appearance; spacing and punctuation are normalised.
NCERT Class 10, Chapter 6: Triangles
CBSE Mathematics syllabus, 2026–27
Get one-to-one help with similarity proofs, proportional sides and clear exam working.
Book a Free Assessment →