NCERT Solutions Class 10 Maths Chapter 6 - Triangles

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles (Updated 2026)

Overview
Important Formulas
Exercise 6.1
Exercise 6.2
Exercise 6.3
FAQs

Important Formulas & Key Concepts

Similar Figures: Two figures having the same shape but not necessarily the same size are called similar figures. All congruent figures are similar, but similar figures need not be congruent.
Similarity of Polygons: Two polygons with the same number of sides are similar if: (i) their corresponding angles are equal, and (ii) their corresponding sides are in the same ratio (proportion).
Basic Proportionality Theorem (BPT / Thales Theorem): 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.
Converse of BPT: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Criteria for Similarity of Triangles:
- AAA / AA Criterion: If corresponding angles of two triangles are equal, their corresponding sides are in proportion, and the triangles are similar.
- SSS Criterion: If corresponding sides of two triangles are in the same ratio, their corresponding angles are equal, and the triangles are similar.
- SAS Criterion: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the triangles are similar.

Exercise 6.1 Solutions

Question 1: Fill in the blanks using the correct word given in brackets:

(i) All circles are _________. (congruent, similar)

(ii) All squares are _________. (similar, congruent)

(iii) All _________ triangles are similar. (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are _________ and (b) their corresponding sides are _________. (equal, proportional)

Answer:

(i) similar

(ii) similar

(iii) equilateral

(iv) (a) equal, (b) proportional

Question 2: Give two different examples of pair of:

(i) Similar figures:

(ii) Non-similar figures:

Answer:

(i) Similar figures:
1. Any two circles of different radii.
2. Any two equilateral triangles of different side lengths.

(ii) Non-similar figures:
1. A triangle and a square.
2. A rectangle and a rhombus.

Question 3: State whether the following quadrilaterals are similar or not:

Answer: The given quadrilaterals PQRS (rhombus with side 1.5 cm) and ABCD (square with side 3 cm) are not similar. Although the ratio of their corresponding sides is equal (1.5/3 = 1/2), their corresponding angles are not equal (angles of PQRS are not 90°, while angles of ABCD are all 90°).

Exercise 6.2 Solutions

Question 1: In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i) Given: AD = 1.5 cm, DB = 3 cm, AE = 1 cm.
Since DE || BC, by Basic Proportionality Theorem (BPT):
AD / DB = AE / EC
1.5 / 3 = 1 / EC
1 / 2 = 1 / EC
EC = 2 cm

(ii) Given: DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm.
Since DE || BC, by BPT:
AD / DB = AE / EC
AD / 7.2 = 1.8 / 5.4
AD / 7.2 = 1 / 3
AD = 7.2 / 3
AD = 2.4 cm

Question 2: E and F are points on the sides PQ and PR respectively of a Triangle PQR. For each case, state whether EF || QR:

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
PE / EQ = 3.9 / 3 = 1.3
PF / FR = 3.6 / 2.4 = 1.5
Since PE / EQ ≠ PF / FR, by converse of BPT, EF is NOT parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
PE / QE = 4 / 4.5 = 8 / 9
PF / RF = 8 / 9
Since PE / QE = PF / RF, by converse of BPT, EF || QR.

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm
FR = PR - PF = 2.56 - 0.36 = 2.20 cm
PE / EQ = 0.18 / 1.10 = 18 / 110 = 9 / 55
PF / FR = 0.36 / 2.20 = 36 / 220 = 9 / 55
Since PE / EQ = PF / FR, by converse of BPT, EF || QR.

Question 3: In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.

Proof:
In Triangle ABC, since LM || CB, by BPT:
AM / AB = AL / AC --- (1)
In Triangle ADC, since LN || CD, by BPT:
AN / AD = AL / AC --- (2)
From (1) and (2), we get:
AM / AB = AN / AD. Hence proved.

Question 4: In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.

Proof:
In Triangle ABE, since DF || AE, by BPT:
BF / FE = BD / DA --- (1)
In Triangle ABC, since DE || AC, by BPT:
BE / EC = BD / DA --- (2)
From (1) and (2), we get:
BF / FE = BE / EC. Hence proved.

Question 5: In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

Proof:
In Triangle PQO, since DE || OQ, by BPT:
PE / EQ = PD / DO --- (1)
In Triangle PRO, since DF || OR, by BPT:
PF / FR = PD / DO --- (2)
From (1) and (2), PE / EQ = PF / FR.
In Triangle PQR, since PE / EQ = PF / FR, by converse of BPT, EF || QR. Hence proved.

Question 6: In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Proof:
In Triangle POQ, AB || PQ ⇒ OA / AP = OB / BQ --- (1)
In Triangle POR, AC || PR ⇒ OA / AP = OC / CR --- (2)
From (1) and (2), OB / BQ = OC / CR.
In Triangle QOR, since OB / BQ = OC / CR, by converse of BPT, BC || QR. Hence proved.

Question 7: Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Proof:
Let ABC be a triangle and D be the mid-point of AB, so AD = DB ⇒ AD / DB = 1.
A line l drawn through D parallel to BC meets AC at E (DE || BC).
By BPT (Theorem 6.1), AD / DB = AE / EC.
Since AD / DB = 1, 1 = AE / EC ⇒ AE = EC.
Thus, E is the mid-point of AC, i.e., the line bisects the third side. Hence proved.

Question 8: Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Proof:
In Triangle ABC, let D and E be the mid-points of AB and AC respectively.
So, AD = DB ⇒ AD / DB = 1, and AE = EC ⇒ AE / EC = 1.
Therefore, AD / DB = AE / EC = 1.
By Converse of BPT (Theorem 6.2), since the line DE divides AB and AC in the same ratio, DE || BC. Hence proved.

Question 9: ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Show that AO/BO = CO/DO.

Proof:
Draw a line OE parallel to AB through O, meeting AD at E.
In Triangle ADC, OE || DC (since OE || AB and AB || DC):
AE / ED = AO / OC --- (1) (by BPT)
In Triangle ABD, OE || AB:
AE / ED = BO / OD --- (2) (by BPT)
From (1) and (2), AO / OC = BO / OD.
Rearranging the terms: AO / BO = CO / DO. Hence proved.

Question 10: The diagonals of a quadrilateral ABCD intersect each other at O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Proof:
Given: AO / BO = CO / DO ⇒ AO / CO = BO / DO.
Draw OE || AB meeting AD at E.
In Triangle ABD, OE || AB ⇒ AE / ED = BO / DO (by BPT).
Since BO / DO = AO / CO, we get AE / ED = AO / CO.
In Triangle ADC, since AE / ED = AO / CO, by converse of BPT, OE || DC.
Since OE || AB and OE || DC, we have AB || DC.
Therefore, quadrilateral ABCD is a trapezium. Hence proved.

Exercise 6.3 Solutions

Question 1: State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion and in symbolic form:

(i) Triangle ABC: ∠A = 60°, ∠B = 80°, ∠C = 40°. Triangle PQR: ∠P = 60°, ∠Q = 80°, ∠R = 40°.
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R. By AAA Similarity Criterion, Triangle ABC ~ Triangle PQR.

(ii) Triangle ABC: AB = 2, BC = 2.5, CA = 3. Triangle PQR: PQ = 6, QR = 4, RP = 5.
AB / QR = 2 / 4 = 1/2
BC / RP = 2.5 / 5 = 1/2
CA / PQ = 3 / 6 = 1/2
Since AB/QR = BC/RP = CA/PQ, by SSS Similarity Criterion, Triangle ABC ~ Triangle QRP.

(iii) Triangle LMP: MP = 2, LP = 3, LM = 2.7. Triangle DEF: DE = 4, EF = 5, DF = 6.
MP / DE = 2 / 4 = 1/2, LP / DF = 3 / 6 = 1/2, LM / EF = 2.7 / 5 ≠ 1/2.
Ratios are not equal, so these triangles are NOT similar.

(iv) Triangle MNL: MN = 2.5, ∠M = 70°, ML = 5. Triangle PQR: PQ = 5, ∠P = 70°, PR = 10.
MN / PQ = 2.5 / 5 = 1/2, ML / PR = 5 / 10 = 1/2, and ∠M = ∠P = 70°.
By SAS Similarity Criterion, Triangle MNL ~ Triangle PQR.

(v) Triangle ABC: AB = 2.5, ∠A = 80°, BC = 3. Triangle DEF: ∠F = 80°, EF = 6, DF = 5.
In Triangle ABC, side AC is not given (angle A is not included between AB and BC). Thus, SAS criterion cannot be applied. These triangles are NOT similar.

(vi) Triangle DEF: ∠D = 70°, ∠E = 80° ⇒ ∠F = 180° - (70° + 80°) = 30°.
Triangle PQR: ∠Q = 80°, ∠R = 30° ⇒ ∠P = 180° - (80° + 30°) = 70°.
∠D = ∠P = 70°, ∠E = ∠Q = 80°, ∠F = ∠R = 30°.
By AAA Similarity Criterion, Triangle DEF ~ Triangle PQR.

Question 2: In Fig. 6.35, Triangle ODC ~ Triangle OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:
1. DOB is a straight line, so ∠DOC + ∠BOC = 180°
∠DOC + 125° = 180° ⇒ ∠DOC = 55°
2. In Triangle ODC, ∠DOC + ∠CDO + ∠DCO = 180°
55° + 70° + ∠DCO = 180° ⇒ ∠DCO = 55°
3. Since Triangle ODC ~ Triangle OBA, corresponding angles are equal:
∠OAB = ∠OCD ⇒ ∠OAB = 55°.

Question 3: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.

Proof:
In Triangle AOB and Triangle COD:
∠OAB = ∠OCD (Alternate interior angles as AB || DC)
∠OBA = ∠ODC (Alternate interior angles as AB || DC)
∠AOB = ∠COD (Vertically opposite angles)
By AAA Similarity Criterion, Triangle AOB ~ Triangle COD.
Corresponding sides of similar triangles are in proportion:
OA / OC = OB / OD. Hence proved.

Question 4: In Fig. 6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that Triangle PQS ~ Triangle TQR.

Proof:
In Triangle PQR, since ∠1 = ∠2 (∠PQR = ∠PRQ), sides opposite to equal angles are equal:
PR = PQ.
Given: QR / QS = QT / PR ⇒ QR / QS = QT / PQ ⇒ QS / QR = PQ / QT.
Now, in Triangle PQS and Triangle TQR:
PQ / QT = QS / QR, and ∠Q (∠1) is common to both triangles.
By SAS Similarity Criterion, Triangle PQS ~ Triangle TQR. Hence proved.

Question 5: S and T are points on sides PR and QR of Triangle PQR such that ∠P = ∠RTS. Show that Triangle RPQ ~ Triangle RTS.

Proof:
In Triangle RPQ and Triangle RTS:
∠R = ∠R (Common angle)
∠P = ∠RTS (Given)
By AA Similarity Criterion, Triangle RPQ ~ Triangle RTS. Hence proved.

Question 6: In Fig. 6.37, if Triangle ABE ≅ Triangle ACD, show that Triangle ADE ~ Triangle ABC.

Proof:
Given Triangle ABE ≅ Triangle ACD.
By CPCT, AB = AC and AE = AD ⇒ AD = AE.
Dividing the equations: AD / AB = AE / AC.
Now, in Triangle ADE and Triangle ABC:
AD / AB = AE / AC and ∠A is common.
By SAS Similarity Criterion, Triangle ADE ~ Triangle ABC. Hence proved.

Question 7: In Fig. 6.38, altitudes AD and CE of Triangle ABC intersect each other at P. Show that:

(i) Triangle AEP ~ Triangle CDP
∠AEP = ∠CDP = 90°, and ∠APE = ∠CPD (Vertically opposite angles). By AA similarity, Triangle AEP ~ Triangle CDP.

(ii) Triangle ABD ~ Triangle CBE
∠ADB = ∠CEB = 90°, and ∠B is common. By AA similarity, Triangle ABD ~ Triangle CBE.

(iii) Triangle AEP ~ Triangle ADB
∠AEP = ∠ADB = 90°, and ∠A is common. By AA similarity, Triangle AEP ~ Triangle ADB.

(iv) Triangle PDC ~ Triangle BEC
∠PDC = ∠BEC = 90°, and ∠C is common. By AA similarity, Triangle PDC ~ Triangle BEC.

Question 8: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Triangle ABE ~ Triangle CFB.

Proof:
In parallelogram ABCD:
∠A = ∠C (Opposite angles of a parallelogram)
Since AD || BC, line AE || BC. Therefore, ∠ABE = ∠CFB (Alternate interior angles).
In Triangle ABE and Triangle CFB:
∠A = ∠C
∠ABE = ∠CFB
By AA Similarity Criterion, Triangle ABE ~ Triangle CFB. Hence proved.

Question 9: In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) Triangle ABC ~ Triangle AMP
In Triangle ABC and Triangle AMP:
∠ABC = ∠AMP = 90°
∠A = ∠A (Common)
By AA Similarity Criterion, Triangle ABC ~ Triangle AMP.

(ii) CA/PA = BC/MP
Since Triangle ABC ~ Triangle AMP, their corresponding sides are proportional:
CA / PA = BC / MP. Hence proved.

Question 10: CD and GH are bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of Triangle ABC and Triangle EFG respectively. If Triangle ABC ~ Triangle FEG, show that:

Solution:
Since Triangle ABC ~ Triangle FEG ⇒ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠EGF.
CD and GH bisect ∠ACB and ∠EGF ⇒ ∠ACD = 1/2 ∠ACB and ∠FGH = 1/2 ∠EGF ⇒ ∠ACD = ∠FGH, and ∠DCB = ∠HGE.

(i) Show CD/GH = AC/FG:
In Triangle DCA and Triangle HGF: ∠A = ∠F and ∠ACD = ∠FGH. By AA similarity, Triangle DCA ~ Triangle HGF.
Therefore, CD / GH = AC / FG.

(ii) Triangle DCB ~ Triangle HGE:
In Triangle DCB and Triangle HGE: ∠B = ∠E and ∠DCB = ∠HGE. By AA similarity, Triangle DCB ~ Triangle HGE.

(iii) Triangle DCA ~ Triangle HGF:
As proved in part (i), Triangle DCA ~ Triangle HGF.

Question 11: In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Triangle ABD ~ Triangle ECF.

Proof:
In Triangle ABC, AB = AC ⇒ ∠ABC = ∠ACB ⇒ ∠ABD = ∠ECF.
In Triangle ABD and Triangle ECF:
∠ADB = ∠EFC = 90° (Since AD ⊥ BC and EF ⊥ AC)
∠ABD = ∠ECF
By AA Similarity Criterion, Triangle ABD ~ Triangle ECF. Hence proved.

Question 12: Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Triangle PQR. Show that Triangle ABC ~ Triangle PQR.

Proof:
Given: AB / PQ = BC / QR = AD / PM.
Since AD is median, BD = BC / 2 ⇒ BC = 2 BD.
Since PM is median, QM = QR / 2 ⇒ QR = 2 QM.
So, BC / QR = (2 BD) / (2 QM) = BD / QM.
Substitute in given relation: AB / PQ = BD / QM = AD / PM.
By SSS Similarity Criterion, Triangle ABD ~ Triangle PQM.
This implies ∠B = ∠Q (Corresponding angles of similar triangles).
Now in Triangle ABC and Triangle PQR:
AB / PQ = BC / QR (Given) and ∠B = ∠Q.
By SAS Similarity Criterion, Triangle ABC ~ Triangle PQR. Hence proved.

Question 13: D is a point on side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB · CD.

Proof:
In Triangle ABC and Triangle DAC:
∠BAC = ∠ADC (Given)
∠C = ∠C (Common angle)
By AA Similarity Criterion, Triangle ABC ~ Triangle DAC.
Corresponding sides are in proportion:
CB / CA = CA / CD ⇒ CA² = CB · CD. Hence proved.

Question 14: Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that Triangle ABC ~ Triangle PQR.

Proof:
Produce AD to E such that AD = DE, join EC.
Produce PM to L such that PM = ML, join LR.
In quadrilateral ABEC, diagonals AE and BC bisect each other at D, so ABEC is a parallelogram ⇒ BE = AC and EC = AB.
Similarly, PQRL is a parallelogram ⇒ QL = PR and LR = PQ.
Given: AB / PQ = AC / PR = AD / PM.
⇒ EC / LR = BE / QL = (2 AD) / (2 PM) = AE / PL.
In Triangle ABE and Triangle PQL, all corresponding sides are in ratio, so Triangle ABE ~ Triangle PQL (SSS criterion) ⇒ ∠BAD = ∠QPM.
Similarly, Triangle ACE ~ Triangle PRL ⇒ ∠CAD = ∠RPM.
Adding both: ∠BAD + ∠CAD = ∠QPM + ∠RPM ⇒ ∠A = ∠P.
Now in Triangle ABC and Triangle PQR: AB / PQ = AC / PR and ∠A = ∠P.
By SAS Similarity Criterion, Triangle ABC ~ Triangle PQR. Hence proved.

Question 15: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:
Let AB be the pole = 6 m, BC be its shadow = 4 m.
Let PQ be the tower = h m, QR be its shadow = 28 m.
At the same time, the angle of elevation of the sun is equal, so ∠C = ∠R and ∠B = ∠Q = 90°.
By AA Similarity Criterion, Triangle ABC ~ Triangle PQR.
AB / PQ = BC / QR
6 / h = 4 / 28
6 / h = 1 / 7
h = 6 × 7 = 42 m.
The height of the tower is 42 m.

Question 16: If AD and PM are medians of triangles ABC and PQR, respectively, where Triangle ABC ~ Triangle PQR, prove that AB/PQ = AD/PM.

Proof:
Given Triangle ABC ~ Triangle PQR ⇒ AB / PQ = BC / QR and ∠B = ∠Q.
Since AD and PM are medians, BC = 2 BD and QR = 2 QM.
So, BC / QR = (2 BD) / (2 QM) = BD / QM.
Therefore, AB / PQ = BD / QM, and ∠B = ∠Q.
By SAS Similarity Criterion, Triangle ABD ~ Triangle PQM.
Hence, AB / PQ = AD / PM. Hence proved.

FAQs

What is the Basic Proportionality Theorem?

It states that a line parallel to one side of a triangle divides the other two sides proportionally.

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Class 10 Mathematics Triangles Chapter 6 Solutions