**NCERT Solutions for Class 10 Maths Chapter 6** are provided in Chapter 6 Triangles Study Circle, which is considered one of the most important chapters for students studying in CBSE Class 10. **NCERT Solutions for Class 10 Maths Chapter 6**, is well explained accordingly.

All solutions are given along with CBSE syllabus for 2023-24. It covers a vast subject including many rules and theorems. While solving different types of questions, students often get confused about which theorem to use, so all the solutions have been made easy in Study Circle.

The **NCERT Solutions for Class 10 Maths Chapter 6** are prepared by the experts subject teachers to help students prepare better for their board exams.

**NCERT Solutions for Class 10 Maths Chapter 6**

**Exercise 6.1**

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

**(i) All circles are __________. (congruent, similar)**

Answer: Similar

**(ii) All squares are __________. (similar, congruent)**

Answer: Similar

**(iii) All __________ triangles are similar. (isosceles, equilateral)
**

Answer: 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: (a) Equal

(b) Proportional

**Question** 2. Give two different examples of pair of

(i) Similar figures

(ii) Non-similar figures

**Solution:–**

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

**Solution :–**

**NCERT Solutions for Class 10 Maths Chapter 6**

**Exercise 6.2**

**Question 1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).**

**Solution:–**

**(i)** Given, in △ ABC, DE∥BC

∴ AD/DB = AE/EC [Using Basic proportionality theorem]

⇒1.5/3 = 1/EC

⇒EC = 3/1.5

EC = 3×10/15 = 2 cm

Hence, EC = 2 cm.

**(ii)** Given, in △ ABC, DE∥BC

∴ AD/DB = AE/EC [Using Basic proportionality theorem]

⇒ AD/7.2 = 1.8 / 5.4

⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10

⇒ AD = 2.4

Hence, AD = 2.4 cm.

** Question 2. E and F are points on the sides PQ and PR, respectively of a ΔPQR. For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm**

**(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm**

**Solution:**

Given, in ΔPQR, E and F are two points on side PQ and PR, respectively. See the figure below;

**(i)** Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm

Therefore, by using Basic proportionality theorem, we get,

PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3

And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5

So, we get, PE/EQ ≠ PF/FR

Hence, EF is not parallel to QR.

**(ii)** Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm

Therefore, by using Basic proportionality theorem, we get,

PE/QE = 4/4.5 = 40/45 = 8/9

And, PF/RF = 8/9

So, we get here,

PE/QE = PF/RF

Hence, EF is parallel to QR.

**(iii)** Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

From the figure,

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55**…………. (i)**

And, PE/FR = 0.36/2.20 = 36/220 = 9/55**………… (ii)**

So, we get here,

PE/EQ = PF/FR

Hence, EF is parallel to QR.

**Question 3. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD**

**Solution:**

In the given figure, we can see, LM || CB,

By using basic proportionality theorem, we get,

AM/AB = AL/AC**……………………..(i)**

Similarly, given, LN || CD and using basic proportionality theorem,

∴AN/AD = AL/AC**……………………………(ii)**

From equation **(i)** and **(ii)**, we get,

AM/AB = AN/AD

Hence, proved.

**Question 4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC**

**Solution:**

In ΔABC, given as, DE || AC

Thus, by using Basic Proportionality Theorem, we get,

∴BD/DA = BE/EC ………………………………………………**(i)**

In ΔBAE, given as, DF || AE

Thus, by using Basic Proportionality Theorem, we get,

∴BD/DA = BF/FE ………………………………………………**(ii)**

From equation **(i)** and **(ii)**, we get

BE/EC = BF/FE

Hence, proved.

**Question 5. In the figure, DE||OQ and DF||OR, show that EF||QR.**

**Solution:**

Given,

In ΔPQO, DE || OQ

So by using Basic Proportionality Theorem,

PD/DO = PE/EQ……………… **..(i)**

Again given, in ΔPOR, DF || OR,

So by using Basic Proportionality Theorem,

PD/DO = PF/FR………………… **(ii)**

From equation **(i)** and **(ii)**, we get,

PE/EQ = PF/FR

Therefore, by converse of Basic Proportionality Theorem,

EF || QR, in ΔPQR.

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

**Solution:**

Given here,

In ΔOPQ, AB || PQ

By using Basic Proportionality Theorem,

OA/AP = OB/BQ…………….**(i)**

Also given,

In ΔOPR, AC || PR

By using Basic Proportionality Theorem

∴ OA/AP = OC/CR……………(ii)

From equation **(i)** and **(ii)**, we get,

OB/BQ = OC/CR

Therefore, by converse of Basic Proportionality Theorem,

In ΔOQR, BC || QR.

**Question 7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).**

**Solution:**

Given, in ΔABC, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

We have to prove that E is the mid point of AC.

Since, D is the mid-point of AB.

∴ AD=DB

⇒AD/DB = 1 …………………………. **(i)**

In ΔABC, DE || BC,

By using Basic Proportionality Theorem,

Therefore, AD/DB = AE/EC

From equation (i), we can write,

⇒ 1 = AE/EC

∴ AE = EC

Hence, proved, E is the midpoint of AC.

**Question 8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).**

**Solution:**

Given, in ΔABC, D and E are the mid points of AB and AC, respectively, such that,

AD=BD and AE=EC.

Since, D is the midpoint of AB

∴ AD=DB

⇒AD/BD = 1……………………………….. **(i)**

** **

Also given, E is the mid-point of AC.

∴ AE=EC

⇒ AE/EC = 1

From equation **(i)** and **(ii)**, we get,

AD/BD = AE/EC

By converse of Basic Proportionality Theorem,

DE || BC

Hence, proved.

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

**Solution:**

Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.

We have to prove, AO/BO = CO/DO

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

In ΔADC, we have OE || DC

Therefore, by using Basic Proportionality Theorem

AE/ED = AO/CO **……………..(i)**

Now, In ΔABD, OE || AB

Therefore, by using Basic Proportionality Theorem

DE/EA = DO/BO**…………….(ii)**

From equation **(i)** and **(ii)**, we get,

AO/CO = BO/DO

⇒AO/BO = CO/DO

Hence, proved.

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

**Solution:**

Given, Quadrilateral ABCD where AC and BD intersect each other at O such that,

AO/BO = CO/DO.

We have to prove here, ABCD is a trapezium

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

In ΔDAB, EO || AB

Therefore, by using Basic Proportionality Theorem

DE/EA = DO/OB ……………………(i)

Also, given,

AO/BO = CO/DO

⇒ AO/CO = BO/DO

⇒ CO/AO = DO/BO

⇒DO/OB = CO/AO **…………………………..(ii)**

From equation **(i)** and **(ii)**, we get

DE/EA = CO/AO

Therefore, by using converse of Basic Proportionality Theorem,

EO || DC also EO || AB

⇒ AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.

**NCERT Solutions for Class 10 Maths Chapter 6**

**Exercise 6.3**

**Question 1 . State which pairs of triangles in the figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:**

**Solution:**

**(i)** Given, in ΔABC and ΔPQR,

∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, by AAA similarity criterion,

∴ ΔABC ~ ΔPQR

**(ii)** Given, in ΔABC and ΔPQR,

AB/QR = 2/4 = 1/2,

BC/RP = 2.5/5 = 1/2,

CA/PA = 3/6 = 1/2

By SSS similarity criterion,

ΔABC ~ ΔQRP

**(iii)** Given, in ΔLMP and ΔDEF,

LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6

MP/DE = 2/4 = 1/2

PL/DF = 3/6 = 1/2

LM/EF = 2.7/5 = 27/50

Here , MP/DE = PL/DF ≠ LM/EF

Therefore, ΔLMP and ΔDEF are not similar.

**(iv)** In ΔMNL and ΔQPR, it is given,

MN/QP = ML/QR = 1/2

∠M = ∠Q = 70°

Therefore, by SAS similarity criterion

∴ ΔMNL ~ ΔQPR

**(v)** In ΔABC and ΔDEF, given that,

AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°

Here , AB/DF = 2.5/5 = 1/2

And, BC/EF = 3/6 = 1/2

⇒ ∠B ≠ ∠F

Hence, ΔABC and ΔDEF are not similar.

**(vi)** In ΔDEF, by sum of angles of triangles, we know that,

∠D + ∠E + ∠F = 180°

⇒ 70° + 80° + ∠F = 180°

⇒ ∠F = 180° – 70° – 80°

⇒ ∠F = 30°

Similarly, In ΔPQR,

∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)

⇒ ∠P + 80° + 30° = 180°

⇒ ∠P = 180° – 80° -30°

⇒ ∠P = 70°

Now, comparing both the triangles, ΔDEF and ΔPQR, we have

∠D = ∠P = 70°

∠F = ∠Q = 80°

∠F = ∠R = 30°

Therefore, by AAA similarity criterion,

Hence, ΔDEF ~ ΔPQR

**Question 2. In figure 6.35, ΔODC ~ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.**

**Solution:–**

As we can see from the figure, DOB is a straight line.

Therefore, ∠DOC + ∠ COB = 180°

⇒ ∠DOC = 180° – 125° (Given, ∠ BOC = 125°)

= 55°

In ΔDOC, sum of the measures of the angles of a triangle is 180º

Therefore, ∠DCO + ∠ CDO + ∠ DOC = 180°

⇒ ∠DCO + 70º + 55º = 180°(Given, ∠ CDO = 70°)

⇒ ∠DCO = 55°

It is given that, ΔODC ~ ΔOBA,

Therefore, ΔODC ~ ΔOBA.

Hence, corresponding angles are equal in similar triangles

∠OAB = ∠OCD

⇒ ∠ OAB = 55°

∠OAB = ∠OCD

⇒ ∠OAB = 55°

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

**Solution:–**

In ΔDOC and ΔBOA,

AB || CD, thus alternate interior angles will be equal,

∴∠CDO = ∠ABO

Similarly,

∠DCO = ∠BAO

Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;

∴∠DOC = ∠BOA

Hence, by AAA similarity criterion,

ΔDOC ~ ΔBOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

⇒OA/OC = OB/OD

Hence, proved.

**Question 4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.**

**Solution:**

In ΔPQR,

∠PQR = ∠PRQ

∴ PQ = PR ………………………**(i)**

Given,

QR/QS = QT/PRUsing equation **(i)**, we get

QR/QS = QT/QP**……………….(ii)**

In ΔPQS and ΔTQR, by equation (ii),

QR/QS = QT/QP

∠Q = ∠Q

∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

** Question 5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.**

**Solution:**

Given, S and T are point on sides PR and QR of ΔPQR

And ∠P = ∠RTS.

In ΔRPQ and ΔRTS,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ΔRPQ ~ ΔRTS (AA similarity criterion)

**Question** 6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

**Solution:**

Given, ΔABE ≅ ΔACD.

∴ AB = AC [By CPCT] ……………………………….**(i)**

And, AD = AE [By CPCT] ……………………………**(ii)**

In ΔADE and ΔABC, dividing eq.(ii) by eq(i),

AD/AB = AE/AC

∠A = ∠A [Common angle]

∴ ΔADE ~ ΔABC [SAS similarity criterion]

**Question** 7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

**(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC**

**Solution:**

Given, altitudes AD and CE of ΔABC intersect each other at the point P.

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (90° each)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB ( 90° each)

∠ABD = ∠CBE (Common Angles)

Hence, by AA similarity criterion,

ΔABD ~ ΔCBE

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (90° each)

∠PAE = ∠DAB (Common Angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (90° each)

∠PCD = ∠BCE (Common angles)

Hence, by AA similarity criterion,

ΔPDC ~ ΔBEC

**Question** 8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

**Solution:**

Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,

In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of a parallelogram)

∠AEB = ∠CBF (Alternate interior angles as AE || BC)

∴ ΔABE ~ ΔCFB (AA similarity criterion)

**Question** **9. In the figure, ABC and AMP are two right triangles, right angled at B and M, respectively, prove that: **

**(i) ΔABC ~ ΔAMP**

**(ii) CA/PA = BC/MP**

**Solution:**

Given, ABC and AMP are two right triangles, right angled at B and M, respectively.

(i) In ΔABC and ΔAMP, we have,

∠CAB = ∠MAP (common angles)

∠ABC = ∠AMP = 90° (each 90°)

∴ ΔABC ~ ΔAMP (AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (AA similarity criterion)

If two triangles are similar then the corresponding sides are always equal,

Hence, CA/PA = BC/MP

**Question10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:**

**(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF**

**Solution:**

Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG, respectively.

(i) From the given condition,

ΔABC ~ ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

Since, ∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)

In ΔACD and ΔFGH,

∠A = ∠F

∠ACD = ∠FGH

∴ ΔACD ~ ΔFGH (AA similarity criterion)

⇒CD/GH = AC/FG

(ii) In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Already proved)

∠B = ∠E (Already proved)

∴ ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Already proved)

∠A = ∠F (Already proved)

∴ ΔDCA ~ ΔHGF (AA similarity criterion)

**Question 11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.**

**Solution:**

Given, ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Already proved)

∴ ΔABD ~ ΔECF (using AA similarity criterion)

**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 ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.**

**Solution:**

Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR

i.e. AB/PQ = BC/QR = AD/PM

We have to prove: ΔABC ~ ΔPQR

As we know here,

AB/PQ = BC/QR = AD/PM

⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)

⇒ ΔABD ~ ΔPQM [SSS similarity criterion]

∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]

⇒ ∠ABC = ∠PQR

In ΔABC and ΔPQR

AB/PQ = BC/QR ………………………….**(i)**

∠ABC = ∠PQR ……………………………**(ii)**

From equation **(i)** and **(ii)**, we get,

ΔABC ~ ΔPQR [SAS similarity criterion]

**Question 13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA ^{2} = CB.CD**

**Solution:**

Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.

In ΔADC and ΔBAC,

∠ADC = ∠BAC (Already given)

∠ACD = ∠BCA (Common angles)

∴ ΔADC ~ ΔBAC (AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.

∴ CA/CB = CD/CA

⇒ CA^{2} = 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 ΔABC ~ ΔPQR.**

**Solution:**

Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that;

AB/PQ = AC/PR = AD/PM

We have to prove, ΔABC ~ ΔPQR

Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.

In ΔABD and ΔCDE, we have

AD = DE [By Construction.]

BD = DC [Since, AP is the median]

and, ∠ADB = ∠CDE [Vertically opposite angles]

∴ ΔABD ≅ ΔCDE [SAS criterion of congruence]

⇒ AB = CE [By CPCT] …………………………..**(i)**

Also, in ΔPQM and ΔMNR,

PM = MN [By Construction.]

QM = MR [Since, PM is the median]

and, ∠PMQ = ∠NMR [Vertically opposite angles]

∴ ΔPQM = ΔMNR [SAS criterion of congruence]

⇒ PQ = RN [CPCT] ………………………………**(ii)**

Now, AB/PQ = AC/PR = AD/PM

From equation **(i)** and **(ii)**,

⇒CE/RN = AC/PR = AD/PM

⇒ CE/RN = AC/PR = 2AD/2PM

⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]

∴ ΔACE ~ ΔPRN [SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠A = ∠P …………………………………………….**(iii)**

Now, in ΔABC and ΔPQR, we have

AB/PQ = AC/PR (Already given)

From equation (iii),

∠A = ∠P

∴ ΔABC ~ ΔPQR [ SAS similarity criterion]

** Question15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.**

**Solution:**

Given, Length of the vertical pole = 6m

Shadow of the pole = 4 m

Let Height of tower = *h* m

Length of shadow of the tower = 28 m

In ΔABC and ΔDEF,

∠C = ∠E (angular elevation of sum)

∠B = ∠F = 90°

∴ ΔABC ~ ΔDEF (AA similarity criterion)

∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)

∴ 6/h = 4/28

⇒h = (6×28)/4

⇒ *h* = 6 × 7

⇒ *h *= 42 m

Hence, the height of the tower is 42 m.

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

**Solution:**

Given, ΔABC ~ ΔPQR

We know that the corresponding sides of similar triangles are in proportion.

∴AB/PQ = AC/PR = BC/QR**……………………………(i**)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….…..**(ii)**

Since AD and PM are medians, they will divide their opposite sides.

∴ BD = BC/2 and QM = QR/2 ……………..………….**(iii)**

From equations **(i)** and **(iii)**, we get

AB/PQ = BD/QM **……………………….(iv)**

In ΔABD and ΔPQM,

From equation (ii), we have

∠B = ∠Q

From equation **(iv), we have,**

AB/PQ = BD/QM

∴ ΔABD ~ ΔPQM (SAS similarity criterion)

⇒AB/PQ = BD/QM = AD/PM