The CBSE Class 10 Chapter 6 Triangles have a total of 6 exercises. The pdf containing the NCERT Solutions of the third exercise, Exercise 6.3, is available here. The solution, apart from the pdf format, is also available below. At BYJUâ€™S, our subject experts solve the questions in the NCERT Class 10 Solutions with the utmost care, giving explanations for the steps that are difficult to understand.

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## NCERT Solutions for Class 10 Maths Chapter 6- Triangles Exercise 6.3

### Access other exercise solutions of class 10 Maths Chapter 6- Triangles

Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)

Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)

Exercise 6.4 Solutions 9 Questions (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.5 Solutions 17 Questions (15 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)

### Access Answers of Maths NCERT class 10 Chapter 6 – Triangles Exercise 6.3

**1. State which pairs of triangles in 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 Î”ABC and Î”PQR,

âˆ A =Â âˆ P = 60Â°

âˆ B = âˆ Q = 80Â°

âˆ C = âˆ R = 40Â°

Therefore by AAA similarity criterion,

âˆ´ Î”ABC ~ Î”PQR

(ii) Given, in Â Î”ABC and Î”PQR,

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

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

CA/PA = 3/6 = 1/2

By SSS similarity criterion,

Î”ABC ~ Î”QRP

(iii) Given, in Î”LMP and Î”DEF,

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 â‰ LM/EF

Therefore, Î”LMP and Î”DEF are not similar.

(iv) In Î”MNL and Î”QPR, it is given,

MN/QP = ML/QR = 1/2

âˆ M = âˆ Q = 70Â°

Therefore, by SAS similarity criterion

âˆ´ Î”MNL ~ Î”QPR

(v) In Î”ABC and Î”DEF, given that,

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

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

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

â‡’ âˆ B â‰ âˆ F

Hence, Î”ABC and Î”DEF are not similar.

(vi) In Î”DEF, by sum of angles of triangles, we know that,

âˆ DÂ +Â âˆ EÂ +Â âˆ F = 180Â°

â‡’ 70Â°Â + 80Â°Â + âˆ F = 180Â°

â‡’ âˆ F = 180Â° – 70Â° – 80Â°

â‡’ âˆ F = 30Â°

Similarly, In Î”PQR,

âˆ PÂ +Â âˆ QÂ +Â âˆ R = 180 (Sum of angles of Î”)

â‡’ âˆ PÂ + 80Â°Â + 30Â° = 180Â°

â‡’ âˆ P = 180Â° – 80Â° -30Â°

â‡’ âˆ P = 70Â°

Now, comparing both the triangles, Î”DEF and Î”PQR, we have

âˆ D = âˆ P = 70Â°

âˆ F = âˆ Q = 80Â°

âˆ F = âˆ R = 30Â°

Therefore, by AAA similarity criterion,

Hence, Î”DEF ~ Î”PQR

**2.Â Â In figure 6.35, Î”ODC ~ Î”OBA, âˆ BOC = 125Â° and âˆ CDO = 70Â°. Find âˆ DOC, âˆ DCO and âˆ OAB.**

**Solution:**

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

Therefore, âˆ DOC + âˆ COB = 180Â°

â‡’ âˆ DOC = 180Â° – 125Â° (Given, âˆ BOC = 125Â°)

= 55Â°

In Î”DOC, sum of the measures of the angles of a triangle is 180Âº

Therefore, âˆ DCO + âˆ CDO + âˆ DOC = 180Â°

â‡’ âˆ DCO + 70Âº + 55Âº = 180Â°(Given, âˆ CDO = 70Â°)

â‡’ âˆ DCO = 55Â°

It is given that, Î”ODC ~Â Î”OBA,

Therefore, Î”ODC ~ Î”OBA.

Hence, Corresponding angles are equal in similar triangles

âˆ OAB = âˆ OCD

â‡’ âˆ OAB = 55Â°

âˆ OAB = âˆ OCD

â‡’ âˆ OAB = 55Â°

**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 Î”DOC and Î”BOA,

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

âˆ´âˆ CDO = âˆ ABO

Similarly,

âˆ DCO = âˆ BAO

Also, for the two triangles Î”DOC and Î”BOA, vertically opposite angles will be equal;

âˆ´âˆ DOC = âˆ BOA

Hence, by AAA similarity criterion,

Î”DOC ~ Î”BOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

â‡’OA/OC = OB/OD

Hence, proved.

**4.Â In the fig.6.36, QR/QS = QT/PR andÂ âˆ 1 =Â âˆ 2. Show thatÂ Î”PQS ~Â Î”TQR.**

**Solution:**

In Î”PQR,

âˆ PQR = âˆ PRQ

âˆ´ PQ = PR â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦…**(i)**

Given,

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

QR/QS = QT/QP**â€¦â€¦â€¦â€¦â€¦â€¦.(ii)**

In Î”PQS and Î”TQR, by equation (ii),

QR/QS = QT/QP

âˆ Q = âˆ Q

âˆ´ Î”PQS ~ Î”TQR [By SAS similarity criterion]

**5. S and T are point on sides PR and QR of Î”PQR such that âˆ P = âˆ RTS. Show that Î”RPQ ~ Î”RTS.**

**Solution: **

Given, S and T are point on sides PR and QR of Î”PQR

And âˆ P = âˆ RTS.

In Î”RPQ and Î”RTS,

âˆ RTS = âˆ QPS (Given)

âˆ R = âˆ R (Common angle)

âˆ´ Î”RPQ ~ Î”RTS (AA similarity criterion)

**6. In the figure, if Î”ABE â‰… Î”ACD, show that Î”ADE ~ Î”ABC.**

**Solution: **

Given, Î”ABE â‰… Î”ACD.

âˆ´ AB = AC [By CPCT] â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦….**(i)**

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

In Î”ADE and Î”ABC, dividing eq.(ii) by eq(i),

AD/AB = AE/AC

âˆ A = âˆ A [Common angle]

âˆ´ Î”ADE ~ Î”ABC [SAS similarity criterion]

**7.Â In the figure, altitudes AD and CE of Î”ABC intersect each other at the point P. Show that:**

**(i) Î”AEP ~ Î”CDP
(ii) Î”ABD ~ Î”CBE
(iii) Î”AEP ~ Î”ADB
(iv) Î”PDC ~ Î”BEC**

**Solution: **

Given, altitudes AD and CE of Î”ABC intersect each other at the point P.

(i) In Î”AEP and Î”CDP,

âˆ AEP = âˆ CDP (90Â° each)

âˆ APE = âˆ CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

Î”AEP ~ Î”CDP

(ii) In Î”ABD and Î”CBE,

âˆ ADB = âˆ CEB ( 90Â° each)

âˆ ABD = âˆ CBE (Common Angles)

Hence, by AA similarity criterion,

Î”ABD ~ Î”CBE

(iii)Â In Î”AEP and Î”ADB,

âˆ AEP = âˆ ADB (90Â° each)

âˆ PAE = âˆ DAB (Common Angles)

Hence, by AA similarity criterion,

Î”AEP ~ Î”ADB

(iv) In Î”PDC and Î”BEC,

âˆ PDC = âˆ BEC (90Â° each)

âˆ PCD = âˆ BCE (Common angles)

Hence, by AA similarity criterion,

Î”PDC ~ Î”BEC

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

**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 Î”ABE and Î”CFB,

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

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

âˆ´ Î”ABE ~ Î”CFB (AA similarity criterion)

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

**(i) Î”ABC ~ Î”AMP**

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

**Solution:**

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

(i) In Î”ABC and Î”AMP, we have,

âˆ CAB = âˆ MAP (common angles)

âˆ ABC = âˆ AMP = 90Â° (each 90Â°)

âˆ´ Î”ABC ~ Î”AMP (AA similarity criterion)

(ii) As, Î”ABC ~ Î”AMP (AA similarity criterion)

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

Hence, CA/PA = BC/MP

**10. CD and GH are respectively the bisectors of âˆ ACB and âˆ EGF such that D and H lie on sides AB and FE of Î”ABC and Î”EFG respectively. If Î”ABC ~ Î”FEG, Show that:**

**(i) CD/GH = AC/FG
(ii) Î”DCBÂ ~Â Î”HGE
(iii) Î”DCA ~ Î”HGF**

**Solution:**

Given, CD and GH are respectively the bisectors of âˆ ACB and âˆ EGF such that D and H lie on sides AB and FE of Î”ABC and Î”EFG respectively.

(i) From the given condition,

Î”ABC ~ Î”FEG.

âˆ´ âˆ A = âˆ F, âˆ B = âˆ E, and âˆ ACB = âˆ FGE

Since, âˆ ACB = âˆ FGE

âˆ´ âˆ ACD = âˆ FGH (Angle bisector)

And, âˆ DCB = âˆ HGE (Angle bisector)

In Î”ACD and Î”FGH,

âˆ A = âˆ F

âˆ ACD = âˆ FGH

âˆ´ Î”ACD ~ Î”FGH (AA similarity criterion)

â‡’CD/GH = AC/FG

(ii) In Î”DCB and Î”HGE,

âˆ DCB = âˆ HGE (Already proved)

âˆ B = âˆ E (Already proved)

âˆ´ Î”DCB ~ Î”HGE (AA similarity criterion)

(iii) In Î”DCA and Î”HGF,

âˆ ACD = âˆ FGH (Already proved)

âˆ A = âˆ F (Already proved)

âˆ´ Î”DCA ~ Î”HGF (AA similarity criterion)

**11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD âŠ¥ BC and EF âŠ¥ AC, prove that Î”ABD ~ Î”ECF.**

**Solution: **

Given, ABC is an isosceles triangle.

âˆ´ AB = AC

â‡’ âˆ ABD = âˆ ECF

In Î”ABD and Î”ECF,

âˆ ADB = âˆ EFC (Each 90Â°)

âˆ BAD = âˆ CEF (Already proved)

âˆ´ Î”ABD ~ Î”ECF (using AA similarity criterion)

**12.Â Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Î”PQR (see Fig 6.41). Show that Î”ABC ~ Î”PQR.**

**Solution: **

Given, Î”ABC and Î”PQR, AB, BC and median AD of Î”ABC are proportional to sides PQ, QR and median PM of Î”PQR

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

We have to prove: Î”ABC ~ Î”PQR

As we know here,

AB/PQ = BC/QR = AD/PM

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

â‡’ Î”ABD ~ Î”PQM [SSS similarity criterion]

âˆ´ âˆ ABD = âˆ PQM [Corresponding angles of two similar triangles are equal]

â‡’ âˆ ABC = âˆ PQR

In Î”ABC and Î”PQR

AB/PQ = BC/QR â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.**(i)**

âˆ ABC = âˆ PQR â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦…**(ii)**

From equationÂ **(i)**Â andÂ **(ii)**, we get,

Î”ABC ~ Î”PQR [SAS similarity criterion]

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

**Solution: **

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

In Î”ADC and Î”BAC,

âˆ ADC = âˆ BAC (Already given)

âˆ ACD = âˆ BCA (Common angles)

âˆ´ Î”ADC ~ Î”BAC (AA similarity criterion)

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

âˆ´ CA/CB = CD/CA

â‡’Â CA^{2}Â = CB.CD.

Hence, proved.

**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 thatÂ Î”ABC ~ Î”PQR.**

**Solution: **

Given: Two triangles Î”ABC and Î”PQR in which AD and PM are medians such that;

AB/PQ = AC/PR = AD/PM

We have to prove, Î”ABC ~ Î”PQR

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 Î”ABD and Î”CDE, we have

AD = DE Â [By Construction.]

BD = DC [Since, AP is the median]

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

âˆ´ Î”ABDÂ â‰…Â Î”CDE [SAS criterion of congruence]

â‡’ AB = CE [By CPCT] â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..**(i)**

Also, in Î”PQM and Î”MNR,

PM = MN [By Construction.]

QM = MR [Since, PM is the median]

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

âˆ´ Î”PQM = Î”MNR [SAS criterion of congruence]

â‡’ PQ = RN [CPCT] â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦…**(ii)**

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

From equation **(i)**Â andÂ **(ii)**,

â‡’CE/RN = AC/PR = AD/PM

â‡’ CE/RN = AC/PR = 2AD/2PM

â‡’ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]

âˆ´ Î”ACE ~ Î”PRN [SSS similarity criterion]

Therefore, âˆ 2 = âˆ 4

Similarly, âˆ 1 = âˆ 3

âˆ´ âˆ 1Â +Â âˆ 2Â =Â âˆ 3Â +Â âˆ 4

â‡’ âˆ A = âˆ P …………………………………………….**(iii)**

Now, in Î”ABC and Î”PQR, we have

AB/PQ = AC/PR (Already given)

From equation (iii),

âˆ A = âˆ P

âˆ´ Î”ABC ~ Î”PQR [ SAS similarity criterion]

**15. 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 =Â *h*Â m

Length of shadow of the tower = 28 m

In Î”ABC and Î”DEF,

âˆ C = âˆ E (angular elevation of sum)

âˆ B = âˆ F = 90Â°

âˆ´ Î”ABC ~ Î”DEF (AA similarity criterion)

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

âˆ´ 6/h = 4/28

â‡’h = (6Ã—28)/4

â‡’Â *h*Â =Â 6 Ã— 7

â‡’Â *hÂ *= 42 m

Hence, the height of the tower is 42 m.

**16. If AD and PM are medians of triangles ABC and PQR, respectively whereÂ Î”ABC ~Â Î”PQR prove that AB/PQ = AD/PM.**

**Solution: **

Given, Î”ABC ~ Î”PQR

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

âˆ´AB/PQ = AC/PR = BC/QR**â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i**)

Also, âˆ A = âˆ P, âˆ B = âˆ Q, âˆ C = âˆ R â€¦â€¦â€¦â€¦.â€¦..**(ii)**

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

âˆ´ BD = BC/2 and QM = QR/2 â€¦â€¦â€¦â€¦â€¦..â€¦â€¦â€¦â€¦.**(iii)**

From equationsÂ **(i)**Â andÂ **(iii)**, we get

AB/PQ = BD/QM **â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(iv)**

In Î”ABD and Î”PQM,

From equation (ii), we have

âˆ B = âˆ Q

From equationÂ **(iv), we have,**

AB/PQ = BD/QM

âˆ´ Î”ABD ~ Î”PQM (SAS similarity criterion)

â‡’AB/PQ = BD/QM = AD/PM

The topic similarity of Triangles is the base of Exercise 6.3, Class 10 NCERT Maths Chapter 6, Triangles. There are 16 Questions in this exercise, which include 1 main question with 6 sub-questions, 12 Short Answer Questions and 3 Long Answer Questions. Exercise 6.3 deals with 3 theorems, which contain AAA, SSS, SAS criterion. The three theorems that form the base of the third exercise of Chapter 6 are:

**Theorem 6.3**: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This criterion is referred to as the AAA (Angleâ€“Angleâ€“Angle) criterion of similarity of two triangles.- If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. This may be referred to as the AA similarity criterion for two triangles
**Theorem 6.4**: If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This criterion is referred to as the SSS (Sideâ€“Sideâ€“Side) similarity criterion for two triangles.**Theorem 6.5**: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This criterion is referred to as the SAS (Sideâ€“Angleâ€“Side) similarity criterion for two triangles.

NCERT Solutions are one of the best study materials a student can use to score great marks in the Class 10 Maths examination. Students who are thorough with the concepts and are well versed in solving the problems present in the NCERT textbook can easily score great marks in the Class 10 board exam as CBSE often asks questions directly or indirectly from the NCERT book.

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