NCERT Solutions for class 10 Maths Chapter 6- Triangles Exercise 6.2

The NCERT solutions for the Exercise 6.2 of NCERT Class 10 Chapter 6 Triangles are available here. The solutions are also available in the downloadable PDF format below. These solutions are prepared by the subject experts at BYJU’S, carefully, giving an explanation for each step alongside. Exercise 6.2 deals with a variety of concepts related to the similarity of the triangle.

The NCERT textbook provides plenty of questions for the students to solve and practice. Solving the NCERT Solutions for Class 10 Maths and practising is more than enough to score high in the Class 10 first term and second term examinations. However, the students should make sure that they practise one problem repeatedly until the concept becomes thorough and clear.

Download PDF of NCERT Solutions for Class 10 Maths Chapter 6- Triangles Exercise 6.2

Access Answers of Maths NCERT Class 10 Chapter 6- Triangles Exercise 6.2

ncert solutions oct14 for class 10 maths chapter 6 triangles 03
ncert solutions oct14 for class 10 maths chapter 6 triangles 04
ncert solutions oct14 for class 10 maths chapter 6 triangles 05
ncert solutions oct14 for class 10 maths chapter 6 triangles 06
ncert solutions oct14 for class 10 maths chapter 6 triangles 07
ncert solutions oct14 for class 10 maths chapter 6 triangles 08
ncert solutions oct14 for class 10 maths chapter 6 triangles 09
ncert solutions oct14 for class 10 maths chapter 6 triangles 10

Access other exercise solutions of Class 10 Maths Chapter 6- Triangles

Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)

Exercise 6.3 Solutions 16 Questions (1 main question with 6 sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)

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 6.2

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

Ncert solutions class 10 chapter 6-4

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.

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;

 

Triangles Exercise 6.2 Answer 3

(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.

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

Ncert solutions class 10 chapter 6-6

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.

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

Ncert solutions class 10 chapter 6-7

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.

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

Ncert solutions class 10 chapter 6-8

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.

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.

Ncert solutions class 10 chapter 6-9

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.

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).

Ncert solutions class 10 chapter 6-10

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.

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.

Ncert solutions class 10 chapter 6-11

We have to prove that: DE || BC.

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.

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.

Ncert solutions class 10 chapter 6-12

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.

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 intersects each other at O such that,

AO/BO = CO/DO.

Ncert solutions class 10 chapter 6-13

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.


The second exercise, Exercise 6.2, of Class 10 NCERT Maths Chapter 6, Triangles includes questions that are based on the topic Similarity of Triangles. There are 10 Questions in Exercise 6.2 of Class 10 out of which 9 are Short Answer Questions and one is a Long Answer Question. The chapter includes different theorems, along with their proofs, related to “Triangles” out of which the Exercise 6.2 deals with mainly 2 theorems. These theorems are used to prove similarities of the triangles. Therefore, the two theorems that form the base of the second exercise of Chapter 6 are:

  • Theorem 6.1 : 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.
  • Theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

If perfect study material is what you are looking for, then, NCERT Solutions is one of the best study materials a student can have in order to score great marks in the first and second term examination. CBSE often asks questions either directly or indirectly from the NCERT textbook. Therefore, the students who are thorough with the concepts and problem-solving method can easily score exceptionally well in the Class 10 term 1 and term 2 exam.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class