NCERT Solutions, Chapter 6: Triangles, Class 10 (Mathematics) Class 10 Notes | EduRev

Class 10 : NCERT Solutions, Chapter 6: Triangles, Class 10 (Mathematics) Class 10 Notes | EduRev

 Page 1


Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
1 
Exercise 6.2 
Question 1:  
In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).  
(i)  
 
(ii)  
 
 
Answer 1:  
 (i)  
 
Let EC = x cm  
It is given that DE || BC.  
By using basic proportionality theorem, we obtain  
Page 2


Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
1 
Exercise 6.2 
Question 1:  
In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).  
(i)  
 
(ii)  
 
 
Answer 1:  
 (i)  
 
Let EC = x cm  
It is given that DE || BC.  
By using basic proportionality theorem, we obtain  
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
2 
  
 
 
(ii)   
 
 
Let AD = x cm  
It is given that DE || BC.  
By using basic proportionality theorem, we obtain  
  
  
 
 
Page 3


Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
1 
Exercise 6.2 
Question 1:  
In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).  
(i)  
 
(ii)  
 
 
Answer 1:  
 (i)  
 
Let EC = x cm  
It is given that DE || BC.  
By using basic proportionality theorem, we obtain  
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
2 
  
 
 
(ii)   
 
 
Let AD = x cm  
It is given that DE || BC.  
By using basic proportionality theorem, we obtain  
  
  
 
 
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
3 
Question 2:  
E and F are points on the sides PQ and PR respectively of a ?PQR. 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   
 
Answer 2:  
 (i)  
 
 
 
Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm  
  
 
 
Page 4


Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
1 
Exercise 6.2 
Question 1:  
In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).  
(i)  
 
(ii)  
 
 
Answer 1:  
 (i)  
 
Let EC = x cm  
It is given that DE || BC.  
By using basic proportionality theorem, we obtain  
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
2 
  
 
 
(ii)   
 
 
Let AD = x cm  
It is given that DE || BC.  
By using basic proportionality theorem, we obtain  
  
  
 
 
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
3 
Question 2:  
E and F are points on the sides PQ and PR respectively of a ?PQR. 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   
 
Answer 2:  
 (i)  
 
 
 
Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm  
  
 
 
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
4 
(ii)   
  
PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm  
  
 
(iii)   
  
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm  
Page 5


Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
1 
Exercise 6.2 
Question 1:  
In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).  
(i)  
 
(ii)  
 
 
Answer 1:  
 (i)  
 
Let EC = x cm  
It is given that DE || BC.  
By using basic proportionality theorem, we obtain  
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
2 
  
 
 
(ii)   
 
 
Let AD = x cm  
It is given that DE || BC.  
By using basic proportionality theorem, we obtain  
  
  
 
 
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
3 
Question 2:  
E and F are points on the sides PQ and PR respectively of a ?PQR. 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   
 
Answer 2:  
 (i)  
 
 
 
Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm  
  
 
 
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
4 
(ii)   
  
PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm  
  
 
(iii)   
  
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm  
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 6) (Triangles)  
(Class – X) 
 
www.tiwariacademy.com  
5 
  
  
Question 3:  
In the following figure, if LM || CB and LN || CD, prove that   
  
 
 
 
 
 
 
 
 
Answer 3:  
  
 
In the given figure, LM || CB  
By using basic proportionality theorem, we obtain  
  
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Summary

,

practice quizzes

,

NCERT Solutions

,

Class 10 (Mathematics) Class 10 Notes | EduRev

,

study material

,

Chapter 6: Triangles

,

Chapter 6: Triangles

,

Semester Notes

,

Class 10 (Mathematics) Class 10 Notes | EduRev

,

Chapter 6: Triangles

,

Free

,

Viva Questions

,

Important questions

,

mock tests for examination

,

MCQs

,

Exam

,

video lectures

,

Extra Questions

,

Previous Year Questions with Solutions

,

shortcuts and tricks

,

NCERT Solutions

,

pdf

,

NCERT Solutions

,

Class 10 (Mathematics) Class 10 Notes | EduRev

,

past year papers

,

Objective type Questions

,

Sample Paper

,

ppt

;