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# 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

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Mathematics
(Chapter – 6) (Triangles)
(Class – X)

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)

(i)

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
Page 2

Mathematics
(Chapter – 6) (Triangles)
(Class – X)

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)

(i)

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
Mathematics
(Chapter – 6) (Triangles)
(Class – X)

2

(ii)

It is given that DE || BC.
By using basic proportionality theorem, we obtain

Page 3

Mathematics
(Chapter – 6) (Triangles)
(Class – X)

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)

(i)

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
Mathematics
(Chapter – 6) (Triangles)
(Class – X)

2

(ii)

It is given that DE || BC.
By using basic proportionality theorem, we obtain

Mathematics
(Chapter – 6) (Triangles)
(Class – X)

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

(i)

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

Page 4

Mathematics
(Chapter – 6) (Triangles)
(Class – X)

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)

(i)

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
Mathematics
(Chapter – 6) (Triangles)
(Class – X)

2

(ii)

It is given that DE || BC.
By using basic proportionality theorem, we obtain

Mathematics
(Chapter – 6) (Triangles)
(Class – X)

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

(i)

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

Mathematics
(Chapter – 6) (Triangles)
(Class – X)

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
(Chapter – 6) (Triangles)
(Class – X)

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)

(i)

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
Mathematics
(Chapter – 6) (Triangles)
(Class – X)

2

(ii)

It is given that DE || BC.
By using basic proportionality theorem, we obtain

Mathematics
(Chapter – 6) (Triangles)
(Class – X)

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

(i)

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

Mathematics
(Chapter – 6) (Triangles)
(Class – X)

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
(Chapter – 6) (Triangles)
(Class – X)

5

Question 3:
In the following figure, if LM || CB and LN || CD, prove that

In the given figure, LM || CB
By using basic proportionality theorem, we obtain

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