Class 10 Math Chapter 6 Triangles Ex 6.2 NCERT Solution

Exercise 6.2 of Class 10 Maths Chapter 6 – Triangles focuses on the Basic Proportionality Theorem (Thales’ Theorem) and its applications. In this exercise, students learn how a line drawn parallel to one side of a triangle divides the other two sides in the same ratio. The problems in this exercise help build a clear understanding of proportional relationships within triangles and how to apply the theorem to solve geometry-based questions. These NCERT solutions are explained in a step-by-step manner to help students grasp the concept easily and improve their problem-solving skills.

Class 10 Math Chapter 6 Triangles Ex 6.2 NCERT

Ex 6.2 Class 10 Maths Question 1.
In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:
Triangles Class 10 Exercise 6.2

Ex 6.2 Class 10 Maths 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

Solution:

Ex 6.2 Class 10 Maths Question 3.
In the given figure, if LM || CB and LN || CD.
Prove that AM/AB=AN/AD
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q3
Solution:
exercise 6.2 class 10

Ex 6.2 Class 10 Maths Question 4.
In the given figure, DE || AC and DF || AE.
Prove that BE/EC=BF/BE
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q4
Solution:
class 10 maths triangles

Ex 6.2 Class 10 Maths Question 5.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.
NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q5
Solution:
Class 10 Triangles Ex 6.2

Ex 6.2 Class 10 Maths Question 6.
In the given 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 For Class 10 Maths Chapter 6 Triangles Ex 6.1 Q6
Solution:
triangles class 10

Ex 6.2 Class 10 Maths Question 7.
Using B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that your have proved it in class IX)
Solution:
ncert solutions for class 10 maths chapter 6

Ex 6.2 Class 10 Maths Question 8.
Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that your have done it in class IX)
Solution:
ch 6 maths class 10

Ex 6.2 Class 10 Maths 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/OD
Solution:
triangles class 10 ncert solutions

Ex 6.2 Class 10 Maths Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/OC=AE/ED Show that ABCD is a trapezium.
Solution:
similar triangles class 10

New Syllabus- Class 10 Math Chapter 6 Triangles Ex 6.2 NCERT

Question 1:
In Fig. 6.17 (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i) AE = 1 cm, AB = 3 cm, AC = 6 cm
By Basic Proportionality Theorem:
AE / AB = EC / AC
1 / 3 = EC / 6
EC = 2 cm

(ii) EC = 3 cm, AC = 4 cm, AB = 6 cm
Let AE = x
Then EC = 3, so AC = AE + EC = x + 3 = 4 ⇒ x = 1
So, AE = 1 cm
By BPT:
AD / DB = AE / EC
Let AD = x, then DB = 6 – x
x / (6 – x) = 1 / 3
3x = 6 – x
4x = 6 ⇒ x = 1.5
AD = 1.5 cm

Question 2:
E and F are points on sides PQ and PR of triangle PQR. For each case, state whether EF || QR.

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
PE / EQ = 3.9 / 3 = 1.3
PF / FR = 3.6 / 2.4 = 1.5
Not equal ⇒ EF is not parallel to QR

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
PE / QE = 4 / 4.5 = 8 / 9
PF / RF = 8 / 9
Equal ⇒ EF || QR

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
PE / PQ = 0.18 / 1.28 ≈ 0.1406
PF / PR = 0.36 / 2.56 ≈ 0.1406
Equal ⇒ EF || QR

Question 3:
In Fig. 6.18, LM || CB and LN || CD
Prove that AM / AB = AN / AD

From LM || CB, by BPT in triangle ABC:
AM / MB = AL / LC ⇒ AM / AB = AL / AC (1)
From LN || CD, by BPT in triangle ACD:
AN / ND = AL / LC ⇒ AN / AD = AL / AC (2)
From (1) and (2), AM / AB = AN / AD

Question 4:
In Fig. 6.19, DE || AC and DF || AE
Prove that BF / FE = BE / EC

From DE || AC in triangle ABC:
BD / DA = BE / EC (1)
From DF || AE in triangle ABE:
BD / DA = BF / FE (2)
From (1) and (2): BF / FE = BE / EC

Question 5:
In Fig. 6.20, DE || OQ and DF || OR
Show that EF || QR

From DE || OQ in triangle POQ:
PE / EQ = PD / DO (1)
From DF || OR in triangle POR:
PF / FR = PD / DO (2)
From (1) and (2):
PE / EQ = PF / FR ⇒ EF || QR

Question 6:
In Fig. 6.21, A, B, and C are points on OP, OQ and OR such that AB || PQ and AC || PR
Show that BC || QR

From AB || PQ in triangle POQ:
OA / AP = OB / BQ (1)
From AC || PR in triangle POR:
OA / AP = OC / CR (2)
From (1) and (2):
OB / BQ = OC / CR ⇒ BC || QR

Question 7:
Using Theorem 6.1, prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.

Let D be midpoint of AB, and line DE || BC
Using BPT in triangle ABC:
AD / DB = AE / EC
Since AD = DB ⇒ AE = EC
So, E is midpoint of AC

Question 8:
Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side.

Let D and E be midpoints of AB and AC
So, AD = DB and AE = EC ⇒ AD / DB = AE / EC
By Converse of BPT ⇒ DE || BC

Question 9:
ABCD is a trapezium in which AB || DC and diagonals intersect at O
Show that AO / CO = BO / DO

In triangles ABO and CDO:
AB || DC ⇒ ∠ABO = ∠DCO and ∠BAO = ∠CDO
So triangles ABO ~ CDO (AA similarity)
⇒ AO / CO = BO / DO

Question 10:
Diagonals of quadrilateral ABCD intersect at O such that AO / CO = BO / DO
Show that ABCD is a trapezium

In triangles ABO and CDO:
AO / CO = BO / DO and ∠AOB = ∠COD (vertically opposite)
⇒ △ABO ~ △CDO (SAS similarity)
⇒ ∠ABO = ∠DCO and ∠BAO = ∠CDO
⇒ AB || DC ⇒ ABCD is a trapezium