# E and F are points on the sides PQ and PR respectively of a Δ 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 = 0.36 cm

**Solution:**

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

As we know that a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (converse of Basic Proportionality theorem)

Here,

PE/EQ = 3.9/3 = 1.3

and

PF/FR = 3.6/2.4 = 1.5

Hence,

PE/EQ ≠ PF/FR

According to the converse of Basic Proportionality theorem, EF is not parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

As we know that a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (converse of Basic Proportionality theorem)

Here,

PE/EQ = 4/4.5 = 8/9

PF/FR = 8/9

Hence,

PE/EQ = PF/FR

According to converse of Basic Proportionality theorem, EF || QR

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

As we know that a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (converse of Basic Proportionality theorem)

Here,

PQ = 1.28cm and PE = 0.18cm

EQ = PQ - PE

= (1.28 - 0.18)cm

= 1.10cm

PR = 2.56cm and PF = 0.36cm

FR = PR - PF

= (2.56 - 0.36)cm

= 2.20cm

Now,

PE/EQ = 0.18cm/1.10cm = 18/110 = 9/55

PF/FR = 0.36cm/2.20cm = 36/220 = 9/55

⇒ PE/EQ = PF/FR

According to converse of Basic Proportionality theorem, EF* || QR*

**Video Solution:**

## E and F are points on the sides PQ and PR respectively of a Δ 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 = 0.36 cm

### Class 10 Maths NCERT Solutions - Chapter 6 Exercise 6.2 Question 2:

E and F are points on the sides PQ and PR respectively of a triangle PQR. Then for the first case, EF is not parallel to QR. For the second case EF || QR and for the third case EF || QR.