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

We know that if 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

We know that if 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

We know that if 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 = 1.28 cm and PE = 0.18 cm

EQ = PQ - PE

= (1.28 - 0.18) cm

= 1.1 cm

PR = 2.56 cm and PF = 0.36 cm

FR = PR - PF

= (2.56 - 0.36)cm

= 2.2 cm

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

☛ Check: NCERT Solutions for Class 10 Maths Chapter 6

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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

Summary:

E and F are points on the sides PQ and PR respectively of a Δ PQR. Then EF || QR for (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm and (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

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☛ Related Questions:

• In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
• In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
• In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
• In Fig. 6.21, A, B and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.