 # E and F are points on the sides PQ and PR respectively of a ΔPQR. 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

Given that in ΔPQR, E and F are two points on side PQ and PR respectively. The figure shows

(i) As per the given data PE = 3.9 cm,

EQ = 3 cm, PF = 3.6 cm

and FR = 2,4 cm

B using Basic proportionality theorem, we get,

PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3

PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5

So, we get, PE/EQ ≠ PF/FR

Thus, EF is not parallel to QR.

(ii) As per the given data PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm

By Basic proportionality theorem, we get,

PE/QE = 4/4.5 = 40/45 = 8/9 And, PF/RF = 8/9

So, Thus, PE/QE = PF/RF

Therefore, EF is parallel to QR.

(iii) As per the given data PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

From the diagram, EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

Also, FR = PR – PF = 2.56 – 0.36 = 2.20 cm

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55…………. (i)

And, PE/FR = 0.36/2.20 = 36/220 = 9/55………… (ii)

So, we get here, PE/EQ = PF/FR

Thus, EF is parallel to QR.