Question

pease help solve this
E and F are points on the side PQ and PR respectively of Δ PQR for each of the following cases, state whather EF ║QR
i)PE=4cm, QE=6.5 cm, PF=8cm, KF=9cm
ii)PE=3.9cm, PF=3.6cm, EQ=1.3cm and ER=2.4
iii)PQ1.28, PR=2.56cm, PE=0.18 and PE=3.6

Answer 1

ΔPQR, E and F are two points on side PQ and PR respectively.

(i) PE = 3.9 cm, EQ = 3 cm (Given)

PF = 3.6 cm, FR = 2,4 cm (Given)

∴ PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3 [By using Basic proportionality theorem]

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

So, PE/EQ ≠ PF/FR

Hence, EF is not parallel to QR.

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

∴ PE/QE = 4/4.5 = 40/45 = 8/9 [By using Basic proportionality theorem]

And, PF/RF = 8/9

So, PE/QE = PF/RF

Hence, EF is parallel to QR.

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

Here, EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm

And, 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)

∴ PE/EQ = PF/FR.

Hence, EF is parallel to QR.