(i)
Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
\(\frac{PE}{EQ}=\frac{3.9}{3}\)=1.3
\(\frac{PF}{FR}\)=\(\frac{3.6}{2.6}\)=1.5
Hence, \(\frac{PE}{EQ}\neq \frac{PF}{FR}\)
Therefore, EF is not parallel to QR
(ii)
Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
\(\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}\)
\(\frac{PF}{FR}=\frac{8}{9}\)
Hence, \(\frac{PE}{EQ}=\frac{PF}{FR}\)
Therefore, EF is parallel to QR.
(iii)
Given that: PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
\(\frac{PE}{EQ}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64}\)
\(\frac{PF}{PR}=\frac{0.36}{2.56}=\frac{9}{64}\)
Hence, \(\frac{PE}{PQ}=\frac{PF}{PR}\)
Therefore, EF is parallel to QR.