Question

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $△PQR$. 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

Answer 1

$E$ and $F$ are two points on side $PQ$ and $PR$ in $△PQR.$

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

Using Basic proportionality theorem,

$\therefore \frac{PE}{EQ}=\frac{3.9}{3}=\frac{39}{30}=\frac{13}{10}=1.3$

$\frac{PF}{FR}=\frac{3.6}{2.4}=\frac{36}{24}=\frac{3}{2}=1.5$

$\frac{PE}{EQ}\ne \frac{PF}{FR}$

So, $EF$ is not parallel to $QR.$

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

Using Basic proportionality theorem,

$\therefore \frac{PE}{QE}=\frac{4}{4.5}=\frac{40}{45}=\frac{8}{9}$

$\frac{PF}{RF}=\frac{8}{9}$

$\frac{PE}{QE}=\frac{PF}{RF}$

So, $EF$ is parallel to $QR.$

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

Using Basic proportionality theorem,

$EQ=PQ-PE=1.28-0.18=1.10$ cm

$FR=PR-PF=2.56-0.36=2.20$ cm

$\frac{PE}{EQ}=\frac{0.18}{1.10}=\frac{18}{110}=\frac{9}{55}$... (i)

$\frac{PE}{FR}=\frac{0.36}{2.20}=\frac{36}{220}=\frac{9}{55}$ ... (ii)

$\therefore \frac{PE}{EQ}=\frac{PF}{FR.}$

So, $EF$ is parallel to $QR.$