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

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In the given question we are to prove that ratio of length of the sides of triangle are equal by using the properties of similarities

ML Aggarwal Avichal Publication, chapter 13.2, question no 3

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, ∴PE/EQ=3.9/3=39/30=13/10=1.3

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

EQPE=FRPF

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, ∴PE/QE=4/4.5=40/45=8/9 PF/RF=8/9 PE/QE=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 PE/EQ=0.18/1.10=18/110=9/55… (i) PE/FR=0.36/2.20=36/220=9/55 … (ii) ∴PE/EQ=PF/FR

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,

∴PE/EQ=3.9/3=39/30=13/10=1.3

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

EQPE=FRPF

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,

∴PE/QE=4/4.5=40/45=8/9

PF/RF=8/9

PE/QE=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

PE/EQ=0.18/1.10=18/110=9/55… (i)

PE/FR=0.36/2.20=36/220=9/55 … (ii)

∴PE/EQ=PF/FR