Question

E and F are point on the side PQ and respectively of a triangle PQR . Show that EF ll QR if PE = 4cm , QE = 4.5 cm , PF = 8 cm and RF = 9 cm

Answer 1

Step-by-step explanation:

(i)Given :

PE = 3.9 cm, EQ = 3 cm ,PF = 3.6 cm, FR = 2,4 cm  

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

∴ 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 || QR.

[By Converse of basic proportionality theorem]

(ii) Given :  

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 || QR.

[By Converse of basic proportionality theorem]

(iii) Given :

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm  

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  

And, PE/FR = 0.36/2.20 = 36/220 = 9/55  

∴ PE/EQ = PF/FR.

Hence, EF || QR.

[By Converse of basic proportionality theorem]

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Answer 2

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Given :

PE = 3.9 cm,

EQ = 3 cm,

PF = 3.6 cm,

FR = 2.4 cm

Now we know,

Triangle Proportionality Theorem: 

If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

So, if the lines EF and QR are to be parallel, then ratio PE:EQ should be proportional to PF:PR

e.g., PE/EQ = 3.9cm/3cm = 1.3

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

hence, PE/EQ ≠ PF/FR

therefore EF is not parallel to QR .

similarly,

(ii) Given :

PE = 4cm

EQ =4.5cm

PF = 8cm

FR = 9cm

PE/EQ = 4/4.5 = 8/9

PF/FR = 8/9

here we see , PE/EQ = PF/FR

therefore, EF || QR

(iii) Given :

PQ = 1.28 cm

PR = 2.56 cm

PE = 1.8cm

PF = 3.6cm

PE/PQ = 1.8/3.6 = 1/2

PQ/PR = 1.28/2.56 = 1/2

we can see that PE/PQ = PQ/PR

therefore , EF || PQ

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