Question

given in figure, AD= 3cm , AE= 5cm, BD= 4cm, CE=4cm, CF= 2cm, BF= 2.5cm. find the pair of parallel lines and their length

Answer 1

Answer:

The lines EF and AB are parallel. The lenght of EF is 3.11 cm.

Step-by-step explanation:

In triangle ABC and EFC,

$\angle ACB=\angle ECF$                    (Common angle)

$\frac{CF}{CB}=\frac{2}{2+2.5}=\frac{2}{4.5}$

$\frac{CE}{CA}=\frac{4}{4+5}=\frac{4}{9}=\frac{2}{4.5}$

$\frac{CF}{CB}=\frac{CE}{CA}$

Since two corresponding sides are proportional and including angle are equal.

By SAS rule of similarity, we get

$\triangle ABC\sim \triangle EFC$

$\angle ABC=\angle EFC$                    (CPCTC)

$\angle BAC=\angle FEC$                    (CPCTC)

Since the corresponding angles are equal, therefore we can say that the lines EF and AB are parallel.

$\frac{EF}{AB}=\frac{2}{4.5}$

$\frac{EF}{7}=\frac{2}{4.5}$

$EF=\frac{14}{4.5}=3.11$

Therefore the lines EF and AB are parallel. The lenght of EF is 3.11 cm.

Answer 2

Given : In given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm,

CE = 4 cm, CF = 2 cm, BF = 2.5 cm,

To Find : the pair of parallel lines and hence their lengths.

Solution:

if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

2.5/2  = 5/4

=> BF/FC  = AE/EC

Hence EF || AB

EF/AB =  CE/AC = CF/BC  = 4/9

EF/(3 + 4) = 4/9

=> EF = 28/9

Check other

3/4  ≠ 5/4

3/4 ≠ 2.5/2

Hence EF || AB

EF = 28/9

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