Question

Three parallel lines AD, BE and CF are intersected by two transversals at points A, B.
Cand D, E, F. IfAB = BC = 4 cm, then find the value of DE:EF​

Answer 1

Given:

Three parallel lines AD, BE and CF are intersected by two transversals at points A, B, C and D, E, F.

AB = BC = 4 cm

To find:

The value of DE:EF

Solution:

We know that,

If three or more parallel lines are cut off by two transversal then, they divide the transversals proportionally.

Here we have

Two transversal lines cutting lines AD, BE & CF at points A, B & C

Therefore, according to the above theorem, we get

$\frac{AB}{BC} = \frac{DE}{EF}$

on substituting AB = 4 cm & BC = 4 cm, we get

⇒ $\frac{4}{4} = \frac{DE}{EF}$

⇒ $\frac{1}{1} = \frac{DE}{EF}$

⇒ $\boxed{\bold{DE:EF = 1 : 1 }}$

Thus, the value of DE : FE is → 1 : 1.

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Also View:

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Three parallel lines are cut by two transversals as shown  in the given figure. If AB = 2 cm, BC = 4 cm and DE 1.5 cm, then the length of EF is=?

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Answer 2

Answer:

1:1 _________________