Question

In the following figure, FD || BC || AE and AC || ED. Find the value of x.


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Answer 1

Step-by-step explanation:

This is your answer OK bro

Answer 2

Concept:

Parallel lines are those lines which do not intersect each other.

Transversal is the line that cut two lines at two distinct points.

Given:

We are given that FD is parallel to BC is parallel to AE that is FD||BC||AE and AC is parallel to ED that is AC||ED.

Find:

We need to find the value of x.

Solution:

In ΔABC,

∠ABC+∠ BCA+∠ CAB=180° [Angle sum property of triangle]

64°+∠ BCA+52°=180°

∠ BCA=180°-64°-52°=64°

∠ FAE=∠BCA [because Alternate angles; A E || B C, AC is the

transversal]

∠ FAE=64°

Now,∠ FAE +x=180° [Adjacent angles in a parallelogram are supplementary]

x=180°-64°

x=116°

Therefore, the value of x is 116°

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