Final answer:
Both parallel sides AB and DE of the isosceles trapezoid are 6 units long, which is determined by solving the equations AB=4x+2 and DE=10x-4 and setting them equal to each other.
Step-by-step explanation:
To find the lengths of the parallel lines AB and DE of an isosceles trapezoid when given AB=4x+2, DE=10x-4, and FC=34, we will utilize the property that the legs (sides) of an isosceles trapezoid are equal.
Since FC is given and corresponds to the length of the leg, it implies that the other leg is equal in length to FC.
However, the information about FC is not directly relevant to finding the lengths of AB and DE.
The key to solving for the lengths of the parallel sides is recognizing that in an isosceles trapezoid, AB and DE must be the same length due to the property that the bases of an isosceles trapezoid are congruent.
Therefore, we can set AB equal to DE:
4x + 2 = 10x - 4
By solving for x, we get:
x = 1
Plugging x back into the expressions for AB and DE:
AB = 4(1) + 2 = 6
DE = 10(1) - 4 = 6
Thus, the lengths of the parallel sides AB and DE are both 6 units.