In the figure shown, line AB is parallel to line CD. Part A what is the measure of angle x. Part b explain how you found the measurr of angle x by identifying the angle relationship that you used along the traversal

Part A

x is 46°

Part B

Alternate angles are angles that are in relatively opposite locations relative to a transversal

Please see attached diagram showing alternate angles

Step-by-step explanation:

Part A

∠DRP  = 110° (Given)

∠QPA = 64° (Given)

∠QPR =

Given that AB is parallel to CD, we have;

∠DRP is congruent to ∠APR (Alternate angles to a transversal RP of parallel lines AB and CD)

Therefore, ∠APR = 115°

∠APR = ∠QPA + ∠QPR (Angle addition postulate)

∴ 115° = 64° +  ∠QPR

∠QPR = 110° - 64° = 46°

x = 46°

Part B.

Given that AB is parallel to CD, the lines common (that intersects) both lines are the transversal lines

The angles formed between the parallel lines and the transversal lines have special relationships based on their position with respect to each other

In the question, the angle 110° given between CD and the transversal RP, is found to at an alternate position to the angle ∠APR between the same transversal RP and AB and given that alternate angles are always equal, angle ∠APR is therefore also equal to 110°.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet