Find the length of segments FD & FE? Show the work.

FD = 8

FE = 21.6

Step-by-step explanation:

By applying Pythagoras theorem in ΔCDF,

CD² = CF² + FD²

(17)² = (15)² + FD²

289 = 225 + FD²

FD =

     =

FD = 8 units

Since AB║DE and CD is a transversal line,

∠BCD ≅ ∠CDF [Alternate interior angles]

m∠CDF = m∠BCD = 55°

By using cosine rule in the right triangle CDE,

cos(55)° = =

cos(55)° =

EF + 8 =

FE = 29.64 - 8

     ≈ 21.6 units  

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

none cuz there are no "following statements" regarding the question

step-by-step explanation: