Me with this geometry problem asap

DE ≈ MN

Step-by-step explanation:

Match the vertices in both triangles, that is

given DEF and MNO , then

The first 2 sides are congruent , that is DE and MN

Also EF and NO ( second/third ), and

DF and MO ( first and third )

(B) DE ≅ MN

Step-by-step explanation:

I always find it best to write one under the other because it makes it easier to see the matches:

Δ D E F

Δ MNO                        

                  Angles                      Sides                            

                ∠D ≅ ∠M                  DE ≅ MN                      

                 ∠E ≅ ∠N                  EF ≅ NO                        

                 ∠F ≅ ∠O                  DF ≅ MO                          

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

i hope this is what you were looking for.

see attached graph