HELP PLEASE!!!

1) - The common difference between the angles =44/(5-1) =11 degrees.
Sum =N/2[2F + (N - 1)*D]
540 =5/2[2F + (5 -1)*11], solve for F
F =86 degrees - the smallest angle.
86+44 =130 degrees - the largest angle.

There are probably easier ways to do this...... but

We have three "outside" triangles and one inner one

Area  of the the three "outside" triangles  =

(1/2)(√3/2)  [ 6*4 + 2 * 4  + 6 * 2)  ]  =  11√3   units^2

And we an construct an inner triangle with sides BF, BD and FD

Length of BF  =  √[ 6^2 + 2^2 – 2(6)(2)(1/2)]  = √28  

Length of BD  =  √[ 6^2 + 4^2 – 2(6)(4)(1/2)]  = √28

Length of FD  =  √[ 2^2 + 4^2 – 2(2)(4)(1/2)]  = √12

Deriving the semi-perimeter of this triangle →     s  =  [2√28 + √12] /  2  =

√3 + 2√7  = √3 + √28     →  s^2  =  31 + 2√84  = 31 + √336

And using Heron's formula  ...

Area of inner triangle = √ [ (√3 + √28)(√3 + √28 – √28)^2 (√3 + √28 – √12)  ]    =

√ [ (√3 + √28)(√3 )^2 (√3 + √28 – √12)  ]  =

√3   *  √ [ (√3 + √28) ((√3 + √28  – √12)  ]  =

√3   *  √ [ (s) ((s  – √12)  ]  =

√3   *  √ [ s^2  – √12 s)  ]  =

√3   *  √ [ 31 + √336 – √12 [ √3 + √28)  ]  =

√3   *  √ [ 31 + √336 – √36 - √336)  ]  =

√3   *  √ [ 31  – 6   ]  =

√3   *  √ [ 25   ]  =

√3   *  5   =

5√3  units^2

So....the total  area  = [11√3  + 5√3]   units^2   = 16√3   units^2

Opened by request.    

In hexagon ABCDEF, AB = DE = 2, BC = EF = 4, CD = FA = 6, and all the interior angles are equal. Find the area of hexagon ABCDEF.

This question makes absolutely no sense to me.

If all the interior angles are equal then it is a regular hexagon and all the sides must be equal too.

Edited:

Oh now I see, the question has just be worded badly. 

The writer did not want to waste written characters, they must have been in short supply. 

Like the trees in a rainforest perhaps!

Tom...riddle    asked me to open this questiuon. 

She thinks this is the answer

https://www.algebra.com/algebra/homework/Surface-area/Surface-area.faq.question.1068016.html