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Using the Law of Sines....we have that

(1/2) side length of  the square  / sin (67.5°)   = 2 / sin (90°)          .....multiply both sides by 2.....

side length of the square =  4 sin (67.5°)  =  4 √ [  (1  - cos (135°)  ) / 2 ] =  4 √  [ (1 + √2 /2) / 2 ]  =  4 √  [ (2 + √2)/4 ] 

So....the area of the square is the square of this  =   16 (2 + √2)/4  =  4 (2 + √2) units^2

Note that the max value for the sine  = 1  and the minimum value is -1.....therefore....

The high point on the graph  is  12 (1)  + 6  = 18

And the low point is 12(-1) + 6  =   -6

To find the midline.....add these values and divide by  2  =   [ 18 + -6 ] / 2 =  12 /  2  =  6

So.....the midline is y = f(x)  = 6

I think you are contradicting yourself.

You said you tried it my way but then you say you only use files on your computer.

My way involves Gyazo. If you have not used Gyazo then you have not tried my way.

It is presented as a ratio.

Nicely done, heureka!!!!!!

You have to prove that the parallelogram exists first

is it a ratio or is a question number?

The question itself is confusing.

I am talking about the hexagon problem.

I have posted pictures before.

I know how to post a picture.

This computer is not cooperating with me. 

I do not use any programs but the files one at the bottom of my computer.

I can't access that site but

8:8

is

1:1

I did not look at it thoroughly but it seems reasonable to me.   

What's wrong with the way guest worded it?

I am not a computer person but that does not make sense to me.  ( Like I said, I am not a computer person. )

There are many programs like Gyazo, including Tinypics which is what Web 2 uses (but you can use it independantly.)

If for some odd reason Gyazo does not work than one of the others problably will.

Are you sure you just have not worked out how to use it yet.

Maybe you should persist a little longer.

My goodness. That person has taken way too many psychedelic drugs!

Thanks Alan :))

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

She thinks this is the answer

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!

Opened by request.    

Here's the picture:

15 is the first answer just like Rom says.

I also got 36 for the second one. 

(a)

Compute the sums of the squares of Rows 1-4 of Pascal's Triangle.

Central binomial coefficients: \(\dbinom{2n}{n}\)

I think your proof is fine.

If the 8 is a question number:

I would solve it like this:

2(2+2)

2(4)

2*4

8

ANSWER:8

If the 8 is part of a ratio/proportion thingy:

8/(2(2+2))

8/(2(4))

8/8

1

ANSWER:1

I tried using your way, and it doesn't work with my HP moniter.