In the given rectangle, diagonal AC is 10.
DX forms two triangles DXA and DXC.
Since these two triangles are not 45-45-90, which would be the case if other diagonal (DB) would've intersected from the center. Thus, we get a 30-60-90 triangle.
Now, X is 90 and therefore DC and DA are the hypotenuse.
Since we don't know which side is larger or smaller, lets consider that DX is larger side.
To find the longer side in triangle DXC, we get 8/2 = 4 * sqrt 3 = 6.928
To find the longer side triangle DXA, we get 6/2 = 3 *sqrt 3 = 5.196
Hence, the answer is C,D, E.

Let's take both extremes.

DX will be at its max when you slide X all the way to C.
So DX has to be less than 8

The smallest size of DX is when it is perpendicular to diagonal AC.
At that moment then angles ADX and DCX are equal and we get triangles DAX and DXC as similar.

Therefore DA/DC = XA/DX
6/8= XA/DX


Through pythagorean theorem
6^2= XA^2 + DX^2
36= ((3/4)DX)^2 + DX^2
36= 9/16*DX^2 + DX^2
36= 25/16 DX^2
DX= sqrt(36*16/25)
DX= 24/5= 4.8

Therefore DX is greater than or equal to 4.8

DX is between 4.8 and 8

Answer: C, D, E

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