A \(\displaystyle 135^{\circ }\) rotation is equivalent to \(\displaystyle \frac{135}{360} = \frac{3}{8}\) of a complete rotation, so rotate as follows:

The image of \(\displaystyle A\) under this rotation, which we will call \(\displaystyle A''\), is at \(\displaystyle F\).

Now, locate the midpoints \(\displaystyle M\) and \(\displaystyle N\), and construct the line \(\displaystyle \overleftrightarrow{MN}\) as described and shown below. Perform the reflection:

The image of \(\displaystyle A ''\) under this transformation - the desired \(\displaystyle A'\) - is located at \(\displaystyle G\).

A \(\displaystyle 120^{\circ }\) rotation is equivalent to \(\displaystyle \frac{120}{360} = \frac{1}{3}\) of a complete rotation, so rotate as follows:

The image of \(\displaystyle A\) under this rotation, which we will call \(\displaystyle A''\), is at \(\displaystyle E\).

Now, construct \(\displaystyle \overleftrightarrow{CF}\), and reflect the hexagon about this line:

The image of \(\displaystyle A ''\) under this reflection - the desired \(\displaystyle A'\) - is located at \(\displaystyle A\) itself.

Between 6:15 and 6:40,

\(\displaystyle 40 - 15 = 25\)

minutes elapse.

The minute hand of a clock rotates one complete \(\displaystyle 360^{\circ }\) clockwise turn about its mount over the course of 60 minutes. Therefore, over 25 minutes, the minute hand rotates \(\displaystyle \frac{25}{60} \times 360 ^{\circ }= 150^{\circ }\) clockwise.

The hour hand of a clock makes one complete \(\displaystyle 360^{\circ }\) clockwise rotation over the course of 12 hours, or

\(\displaystyle 12 \times 60 = 720\)

minutes.

Therefore, over the course of 20 minutes, the hour hand rotates

\(\displaystyle \frac{20}{720} \times 360 ^{\circ } = 10^{\circ }\).

The minute hand of a clock rotates one complete \(\displaystyle 360^{\circ }\) clockwise turn about its mount over the course of 60 minutes, or, equivalently, since there are 60 seconds in a minute, every

\(\displaystyle 60 \times 60 = 3,600\) seconds.

Over the course of 1 minute 40 seconds - or, since one minute is equal to 60 seconds,  \(\displaystyle 1 \times 60 + 40 = 100\) seconds - the minute hand rotates clockwise

\(\displaystyle \frac{100}{3,600} \times 360^{\circ } = 10^{\circ }\).

Examine the figure below:

If we connect the horizontal line with the line along the rotated nine at right, we see that it is the result of a one-third turn clockwise; the angle between them \(\displaystyle \frac{1}{3} \times 360^{\circ } = 120^{\circ }\)

Examine the figure below:

If we connect the horizontal line with the line along the rotated nine at right, we see that it is the result of a one-eighth turn clockwise; the angle between them \(\displaystyle \frac{1}{8} \times 360^{\circ } = 45^{\circ }\).

Examine the figure below. 

If we connect the horizontal line with the line along the rotated "omega" at right, we see that it is the result of a one-sixth turn counterclockwise; the angle between them is one sixth of \(\displaystyle 360^{\circ }\), or  \(\displaystyle \frac{1}{6} \times 360^{\circ } = 60^{\circ }\).

The hour hand of a clock makes one complete \(\displaystyle 360^{\circ }\) clockwise rotation over the course of 12 hours, or, since one hour is equal to 60 minutes,

\(\displaystyle 12 \times 60 = 720\textup{ minutes}\)

Therefore, over the course of 1 hour 20 minutes - which is equal to \(\displaystyle 1 \times 60 + 20 = 80\) minutes - the hour hand rotates

\(\displaystyle \frac{80}{720} \times 360 ^{\circ } = 40^{\circ }\).

The circumcenter of a triangle can be located by finding the intersection of the perpendicular bisectors of the three sides of the triangle. The perpendicular bisectors are shown below, with point of intersection  \(\displaystyle O\):

It can be seen that, as is characteristic of a right triangle, this point is the midpoint of the hypotenuse. Construct \(\displaystyle \overline{BO}\). A dilation of scale factor \(\displaystyle \frac{1}{2}\)  with center \(\displaystyle O\) can be performed by letting \(\displaystyle A '\), \(\displaystyle B'\), and \(\displaystyle C'\) be the midpoints of \(\displaystyle \overline{AO}\), \(\displaystyle \overline{BO}\), and \(\displaystyle \overline{CO}\), respectively: 

Removing the perpendicular bisectors and \(\displaystyle O\), we see that the correct choice is the figure