HiSET - Rotations | Practice Hub

Letters

Examine the figures in the above diagram. The figure at right is the result of performing which of the following transformations on the figure at left?

A $45^{\circ }$ clockwise rotation

A $90^{\circ }$ clockwise rotation

A $45^{\circ }$ counterclockwise rotation

A $90^{\circ }$ counterclockwise rotation

None of the other choices gives the correct response.

Explanation

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 $\frac{1}{8} \times 360^{\circ } = 45^{\circ }$ .

Nonagons

Examine the figures in the above diagram. Figure 2 is the result of performing which of the following transformations on Figure 1?

$72^{\circ }$

$84^{\circ }$

$36^{\circ }$

$48^{\circ }$

$60^{\circ }$

Explanation

The diagram below superimposes the two figures:

Nonagons

The transformation moves the black diagonal to the position of the red diagonal, and, consequently, points and to points and , respectively. This constitutes two-tenths of a complete turn clockwise, or a clockwise rotation of $\frac{2}{10} \times 360^{\circ } = 72^{\circ }$

Letters

Examine the figures in the above diagram. The figure at right is the result of performing which of the following transformations on the figure at left?

A $60^{\circ }$ counterclockwise rotation

A $60^{\circ }$ clockwise rotation

A $120^{\circ }$ clockwise rotation

A $120^{\circ }$ counterclockwise rotation

A $180^{\circ }$ rotation

Explanation

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 $360^{\circ }$ , or $\frac{1}{6} \times 360^{\circ } = 60^{\circ }$ .

What is the result of rotating the point $(2,\frac{1}{3})$ about the origin in the plane by $180 \degree$ ?

$(-2,-\frac{1}{3})$

$(-2,\frac{1}{3})$

$(2,-\frac{1}{3})$

$(2,\frac{1}{3})$

$(\frac{1}{3},2)$

Explanation

Rotating a point

geometrically in the plane about the origin $180\degree$ is equivalent to negating the coordinates of the point algebraically to obtain

Thus, since our initial point was

$(2,\frac{1}{3})$

we negate both coordinates to get

$(-2,-\frac{1}{3})$

as the rotation about the origin by $180\degree$ .

Letters

Examine the figures in the above diagram. The figure at right is the result of performing which of the following transformations on the figure at left?

A $120^{\circ }$ clockwise rotation

A $60^{\circ }$ clockwise rotation

A $60^{\circ }$ counterclockwise rotation

A $120^{\circ }$ counterclockwise rotation

A $180^{\circ }$ rotation

Explanation

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 $\frac{1}{3} \times 360^{\circ } = 120^{\circ }$

Hexagon

Let and be the midpoints of $\overline{AF}$ and $\overline{CD}$ , respectively.

Rotate the above hexagon $120^{\circ }$ clockwise, then reflect it about the line through $\overleftrightarrow{MN}$ . Call the image of after these transformations.

will be located in the same position as which of the following points?

Explanation

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

Hexagon

The image of under this rotation, which we will call , is at .

Now, locate the midpoints and , and construct the line $\overleftrightarrow{MN}$ as described and shown below. Perform the reflection:

Hexagon

It can be seen that the image of under this transformation - the desired - is located at .

Hexagon

Rotate the above hexagon $120^{\circ }$ counterclockwise, then reflect it about $\overleftrightarrow{CF}$ . Call the image of after these transformations.

will be located in the same position as which of the following points?

Explanation

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

Hexagon

The image of under this rotation, which we will call , is at .

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

Hexagon

The image of under this reflection - the desired - is located at itself.

Octagon

In the above octagon, let and be the midpoints of $\overline{BC}$ and $\overline{FG}$ , respectively.

Rotate the above octagon $135^{\circ }$ counterclockwise, then reflect it about $\overleftrightarrow{MN}$ . Call the image of after these transformations.