Measurement and Geometry

Help Questions

HiSET › Measurement and Geometry

Questions 1 - 10

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

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

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 $\frac{2}{10} \times 360^{\circ } = 72^{\circ }$

Consider regular Hexagon ; let and be the midpoints of $\overline{AB}$ and $\overline{DE}$ . Reflect the hexagon about $\overleftrightarrow{AD}$ , then again about $\overleftrightarrow{MN}$ . Which of the following clockwise rotations about the center would result in each point being its own image under this series of transformations?

$300^{\circ }$

$240^{\circ }$

$180^{\circ }$

$120^{\circ }$

$60^{\circ }$

Explanation

Refer to the figure below, which shows the reflection of the given hexagon about $\overleftrightarrow{AD}$ ; we will call the image of , call the image of , and so forth.

Hexagon

Now, refer to the figure below, which shows the reflection of the image about $\overleftrightarrow{MN}$ ; we will call the image of , call the image of , and so forth.

Hexagon

Note that the vertices coincide with those of the original hexagon, and that the images of the points are in the same clockwise order as the original points. Since coincides with , coincides with , and so forth, a clockwise rotation of five-sixth of a complete turn - that is, $\frac{5}{6} \times 360 ^{\circ } = 300 ^{\circ }$ ,

is required to make each point its own image under the three transformations.

$\bigtriangleup ABC \sim \bigtriangleup DEF$ with scale factor 5:4, with $\bigtriangleup ABC$ the larger triangle.

Complete the sentence: the area of $\bigtriangleup ABC$ is % greater than that of $\bigtriangleup DEF$ .

(Select the closest whole percent)

Explanation

The ratio of the areas of two similar triangles is equal to the square of their scale factor. The scale factor is equal to $\frac{5}{4}$ , so the ratio of the areas is the square of this, or $\left ( \frac{5}{4} \right ) ^{2} = \frac{25}{16}$ .

This makes the area of larger triangle $\bigtriangleup ABC$ equal to $\frac{25}{16} \times 100 % = 156 \frac{1}{4} %$ of that of smaller triangle $\bigtriangleup DEF$ —or, equivalently, $56 \frac{1}{4} %$ greater.

Two of the angles of a triangle are congruent; the third has measure ten degrees greater than either one of the first two. What is the measure of the third angle?

$66 \frac{2}{3} ^{\circ }$

$63 \frac{1}{3} ^{\circ }$

$53 \frac{1}{3} ^{\circ }$

$56 \frac{2}{3} ^{\circ }$

$73\frac{1}{3}^{\circ }$

Explanation

Let be the measure of the third angle. Since its measure is ten degrees greater than either of the others, then the common measure of the other two is . The sum of the measures of the angles of a triangle is 180 degrees, so

To solve for , first ungroup and collect like terms:

Isolate ; first add 20:

Divide by 3:

$3x \div 3 = 200 \div 3$

Since $200 \div 3 = 66 \textup{ R }2$ ,

$x = 66 \frac{2}{3}$ .

The third angle measures $66 \frac{2}{3} ^{\circ }$ .

What is the result of reflecting the point over the y-axis in the coordinate plane?

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

Explanation

Reflecting a point

over the y-axis geometrically is the same as negating the x-coordinate of the ordered pair to obtain

Thus, since our initial point was

and we want to reflect it over the y-axis, we obtain the reflection by negating the first term of the ordered pair to get

Consider regular Hexagon ; let and be the midpoints of $\overline{AB}$ and $\overline{DE}$ . Reflect the hexagon about $\overleftrightarrow{BE}$ , then again about $\overleftrightarrow{MN}$ . With which of the following points does the image of under these reflections coincide?

Explanation

Refer to the figure below, which shows the reflection of about $\overleftrightarrow{BE}$ ; we will call this image .

Hexagon

Note that coincides with . Now, refer to the figure below, which shows the reflection of about $\overleftrightarrow{MN}$ ; we will call this image - the final image -