Measurement and Geometry - HiSET
Card 1 of 536
The graph on the left shows an object in the Cartesian plane. A transformation is performed on it, resulting in the graph on the right.
Which of the following transformations best fits the graphs?
The graph on the left shows an object in the Cartesian plane. A transformation is performed on it, resulting in the graph on the right.
Which of the following transformations best fits the graphs?
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A dilation is the stretching or shrinking of a figure.
A rotation is the turning of a a figure about a point.
A reflection is the flipping of a figure across a line.
A translation is is the sliding of a figure in a direction.
With a translation, the image is not only congruent to its original size and shape, but its orientation remains the same. A translation fits this figure best because the shape seems to move upward and rightward without changing size, shape, or orientation.
A dilation is the stretching or shrinking of a figure.
A rotation is the turning of a a figure about a point.
A reflection is the flipping of a figure across a line.
A translation is is the sliding of a figure in a direction.
With a translation, the image is not only congruent to its original size and shape, but its orientation remains the same. A translation fits this figure best because the shape seems to move upward and rightward without changing size, shape, or orientation.
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Consider regular Hexagon 
.
On this hexagon, perform the translation 
. Then perform a 
clockwise rotation on the image with center at 
.
Let 
be the image of 
under these transformations, 
be the image of 
, and so forth. Under these images, which point on the original hexagon does 
fall?
Consider regular Hexagon 
On this hexagon, perform the translation 
Let 
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The translation 
on a figure is the translation that shifts a figure so that the image of 
coincides with 
. All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation, with the image of 
marked as 
:
If this new hexagon is rotated clockwise 
- one third of a turn - about 
- the image is the original hexagon, but the vertices can be relabeled. Letting 
be the image of 
under this rotation, and so forth:
coincides with 
in the original hexagon, making 
the correct response.
The translation 
If this new hexagon is rotated clockwise 
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Consider regular Hexagon 
.
On this hexagon, perform the translation 
. Then perform a 
rotation on the image with center at 
.
Let 
be the image of 
under these transformations, 
be the image of 
, and so forth. Under these images, which point on the original hexagon does 
fall?
Consider regular Hexagon 
On this hexagon, perform the translation 
Let 
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The translation 
on a figure is the translation that shifts a figure so that the image of coincides with 
. All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation, with 
the image of 
:
If this new hexagon is rotated 
- one half of a turn - about 
- the image is the original hexagon, but the vertices can be relabeled. Letting 
be the image of 
under this rotation, and so forth:
coincides with 
in the original hexagon, making 
the correct response.
The translation 
If this new hexagon is rotated 
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Consider regular Hexagon 
.
On this hexagon, perform the translation 
. Then perform a 
rotation on the image with center at 
. Let 
be the image of 
under these transformations, and so forth.
Which of the following correctly shows Hexagon 
relative to Hexagon ?
Consider regular Hexagon
On this hexagon, perform the translation 
Which of the following correctly shows Hexagon 
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The translation 
on a figure is the translation that shifts a figure so that the image of 
, which we will call 
, coincides with 
. All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation:
If this new hexagon is rotated clockwise - one third of a turn - about , and call 
the image of 
, and so forth, the result is as follows:
Removing the intermediate markings, we see that the correct response is
The translation 
If this new hexagon is rotated clockwise 
Removing the intermediate markings, we see that the correct response is
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Consider regular Hexagon .
On this hexagon, perform the translation 
. Then reflect the hexagon about 
. Let 
be the image of 
under these transformations, and so forth.
Which point on Hexagon is the image of 
under these transformations?
Consider regular Hexagon
On this hexagon, perform the translation 
Which point on Hexagon 
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The translation 
on a figure is the translation that shifts a figure so that the image of 
, which we will call 
, coincides with 
. All other points shift the same distance in the same direction. Below shows the image of the given hexagon under this translation, with the image of 
marked as 
:
If the image is reflected about 
, the new image is the original hexagon. Calling 
the image of 
under this reflection, we get the following:
, the image of 
under these two transformations, coincides with 
.
The translation 
If the image is reflected about 
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Translate the graph of the equation
left four units and down six units. Give the equation of the image.
Translate the graph of the equation
left four units and down six units. Give the equation of the image.
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If the graph of an equation is translated to the right 
units, and upward 
units, the equation of the image can be found by replacing 
with 
and 
with 
in the equation of the original graph.
Since we are moving the graph of the equation
left four units and down six units, we set 
and 
; we can replace 
with 
, or 
, and 
with 
, or 
. The equation of the image can be written as
Simplify by distributing:
Collect like terms:
Add 26 to both sides:
,
the correct choice.
If the graph of an equation is translated to the right 
Since we are moving the graph of the equation
left four units and down six units, we set 
Simplify by distributing:
Collect like terms:
Add 26 to both sides:
the correct choice.
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Translate the graph of the equation
right two units and up five units. Give the equation of the image.
Translate the graph of the equation
right two units and up five units. Give the equation of the image.
Tap to reveal answer
If the graph of an equation is translated to the right 
units, and upward 
units, the equation of the image can be found by replacing 
with 
and 
with 
in the equation of the original graph.
Since we are moving the graph of the equation
right two units and up five units, we set 
and 
; we can therefore replace 
with 
and 
with 
. The equation of the image can be written as
This can be rewritten by applying the binomial square pattern as follows:
Collect like terms; the equation becomes
Subtract 100 from both sides:
If the graph of an equation is translated to the right 
Since we are moving the graph of the equation
right two units and up five units, we set 
This can be rewritten by applying the binomial square pattern as follows:
Collect like terms; the equation becomes
Subtract 100 from both sides:
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Translate the graph of the equation
right four units and down two units. Give the equation of the image.
Translate the graph of the equation
right four units and down two units. Give the equation of the image.
Tap to reveal answer
If the graph of an equation is translated to the right 
units, and upward 
units, the equation of the image can be found by replacing 
with 
and 
with 
in the equation of the original graph.
Since we are moving the graph of the equation
right four units and down two units, we set 
and 
; we can therefore replace 
with 
, and 
with 
, or 
. The equation of the image can be written as
The expression at right can be simplified. First, use the distributive property on the middle expression:
Now, simplify the first expression by using the binomial square pattern:
Collect like terms on the right:
Subtract 2 from both sides:
,
the equation of the image.
If the graph of an equation is translated to the right 
Since we are moving the graph of the equation
right four units and down two units, we set 
The expression at right can be simplified. First, use the distributive property on the middle expression:
Now, simplify the first expression by using the binomial square pattern:
Collect like terms on the right:
Subtract 2 from both sides:
the equation of the image.
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Translate the graph of the equation
left three units and down five units. Give the equation of the image.
Translate the graph of the equation
left three units and down five units. Give the equation of the image.
Tap to reveal answer
If the graph of an equation is translated to the right 
units, and upward 
units, the equation of the image can be found by replacing 
with 
and 
with 
in the equation of the original graph.
Since we are moving the graph of the equation
left three units and down five units, we set 
and 
; we can therefore replace 
with 
, or 
, and 
with 
, or 
. The equation of the image can be written as
We can simplify the expression on the right by distributing:
Collect like terms:
Subtract 5 from both sides:
,
the correct equation of the image.
If the graph of an equation is translated to the right 
Since we are moving the graph of the equation
left three units and down five units, we set 
We can simplify the expression on the right by distributing:
Collect like terms:
Subtract 5 from both sides:
the correct equation of the image.
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On the coordinate plane, let 
, 
, and 
be located at the origin, 
, and 
. Construct the median of 
from 
and let the foot of the median be 
. On the triangle, perform the translation 
. Where is the image of 
?
On the coordinate plane, let 
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By definition, a median of a triangle has as its endpoints one vertex and the midpoint of the opposite side. Therefore, the endpoints of the median from 
are 
itself, which is at 
, and 
, which itself is the midpoint of the side with origin 
and 
, which is 
, as its endpoints.
The midpoint of a segment with endpoints at 
and 
is located at
,
so, substituting the coordinates of 
and 
in the formula, we see that 
is
, or 
.
See the figure below:
To perform the translation 
, or, equivalently,
,
on a point, it is necessary to add
and
to the 
- and 
- coordinates, respectively. Therefore, the image of 
is located at
,
or
.
By definition, a median of a triangle has as its endpoints one vertex and the midpoint of the opposite side. Therefore, the endpoints of the median from 
The midpoint of a segment with endpoints at
so, substituting the coordinates of 
See the figure below:
To perform the translation 
on a point, it is necessary to add
and
to the 
or
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On the coordinate plane, let 
and 
be located at 
and 
, respectively. Let 
be the midpoint of 
and let 
be the midpoint of 
. On the segment, perform the translation 
. Where is the image of 
located?
On the coordinate plane, let 
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The midpoint of a segment with endpoints at 
and 
is located at
Substitute the coordinates of 
and 
in this formula to find that midpoint 
of 
is located at
, or 
.
Substitute the coordinates of 
and 
to find that midpoint 
of 
is located at
, or 
.
To perform the translation 
, or, equivalently,
,
on a point, it is necessary to add
to its 
-coordinate, and
to its 
-coordinate.
Therefore, the 
-coordinate of the image of 
under this translation is
;
its 
-coordinate is
The image of 
is located at 
.
The midpoint of a segment with endpoints at 
Substitute the coordinates of 
Substitute the coordinates of 
To perform the translation 
on a point, it is necessary to add
to its 
to its 
Therefore, the 
its 
The image of 
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Find the area of a square with the following side length:
Find the area of a square with the following side length:
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We can find the area of a circle using the following formula:
In this equation the variable, 
, represents the length of a single side.
Substitute and solve.
We can find the area of a circle using the following formula:
In this equation the variable,
Substitute and solve.
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The perimeter of a square is 
. In terms of 
, give the area of the square.
The perimeter of a square is 
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Since a square comprises four segments of the same length, the length of one side is equal to one fourth of the perimeter of the square, which is 
. The area of the square is equal to the square of this sidelength, or
.
Since a square comprises four segments of the same length, the length of one side is equal to one fourth of the perimeter of the square, which is 
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The volume of a sphere is equal to 
. Give the surface area of the sphere.
The volume of a sphere is equal to 
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The volume of a sphere can be calculated using the formula
Solving for 
:
Set 
. Multiply both sides by 
:
Divide by 
:
Take the cube root of both sides:
Now substitute for 
in the surface area formula:
,
the correct response.
The volume of a sphere can be calculated using the formula
Solving for 
Set 
Divide by 
Take the cube root of both sides:
Now substitute for 
the correct response.
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Express the area of a square plot of land 60 feet in sidelength in square yards.
Express the area of a square plot of land 60 feet in sidelength in square yards.
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One yard is equal to three feet, so convert 60 feet to yards by dividing by conversion factor 3:
Square this sidelength to get the area of the plot:
,
the correct response.
One yard is equal to three feet, so convert 60 feet to yards by dividing by conversion factor 3:
Square this sidelength to get the area of the plot:
the correct response.
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A square has perimeter 
. Give its area in terms of 
.
A square has perimeter 
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Divide the perimeter to get the length of one side of the square.
Divide each term by 4:
Square this sidelength to get the area of the square. The binomial can be squared by using the square of a binomial pattern:
Divide the perimeter to get the length of one side of the square.
Divide each term by 4:
Square this sidelength to get the area of the square. The binomial can be squared by using the square of a binomial pattern:
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A cube has surface area 6. Give the surface area of the sphere that is inscribed inside it.
A cube has surface area 6. Give the surface area of the sphere that is inscribed inside it.
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A cube with surface area 6 has six faces,each with area 1. As a result, each edge of the cube has length the square root of this, which is 1.
This is the diameter of the sphere inscribed in the cube, so the radius of the sphere is half this, or 
. Substitute this for 
in the formula for the surface area of a sphere:
,
the correct choice.
A cube with surface area 6 has six faces,each with area 1. As a result, each edge of the cube has length the square root of this, which is 1.
This is the diameter of the sphere inscribed in the cube, so the radius of the sphere is half this, or 
the correct choice.
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The graphic below shows a blueprint for a swimming pool.
If the pool is going to be 66 inches deep, how many cubic feet of water will it be able to hold? (1 ft = 12 in)
The graphic below shows a blueprint for a swimming pool.
If the pool is going to be 66 inches deep, how many cubic feet of water will it be able to hold? (1 ft = 12 in)
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Notice that the outer dimensions of the blueprint are the dimensions for the entire pool, including the concrete, while the inner dimensions are for the part of the pool that will be filled with water. Therefore, we want to focus on just the inner dimensions.
Notice that the depth is given in inches, while the dimensions are in feet. Convert 66 inches to feet by dividing 66 by 12, since 12 inches makes a foot:
The inch units cancel out and leave us with just the feet units. 66 in is 5.5 ft.
Now we have all of the information we need to solve for the volume of the pool. The pool is a rectangular prism, and the formula for volume of a rectangular prism is
(In this case, the "height" of the swimming pool is its depth.)
The blueprint shows that the pool is 40 ft long and 30 ft wide. Plugging in the measurements from the problem, we get
Multiplying this out, we get 
.
Notice that the outer dimensions of the blueprint are the dimensions for the entire pool, including the concrete, while the inner dimensions are for the part of the pool that will be filled with water. Therefore, we want to focus on just the inner dimensions.
Notice that the depth is given in inches, while the dimensions are in feet. Convert 66 inches to feet by dividing 66 by 12, since 12 inches makes a foot:
The inch units cancel out and leave us with just the feet units. 66 in is 5.5 ft.
Now we have all of the information we need to solve for the volume of the pool. The pool is a rectangular prism, and the formula for volume of a rectangular prism is
(In this case, the "height" of the swimming pool is its depth.)
The blueprint shows that the pool is 40 ft long and 30 ft wide. Plugging in the measurements from the problem, we get
Multiplying this out, we get 
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Given: 
and 
such that 
and 
.
Does sufficient information exist to prove that 
, and if so, by what postulate or theorem?
Given: 
Does sufficient information exist to prove that
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We are given that, between the triangles, two pairs of corresponding sides are proportional, and that a pair of corresponding angles are congruent. The angles that are congruent are the included angles of their respective sides. By the SAS Similarity Postulate, this is enough to prove that 
.
We are given that, between the triangles, two pairs of corresponding sides are proportional, and that a pair of corresponding angles are congruent. The angles that are congruent are the included angles of their respective sides. By the SAS Similarity Postulate, this is enough to prove that 
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Given: 
and 
such that 
and 
.
Does sufficient information exist to prove that 
, and if so, by what postulate or theorem?
Given: 
Does sufficient information exist to prove that 
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We are given that, between the triangles, two pairs of corresponding angles are congruent. By the AA Similarity Postulate, this is enough to prove that 
.
We are given that, between the triangles, two pairs of corresponding angles are congruent. By the AA Similarity Postulate, this is enough to prove that 
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