Measurement and Geometry - HiSET

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Question

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?

Answer

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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Question

Consider regular Hexagon .

On this hexagon, perform the translation $F\rightarrow B$ . Then perform a $120^{\circ }$ 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?

Answer

The translation $F\rightarrow B$ 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 $120 ^{\circ }$ - 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.

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Question

Consider regular Hexagon .

On this hexagon, perform the translation $F\rightarrow C$ . Then perform a $180^{\circ }$ 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?

Answer

The translation $F\rightarrow C$ 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 $180 ^{\circ }$ - 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.

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Question

Consider regular Hexagon .

On this hexagon, perform the translation $A\rightarrow B$ . Then perform a $120^{\circ }$ 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 ?

Answer

The translation $A\rightarrow B$ 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 $120 ^{\circ }$ - 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

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Question

Consider regular Hexagon .

On this hexagon, perform the translation $A\rightarrow B$ . Then reflect the hexagon about $\overleftrightarrow{BD}$ . Let be the image of under these transformations, and so forth.

Which point on Hexagon is the image of under these transformations?

Answer

The translation $A\rightarrow B$ 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 $\overleftrightarrow{BD}$ , 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 .

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Question

Translate the graph of the equation

left four units and down six units. Give the equation of the image.

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 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.

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Question

Translate the graph of the equation

$x^{2}+ y^{2}= 100$

right two units and up five units. Give the equation of the image.

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

$x^{2}+ y^{2}= 100$

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

$(x-2)^{2}+( y-5)^{2}= 100$

This can be rewritten by applying the binomial square pattern as follows:

$x^{2}- 2 (x) (2)+ 2^{2}+ y^{2}- 2 (5)(x)+ 5^{2} = 100$

$x^{2}- 4x+4+ y^{2}-10y+2 5 = 100$

Collect like terms; the equation becomes

$x^{2}- 4x + y^{2}-10y+29 = 100$

Subtract 100 from both sides:

$x^{2}- 4x + y^{2}-10y+29- 100 = 100 - 100$

$x^{2}- 4x + y^{2}-10y-71=0$

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Question

Translate the graph of the equation

$y = x^{2}+ 5x - 7$

right four units and down two units. Give the equation of the image.

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

$y = x^{2}+ 5x - 7$

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

$y+2 = (x-4)^{2}+ 5(x-4) - 7$

The expression at right can be simplified. First, use the distributive property on the middle expression:

$y+2 = (x-4)^{2}+ 5x-5 (4) - 7$

$y+2 = (x-4)^{2}+ 5x-20 - 7$

Now, simplify the first expression by using the binomial square pattern:

$y+2 = x^{2} -2 (x)(4) +4^{2}+ 5x-20 - 7$

$y+2 = x^{2} -8x+16+ 5x-20 - 7$

Collect like terms on the right:

$y+2 = x^{2} -8x+ 5x+16 -20 - 7$

$y+2 = x^{2} -3x-11$

Subtract 2 from both sides:

$y+2 -2= x^{2} -3x-11-2$

$y= x^{2} -3x-13$ ,

the equation of the image.

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Question

Translate the graph of the equation

left three units and down five units. Give the equation of the image.

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:

$y+5 = | 3x+3 \cdot 3 - 8 |$

Collect like terms:

Subtract 5 from both sides:

,

the correct equation of the image.

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Question

On the coordinate plane, let , , and be located at the origin, , and . Construct the median of $\bigtriangleup OAB$ from and let the foot of the median be . On the triangle, perform the translation $A \rightarrow M$ . Where is the image of ?

Answer

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 $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is located at

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$ ,

so, substituting the coordinates of and in the formula, we see that is

$\left ( \frac{0+6}{2}, \frac{0+4}{2} \right )$ , or .

See the figure below:

To perform the translation $A \rightarrow M$ , or, equivalently,

$(8, 0) \rightarrow (3,2)$ ,

on a point, it is necessary to add

and

to the - and - coordinates, respectively. Therefore, the image of is located at

,

or

.

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Question

On the coordinate plane, let and be located at and , respectively. Let be the midpoint of $\overline{AB}$ and let be the midpoint of $\overline{AM}$ . On the segment, perform the translation $N \rightarrow B$ . Where is the image of located?

Answer

The midpoint of a segment with endpoints at $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is located at

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

Substitute the coordinates of and in this formula to find that midpoint of $\overline{AB}$ is located at

$\left ( \frac{3+9}{2}, \frac{9+(-5)}{2} \right )$ , or .

Substitute the coordinates of and to find that midpoint of $\overline{AM}$ is located at

$\left ( \frac{3+6}{2}, \frac{9+2}{2} \right )$ , or $\left ( \frac{9}{2}, \frac{11}{2} \right )$ .

To perform the translation $N \rightarrow B$ , or, equivalently,

$\left ( \frac{9}{2}, \frac{11}{2} \right ) \rightarrow (9, -5)$ ,

on a point, it is necessary to add

$9 - \frac{9}{2} = \frac{18}{2} - \frac{9}{2} = \frac{9}{2}$

to its -coordinate, and

$-5 - \frac{11}{2} = - \frac{10}{2} - \frac{11}{2} = - \frac{21}{2}$

to its -coordinate.

Therefore, the -coordinate of the image of under this translation is

$6+ \frac{9}{2} = \frac{12}{2}+ \frac{9}{2} = \frac{21}{2}$ ;

its -coordinate is

$2 + \left (- \frac{21}{2} \right ) = \frac{4}{2}- \frac{21}{2}=-\frac{17}{2}$

The image of is located at $\left (\frac{21}{2},-\frac{17}{2} \right )$ .

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Question

Find the area of a square with the following side length:

Answer

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.

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Question

The perimeter of a square is . In terms of , give the area of the square.

Answer

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 $\frac{1}{4}P$ . The area of the square is equal to the square of this sidelength, or

$\left (\frac{1}{4}P \right )^{2} = \left (\frac{1}{4} \right )^{2} P^{2} = \frac{1}{16}P^{2}$ .

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Question

The volume of a sphere is equal to $108 \pi$ . Give the surface area of the sphere.

Answer

The volume of a sphere can be calculated using the formula

$V= \frac{4}{3} \pi r^{3}$

Solving for :

Set $V = 108 \pi$ . Multiply both sides by $\frac{3}{4}$ :

$\frac{3}{4} \cdot \frac{4}{3} \pi r^{3} = \frac{3}{4} \cdot 108 \pi$

$\pi r^{3} = 81\pi$

Divide by $\pi$ :

$\frac{\pi r^{3} }{\pi}= \frac{81\pi}{\pi}$

$r^{3} = 81$

Take the cube root of both sides:

$r = \sqrt[3]{81} = \sqrt[3]{27(3)} = \sqrt[3]{27} \cdot \sqrt[3]{3} = 3 \sqrt[3]{3}$

Now substitute for in the surface area formula:

$A = 4 \pi r^{2}$

$A = 4 \pi (3 \sqrt[3]{3})^{2}= 4 \pi (3 )^{2}( \sqrt[3]{3})^{2} = 4 \pi (9) ( \sqrt[3]{3^{2}}) = 36 \pi \sqrt[3]{9}$ ,

the correct response.

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Question

Express the area of a square plot of land 60 feet in sidelength in square yards.

Answer

One yard is equal to three feet, so convert 60 feet to yards by dividing by conversion factor 3:

$60 \textup{ ft} \div 3 \textup{ ft / yd}= 20 \textup{ yd}$

Square this sidelength to get the area of the plot:

$20 \textup{ yd} \times 20 \textup{ yd} = 400 \textup{ yd} ^{2}$ ,

the correct response.

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Question

A square has perimeter . Give its area in terms of .

Answer

Divide the perimeter to get the length of one side of the square.

$s= \frac{P}{4}$

$s= \frac{12y+4}{4}$

Divide each term by 4:

$s= \frac{12y}{4}+ \frac{4}{4} = 3y+ 1$

Square this sidelength to get the area of the square. The binomial can be squared by using the square of a binomial pattern:

$A= s^{2}$

$=( 3y+ 1)^{2}$

$=( 3y )^{2}+ 2 (3y)(1)+ 1^{2}$

$= 3^{2}y ^{2}+ 2 (3 )(1)y+ 1^{2}$

$=9y ^{2}+ 6y+ 1$

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Question

A cube has surface area 6. Give the surface area of the sphere that is inscribed inside it.

Answer

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 $r= \frac{1}{2}$ . Substitute this for in the formula for the surface area of a sphere:

$A = 4\pi r^{2} = 4\pi \left ( \frac{1}{2} \right )^{2} = 4\pi \left ( \frac{1}{4} \right ) = \pi$ ,

the correct choice.

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Question

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)

Answer

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:

$66;in\cdot\frac{1 ft}{12 in}=5.5 ft$

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

$length\cdot width\cdot height$

(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

$40,ft\cdot30,ft\cdot5.5,ft$

Multiplying this out, we get .

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ such that $\frac{AB}{DE} = \frac{BC}{EF}$ and $\angle B \cong \angle E$ .

Does sufficient information exist to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , and if so, by what postulate or theorem?

Answer

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 $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

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Question

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ such that $\angle C \cong \angle F$ and $\angle A \cong \angle D$ .

Does sufficient information exist to prove that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , and if so, by what postulate or theorem?

Answer

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 $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

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