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Common Core: High School - Geometry : High School: Geometry

Study concepts, example questions & explanations for Common Core: High School - Geometry

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6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #5 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Find the perimeter of the following polygon with the points (-4, 11) , (-12, -12) , (4, 7) , and (1, 9) to two decimal places.

Possible Answers:

P = 58.18

P = 48.18

P = 53.33

P = 106.6

P = 59.18

Correct answer:

P = 58.18

Explanation:

In order to find the perimeter, we need to find the sum of the edges of the polygon.

In order to do this, we need to find the distance between the points.

Let's recall the distance formula.

$D = \sqrt{\left(x_{0} - x_{1}\right)^{2} + \left(y_{0} - y_{1}\right)^{2}}$

Lets say that

$\\p_{1} = \left ( -4, \quad 11\right )\\ p_{2} = \left ( -12, \quad -12\right ) \\p_{3} = \left ( 4, \quad 7\right ) \\ p_{4} = \left ( 1, \quad 9\right )$

Here is a picture of the polygon

Plot5

Let's first find the distance between $p_{1}$ and $p_{2}$

$\\D = \sqrt{\left(-4- -12\right)^{2} + \left(11- - 12\right)^{2}} \\D = \sqrt{\left(8)^{2} + \left(23\right)^{2}}$

$\\D = \sqrt{64 + 529}$

$D{\left (p_{1},p_{2} \right )} = \sqrt{593}$

Now let's find the distance between $p_{2}$ and $p_{3}$

$\\D = \sqrt{\left(-12-4\right)^{2} + \left(- 12-7\right)^{2}} \\D = \sqrt{\left(-16)^{2} + \left(-19\right)^{2}}$

$\\D = \sqrt{256 + 361}$

$D{\left (p_{2},p_{3} \right )} = \sqrt{617}$

Now let's find the distance between $p_{3}$ and $p_{4}$

$\\D = \sqrt{\left(4-1\right)^{2} + \left(7-9\right)^{2}} \\D = \sqrt{\left(3)^{2} + \left(-2\right)^{2}}$

$\\D = \sqrt{9 + 4}$

$D{\left (p_{3},p_{4} \right )} = \sqrt{13}$

Now let's find the distance between $p_{4}$ and $p_{1}$

$\\D = \sqrt{\left(1--4\right)^{2} + \left(9-11\right)^{2}} \\D = \sqrt{\left(5)^{2} + \left(-2\right)^{2}}$

$\\D = \sqrt{25 + 4}$

$D{\left (p_{4},p_{1} \right )} = \sqrt{29}$

Now since we have the distances, we can simply sum up each edge to get the perimeter.

$\\P = \sqrt{13} + \sqrt{29} + \sqrt{593} + \sqrt{617} \\P = 58.18$

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Example Question #6 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Find the perimeter of the following polygon with the points (0, -7) , (-6, -2) , (-5, -5) , and (-1, 1) to two decimal places.

Possible Answers:

P = 18.99

P = 27.25

P = 26.25

P = 16.25

P = 98.81

Correct answer:

P = 26.25

Explanation:

In order to find the perimeter, we need to find the sum of the edges of the polygon.

In order to do this, we need to find the distance between the points.

Let's recall the distance formula.

$D = \sqrt{\left(x_{0} - x_{1}\right)^{2} + \left(y_{0} - y_{1}\right)^{2}}$

Lets say that

$\\p_{1} = \left ( 0, \quad -7\right )\\ p_{2} = \left ( -6, \quad -2\right ) \\p_{3} = \left ( -5, \quad -5\right ) \\ p_{4} = \left ( -1, \quad 1\right )$

Here is a picture of the polygon

Plot6

Let's first find the distance between $p_{1}$ and $p_{2}$

$\\D = \sqrt{\left(0--6\right)^{2} + \left(-7--2}\right)^{2}} \\D = \sqrt{\left(6\right)^{2} + \left(-5}\right)^{2}} \\D = \sqrt{36 + 25}$

$D{\left (p_{1},p_{2} \right )} = \sqrt{61}$

Now let's find the distance between $p_{2}$ and $p_{3}$

$\\D = \sqrt{\left(-6--5\right)^{2} + \left(-2--5}\right)^{2}} \\D = \sqrt{\left(-1\right)^{2} + \left(3}\right)^{2}} \\D = \sqrt{1 + 9}$

$D{\left (p_{2},p_{3} \right )} = \sqrt{10}$

Now let's find the distance between $p_{3}$ and $p_{4}$

$\\D = \sqrt{\left(-5--1\right)^{2} + \left(-5-1}\right)^{2}} \\D = \sqrt{\left(-4\right)^{2} + \left(-6}\right)^{2}} \\D = \sqrt{16 + 36}$

$D{\left (p_{3},p_{4} \right )} = 2 \sqrt{13}$

Now let's find the distance between $p_{4}$ and $p_{1}$

$\\D = \sqrt{\left(-1-0\right)^{2} + \left(1--7}\right)^{2}} \\D = \sqrt{\left(-1\right)^{2} + \left(8}\right)^{2}} \\D = \sqrt{1 + 64}$

$D{\left (p_{4},p_{1} \right )} = \sqrt{65}$

Now since we have the distances, we can simply sum up each edge to get the perimeter.

$\\P = \sqrt{10} + 2 \sqrt{13} + \sqrt{61} + \sqrt{65} \\P = 26.25$

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Example Question #7 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Find the perimeter of the following polygon with the points (-7, 12) , (9, 7) , (-14, 5) , and (1, 9) to two decimal places.

Possible Answers:

P = 63.92

P = 140.8

P = 56.23

P = 53.92

P = 64.92

Correct answer:

P = 63.92

Explanation:

In order to find the perimeter, we need to find the sum of the edges of the polygon.

In order to do this, we need to find the distance between the points.

Let's recall the distance formula.

$D = \sqrt{\left(x_{0} - x_{1}\right)^{2} + \left(y_{0} - y_{1}\right)^{2}}$

Lets say that

$\\p_{1} = \left ( -7, \quad 12\right ) \\p_{2} = \left ( 9, \quad 7\right ) \\p_{3} = \left ( -14, \quad 5\right ) \\ p_{4} = \left ( 1, \quad 9\right )$

Here is a picture of the polygon

Plot7

Let's first find the distance between $p_{1}$ and $p_{2}$

$\\D = \sqrt{\left(-7 - 9\right)^{2} + \left(12 - 7\right)^{2}} \\D= \sqrt{\left(-16\right)^{2} + \left(5\right)^{2}} \\D= \sqrt{256 + 25}$

$D{\left (p_{1},p_{2} \right )} = \sqrt{281}$

Now let's find the distance between $p_{2}$ and $p_{3}$

$\\D = \sqrt{\left( 9--14\right)^{2} + \left( 7-5\right)^{2}} \\D= \sqrt{\left(23\right)^{2} + \left(2\right)^{2}} \\D= \sqrt{529 + 4}$

$D{\left (p_{2},p_{3} \right )} = \sqrt{533}$

Now let's find the distance between $p_{3}$ and $p_{4}$

$\\D = \sqrt{\left( -14-1\right)^{2} + \left( 5-9\right)^{2}} \\D= \sqrt{\left(-15\right)^{2} + \left(-4\right)^{2}} \\D= \sqrt{225 + 16}$

$D{\left (p_{3},p_{4} \right )} = \sqrt{241}$

Now let's find the distance between $p_{4}$ and $p_{1}$

$\\D = \sqrt{\left( 1--7\right)^{2} + \left( 9-12\right)^{2}} \\D= \sqrt{\left(8\right)^{2} + \left(-3\right)^{2}} \\D= \sqrt{64 + 9}$

$D{\left (p_{4},p_{1} \right )} = \sqrt{73}$

Now since we have the distances, we can simply sum up each edge to get the perimeter.

$\\P = \sqrt{73} + \sqrt{241} + \sqrt{281} + \sqrt{533} \\P = 63.92$

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Example Question #8 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Find the perimeter of the following polygon with the points (2, 1) , (19, -7) , (11, 11) , and (4, 8) to two decimal places.

Possible Answers:

P = 46.83

P = 43.38

P = 54.38

P = 118.9

P = 53.38

Correct answer:

P = 53.38

Explanation:

In order to find the perimeter, we need to find the sum of the edges of the polygon.

In order to do this, we need to find the distance between the points.

Let's recall the distance formula.

$D = \sqrt{\left(x_{0} - x_{1}\right)^{2} + \left(y_{0} - y_{1}\right)^{2}}$

Lets say that

$\\p_{1} = \left ( 2, \quad 1\right ) \\p_{2} = \left ( 19, \quad -7\right ) \\p_{3} = \left ( 11, \quad 11\right ) \\ p_{4} = \left ( 4, \quad 8\right )$

Here is a picture of the polygon

Plot8

Let's first find the distance between $p_{1}$ and $p_{2}$

$\\D = \sqrt{\left(2 - 19\right)^{2} + \left(1- - 7\right)^{2}} \\D= \sqrt{\left(-17\right)^{2} + \left(8\right)^{2}} \\D= \sqrt{289 + 64}$

$D{\left (p_{1},p_{2} \right )} = \sqrt{353}$

Now let's find the distance between $p_{2}$ and $p_{3}$

$\\D = \sqrt{\left( 19-11\right)^{2} + \left( -7-11\right)^{2}} \\D= \sqrt{\left(8\right)^{2} + \left(-18\right)^{2}} \\D= \sqrt{64 + 324}\\D=\sqrt{388}$

$D{\left (p_{2},p_{3} \right )} = 2 \sqrt{97}$

Now let's find the distance between $p_{3}$ and $p_{4}$

$\\D = \sqrt{\left( 11-4\right)^{2} + \left( 11-8\right)^{2}} \\D= \sqrt{\left(7\right)^{2} + \left(3\right)^{2}} \\D= \sqrt{49+ 9}$

$D{\left (p_{3},p_{4} \right )} = \sqrt{58}$

Now let's find the distance between $p_{4}$ and $p_{1}$

$\\D = \sqrt{\left( 4-2\right)^{2} + \left( 8-1\right)^{2}} \\D= \sqrt{\left(2\right)^{2} + \left(7\right)^{2}} \\D= \sqrt{4 + 49}$

$D{\left (p_{4},p_{1} \right )} = \sqrt{53}$

Now since we have the distances, we can simply sum up each edge to get the perimeter.

$\\P = \sqrt{53} + \sqrt{58} + \sqrt{353} + 2 \sqrt{97} \\P = 53.38$

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Example Question #9 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Find the perimeter of the following polygon with the points (-9, -10) , (0, -19) , (11, -6) , and (0, -6) to two decimal places.

Possible Answers:

P = 50.61

P = 40.61

P = 139.2

P = 51.61

P = 41.74

Correct answer:

P = 50.61

Explanation:

In order to find the perimeter, we need to find the sum of the edges of the polygon.

In order to do this, we need to find the distance between the points.

Let's recall the distance formula.

$D = \sqrt{\left(x_{0} - x_{1}\right)^{2} + \left(y_{0} - y_{1}\right)^{2}}$

Lets say that

$\\p_{1} = \left ( -9, \quad -10\right ) \\p_{2} = \left ( 0, \quad -19\right ) \\p_{3} = \left ( 11, \quad -6\right ) \\ p_{4} = \left ( 0, \quad -6\right )$

Here is a picture of the polygon

Plot9

Let's first find the distance between $p_{1}$ and $p_{2}$

$\\D = \sqrt{\left(-9 - 0\right)^{2} + \left(-10--19\right)^{2}} \\D=\sqrt{\left(-9\right)^{2} + \left(9\right)^{2}} \\D=\sqrt{81 + 81}$

$D{\left (p_{1},p_{2} \right )} = 9 \sqrt{2}$

Now let's find the distance between $p_{2}$ and $p_{3}$

$\\D = \sqrt{\left(0-11\right)^{2} + \left(-19--6\right)^{2}} \\D=\sqrt{\left(-11\right)^{2} + \left(-13\right)^{2}} \\D=\sqrt{121 + 169}$

$D{\left (p_{2},p_{3} \right )} = \sqrt{290}$

Now let's find the distance between $p_{3}$ and $p_{4}$

$\\D = \sqrt{\left(11-0\right)^{2} + \left(-6--6\right)^{2}} \\D=\sqrt{\left(11\right)^{2} + \left(0\right)^{2}} \\D=\sqrt{121 + 0}$

$D{\left (p_{3},p_{4} \right )} = 11$

Now let's find the distance between $p_{4}$ and $p_{1}$

$\\D = \sqrt{\left(0--9\right)^{2} + \left(-6--10\right)^{2}} \\D=\sqrt{\left(9\right)^{2} + \left(4\right)^{2}} \\D=\sqrt{81 + 16}$

$D{\left (p_{4},p_{1} \right )} = \sqrt{97}$

Now since we have the distances, we can simply sum up each edge to get the perimeter.

$\\P = \sqrt{97} + 11 + 9 \sqrt{2} + \sqrt{290} \\P = 50.61$

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Example Question #10 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Find the perimeter of the following polygon with the points (-1, 13) , (-3, -2) , (1, -7) , and (-2, -9) to two decimal places.

Possible Answers:

P = 245.4

P = 37.16

P = 48.16

P = 47.16

P = 27.34

Correct answer:

P = 47.16

Explanation:

In order to find the perimeter, we need to find the sum of the edges of the polygon.

In order to do this, we need to find the distance between the points.

Let's recall the distance formula.

$D = \sqrt{\left(x_{0} - x_{1}\right)^{2} + \left(y_{0} - y_{1}\right)^{2}}$

Lets say that

$\\p_{1} = \left ( -1, \quad 13\right ) \\p_{2} = \left ( -3, \quad -2\right ) \\p_{3} = \left ( 1, \quad -7\right ) \\ p_{4} = \left ( -2, \quad -9\right )$

Here is a picture of the polygon

Plot10

Let's first find the distance between $p_{1}$ and $p_{2}$

$\\D = \sqrt{\left(-1--3\right)^{2} + \left(13--2\right)^{2}} \\D=\sqrt{\left(2\right)^{2} + \left(15\right)^{2}} \\D=\sqrt{4+225 }$

$D{\left (p_{1},p_{2} \right )} = \sqrt{229}$

Now let's find the distance between $p_{2}$ and $p_{3}$

$\\D = \sqrt{\left(-3-1\right)^{2} + \left(-2--7\right)^{2}} \\D=\sqrt{\left(-4\right)^{2} + \left(5\right)^{2}} \\D=\sqrt{16+25}$

$D{\left (p_{2},p_{3} \right )} = \sqrt{41}$

Now let's find the distance between $p_{3}$ and $p_{4}$

$\\D = \sqrt{\left(1--2\right)^{2} + \left(-7--9\right)^{2}} \\D=\sqrt{\left(3\right)^{2} + \left(2\right)^{2}} \\D=\sqrt{9+4 }$

$D{\left (p_{3},p_{4} \right )} = \sqrt{13}$

Now let's find the distance between $p_{4}$ and $p_{1}$

$\\D = \sqrt{\left(-2--1\right)^{2} + \left(-9-13\right)^{2}} \\D=\sqrt{\left(-1\right)^{2} + \left(-22\right)^{2}} \\D=\sqrt{1+484 }$

$D{\left (p_{4},p_{1} \right )} = \sqrt{485}$

Now since we have the distances, we can simply sum up each edge to get the perimeter.

$\\P = \sqrt{13} + \sqrt{41} + \sqrt{229} + \sqrt{485} \\P = 47.16$

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Example Question #11 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Find the perimeter of the following polygon with the points (8, 2) , (0, -10) , (-11, 7) , and (0, -14) to two decimal places.

Possible Answers:

P = 76.27

P = 60.17

P = 66.27

P = 77.27

P = 237.3

Correct answer:

P = 76.27

Explanation:

In order to find the perimeter, we need to find the sum of the edges of the polygon.

In order to do this, we need to find the distance between the points.

Let's recall the distance formula.

$D = \sqrt{\left(x_{0} - x_{1}\right)^{2} + \left(y_{0} - y_{1}\right)^{2}}$

Lets say that

$\\p_{1} = \left ( 8, \quad 2\right ) \\p_{2} = \left ( 0, \quad -10\right ) \\p_{3} = \left ( -11, \quad 7\right ) \\ p_{4} = \left ( 0, \quad -14\right )$

Here is a picture of the polygon

Plot11

Let's first find the distance between $p_{1}$ and $p_{2}$

$\\D = \sqrt{\left(8-0\right)^{2} + \left(2--10\right)^{2}} \\D=\sqrt{\left(8\right)^{2} + \left(12\right)^{2}} \\D=\sqrt{64+144}$

$D{\left (p_{1},p_{2} \right )} = 4 \sqrt{13}$

Now let's find the distance between $p_{2}$ and $p_{3}$

$\\D = \sqrt{\left(0--11\right)^{2} + \left(-10-7\right)^{2}} \\D=\sqrt{\left(11\right)^{2} + \left(-17\right)^{2}} \\D=\sqrt{121+289}$

$D{\left (p_{2},p_{3} \right )} = \sqrt{410}$

Now let's find the distance between $p_{3}$ and $p_{4}$

$\\D = \sqrt{\left(-11-0\right)^{2} + \left(7--14\right)^{2}} \\D=\sqrt{\left(-11\right)^{2} + \left(21\right)^{2}} \\D=\sqrt{121+441}$

$D{\left (p_{3},p_{4} \right )} = \sqrt{562}$

Now let's find the distance between $p_{4}$ and $p_{1}$

$\\D = \sqrt{\left(0-8\right)^{2} + \left(-14-2\right)^{2}} \\D=\sqrt{\left(-8\right)^{2} + \left(-16\right)^{2}} \\D=\sqrt{64+256}$

$D{\left (p_{4},p_{1} \right )} = 8 \sqrt{5}$

Now since we have the distances, we can simply sum up each edge to get the perimeter.

$\\P = 4 \sqrt{13} + 8 \sqrt{5} + \sqrt{410} + \sqrt{562} \\P = 76.27$

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Example Question #12 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Find the perimeter of the following polygon with the points (7, 2) , (-4, -18) , (8, 4) , and (-3, -8) to two decimal places.

Possible Answers:

P = 79.31

P = 78.31

P = 65.58

P = 205.6

P = 68.31

Correct answer:

P = 78.31

Explanation:

In order to find the perimeter, we need to find the sum of the edges of the polygon.

In order to do this, we need to find the distance between the points.

Let's recall the distance formula.

$D = \sqrt{\left(x_{0} - x_{1}\right)^{2} + \left(y_{0} - y_{1}\right)^{2}}$

Lets say that

$\\p_{1} = \left ( 7, \quad 2\right ) \\p_{2} = \left ( -4, \quad -18\right ) \\p_{3} = \left ( 8, \quad 4\right ) \\p_{4} = \left ( -3, \quad -8\right )$

Here is a picture of the polygon

Plot12

Let's first find the distance between $p_{1}$ and $p_{2}$

$\\D = \sqrt{\left(7--4\right)^{2} + \left(2--18\right)^{2}} \\D=\sqrt{\left(11\right)^{2} + \left(20\right)^{2}} \\D=\sqrt{121+400 }$

$D{\left (p_{1},p_{2} \right )} = \sqrt{521}$

Now let's find the distance between $p_{2}$ and $p_{3}$

$\\D = \sqrt{\left(-4-8\right)^{2} + \left(-18-4\right)^{2}} \\D=\sqrt{\left(-12\right)^{2} + \left(-22\right)^{2}} \\D=\sqrt{144+484 }$

$D{\left (p_{2},p_{3} \right )} = 2 \sqrt{157}$

Now let's find the distance between $p_{3}$ and $p_{4}$

$\\D = \sqrt{\left(8--3\right)^{2} + \left(4--8\right)^{2}} \\D=\sqrt{\left(11\right)^{2} + \left12(\right)^{2}} \\D=\sqrt{121+144 }$

$D{\left (p_{3},p_{4} \right )} = \sqrt{265}$

Now let's find the distance between $p_{4}$ and $p_{1}$

$\\D = \sqrt{\left(-3-7\right)^{2} + \left(-8-2\right)^{2}} \\D=\sqrt{\left(-10\right)^{2} + \left(-10\right)^{2}} \\D=\sqrt{100+100 }$

$D{\left (p_{4},p_{1} \right )} = 10 \sqrt{2}$

Now since we have the distances, we can simply sum up each edge to get the perimeter.

$\\P = 10 \sqrt{2} + \sqrt{265} + \sqrt{521} + 2 \sqrt{157} \\P = 78.31$

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Example Question #1 : Circumference And Area Of A Circle, Volume Of A Cylinder, Pyramid, And Cone Formulas: Ccss.Math.Content.Hsg Gmd.A.1

If a circle has a circumference of what is the radius?

Possible Answers:

$\frac{1089}{4 \pi^{2}}$

$\frac{33}{2 \pi}$

$\frac{33}{4 \pi}$

Correct answer:

$\frac{33}{2 \pi}$

Explanation:

In order to find the radius of a circle, we need to recall the equation that involves both the radius and circumference.

$C = 2 \pi r$

Since we are given the circumference, we simply substitute 33 for $\uptext{C}$ , and solve for $\uptext{r}$ .

$33 = 2 \pi r$

Divide by $2 \pi$ on each side to get

$r = \frac{33}{2 \pi}$

Thus the radius is

$\frac{33}{2 \pi}$

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Example Question #2 : Circumference And Area Of A Circle, Volume Of A Cylinder, Pyramid, And Cone Formulas: Ccss.Math.Content.Hsg Gmd.A.1

If a cylinder has a radius of and a height of what is the volume?

Possible Answers:

$50 \pi$

375

$25 \pi^{2}$

5625

$5625 \pi$

Correct answer:

$5625 \pi$

Explanation:

In order to find the volume, we need to recall the equation for the volume of a cylinder.

$V = \pi h r^{2}$

Since we are given the radius, and the height, we can simply plug in those values into the equation.

$\\r=15\rightarrow r^2=225 \\h=25$

$V = 5625 \pi$

Thus the volume is

$5625 \pi$

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Common Core: High School - Geometry : High School: Geometry

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

Example Question #5 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Example Question #6 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Example Question #7 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Example Question #8 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Example Question #9 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Example Question #10 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Example Question #11 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Example Question #12 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Example Question #1 : Circumference And Area Of A Circle, Volume Of A Cylinder, Pyramid, And Cone Formulas: Ccss.Math.Content.Hsg Gmd.A.1

Example Question #2 : Circumference And Area Of A Circle, Volume Of A Cylinder, Pyramid, And Cone Formulas: Ccss.Math.Content.Hsg Gmd.A.1

All Common Core: High School - Geometry Resources

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