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Common Core: High School - Geometry : Expressing Geometric Properties with Equations

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

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All Common Core: High School - Geometry Resources

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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Common Core: High School - Geometry : Expressing Geometric Properties with Equations

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

All Common Core: High School - Geometry Resources

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