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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 #2 : Points That Partitions Segments In A Given Ratio: Ccss.Math.Content.Hsg Gpe.B.6

Find the midpoint of a line segment with endpoints of (10, 12) and (-4, -13) .

Possible Answers:

$\left ( 6, \quad -1\right )$

$\left ( 3.0, \quad -0.5\right )$

$\left ( 7.0, \quad 12.5\right )$

$\left ( -14, \quad -25\right )$

$\left ( -2.5, \quad -0.9230769230769231\right )$

Correct answer:

$\left ( 3.0, \quad -0.5\right )$

Explanation:

In order to find the midpoint between these end points, we need to recall the midpoint formula.

$\left ( x_{m} = \frac{x_{0}}{2} + \frac{x_{1}}{2}, \quad y_{m} = \frac{y_{0}}{2}+\frac{y_1}{2}\right )$

Where $x_{m}$ is the mid point for and $y_{m}$ is the midpoint for

Now we simply substitute values for $x_{0}$ , $x_{1}$ , $y_{0}$ , $y_{1}$

$\left ( x_{m} = 3, \quad y_{m} = - \frac{1}{2}\right )$

Now plot the points to verify the results.

Plot7

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Example Question #3 : Points That Partitions Segments In A Given Ratio: Ccss.Math.Content.Hsg Gpe.B.6

Find the midpoint of a line segment with endpoints of (-9, 2) and (-18, -8) .

Possible Answers:

$\left ( -9, \quad -10\right )$

$\left ( 0.5, \quad -0.25\right )$

$\left ( 4.5, \quad 5.0\right )$

$\left ( -27, \quad -6\right )$

$\left ( -13.5, \quad -3.0\right )$

Correct answer:

$\left ( -13.5, \quad -3.0\right )$

Explanation:

In order to find the midpoint between these end points, we need to recall the midpoint formula.

$\left ( x_{m} = \frac{x_{0}}{2} + \frac{x_{1}}{2}, \quad y_{m} = \frac{y_{0}}{2}+\frac{y_1}{2}\right )$

Where $x_{m}$ is the mid point for and $y_{m}$ is the midpoint for

Now we simply substitute values for $x_{0}$ , $x_{1}$ , $y_{0}$ , $y_{1}$

$\left ( x_{m} = - \frac{27}{2}, \quad y_{m} = -3\right )$

Now, plot the points to verify the result.

Plot9

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Example Question #11 : Points That Partitions Segments In A Given Ratio: Ccss.Math.Content.Hsg Gpe.B.6

Find the midpoint of a line segment with endpoints of (0, -5) and (-20, -9) .

Possible Answers:

$\left ( -20, \quad -14\right )$

$\left ( 10.0, \quad 2.0\right )$

$\left ( -10.0, \quad -7.0\right )$

$\left ( -0.0, \quad 0.5555555555555556\right )$

$\left ( -20, \quad -4\right )$

Correct answer:

$\left ( -10.0, \quad -7.0\right )$

Explanation:

In order to find the midpoint between these end points, we need to recall the midpoint formula.

$\left ( x_{m} = \frac{x_{0}}{2} + \frac{x_{1}}{2}, \quad y_{m} = \frac{y_{0}}{2}+\frac{y_1}{2}\right )$

Where $x_{m}$ is the mid point for and $y_{m}$ is the midpoint for

Now we simply substitute values for $x_{0}$ , $x_{1}$ , $y_{0}$ , $y_{1}$

$\left ( x_{m} = -10, \quad y_{m} = -7\right )$

Now plot the points to verify the result.

Plot10

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Example Question #11 : Points That Partitions Segments In A Given Ratio: Ccss.Math.Content.Hsg Gpe.B.6

Find the midpoint of a line segment with endpoints of (9, 9) and (-10, -4) .

Possible Answers:

$\left ( -19, \quad -13\right )$

$\left ( -1, \quad 5\right )$

$\left ( -0.9, \quad -2.25\right )$

$\left ( 9.5, \quad 6.5\right )$

$\left ( -0.5, \quad 2.5\right )$

Correct answer:

$\left ( -0.5, \quad 2.5\right )$

Explanation:

In order to find the midpoint between these end points, we need to recall the midpoint formula.

$\left ( x_{m} = \frac{x_{0}}{2} + \frac{x_{1}}{2}, \quad y_{m} = \frac{y_{0}}{2}+\frac{y_1}{2}\right )$

Where $x_{m}$ is the mid point for and $y_{m}$ is the midpoint for

Now we simply substitute values for $x_{0}$ , $x_{1}$ , $y_{0}$ , $y_{1}$

$\left ( x_{m} = - \frac{1}{2}, \quad y_{m} = \frac{5}{2}\right )$

From here, plot the points to verify the midpoint.

Plot11

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Example Question #11 : Points That Partitions Segments In A Given Ratio: Ccss.Math.Content.Hsg Gpe.B.6

Find the midpoint of a line segment with endpoints of (13, -9) and (-14, -14) .

Possible Answers:

$\left ( 13.5, \quad 2.5\right )$

$\left ( -27, \quad -5\right )$

$\left ( -1, \quad -23\right )$

$\left ( -0.5, \quad -11.5\right )$

$\left ( -0.9285714285714286, \quad 0.6428571428571429\right )$

Correct answer:

$\left ( -0.5, \quad -11.5\right )$

Explanation:

In order to find the midpoint between these end points, we need to recall the midpoint formula.

$\left ( x_{m} = \frac{x_{0}}{2} + \frac{x_{1}}{2}, \quad y_{m} = \frac{y_{0}}{2}+\frac{y_1}{2}\right )$

Where $x_{m}$ is the mid point for and $y_{m}$ is the midpoint for

Now we simply substitute values for $x_{0}$ , $x_{1}$ , $y_{0}$ , $y_{1}$

$\left ( x_{m} = - \frac{1}{2}, \quad y_{m} = - \frac{23}{2}\right )$

From here, plot the points to verify the result.

Plot12

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Example Question #351 : High School: Geometry

Find the midpoint of a line segment with endpoints of (-8, 5) and (-7, 3) .

Possible Answers:

$\left ( 1, \quad -2\right )$

$\left ( -0.5, \quad 1.0\right )$

$\left ( -7.5, \quad 4.0\right )$

$\left ( 1.1428571428571428, \quad 1.6666666666666667\right )$

$\left ( -15, \quad 8\right )$

Correct answer:

$\left ( -7.5, \quad 4.0\right )$

Explanation:

In order to find the midpoint between these end points, we need to recall the midpoint formula.

$\left ( x_{m} = \frac{x_{0}}{2} + \frac{x_{1}}{2}, \quad y_{m} = \frac{y_{0}}{2}+\frac{y_1}{2}\right )$

Where $x_{m}$ is the mid point for and $y_{m}$ is the midpoint for

Now we simply substitute values for $x_{0}$ , $x_{1}$ , $y_{0}$ , $y_{1}$

$\left ( x_{m} = - \frac{15}{2}, \quad y_{m} = 4\right )$

Now, plot the points to verify the results.

Plot4

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Example Question #1 : 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 (10, 14) , (7, -20) , ( 1, 8) and (1, -15) to two decimal places.

Possible Answers:

P = 116.1

P = 389.4

P = 117.1

P = 88.8

P = 106.1

Correct answer:

P = 116.1

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

$\\p_{1} = \left ( 10, \quad 14\right ) \\p_{2} = \left ( 7, \quad -20\right ) \\p_{3} = \left ( 1, \quad 8\right ) \\p_{4} = \left ( 1, \quad -15\right )$

Here is a picture of the polygon

Plot1

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

$\\D = \sqrt{\left(10 - 7\right)^{2} + \left(14- - 20\right)^{2}} \\D=\sqrt{\left(3\right)^{2} + \left(34\right)^{2}} \\D=\sqrt{9 + 1156}$

$D{\left (p_{1},p_{2} \right )} = \sqrt{1165}$

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

$\\D = \sqrt{\left(7 - 1\right)^{2} + \left(-20 - 8\right)^{2}} \\D=\sqrt{\left(6\right)^{2} + \left(-28\right)^{2}} \\D=\sqrt{36 + 784}\\D=\sqrt{820}$

$D{\left (p_{2},p_{3} \right )} = 2 \sqrt{205}$

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

$\\D = \sqrt{\left(1 - 1\right)^{2} + \left(8- - 15\right)^{2}} \\D=\sqrt{\left(0\right)^{2} + \left(23\right)^{2}} \\D=\sqrt{0 + 529}$

$D{\left (p_{3},p_{4} \right )} = 23$

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

$\\D = \sqrt{\left(1 - 10\right)^{2} + \left(-15 - 14\right)^{2}} \\D=\sqrt{\left(-9\right)^{2} + \left(-29\right)^{2}} \\D=\sqrt{81 + 841}$

$D{\left (p_{4},p_{1} \right )} = \sqrt{922}$

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

$\\P = 23 + 2 \sqrt{205} + \sqrt{922} + \sqrt{1165} \\P = 116.1$

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Example Question #351 : High School: Geometry

Find the perimeter of the following polygon with the points (0, 12) , (1, 4) , (-14, -4) and (-1, -19) to two decimal places.

Possible Answers:

P = 48.01

P = 75.93

P = 355.1

P = 65.93

P = 76.93

Correct answer:

P = 75.93

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

$\\p_{1} = \left ( 0, \quad 12\right ) \\p_{2} = \left ( 1, \quad 4\right ) \\p_{3} = \left ( -14, \quad -4\right ) \\ p_{4} = \left ( -1, \quad -19\right )$

Here is a picture of the polygon

Plot2

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

$\\D = \sqrt{(0) - 1)^{2} + \left(12 - 4\right)^{2}} \\D=\sqrt{1+ \left(8\right)^{2}} \\D=\sqrt{1+64}$

$D{\left (p_{1},p_{2} \right )} = \sqrt{65}$

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

$\\D = \sqrt{(1) --14)^{2} + \left(4- - 4\right)^{2}} \\D=\sqrt{15^2+ \left(8\right)^{2}} \\D=\sqrt{225+64}\\D=\sqrt{289}$

$D{\left (p_{2},p_{3} \right )} = 17$

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

$\\D = \sqrt{(-14) -- 1)^{2} + \left(-4 - -19\right)^{2}} \\D=\sqrt{(-13)^2+ \left(15\right)^{2}} \\D=\sqrt{169+225}$

$D{\left (p_{3},p_{4} \right )} = \sqrt{394}$

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

$\\D = \sqrt{(-1) - 0)^{2} + \left(-19 - 12\right)^{2}} \\D=\sqrt{1+ \left(-31\right)^{2}} \\D=\sqrt{1+961}$

$D{\left (p_{4},p_{1} \right )} = \sqrt{962}$

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

$\\P = \sqrt{65} + 17 + \sqrt{394} + \sqrt{962} \\P = 75.93$

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Example Question #2 : 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, 13) , (5, -18) , (-6, -6) , and (-2, -5) to two decimal places.

Possible Answers:

P = 257.5

P = 72.14

P = 62.14

P = 73.14

P = 53.61

Correct answer:

P = 72.14

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

$\\p_{1} = \left ( 8, \quad 13\right ) \\p_{2} = \left ( 5, \quad -18\right ) \\p_{3} = \left ( -6, \quad -6\right ) \\p_{4} = \left ( -2, \quad -5\right )$

Here is a picture of the polygon

Plot3

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

$\\D = \sqrt{\left(8 - 5\right)^{2} + \left(13- - 18\right)^{2}} \\D=\sqrt{\left(3\right)^{2} + \left(31\right)^{2}} \\D=\sqrt{9+ 961}$

$D{\left (p_{1},p_{2} \right )} = \sqrt{970}$

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

$\\D = \sqrt{\left(5- -6\right)^{2} + \left(-18- - 6\right)^{2}} \\D=\sqrt{\left(11\right)^{2} + \left(-12\right)^{2}} \\D=\sqrt{121+144}$

$D{\left (p_{2},p_{3} \right )} = \sqrt{265}$

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

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

$D{\left (p_{3},p_{4} \right )} = \sqrt{17}$

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

$\\D = \sqrt{\left(-2- 8\right)^{2} + \left(-5- 13\right)^{2}} \\D=\sqrt{\left(-10\right)^{2} + \left(-18\right)^{2}} \\D=\sqrt{100+324}\\D=\sqrt{424}$

$D{\left (p_{4},p_{1} \right )} = 2 \sqrt{106}$

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

$\\P = \sqrt{17} + \sqrt{265} + 2 \sqrt{106} + \sqrt{970} \\P = 72.14$

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Example Question #3 : 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, 10) , (-7, -8) , (-18, -10) , and (-1, -14) to two decimal places.

Possible Answers:

P = 77.71

P = 67.71

P = 54.64

P = 308.4

P = 78.71

Correct answer:

P = 77.71

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

$\\p_{1} = \left ( 8, \quad 10\right ) \\p_{2} = \left ( -7, \quad -8\right ) \\p_{3} = \left ( -18, \quad -10\right ) \\ p_{4} = \left ( -1, \quad -14\right )$

Here is a picture of the polygon

Plot4

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

$\\D = \sqrt{\left(8 - -7\right)^{2} + \left(10- - 8\right)^{2}} \\D= \sqrt{\left(15\right)^{2} + \left(18\right)^{2}} \\D=\sqrt{225+324}$

$D{\left (p_{1},p_{2} \right )} = 3 \sqrt{61}$

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

$\\D = \sqrt{\left( -7--18\right)^{2} + \left(- 8--10\right)^{2}} \\D= \sqrt{\left(11\right)^{2} + \left(2\right)^{2}} \\D=\sqrt{121+4}$

$D{\left (p_{2},p_{3} \right )} = 5 \sqrt{5}$

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

$\\D = \sqrt{\left( -18--1\right)^{2} + \left(-10--14\right)^{2}} \\D= \sqrt{\left(-17\right)^{2} + \left(4\right)^{2}} \\D=\sqrt{289+16}$

$D{\left (p_{3},p_{4} \right )} = \sqrt{305}$

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

$\\D = \sqrt{\left( -1-8\right)^{2} + \left(-14-10\right)^{2}} \\D= \sqrt{\left(-9\right)^{2} + \left(-24\right)^{2}} \\D=\sqrt{81+576}$

$D{\left (p_{4},p_{1} \right )} = 3 \sqrt{73}$

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

$\\P = 5 \sqrt{5} + \sqrt{305} + 3 \sqrt{61} + 3 \sqrt{73} \\P = 77.71$

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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 #2 : Points That Partitions Segments In A Given Ratio: Ccss.Math.Content.Hsg Gpe.B.6

Example Question #3 : Points That Partitions Segments In A Given Ratio: Ccss.Math.Content.Hsg Gpe.B.6

Example Question #11 : Points That Partitions Segments In A Given Ratio: Ccss.Math.Content.Hsg Gpe.B.6

Example Question #11 : Points That Partitions Segments In A Given Ratio: Ccss.Math.Content.Hsg Gpe.B.6

Example Question #11 : Points That Partitions Segments In A Given Ratio: Ccss.Math.Content.Hsg Gpe.B.6

Example Question #351 : High School: Geometry

Example Question #1 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Example Question #351 : High School: Geometry

Example Question #2 : Use Distance Formula To Compute And Compare Perimeters And Areas Of Polygons: Ccss.Math.Content.Hsg Gpe.B.7

Example Question #3 : 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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