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All Common Core: High School - Geometry Resources

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Flashcards: 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.

P = 117.1

P = 389.4

P = 116.1

P = 106.1

P = 88.8

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Flashcards: 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.

Your answer:

P = 117.1

Your answer:

P = 389.4

Your answer:

P = 106.1

Your answer:

P = 88.8

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.

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

Flashcards: 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, 12) , (1, 4) , (-14, -4) and (-1, -19) to two decimal places.

P = 75.93

P = 48.01

P = 355.1

P = 65.93

P = 76.93

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Flashcards: 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, 12) , (1, 4) , (-14, -4) and (-1, -19) to two decimal places.

Your answer:

P = 48.01

Your answer:

P = 355.1

Your answer:

P = 65.93

Your answer:

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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All Common Core: High School - Geometry Resources

Flashcards: Use Distance Formula to Compute and Compare Perimeters and Areas of Polygons: CCSS.Math.Content.HSG-GPE.B.7

Flashcards: Use Distance Formula to Compute and Compare Perimeters and Areas of Polygons: CCSS.Math.Content.HSG-GPE.B.7

Flashcards: Use Distance Formula to Compute and Compare Perimeters and Areas of Polygons: CCSS.Math.Content.HSG-GPE.B.7

Flashcards: Use Distance Formula to Compute and Compare Perimeters and Areas of Polygons: CCSS.Math.Content.HSG-GPE.B.7

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