Use Distance Formula to Compute and Compare Perimeters and Areas of Polygons: CCSS.Math.Content.HSG-GPE.B.7 - Common Core: High School - Geometry

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Question

Find the perimeter of the following polygon with the points , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , , and to two decimal places.

Answer

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

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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Question

Find the perimeter of the following polygon with the points , , , and to two decimal places.

Answer

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

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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