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consider polygon abcde on the coordinate grid what is the best approximation for the perimeter of polygon ABCDE

consider polygon abcde on the coordinate grid what is the best approximation for the-example-1
User Luca Murra
by
2.9k points

1 Answer

7 votes
7 votes

24.28 units

Step-by-step explanation

Perimeter is the distance around the edge of a shape, to find the perimeter of the given figure we can use the distance bewteen 2 points formula,it says


\begin{gathered} d=√((x_2-x_1)^2+(y_2-y_1)^2) \\ where \\ P1(x_1,y_1) \\ P2(x_2,y_2) \end{gathered}

so

Step 1

identify the vertices ;


\begin{gathered} A(-2,3) \\ B(0,6) \\ C(5,4) \\ D(4,-1) \\ E(-1,-2) \end{gathered}

Step 2

now, find the length of each segment

a)AB

replace in the formula


\begin{gathered} d=√((x_2-x_1)^2+(y_2-y_1)^2) \\ d_(AB)=√((0-(-2))^2+(6-3)^2) \\ d_(AB)=√(4+9) \\ d_(AB)=√(13) \end{gathered}

b)BC


\begin{gathered} d=√((x_2-x_1)^2+(y_2-y_1)^2) \\ d_(BC)=√((5-0)^2+(4-6)^2) \\ d_(BC)=√(25+4) \\ d_(BC)=√(29) \end{gathered}

c)CD


\begin{gathered} d=√((x_2-x_1)^2+(y_2-y_1)^2) \\ d_(CD)=√((4-5)^2+(-1-4)^2) \\ d_(CD)=√(1+25) \\ d_(CD)=√(26) \end{gathered}

d)DE


\begin{gathered} d=√((x_2-x_1)^2+(y_2-y_1)^2) \\ d_(DE)=√((-1-4)^2+(-2-(-1))^2) \\ d_(DE)=√(25+1) \\ d_(DE)=√(26) \end{gathered}

e) EA


\begin{gathered} d=√((x_2-x_1)^2+(y_2-y_1)^2) \\ d_(EA)=√((-2-(-1))^2+(3-(-2))^2) \\ d_(EA)=√(1+25) \\ d_(EA)=√(26) \end{gathered}

Step 3

finally, the perimeter is the sum of the sides length, so


\begin{gathered} perimeter=AB+BC+CD+DE+EA \\ replacing \\ perimeter=√(13)+√(29)+√(26)+√(26)+√(26) \\ perimeter=24.28 \end{gathered}

so, the perimeter is 24.28 units

consider polygon abcde on the coordinate grid what is the best approximation for the-example-1
User Jarred Sumner
by
2.8k points
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