QUESTION IMAGE
Question
type the correct answer in the box. use numerals instead of words. if necessary, use / for the fraction bar.
the school puts up fencing around an area in the shape of a hexagon for an outdoor concert. how much fencing does the school use? round your answer to the nearest whole yard.
graph of a hexagon on a coordinate grid with x and y axes labeled in yards
the school uses about \boxed{} yards of fencing.
Step1: Identify the vertices of the hexagon
From the graph, the vertices (in order) are: \((-6, 4)\), \((-5, -2)\), \((0, -4)\), \((5, -2)\), \((6, 4)\), \((0, 6)\)
Step2: Calculate the distance between consecutive vertices using the distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Distance between \((-6, 4)\) and \((-5, -2)\):
\(d_1=\sqrt{(-5 - (-6))^2 + (-2 - 4)^2}=\sqrt{(1)^2 + (-6)^2}=\sqrt{1 + 36}=\sqrt{37}\approx 6.08\)
Distance between \((-5, -2)\) and \((0, -4)\):
\(d_2=\sqrt{(0 - (-5))^2 + (-4 - (-2))^2}=\sqrt{(5)^2 + (-2)^2}=\sqrt{25 + 4}=\sqrt{29}\approx 5.39\)
Distance between \((0, -4)\) and \((5, -2)\):
\(d_3=\sqrt{(5 - 0)^2 + (-2 - (-4))^2}=\sqrt{(5)^2 + (2)^2}=\sqrt{25 + 4}=\sqrt{29}\approx 5.39\)
Distance between \((5, -2)\) and \((6, 4)\):
\(d_4=\sqrt{(6 - 5)^2 + (4 - (-2))^2}=\sqrt{(1)^2 + (6)^2}=\sqrt{1 + 36}=\sqrt{37}\approx 6.08\)
Distance between \((6, 4)\) and \((0, 6)\):
\(d_5=\sqrt{(0 - 6)^2 + (6 - 4)^2}=\sqrt{(-6)^2 + (2)^2}=\sqrt{36 + 4}=\sqrt{40}\approx 6.32\)
Distance between \((0, 6)\) and \((-6, 4)\):
\(d_6=\sqrt{(-6 - 0)^2 + (4 - 6)^2}=\sqrt{(-6)^2 + (-2)^2}=\sqrt{36 + 4}=\sqrt{40}\approx 6.32\)
Step3: Sum up all the distances
\(Perimeter = d_1 + d_2 + d_3 + d_4 + d_5 + d_6\)
\(Perimeter\approx 6.08 + 5.39 + 5.39 + 6.08 + 6.32 + 6.32\)
\(Perimeter\approx 6.08\times2 + 5.39\times2 + 6.32\times2\)
\(Perimeter\approx 12.16 + 10.78 + 12.64\)
\(Perimeter\approx 35.58\approx 36\)
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