Question

determine the perimeter of the given figure on the coordinate plane. round your answer to the nearest hundredth, if necessary. figure abcdef > enter the answer in the space provided. use numbers instead of words. units

Explanation:

Step1: Identify coordinates

Assume \(A(-6, 2)\), \(B(0, 0)\), \(C(3, 4)\), \(D(3, - 6)\), \(E(0,-3)\), \(F(-6,-3)\)

Step2: Use distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) for non - horizontal/vertical sides

For \(AB\): \(d_{AB}=\sqrt{(-6 - 0)^2+(2 - 0)^2}=\sqrt{36 + 4}=\sqrt{40}=2\sqrt{10}\approx6.32\)
For \(BC\): \(d_{BC}=\sqrt{(3 - 0)^2+(4 - 0)^2}=\sqrt{9 + 16}=\sqrt{25}=5\)
For \(CD\): \(d_{CD}=\sqrt{(3 - 3)^2+(-6 - 4)^2}=\sqrt{0 + 100}=10\)
For \(DE\): \(d_{DE}=\sqrt{(0 - 3)^2+(-3+6)^2}=\sqrt{9 + 9}=\sqrt{18}=3\sqrt{2}\approx4.24\)
For \(EF\): Since it's a horizontal line, \(d_{EF}=\vert0+6\vert = 6\)
For \(FA\): Since it's a vertical line, \(d_{FA}=\vert2 + 3\vert=5\)

Step3: Calculate perimeter

\(P=d_{AB}+d_{BC}+d_{CD}+d_{DE}+d_{EF}+d_{FA}\)
\(P\approx6.32+5 + 10+4.24+6+5=36.56\)

Answer:

\(36.56\)