Question

find the perimeter of the trapezoid with these vertices (4,6), (-2,6), (4,-1), (1,-1) give an exact answer (not a decimal approximation). simplify your answer as much as possible.

Explanation:

Step1: Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

For the side between \((4,6)\) and \((-2,6)\):
\(x_1 = 4,y_1 = 6,x_2=-2,y_2 = 6\)
\(d_1=\sqrt{(-2 - 4)^2+(6 - 6)^2}=\sqrt{(-6)^2+0^2}=\sqrt{36}=6\)

Step2: For the side between \((4,6)\) and \((4,-1)\)

\(x_1 = 4,y_1 = 6,x_2 = 4,y_2=-1\)
\(d_2=\sqrt{(4 - 4)^2+(-1 - 6)^2}=\sqrt{0^2+(-7)^2}=\sqrt{49}=7\)

Step3: For the side between \((-2,6)\) and \((1,-1)\)

\(x_1=-2,y_1 = 6,x_2 = 1,y_2=-1\)
\(d_3=\sqrt{(1+2)^2+(-1 - 6)^2}=\sqrt{3^2+(-7)^2}=\sqrt{9 + 49}=\sqrt{58}\)

Step4: For the side between \((4,-1)\) and \((1,-1)\)

\(x_1 = 4,y_1=-1,x_2 = 1,y_2=-1\)
\(d_4=\sqrt{(1 - 4)^2+(-1+1)^2}=\sqrt{(-3)^2+0^2}=\sqrt{9}=3\)

Step5: Calculate the perimeter \(P\)

\(P=d_1 + d_2+d_3+d_4=6 + 7+\sqrt{58}+3=16+\sqrt{58}\)

Answer:

\(16+\sqrt{58}\)