Question

find the perimeter of the parallelogram with these vertices. (3, 6), (-5, 6), (6, -1), (-2, -1) give an exact answer (not a decimal approximation). simplify your answer as much as possible.

Explanation:

Step1: Recall distance formula

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Find length of one - side

Let's find the distance between $(3,6)$ and $(-5,6)$. Here $x_1 = 3,y_1 = 6,x_2=-5,y_2 = 6$. Then $d_1=\sqrt{(-5 - 3)^2+(6 - 6)^2}=\sqrt{(-8)^2+0^2}=\sqrt{64}=8$.

Step3: Find length of adjacent - side

Let's find the distance between $(3,6)$ and $(6,-1)$. Here $x_1 = 3,y_1 = 6,x_2 = 6,y_2=-1$. Then $d_2=\sqrt{(6 - 3)^2+(-1 - 6)^2}=\sqrt{3^2+(-7)^2}=\sqrt{9 + 49}=\sqrt{58}$.

Step4: Calculate perimeter of parallelogram

The perimeter $P$ of a parallelogram with adjacent - side lengths $a$ and $b$ is $P = 2(a + b)$. Here $a = 8$ and $b=\sqrt{58}$, so $P=2(8+\sqrt{58})=16 + 2\sqrt{58}$.

Answer:

$16 + 2\sqrt{58}$