Question

answer attempt 1 out of 3 perimeter = units submit answer

Explanation:

Step1: Identify side - lengths using distance formula

For a line segment with endpoints $(x_1,y_1)$ and $(x_2,y_2)$, the distance $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Assume the parallelogram has vertices $U(-1,-5)$, $V(7,-5)$, $W(8,3)$, $X(0,3)$. The horizontal side $UV$: Since the $y -$coordinates of $U$ and $V$ are the same, $d_{UV}=\vert7-(-1)\vert = 8$. The non - horizontal side $UX$: Using the distance formula with $U(-1,-5)$ and $X(0,3)$, $d_{UX}=\sqrt{(0 + 1)^2+(3 + 5)^2}=\sqrt{1 + 64}=\sqrt{65}$.

Step2: Calculate the perimeter of the parallelogram

The perimeter $P$ of a parallelogram with adjacent side - lengths $a$ and $b$ is $P = 2(a + b)$. Here, $a = 8$ and $b=\sqrt{65}$, so $P=2(8+\sqrt{65})=16 + 2\sqrt{65}\approx16+2\times8.062 = 16+16.124=32.124$. Since we can also count the grid units more simply for a parallelogram on a grid. By counting the horizontal and vertical displacements, the two adjacent side - lengths are 8 and $\sqrt{1^2 + 8^2}=\sqrt{65}$. The perimeter $P = 2(8+\sqrt{65})\approx32.12$. If we assume we are working with grid - unit counting precisely, the horizontal side length between two points with the same $y$ - value is 8 units and the non - horizontal side length (by Pythagorean theorem for right - triangle formed by grid - lines) for example between $U(-1,-5)$ and $X(0,3)$ is $\sqrt{(0 + 1)^2+(3 + 5)^2}=\sqrt{65}\approx 8$. The perimeter of the parallelogram $P=2(8 + \sqrt{65})\approx32$ (rounded to the nearest whole number).

Answer:

$32$