Question

find the perimeter of the polygon with the vertices (x(-1,3)), (y(3,0)), and (z(-1,-2)). round your answer to the nearest hundredth. the perimeter is about (square) units.

Explanation:

Step1: Use distance formula

The distance formula 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: Calculate distance $XY$

For points $X(-1,3)$ and $Y(3,0)$, $x_1=-1,y_1 = 3,x_2=3,y_2 = 0$.
$d_{XY}=\sqrt{(3-(-1))^2+(0 - 3)^2}=\sqrt{(4)^2+(-3)^2}=\sqrt{16 + 9}=\sqrt{25}=5$.

Step3: Calculate distance $YZ$

For points $Y(3,0)$ and $Z(-1,-2)$, $x_1 = 3,y_1=0,x_2=-1,y_2=-2$.
$d_{YZ}=\sqrt{(-1 - 3)^2+(-2-0)^2}=\sqrt{(-4)^2+(-2)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}\approx4.47$.

Step4: Calculate distance $ZX$

For points $Z(-1,-2)$ and $X(-1,3)$, $x_1=-1,y_1=-2,x_2=-1,y_2 = 3$.
$d_{ZX}=\sqrt{(-1-(-1))^2+(3-(-2))^2}=\sqrt{(0)^2+(5)^2}=5$.

Step5: Calculate perimeter

$P=d_{XY}+d_{YZ}+d_{ZX}=5 + 2\sqrt{5}+5=10 + 2\sqrt{5}\approx10+4.47=14.47$.

Answer:

$14.47$