Question

Look at the figure XYZ in the coordinate plane. Find the perimeter of the figure rounded to the nearest tenth.

Answer 1

the distance form X to Y is clearly -6 to 0 is 6 units, and 0 to 8 is 8 units, so 6 + 8 = 14 units.

now, for XZ and ZY we can simply use as stated, the distance formula to get those and then add them all to get the perimeter.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ X(\stackrel{x_1}{-6}~,~\stackrel{y_1}{2})\qquad Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ XZ=\sqrt{[5-(-6)]^2+[8-2]^2}\implies XZ=\sqrt{(5+6)^2+(8-2)^2} \\\\\\ XZ=\sqrt{121+36}\implies \boxed{XZ=\sqrt{157}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ Z(\stackrel{x_2}{5}~,~\stackrel{y_2}{8})\qquad Y(\stackrel{x_2}{8}~,~\stackrel{y_2}{2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ ZY=\sqrt{(8-5)^2+(2-8)^2}\implies ZY=\sqrt{9+36}\implies \boxed{ZY=\sqrt{45}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{perimeter}{14+\sqrt{157}+\sqrt{45}}\qquad \approx \qquad 33.2$