Question

What is the area of the space

Answer 1

check the picture below.

since we know is a rectangle, then we can simply find a width and a length, say hmmm CD and CB and get their product, recall that the area of a rectangle is width*length.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ C(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})\qquad D(\stackrel{x_2}{-7}~,~\stackrel{y_2}{1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ CD=\sqrt{[-7-(-4)]^2+[1-(-3)]^2}\implies CD=\sqrt{(-7+4)^2+(1+3)^2} \\\\\\ CD=\sqrt{(-3)^2+4^2}\implies CD=\sqrt{9+16}\implies \boxed{CD=5}\\\\ -------------------------------$

$\bf C(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})\qquad B(\stackrel{x_2}{4}~,~\stackrel{y_2}{3}) \\\\\\ CB=\sqrt{[4-(-4)]^2+[3-(-3)]^2}\implies CB=\sqrt{(4+4)^2+(3+3)^2} \\\\\\ CB=\sqrt{8^2+6^2}\implies CB=\sqrt{64+36}\implies \boxed{CB=10}\\\\ -------------------------------\\\\ \stackrel{\textit{area of rectangle}}{CD\cdot CB}\implies 50$