[tex]\bf \stackrel{\textit{area of a trapezoid}}{A=\cfrac{h(a+b)}{2}}~~ \begin{cases} h=height\\ a,b=bases \end{cases}\qquad therefore\qquad A=\cfrac{AB(AD+BC)}{2}[/tex]
so, let's find how long each of those segments are anyway.
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-1}~,~\stackrel{y_1}{5})\qquad B(\stackrel{x_2}{3}~,~\stackrel{y_2}{2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AB=\sqrt{[3-(-1)]^2+[2-5]^2}\implies AB=\sqrt{(3+1)^2+(2-5)^2} \\\\\\ AB=\sqrt{16+9}\implies \boxed{AB=5}[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_2}{3}~,~\stackrel{y_2}{2})\qquad C(\stackrel{x_1}{0}~,~\stackrel{y_1}{-2})\qquad \qquad \\\\\\ BC=\sqrt{(0-3)^2+(-2-2)^2}\implies BC=\sqrt{9+16}\implies \boxed{BC=5}[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ D(\stackrel{x_2}{-13}~,~\stackrel{y_2}{-11})\qquad A(\stackrel{x_1}{-1}~,~\stackrel{y_1}{5}) \\\\\\ AD=\sqrt{[-1-(-13)]^2+[5-(-11)]^2}\implies AD=\sqrt{(-1+13)^2+(5+11)^2} \\\\\\ AD=\sqrt{144+256}\implies AD=\sqrt{400}\implies \boxed{AD=20}[/tex]
[tex]\bf \stackrel{\textit{area of a trapezoid}}{A=\cfrac{h(a+b)}{2}}~~ \begin{cases} h=height\\ a,b=bases \end{cases}\qquad therefore\qquad A=\cfrac{AB(AD+BC)}{2} \\\\\\ A=\cfrac{5(20+5)}{2}\implies A=\cfrac{5(25)}{2}\implies A=\cfrac{125}{2}\implies \blacktriangleright A=62.5 \blacktriangleleft[/tex]
