What is the area of trapezoid ABCD ? (Enter your answer as a decimal or whole number)

check the picture below.

bear in mind that, the "bases" are the two parallel sides, and the height is the distance between them.

[tex]\bf \textit{area of this trapezoid}\\\\ A=\cfrac{AB(BC+AD)}{2}\\\\ -------------------------------\\\\ \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ -1}}\quad ,&{{ 5}})\quad % (c,d B&({{ 3}}\quad ,&{{ 2}}) \end{array}\qquad % distance value 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 AB=\sqrt{25}\implies \boxed{AB=5}[/tex]

[tex]\bf -------------------------------\\\\ \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) B&({{ 3}}\quad ,&{{ 2}})\quad % (c,d C&({{ 0}}\quad ,&{{ -2}}) \end{array} \\\\\\ BC=\sqrt{(0-3)^2+(-2-2)^2}\implies BC=\sqrt{9+16} \\\\\\ BC=\sqrt{25}\implies \boxed{BC=5}[/tex]

[tex]\bf -------------------------------\\\\ \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ -1}}\quad ,&{{ 5}})\quad % (c,d D&({{ -13}}\quad ,&{{ -11}}) \end{array} \\\\\\ AD=\sqrt{[-13-(-1)]^2+[-11-5]^2} \\\\\\ AD=\sqrt{(-13+1)^2+(-16)^2}\implies AD=\sqrt{144+256} \\\\\\ AD=\sqrt{400}\implies \boxed{AD=\sqrt{20}}[/tex]

so, the area for this trapezoid is then

[tex]\bf A=\cfrac{5(5+20)}{2}\implies A=\cfrac{125}{2}\implies A=62\frac{1}{2}[/tex]