Respuesta :

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]
Ver imagen jdoe0001

Area of the trapezoid ABCD is

[tex]62\frac{1}{2}[/tex]

Given :

The diagram of a trapezoid ABCD

Area of trapezoid is [tex]\frac{base1+base2}{2} \cdot height[/tex]

Lets find the length of base 1  and base 2 using distance formula

Base 1 is AD. find the distance AD

[tex]AD=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\\AD=\sqrt{\left(-13-\left(-1\right)\right)^2+\left(-11-5\right)^2}\\AD=20[/tex]

Now we find out BD that is base 2

[tex]BD=\sqrt{\left(0-3\right)^2+\left(-2-2\right)^2}\\BD=5[/tex]

Now find out height  AB

A is (-1,5) and B(3,2)

[tex]AB=\sqrt{\left(-1-3\right)^2+\left(5-2\right)^2}\\AB=5\\[/tex]

[tex]Area =\frac{AD+BC}{2} \cdot AB\\Area=\frac{20+5}{2} \cdot 5\\Area=\frac{125}{2}\\Area=62\frac{1}{2}[/tex]

Area of the trapezoid ABCD is

[tex]62\frac{1}{2}[/tex]

Learn more : brainly.com/question/11150764

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