ANSWER
Area of the trap-ezoid is [tex]50[/tex] square units
EXPLANATION
The given trap-ezoid has vertices [tex]A(-2,2)[/tex], [tex]B(2,5)[/tex], [tex]C(11,-7)[/tex] and [tex]D(1,-2)[/tex].
Area of a trap-ezoid is given by
[tex]Area =\frac{1}{2}(sum\:of\: paralle\:sides)\times \:vertical\: height[/tex]
We use the distance formula to determine length of all the necessary sides and plug them in to the formula.
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
The vertical height of the trap-ezoid is
[tex]|AB|=\sqrt{(2--2)^2+(5-2)^2}[/tex]
[tex]|AB|=\sqrt{(2+2)^2+(5-2)^2}[/tex]
[tex]|AB|=\sqrt{(4)^2+(3)^2}[/tex]
[tex]|AB|=\sqrt{16+9}[/tex]
[tex]|AB|=\sqrt{25}[/tex]
[tex]|AB|=5[/tex] units
The length of the parallel sides are;
[tex]|AD|=\sqrt{(1--2)^2+(-2-2)^2}[/tex]
[tex]|AD|=\sqrt{(3)^2+(-4)^2}[/tex]
[tex]|AD|=\sqrt{9+16}[/tex]
[tex]|AD|=\sqrt{25}[/tex]
[tex]|AD|=5[/tex]
and
[tex]|BC|=\sqrt{(11-2)^2+(-7-5)^2}[/tex]
[tex]|BC|=\sqrt{(9)^2+(-12)^2}[/tex]
[tex]|BC|=\sqrt{81+144}[/tex]
[tex]|BC|=\sqrt{225}[/tex]
[tex]|BC|=15[/tex] units
we now substitute all these values to obtain,
[tex]Area =\frac{1}{2}(15+5)\times 5[/tex]
[tex]Area =\frac{1}{2}(20)\times 5[/tex]
[tex]Area =(10)\times 5[/tex]
[tex]Area =50[/tex] square units.
