Respuesta :

the picture in the attached figure

we know that
in the right triangle ACD
AD²=AC²+CD²-----> (√75)²+5²-----> AD²=100-----> AD=10 units
sin ∠D=AC/AD----> sin ∠ D=√75/10
cos ∠D=CD/AD----> cos ∠D=5/10-----> cos ∠D=1/2

in the right triangle FDC
sin ∠ D=FC/CD------> FC=CD*sin  ∠ D----> FC=5*(√75)/10
FC=√75/2 units
cos ∠D=FD/CD-----> FD=CD*cos ∠D-----> FD=5*(1/2)---> FD=2.5 units
FD=2.5 units

AE=FD
BC=AD-2*(FD)----> BC=10-2*2.5-----> BC=5 units

 Area of trapezoid  ABCD=(BC+AD)*FC/2
[tex]Area . trapezoid= \frac{1}{2} [(5+10) \sqrt{75}/2] \\ Area.trapezoid= \frac{1}{4} 15 \sqrt{75 } \\ Area.trapezoid= \frac{75}{4} \sqrt{3} [/tex]

the answer is
A=(75/4)√3 units²

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