Question

Given: ABCD is a trapezoid, AC ⊥ CD , AB = CD, AC= square root of 75 , AB = 5 Find: Area of ABCD

Answer 1

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

`Area . trapezoid= (1)/(2) [(5+10) √(75)/2] \\ Area.trapezoid= (1)/(4) 15 √(75 ) \\ Area.trapezoid= (75)/(4) √(3)`

the answer is
A=(75/4)√3 units²