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

Look at the picture.
Use the Pythagorean theorem:
[tex]|AD|^2=5^2+(\sqrt{75})^2\\\\|AD|^2=25+75\\\\|AD|^2=100\to|AD|=\sqrt{100}\to|AD|=10[/tex]
The area of a triangle ADC:
[tex]A_\Delta=\dfrac{|DC|\cdot|AC|}{2}\ and\ A_\Delta=\dfrac{|AD|\cdot h}{2}[/tex]
Substitute:
[tex]\dfrac{10h}{2}=\dfrac{5\sqrt{75}}{2}\\\\5h=\dfrac{5\sqrt{75}}{2}\ \ \ |:5\\\\h=\dfrac{\sqrt{75}}{2}[/tex]
The triangles ADC and DCE are similar. Therefore:
[tex]\dfrac{|ED|}{\frac{\sqrt{75}}{2}}=\dfrac{5}{\sqrt{75}}\\\\\dfrac{2|ED|}{\sqrt{75}}=\dfrac{5}{\sqrt{75}}\ \ \ |\cdot\sqrt{75}\\\\2|ED|=5\ \ \ |:2\\\\|ED|=2.5[/tex]
[tex]|BC|=|AD|-2|ED|\\\\|BC|=10-2\cdot2.5=10-5=5[/tex]
The area of the trapezoid:
[tex]A=\dfrac{|AD|+|BC|}{2}\cdot h[/tex]
Substitute:
[tex]A=\dfrac{10+5}{2}\cdot\dfrac{\sqrt{75}}{2}=\dfrac{15\sqrt{75}}{4}\\\\=\dfrac{15\sqrt{25\cdot3}}{4}=\dfrac{15\cdot5\sqrt3}{4}=\dfrac{75\sqrt3}{4}[/tex]