A semicircle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn, meeting the circumference of the semicircle at D. Given that AC=2cm and CD=6cm, the area of the semicircle in square cm will be ?

Consider that O be center of circle and radius be x.
C can be anywhere on AB.
Now, consider again C is somewhere between A and O.
Therefore,
           \(\begin{array}{l}AC = 2\;{\rm{cm}}\\CO = \left( {x - 2} \right)\\DO = x,CD = 6{\rm{ cm}}\end{array}\)
then,
           CDO form a right angled triangle with right angle at C.

Using Pythagoras theorem,
           \(\begin{array}{c}{x^2} = {6^2} + {\left( {x + 2} \right)^2}\\{x^2} = 36 + {x^2} + 4x + 4\\4x = 40\\x = 10{\rm{ cm}}\end{array}\)
Therefore,
             area=\(\frac{{\pi {x^2}}}{2}\)
             area=50\(\pi \)