a square ABCD,has side length 4cm and M is the midpoint of side CD,a semi circle is drawn inside the square.Also with AB as radius an arc is drawn from B to D. The point of intersection of arc and semicircle is P.What is the distance from P to AD?

Dear Ashwen,
Taking D as origin...
equation of the quadrant=x^2 + (y-4)^2=4^2
for the semicirle..= (x-2)^2 + y^2=2^2
Hence solving them and getting the point of intersection a (16/5,8/5)
Hence distance from AD= distance from y axis=x coordinate=16/5=3.2 units (ANS).
Regards.