Construct a paralelogram /PQRS/ Such that /PQ/=9.5cm,/PQR/=105,diagonal/PR/=12cm,measure circle/QR/
105° = 90° + 15°
It's easy to construct a 15° angle -- just bisect a 60° angle twice.
Draw segment PQ of length 9.
So, at Q, construct the 105° angle
With P as center and a radius of 12, construct a circle.
Where the circle intersects the ray from Q will be vertex R.
Now you have PQR, so either
construct PS parallel to QR or RS parallel to PQ.
Not sure what you mean by "circle/QR/" but inscribed angles are 1/2 the central angle in a circle...