Suresh,
I drew a picture of your truncated cone and extended it to a vertex P.
In the diagram the distance PQ is x mm. Using similar triangles
^{(600 + x)}/_{1475} = ^{x}/_{447}
Solving for x I found that x = 260.89 mm.
Using Pythagoras theorem with the triangles PQT and PRS I found PT = 517.56 mm and PS = 1707.85 mm. Thus the plate can be contructed from a disk of radius
PS = 1707.85 mm by removing a cisk of radius PT = 517.56 mm.
Using the same method as in the answer to a previous problem I found the angle VPS to be 49.09 degrees.
Penny
