Hi Sonny,
I recently answered a very similar question from Wendy. Her cone had a base opening of 4" and a top opening of 2" but it was 12" tall where your truncated cone has a height of 6". I am going to refer to the diagrams I drew for her to answer your question.
For your situation BC = 6" so AC = 12 " and
AD^{2} = AC^{2} + CD^{2} = 144 + 16 = 160
and hence AD = √160 = 12.65 inches.
Now slice the cone from bottom to top and roll it out flat. What you get is a sector of a circle of radius 12.65 inches. The length of the arc is the circumference of the base circle of the cone. This is a circle of radius 4 inches so the length of the arc is
arc length = circle circumference = 2 radius = 2 4 = 25.13 inches.
The relationship among the length of an arc, the radius of the circle and the angle at the centre of the circle is
arc length = radius angle
where the angle is measured in radians. Thus
25.13 = 12.65 angle
and hence the angle is 1.99 radians. There are radians in 180^{o} and thus
1.99 radians = ^{180}/_{} 1.99 = 114.02^{0}
Thus you need to draw a circle of radius 12.65 inches. Cut out a sector with a centre angle of 114.02^{0} . Cut out a sector of radius ^{12.65}/_{2} = 6.33 inches inches and roll up what's left to form your truncated cone.
Penny
