A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diamter and height of the cylinder are 6 cm and 12 cm , respectively. If the slant height of the conical portion is 5 cm, the find the total surface area and volume of the rocket. ( use `pi = 3.14`).

Explanation

Since, rocket is the combination of a right circular cylinder and a cone.
Given, diameter of the cylinder = 6 cm
`therefore` Radius of the cylinder = 12 cm
and height of the cylinder `= pir^(2)h= 3.14 xx (3)^(3) xx 12`
`" "= 339.12 cm^(3)`
and height of the cylinder = 12 cm
`therefore` Volume of the cylinder `= pir^(2)h = 3.14 xx (3)^(3) xx 12`
`= 339.12 cm^(3)`
and curved surface area `= 2pirh`
`= 2 xx 3.14 xx 3 xx 12 = 226.08`
Now, in right angled `DeltaAOc`
`" " h = sqrt(5^(2) -3^(2)) = sqrt(25-9) = sqrt(16) = 4`
` therefore` Height of the cone h = 4 cm
Radius of the cone ,r= 3cm
Now, volume of the cone
`" " =(1)/(3) pir^(2)h = (1)/(3) xx 3.14 xx (3)^(2) xx 4`
`" " = (113.04)/(3) = 37.68 cm^(3)`
and curved surface area `= pirl = 3.14 xx 3xx5 = 47.1`
Hence, total volume of the rocket`=339.12 +37.68 = 376. 8 cm ^(3)`
and total surface area of the rocket = CSA of cone + CSA of cylinder + Area of base of cylinder
`" " = 47.1 + 226.08 + 28.26`
`" " = 301.44 cm^(2)`