Question

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 diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket [Use π = 3.14].

Answer 1

Since, rocket is the combination of a right circular cylinder and a cone.

Given, diameter of the cylinder = 6 cm

∴ Radius of the cylinder = `6/2` = 3 cm

And height of the cylinder = 12 cm

∴ Volume of the cylinder = πr²h

= 3.14 × (3)² × 12

= 339.12 cm³

And curved surface area = 2πrh

= 2 × 3.14 × 3 × 12

= 226.08

Now, In right angled ΔAOC,

h = `sqrt(5^2 - 3^2)`

= `sqrt(25 - 9)`

= `sqrt(16)`

= 4

∴ Height of the cone, h = 4 cm

Radius of the cone, r = 3 cm

Now, volume of the cone

= `1/3 pir^2h`

= `1/3 xx 3.14 xx (3)^2 xx 4`

= `113.04/3`

= 37.68 cm³

And curved surface area = πrl

= 3.14 × 3 × 5

= 47.1

Hence, total volume of the rocket

= 339.12 + 37.68

= 376.8 cm³

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 cm²