Question

4. A rocket is in the form of a right circular
cylinder closed at the
closed at the lower end and
surmounted by a cone of the same radius.
The diameter and height of the cylinder are
6 cm and 12 cm respectively, while the slant
height of the cone is 5 cm. Find the total
surface area and volume of the rocket. (Use
T = 3.14) pls answer the above question​

Answer 1

Answer:

Given diameter of the cylinder = 6 cm

Radius of the cylinder, r = 6/2 = 3 cm

Height of the cylinder, H = 12 cm

Slant height of the cone, l = 5 cm

Radius of the cone, r = 3 cm

Height of the cone, h = √(l2-r2)

h = √(52-32)

h = √(25-9)

h = √16

h = 4 cm

Total surface area of the rocket = curved surface area of cylinder +base area of cylinder+ curved surface area of cone

= 2rH+r2+rl

= r(2H+r+l)

= 3.14×3×(2×12+3+5)

= 3.14×3×(24+3+5)

= 3.14×3×32

= 301.44 cm2

Hence the Total surface area of the rocket is 301.44 cm2.

Volume of the rocket = Volume of the cone + volume of the cylinder

= (1/3)r2h +r2H

= r2((h/3)+H)

= 3.14×32×((4/3)+12)

= 3.14×9×((4+36)/3)

= 3.14×9×(40/3)

= 3.14×3×40

= 376.8 cm3

Step-by-step explanation:

Answer 2

Your answer is in the above of the attachment.............