 # ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.4 in Chapter 17 - Mensuration

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 the 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 the volume of the rocket. (Use π = 3.14).

Solution:

Here, the rocket is a combination of a cone and a cylinder, thus we must sum the volumes of the two to determine its volume. 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

= 2π rH+π r2+π rl

=π 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)π r2h +π r2H

=π 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

Hence the volume of the rocket is 376.8 cm3.

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