In this question we have to find the the total surface area

and the volume of the rocket given that a rocket is in the form

of a circular cylinder closed at the lower end with a

cone of the same radius attached to the top. The cylinder

is of radius 2.5 m andheight 21 m and the cone has the slant height 8 m.

This question is from r d sharma class 10 maths.

I found this question while doing maths class 10.

I need help in getting the solution.

Given,

Radius of the cylindrical portion of the rocket (R) = 2.5 m

Height of the cylindrical portion of the rocket (H) = 21 m

Slant Height of the Conical surface of the rocket (L) = 8 m

Curved Surface Area of the Cone (S1) = Ï€RL = Ï€(2.5)(8)= 20Ï€

And,

Curved Surface Area of the Cone (S2) = 2Ï€RH + Ï€R2

S2 = (2Ï€ Ã— 2.5 Ã— 21) + Ï€ (2.5)2

S2 = (Ï€ Ã— 105) + (Ï€ Ã— 6.25)

Thus, the total curved surface area S is

S = S1 + S2

S = (Ï€20) + (Ï€105) + (Ï€6.25)

S = (22/7)(20 + 105 + 6.25) = 22/7 x 131.25

S = 412.5 m2

Therefore, the total Surface Area of the Conical Surface = 412.5 m2

Now, calculating the volume of the rocket

Volume of the conical part of the rocket (V1) = 1/3 Ã— 22/7 Ã— R2 Ã— h

V1 = 1/3 Ã— 22/7 Ã— (2.5)2 Ã— h

Let, h be the height of the conical portion in the rocket.

We know that,

L2 = R2 + h2

h2 = L2 â€“ R2 = 82 â€“ 2.52

h = 7.6 m

Using the value of h, we will get

Volume of the conical part (V1) = 1/3 Ã— 22/7 Ã— 2.52 Ã— 7.6 m2 = 49.67 m2

Next,

Volume of the Cylindrical Portion (V2) = Ï€R2h

V2 = 22/7 Ã— 2.52 Ã— 21 = 412.5 m2

Thus, the total volume of the rocket = V1 + V2

V = 412.5 + 49.67 = 462.17 m2

Hence, the total volume of the Rocket is 462.17 m2