Math, asked by BrainlyHelper, 1 year ago

A rocket is in the form of a circular cylinder closed at the lower end with the cone of the same base radius attach to the top the cylinder is of radius 2 5 m and height 21 m & the cone has slant height 8cm find the total surfce area of the volume of the rocket .

Answers

Answered by nikitasingh79
5

Answer:

Total surface area of rocket is 412.5 m² and the Volume of the rocket is 462.26 m³.

Step-by-step explanation:

Given :  

Radius of the cylinder and cone , r = 2.5 m

Height of a cylinder, h  = 21 m

Slant height of the cone , l = 8 m  

Curved surface area of cone = πrl

= 22/7 × 2.5 × 8

= 440/7

CSA of cone = 62.86 m²

Curved surface area of cylinder = 2πrh

= 2 × 22/7 × 2.5 × 21

CSA of cylinder  = 330 m²

Area of the base of the cylinder = πr²

= 22/7 × 2.5 × 2.5

= 19.64 m²

Total surface area of rocket = CSA of cone + CSA of cylinder + Area of the base of the cylinder

TSA of rocket  = 62.85 + 330 + 19.64 = 412.5 m²

Total surface area of rocket =  412.5 m²

Volume of cylinder = πr²h

= 22/7 × 2.5 × 2.5 × 21 = 22 × 2.5 × 2.5 × 3

Volume of cylinder = 412.5 m³

Height of the cone, h1 = √l² - r²

h1 = √8² - (2.5)²

h1 = √64 - 6.25

h1 = √57.75

h1 = 7.59 m

Volume of the cone = 1/3πr²h1

= ⅓ × 22/7× 2.5 × 2.5 × 7.59

= 49.76  m³

Volume of the rocket = 412.5 + 49.76 = 462.26 m³

Hence, Total surface area of rocket is 412.5 m² and the Volume of the rocket is 462.26 m³.

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Answered by Kusumsahu7
2

Given radius of the cylindrical portion of the rocket (say, R) = 2.5m

Given height of the cylindrical portion of the rocket (say, H) = 21m

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

Curved Surface Area of the Cone (say S1) = RL

S1 =  m2                                                                                    ……. E.1

                             Curved Surface Area of the Cone (say, S2) = 2RH + R2

                                                                                                                       S2 = (2 π   2.5  21) + (π   (2.5)2 )

S2 = (π  105) + (π  6.25)                                            …….. E.2

So, The total curved surface area = E.1 + E.2

S = S1 + S2

S = (π   20) + (π  105) + (π  6.25)

S = 62.83 + 329.86 + 19.63

S = 412.3 m2

Hence, the total Curved Surface Area of the Conical Surface = 412.3 m2

Volume of the conical surface of the rocket = 13×227×R2×h

V1 = 13×227×(2.5)2×h                          ……. E.3

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

Now,

L2 = R2 + h2

h2 = L2 – R2

                                             h = L2−R2−−−−−−−√

h = 82−2.52−−−−−−−√

h = 23.685 m

Putting the value of h in E.3, we will get

Volume of the conical portion (V1) = 13×227×2.52×23.685 m2                         …….. E.4

Volume of the Cylindrical Portion (V2) = πR2h

V2 = 227×2.52×21

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

V = 461.84 m2

Hence, the total volume of the Rocket (V) is 461.84 m2

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