Question -
Answer -
Given,
Radius of thecylindrical portion of the rocket (R) = 2.5 m
Height of thecylindrical portion of the rocket (H) = 21 m
Slant Height of theConical surface of the rocket (L) = 8 m
Curved Surface Area ofthe Cone (S1) = πRL = π(2.5)(8)= 20π
And,
Curved Surface Area ofthe Cone (S2) = 2πRH + πR2
S2 =(2π × 2.5 × 21) + π (2.5)2
S2 =(π × 105) + (π × 6.25)
Thus, the total curvedsurface area S is
S = S1 +S2
S = (π20) + (π105) +(π6.25)
S = (22/7)(20 + 105 +6.25) = 22/7 x 131.25
S = 412.5 m2
Therefore, the totalSurface Area of the Conical Surface = 412.5 m2
Now, calculating thevolume of the rocket
Volume of the conicalpart of the rocket (V1) = 1/3 × 22/7 × R2 × h
V1 = 1/3× 22/7 × (2.5)2 × h
Let, h be the heightof the conical portion in the rocket.
We know that,
L2 = R2 +h2
h2 = L2 –R2 = 82 – 2.52
h = 7.6 m
Using the value of h,we will get
Volume of the conicalpart (V1) = 1/3 × 22/7 × 2.52 × 7.6 m2 =49.67 m2
Next,
Volume of theCylindrical Portion (V2) = πR2h
V2 = 22/7× 2.52 × 21 = 412.5 m2
Thus, the total volumeof the rocket = V1 + V2
V = 412.5 + 49.67 =462.17 m2
Hence, the totalvolume of the Rocket is 462.17 m2