the total surface area of a solid composed of a cone with hemispherical base is 361.1 cm. (π = 3.14) the dimension are shown in figure. find the total height of the solid
Answers
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.If a solid is composed of a cone with a hemispherical base whose area is 361.1 cm^2 and slant height is 13 cm then the total height of the solid is 17 cm.
Step-by-step explanation:
Required Formulas:
Curved Surface area of the Cone = πrl
Curved Surface area of the hemisphere = 2πr²
It is given that,
A solid is in the shape of a cone with hemispherical base
The total surface area of the solid = 361.1 cm²
The slant height of the cone, l = 13 cm
Step 1:
Let the radius of the hemispherical base be denoted as “r” cm.
Now,
The curved surface area of the conical part of the solid = πrl = 3.14 * r * 13 = [40.82 r] cm²
And,
The curved surface area of the hemisphere part of the solid = 2πr² = 2 * 3.14 * r² = [6.28r²] cm².
We know that,
The total surface area of the solid = [C.S.A of the conical part] + [C.S.A of the hemisphere part]
⇒ 361.1 = [40.82 r] + [6.28r²]
⇒ 6.28r² + 40.82r – 361.1 = 0
⇒ r² + 6.5r – 57.5 = 0
⇒ r² + 11.5r – 5r – 57.5 = 0
⇒ r(r+11.5) – 5(r+11.5) = 0
⇒ (r+11.5)(r-5) = 0
⇒ r = 5 cm …… [neglecting the negative value]
Step 2:
Let the height of the cone be denoted as “h” cm.
We know the formula of the slant height of a cone is given by,
l² = h² + r²
Substituting the value of l and r in the formula, we get
h = √[13² – 5²]
⇒ h = √[144]
⇒ h = 12 cm
Thus,
The total height of the solid is given by,
= [height of the cone] + [radius of the hemisphere base]
= h + r
= 12 + 5
= 17 cm
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Answer:
Solution:
Given radius of the cylindrical portion of the rocket (say, R) = 2.5 m
Given height of the cylindrical portion of the rocket (say, H) = 21 m
Given Slant Height of the Conical surface of the rocket (say, L) = 8 m
Curved Surface Area of the Cone (say S1) = RL
S1 = m2 .... E.1
Curved Surface Area of the Cone (say, S2) = 2RH + R2
S2 = (2π × 2.5 × 21) + π (2.5)2
S2 = (π × 10.5) + (π ×6.25) ... E.2
So, The total curved surface area = E.1 + E.2
S = S1 + S2
S = (π20) + (π105) + (π6.25)
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 = 1/3 × 22/7 × R2 × h
V1 = 1/3 × 22/7 × (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 = 23.685 m
Putting the value of h in E.3, we will get
Volume of the conical portion (V1) = 1/3 × 22/7 × 2.52 × 23.685 m2 .... E.4
Volume of the Cylindrical Portion (V2) = πR2h
V2 = 22/7 × 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