Ram and mehmood are two neighbours ram wears a cap which is in the shape of a hemisphere of radius 7 cm and mehmood wears a turkish cap which is in the shape of a frustum of a cone if the radius of its open side is 10 cm and the radius of the upper base is 4 cm and its slant height is 15 cm find the curved surface area of the both cape
Answers
The curved surface area of hemispherical cap and Turkish caps of Ram and Mahmood respectively is 968 cm².
Step-by-step explanation:
It is given that,
The radius of the hemispherical cap, r = 7 cm
The radius of the lower base of the frustum shaped cap, r₁ = 10 cm
The radius of the upper base of the frustum shaped cap, r₂ = 4 cm
The slant height of the frustum shaped cap = 15 cm
Required Formulas:
- Curved Surface Area of Hemisphere = 2πr²
- Curved Surface Area of Frustum = π(r₁ + r₂)l
Now, substituting the given values in the formulas above we get,
The CSA of both the caps is given by,
= [CAS of hemisphere] + [CSA of frustum]
= [2πr²] + [π(r₁ + r₂)l]
= [2 × (22/7) × 7²] + [(22/7) × (10 + 4) × 15]
= [2 × 22 × 7] + [(22/7) × (14) × 15]
= 308 + [22 × 2 × 15]
= 308 + 660
= 968 cm²
Thus, the curved surface area of the caps of Ram and Mahmood is 968 cm².
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Answer:
hello friend
Step-by-step explanation:
The CSA of both the caps is given by,
=[CAS of hemisphere] + [CSA of frustum]
=[2 pi r^ 2 ]+[ pi(r 1 +r 2 )l]
=[2*(22/7)*7^ 2 ]+[(22/7)*(10+4)*15]
=[2*22*7]+[(22/7)*(14)*15]
=308+[22*2*15]
=308+660
=968 cm^ 2