A right circular cone is placed over a cylinder of the same radius. No...
Cylinder painted area = 2π rh+π r^2
Cone painted area = π rl
2h+r/√ (r^2 +h1^2 ) = 3:1
h = 6
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A right circular cone is placed over a cylinder of the same radius. No...
Problem Analysis:
We have a right circular cone and a cylinder of the same radius with the cone placed over the cylinder. The combined structure is painted on all sides and then separated. We need to find the height of the cylinder if the height of the cone is 4 m and radius is 3 m, and the ratio of the area painted on the cylinder to the cone is 3:1.
Solution:
Let the height of the cylinder be 'h' meters.
We know that the cone and the cylinder have the same radius, so the area of the cylinder painted is only the curved surface area (CSA) of the cylinder, and the area of the cone painted is the total surface area (TSA) of the cone.
The TSA of the cone is given by:
TSA of cone = πr(l + r) square units
where r is the radius of the cone and l is the slant height of the cone.
In this case, r = 3 m and l = 5 m (using Pythagoras theorem)
So, TSA of cone = π3(5 + 3) = 24π sq. m
The CSA of the cylinder is given by:
CSA of cylinder = 2πrh square units
where r is the radius of the cylinder and h is the height of the cylinder.
In this case, r = 3 m and CSA of cylinder/TSA of cone = 3/1
So, CSA of cylinder = 3/4 × TSA of cone = 18π sq. m
Therefore, 2πrh = 18π
h = 3
Therefore, the height of the cylinder is 6 meters.
Answer: Option (b) 6 m.
A right circular cone is placed over a cylinder of the same radius. No...
Cylinder painted area = 2πrh+πr²
Cone painted area = πrl
2h+r/√ (r² +h1² ) = 3:1
h = 6