If the height of the circumscribed cone of a sphere is three times the radius of the sphere, what is the ratio of the side area of the cone to the surface area of the sphere? It's a process. There is also a question: a cylindrical cylinder with a bottom radius of R is filled with an appropriate amount of water. If a solid iron ball with a radius of R is put in, the height of the water surface will just rise R. R/r？ 、

If the height of the circumscribed cone of a sphere is three times the radius of the sphere, what is the ratio of the side area of the cone to the surface area of the sphere? It's a process. There is also a question: a cylindrical cylinder with a bottom radius of R is filled with an appropriate amount of water. If a solid iron ball with a radius of R is put in, the height of the water surface will just rise R. R/r？ 、

1. Tip: the height of the circumscribed cone of the ball is three times of the radius of the ball. It tells you that the bottom of the cone and the cone of the cone are tangent to the ball, the radius of the ball is r, and the height of the cone is 3R;
Let the vertex of the cone be a, the center of the bottom be B, BC be the radius of the bottom, AC be the generatrix, d be the tangent point, o be the center of the ball, OD = R be the radius of the ball,
Because AB = 3od, OD ⊥ AC, Ao = 2od
So ∠ oad = 30 degree, AC = AB / cos30 degree = 3R / cos30 degree = (2 √ 3) r,
BC=AB*tan30°=3R* tan30°=(√3)R
Side area of cone = π (√ 3) r (2 √ 3) r = 6 π R ^ 2
Sphere area = 4 π R ^ 2 [sphere surface area = 4 π R ^ 2, R --- sphere radius]
The ratio of the side area of the cone to that of the sphere is 6 π R ^ 2 / 4 π R ^ 2 = 3 / 2
2. The volume of the sphere V1 = 4 / 3 π R ^ 3
When the water level of the cylinder rises R, the volume of the sphere causes the water level to rise, and the volume V2 is π R ^ 2 * r
V1=V2
4/3πr^3=πR^2*r
R/r=2√3