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?

First of all, you should know: the cone's axial section is an isosceles triangle, its height is equal to the cone's height, the ball's axial section is a circle, and the radius of the circle is equal to the radius of the ball
The height of the isosceles triangle of the cone axis section is equal to 3 times of the radius of its inscribed circle. According to this condition and the above two conditions, we can know that the cross-section triangle is an equilateral triangle. (the proof method is: the area of the triangle is equal to the bottom * height / 2, also equal to the radius of three sides and * inscribed circle / 2, and then from the height to 3 times of the radius, we can get that the bottom and waist are equal.)
Now draw a picture (an isosceles triangle, an inscribed circle, let the radius of the inscribed circle be r, then we can get the side length of the triangle a = 2 √ 3R). The generatrix and bottom length of the cone are both a, and the radius of the ball is R. it's easy to find the following
The result is: 3:2

If the radius of the cone is 1 and the length of the generatrix is 3, the surface area of the inscribed sphere of the cone is calculated

It can be transformed into finding the radius of the inscribed circle (radius of the ball) of an isosceles triangle whose base is 2 and waist is 3

In rectangular paper ABCD, ab = 3, BC = 4, fold along diagonal BD (make triangle abd and triangle EBD in the same plane) to find the weight of angle abd and triangle EBD
Finding the overlapping area of angular abd and triangular EBD

ABCD is a rectangle
∴AD∥BC
■ ∠ FDB = ∠ CBD (AD and be are handed over to f)
∵∠CBD=∠FBD
∴∠FDB=∠FBD
∴DF=BF
∵AF=AD-DF=4-DF
In RT △ ABF:
BF²=AF²+AB²
DF²=(4-DF)²+3²
DF=25/8
The overlap area of ∧ abd and ∧ EBD
=1/2DF×AB
=1/2×25/8×3
=75/16

Fold △ abd along the diagonal BD of rectangle ABCD to get △ a ′ BD, a ′ D intersects BC with F, as shown in the figure, △ BDF is what kind of triangle? Please give reasons

The ∵ quadrilateral ABCD is a rectangle, ∵ ad ∥ BC, ∵ 1 = ∠ 2, ∵ fold △ abd along the diagonal BD of rectangular ABCD to get ∵ a ′ BD, ≌ abd ≌ a ′ BD, ∵ 2 = ∠ 3, ∵ 1 = ∠ 3,

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?