To solve this problem, let's start by finding the individual components of solid A.
Let the radius of the hemisphere in solid A be denoted as r, and the height of the cylinder be denoted as h.
The volume of a hemisphere is given by V_hemisphere = (2/3)πr^3, and the volume of a cylinder is given by V_cylinder = πr^2h.
Given that the volume of solid A is 792π, we can set up the equation:
(2/3)πr^3 + πr^2h = 792π
To simplify the equation, we can divide both sides by π:
(2/3)r^3 + r^2h = 792
Now, let's consider solid B. Since it has a similar shape to solid A, the ratio of their volumes is the same as the ratio of their surface areas.
The volume of solid B is given as 99π, so we can set up the equation:
(2/3)r_b^3 + r_b^2h_b = 99
Given that the ratio of the radius to the height of the cylinder is 1:3, we can express h in terms of r as h = 3r.
Substituting this into the equations, we have:
(2/3)r^3 + r^2(3r) = 792
(2/3)r_b^3 + r_b^2(3r_b) = 99
Simplifying the equations further, we get:
(2/3)r^3 + 3r^3 = 792
(2/3)r_b^3 + 3r_b^3 = 99
Combining like terms:
(8/3)r^3 = 792
(8/3)r_b^3 = 99
To isolate r^3 and r_b^3, we divide both sides by (8/3):
r^3 = 297
r_b^3 = 37.125
Now, let's calculate the surface areas of solid A and solid B.
The surface area of a hemisphere is given by A_hemisphere = 2πr^2, and the surface area of a cylinder is given by A_cylinder = 2πrh.
For solid A, the surface area is:
A_a = 2πr^2 (hemisphere) + 2πrh (cylinder)
A_a = 2πr^2 + 2πrh
A_a = 2πr^2 + 2πr(3r) (substituting h = 3r)
A_a = 2πr^2 + 6πr^2
A_a = 8πr^2
For solid B, the surface area is:
A_b = 2πr_b^2 (hemisphere) + 2πr_bh_b (cylinder)
A_b = 2πr_b^2 + 2πr_b(3r_b) (substituting h_b = 3r_b)
A_b = 2πr_b^2 + 6πr_b^2
A_b = 8πr_b^2
Now, let's calculate the ratio of the surface areas:
Ratio = A_a : A_b
Ratio = 8πr^2 : 8πr_b^2
Ratio = r^2 : r_b^2
Ratio = (297) : (37.125)
Ratio = 8 : 1
Therefore, the ratio of the surface areas is 1:8.