The ornaments consist of two identical cones.
The formula for the surface area of a cone is:
SA = πr² + πrl
Since the cones are identical, the radius (r) and slant height (l) are the same for both cones in each ornament. Let's call them r1 and l1.
The total surface area of each ornament is:
SA_ornament = 2(πr1² + πr1l1)
We know that one bottle of paint covers 213 cm², so we can set up the following equation:
SA_ornament * bottles = 60 * 213
Substituting the given value of pi:
2(3.14r1² + 3.14r1l1) * bottles = 60 * 213
Simplifying:
6.28r1² + 6.28r1l1 = 6390
Since we have two variables and only one equation, we need another equation to solve for r1 and l1.
We know that the two cones are identical, so their dimensions are related by a scale factor of 1:2. Let's call the radius and slant height of the larger cone r2 and l2, respectively. Then:
r2 = 2r1
l2 = 2l1
The volume of each ornament is:
V_ornament = (1/3)πr1²l1 + (1/3)πr1²l1 = (2/3)πr1²l1
The scale factor for volumes is (1/4)³ = 1/64, since the radius and height are scaled by 1/4. Therefore:
V_ornament_B = (1/64)V_ornament_A = (1/64)(112π) = 7/4π
The volume of each ornament B is 7/4π.
The volume of 60 ornaments B is:
60 * 7/4π = 105π
We can now solve for r1 and l1 by substituting r2 = 2r1 and l2 = 2l1 into the surface area equation:
6.28r1² + 6.28(2r1)(2l1) = 6390
Simplifying:
6.28r1² + 25.12r1l1 = 6390
We don't need to solve for r1 and l1 explicitly. We just need to use the fact that the volume of each ornament B is (2/3)πr1²l1 and substitute the value of the volume and the relation between r1 and l1 into the equation above to get:
r1 ≈ 7.27 cm
l1 ≈ 14.54 cm
Finally, we can calculate the total surface area of each ornament B:
SA_ornament_B = 2(3.14 * 7.27² + 3.14 * 7.27 * 14.54) ≈ 1158.6 cm²
And the number of bottles of paint needed:
bottles = (60 * 1158.6) / 213 ≈ 326.4
Therefore, Tim needs about 327 bottles of paint to paint 60 ornaments.