Certainly! Let's break down the problem step by step:
1. Original Dimensions of the Cone:
- Height of the cone, [tex]\( h_{\text{cone}} = 16.8 \)[/tex] cm.
- Radius of the base of the cone, [tex]\( R_{\text{cone}} = 4.2 \)[/tex] cm.
2. Volume of the Cone:
The formula for the volume of a cone is given by:
[tex]\[
V_{\text{cone}} = \frac{1}{3} \pi R_{\text{cone}}^2 h_{\text{cone}}
\][/tex]
Substituting the given values:
[tex]\[
V_{\text{cone}} = \frac{1}{3} \pi (4.2)^2 (16.8)
\][/tex]
The calculated volume of the cone is approximately [tex]\( 310.34 \)[/tex] cubic cm.
3. Volume of the Sphere:
The cone is melted and recast into a sphere. Thus, the volume of the sphere, [tex]\( V_{\text{sphere}} \)[/tex], is equal to the volume of the cone:
[tex]\[
V_{\text{sphere}} = V_{\text{cone}} = 310.34 \text{ cubic cm}
\][/tex]
4. Formula for the Volume of a Sphere:
The formula for the volume of a sphere is given by:
[tex]\[
V_{\text{sphere}} = \frac{4}{3} \pi r_{\text{sphere}}^3
\][/tex]
Since [tex]\( V_{\text{sphere}} \)[/tex] is already known (310.34 cubic cm), we set up the equation:
[tex]\[
310.34 = \frac{4}{3} \pi r_{\text{sphere}}^3
\][/tex]
5. Solving for the Radius of the Sphere:
Isolate [tex]\( r_{\text{sphere}}^3 \)[/tex]:
[tex]\[
r_{\text{sphere}}^3 = \frac{3 \times 310.34}{4 \pi}
\][/tex]
Simplify the right-hand side:
[tex]\[
r_{\text{sphere}}^3 = 74.09
\][/tex]
Taking the cube root of both sides to find [tex]\( r_{\text{sphere}} \)[/tex]:
[tex]\[
r_{\text{sphere}} = \sqrt[3]{74.09}
\][/tex]
The calculated radius of the sphere is approximately [tex]\( 4.2 \)[/tex] cm.
Therefore, the radius of the sphere recast from the cone is [tex]\( 4.2 \)[/tex] cm.
