The tip of a solid metal cone was placed into a cube that has 10 inch edges, as shown.If the water in the cube rose from 6 inches to 8 inches when the cone was placed in the cube, what is theradius of the base of the submerged portion of the cone?

Answer:[tex]r=\sqrt{\dfrac{75}{\pi}}\approx 4.89\ in[/tex]Step-by-step explanation:If the water in the cube rose from 6 inches to 8 inches when the cone was placed in the cube, then the difference in volumes is the volume of the submerged portion of the cone.Initially, 10 in by 10 in by 6 in:[tex]V_{initial}=10\cdot 10\cdot 6=600\ in^2.[/tex]At he end, 10 in by 10 in by 8 in:[tex]V_{final}=10\cdot 10\cdot 8=800\ in^3.[/tex]Thus,[tex]V_{submerged \ cone}=800-600=200\ in^3.[/tex]Use cone's volume formula [tex]V_{cone}=\dfrac{1}{3}\pi r^2 \cdot h,[/tex]where r is the radius of the cone's base, h is the height of the cone.From the diagram, h=8 in, then[tex]200=\dfrac{1}{3}\cdot \pi\cdot r^2\cdot 8\\ \\\pi r^2=75\\ \\r^2=\dfrac{75}{\pi}\\ \\r=\sqrt{\dfrac{75}{\pi}}\approx 4.89\ in[/tex]