Respuesta :
To find the volume of the cube which remains, we will subtract the volume of the cone that was removed from the volume of the original cube, as follows:
[tex]\begin{gathered} \text{V}_{cube\text{ remaining }}=V_{orig\text{ inal cube}}-V_{cone} \\ \text{where:} \\ V_{orig\text{ inal cube}}=s^3 \\ V_{cone}=\frac{1}{3}\pi\times r^2h \end{gathered}[/tex]Where:
s= the length of each side of the cube
r= the base radius of the cone
h= the hight of the cone
According to the question:
s= 6mm
r= 2mm
h= 6mm
Therefore:
[tex]\begin{gathered} V_{orig\text{ inal cube}}=s^3=6^3=6\times6\times6=216mm^3 \\ V_{cone}=\frac{1}{3}\pi\times r^2h\text{ = }\frac{1}{3}\pi\times2^2\times6=\text{ }\frac{1}{3}\pi\text{ }\times4\times6\text{ = }\frac{24}{3}\pi\text{ = }\frac{24\text{ }\times3.142}{3}=25.136^{} \\ V_{cone}=25.136mm^3 \end{gathered}[/tex]Finally:
[tex]\begin{gathered} \text{V}_{cube\text{ remaining }}=V_{orig\text{ inal cube}}-V_{cone} \\ \text{V}_{cube\text{ remaining }}=216_{}-25.136=190.864\operatorname{mm}^3 \end{gathered}[/tex]Therefore, the volume of the remaining cube is 190.864 cubic millimeter
Answer: Option J
