Question 11 Mark
Write True or False and justify your answer in the following:
If a sphere is inscribed in a cube, then the ratio of the volume of the cube to the volume of the sphere will be $6:\pi.$
If a sphere is inscribed in a cube, then the ratio of the volume of the cube to the volume of the sphere will be $6:\pi.$
Answer
View full question & answer→True.Solution:
Volume of cube $(\text{V}_1)=(\text{Side})^ 3$
Radius of sphere $(\text{r})=\frac{\text{Side}}{2}$
Volume of sphere $(\text{V}_1)=\frac{4}{3}\pi(\text{r}^3)$
$=\frac{4}{3}\pi\Big(\frac{\text{Side}}{2}\Big)^3=\frac{4}{3}\pi\times\frac{(\text{Side})^3}{8}$
$\frac{\text{V}_1}{\text{V}_2}=\frac{(\text{Side})^3}{\frac{4}{3}\pi\frac{\text{(Side)}^3}{8}}=\frac{6}{\pi}\text{ or }6:\pi$
Hence, the ratio of the volume of the cube to the volume of the sphere is $6:\pi.$
Volume of cube $(\text{V}_1)=(\text{Side})^ 3$
Radius of sphere $(\text{r})=\frac{\text{Side}}{2}$
Volume of sphere $(\text{V}_1)=\frac{4}{3}\pi(\text{r}^3)$
$=\frac{4}{3}\pi\Big(\frac{\text{Side}}{2}\Big)^3=\frac{4}{3}\pi\times\frac{(\text{Side})^3}{8}$
$\frac{\text{V}_1}{\text{V}_2}=\frac{(\text{Side})^3}{\frac{4}{3}\pi\frac{\text{(Side)}^3}{8}}=\frac{6}{\pi}\text{ or }6:\pi$
Hence, the ratio of the volume of the cube to the volume of the sphere is $6:\pi.$
Clearly, $\angle\text{AOB}=90^\circ$ Let the radius of the base = r unit, height = h units and slant height = l unit, then $\text{l}^2=\text{h}^2+\text{r}^2\Rightarrow\text{l}=\sqrt{\text{h}^2+\text{r}^2}$