A cone with diameter $12$cm and height $8$cm. Find the volume of the inscribed Sphere. A cone with diameter $12\ cm$ and height $8\ cm$. Find the volume of the inscribed Sphere. Can someone help me solve this maths problem?
 A: Notice, let $r$ be the radius of the sphere inscribed in the cone having diameter of base $12\ cm$ (radius, $R=6\ cm$) & the vertical height $H=8\ cm$. 
If $\alpha$ is the semi-apex angle of cone then using geometry in a right triangle we get 
$$\tan \alpha=\frac{R}{H}=\frac{6}{8}=\frac{3}{4}\ \ (\forall\ \ 0<\alpha<\pi/2)$$ 
$$\implies \sin \alpha=\frac{\tan \alpha}{\sqrt{1+\tan^2\alpha}}$$
$$=\frac{\frac{3}{4}}{\sqrt{1+\left(\frac{3}{4}\right)^2}}$$$$\sin \alpha=\frac{3}{5}$$
Using geometry, we also have $$\sin\alpha=\frac{r}{h-r}$$
$$\frac{r}{8-r}=\frac{3}{5}$$ $$8r=24\implies r=3\ $$
Hence, the volume of sphere inscribed into the cone is 
$$V=\frac{4\pi r^3}{3}=\frac{4\pi (3)^3}{3}=36 \pi$$
$$\color{red}{V=36\pi\ cm^3}$$
A: The cone has radius 6, height 8 and slant height 10. I leave it to you to draw a diagram involving Pythagorean triplet proportions of sides 3:4:5 ; 
Now the centre of circle touching two lines (slanted & base) is to be connected to base corner. It must be a bisector... whose property is ... proportion of segments on either side of the bisector to be same:
$$ \frac {6} {10} = \frac {x}  {8-x} $$
which fixes radius of base as 3 units. Now you can take it further to find volume.
