Tetrahedron- Inscribed Sphere

In tetrahedron $ABCD, AB=BC=CA$ and $DA=DB=DC$. Given that the altitude of $ABCD$ from point $D$ is $24$ and that the radius of the inscribed sphere of $ABCD$ is $11$, determine $AB$.

If we call $A1$, $A2$, $A3$, and $A4$ the areas of the faces of a tetrahedron, the radius $R$ of the inscribed sphere can be calculated using the formula

$$V = \frac{1}{3} R (A1 + A2 + A3 + A4)$$

where V is the volume of the tetrahedron.

In this case, calling $s$ the length of $AB=BC=CA$, we have that:

• the area of the base is $s^2 \frac{\sqrt{3}}{4}$;

• the volume is given by $V=\frac{1}{3} s^2 \frac{\sqrt{3}}{4} \cdot 24= 2 \sqrt{3} s^2$;

• the areas of the lateral faces are given by $\frac{s}{2} \sqrt{24^2+(s \frac{\sqrt{3}}{6})^2}=\frac{s}{2} \sqrt{576+s^2/12}$.

We then can write

$$2 \sqrt{3} s^2= \frac{1}{3} \cdot 11 (s^2 \frac{\sqrt{3}}{4}+ 3 \cdot \frac{s}{2} \sqrt{576+s^2/12})$$

which solved for $s$ gives $s=132$.