A tetrahedron has edges extending from one vertex that are pairwise perpendicular to each other. The lengths of the edges are $9 \mathrm{~cm}, 12 \mathrm{~cm}, 16 \mathrm{~cm}$. What is the height of the tetrahedron from this vertex?
We write the volume of the tetrahedron in two ways, and from this, we will calculate the desired height \( m \) of the body.

\[
V=\frac{\frac{9 \cdot 16}{2} \cdot 12}{3}=\frac{T_{D B C \Delta} \cdot m}{3}
\]

The edges starting from vertex \( D \) form three right-angled triangles, whose hypotenuses are:

\[
B A=\sqrt{9^{2}+12^{2}}=15, \quad A C=\sqrt{12^{2}+16^{2}}=20, \quad B C=\sqrt{9^{2}+16^{2}}=\sqrt{337}
\]

The area of triangle \( A B C \) can be calculated using the formula \( T=\frac{b c \sin \alpha}{2} \). For this, we first write the cosine rule using the three sides:

\[
\cos \alpha=\frac{400+225-337}{2 \cdot 20 \cdot 15}=\frac{12}{25}
\]

from which \(\sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\frac{\sqrt{481}}{25}\). Thus,

\[
V=\frac{\frac{9 \cdot 16}{2} \cdot 12}{3}=\frac{\frac{15 \cdot 20 \cdot \frac{\sqrt{481}}{25}}{2} \cdot m}{3}
\]

from which

\[
m=\frac{9 \cdot 8 \cdot 12 \cdot 25}{15 \cdot 10 \cdot \sqrt{481}}=\frac{144}{\sqrt{481}} \approx 6.5658 \text{ cm}
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_915fb3c406ed7d8a7818g-1.jpg?height=868&width=374&top_left_y=1084&top_left_x=865)