11.227 The lengths of the lateral edges of a triangular pyramid are equal to $a, b$, and $c$; the planar angles formed by these edges are right angles. Find the length of the height dropped to the base of the pyramid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-277.jpg?height=295&width=323&top_left_y=699&top_left_x=189)

Let in the given pyramid $FABC$ the lateral edges $FA, FB, FC$ be pairwise perpendicular and equal to $a, b$, and $c$ respectively. Draw the height $AK$ in $\triangle ABC$. $FA \perp FBC$ (Theorem 11), $BC \perp FK$ (Theorem 12), $BC \perp AFK$ (Theorem 11) $\Rightarrow AFK \perp ABC$ (Theorem 14) $\Rightarrow$ the height $FM$ of the pyramid $FABC$ lies in the plane $AFK$.

Obviously, $V_{FABC} = \frac{abc}{6}$. On the other hand,

$V_{FABC} = \frac{1}{3} S_{ABC} \cdot FM = \frac{AK \cdot BC \cdot FM}{2 \cdot 3}$. But $BC = \sqrt{b^2 + c^2}$, $AK = \sqrt{a^2 + FK^2}$,

$FK \cdot BC = FC \cdot FB = 2 S_{FBC} \Rightarrow FK = \frac{bc}{\sqrt{b^2 + c^2}} \Rightarrow$

$\Rightarrow AK = \frac{\sqrt{a^2 b^2 + a^2 c^2 + b^2 c^2}}{\sqrt{b^2 + c^2}} = \frac{\sqrt{a^2 b^2 + a^2 c^2 + b^2 c^2}}{BC} \Rightarrow \frac{1}{6} FM \cdot AK \cdot BC = $

$= \frac{1}{6} FM \cdot \sqrt{a^2 b^2 + a^2 c^2 + b^2 c^2} = \frac{abc}{6} \Rightarrow FM = \frac{abc}{\sqrt{a^2 b^2 + a^2 c^2 + b^2 c^2}}$.

Answer:

$$
\frac{abc}{\sqrt{a^2 b^2 + a^2 c^2 + b^2 c^2}}
$$