4A. Let $ABCD$ be a triangular pyramid where all three angles at vertex $D$ are right angles. Calculate the height of the pyramid dropped from vertex $D$, given the lengths of the edges $\overline{AB}=\overline{AC}=5$ and $\overline{BC}=4\sqrt{2}$.
Solution. Triangles $ABD$ and $ACD$ are congruent, as they are right triangles with vertex at $D$, a common leg $AD$, and equal hypotenuses $\overline{AB}=\overline{AC}$. From the congruence of these triangles, we get $\overline{BD}=\overline{CD}$. We will denote this length by $x$. Triangle $BCD$ is a right triangle with vertex at $D$. According to the Pythagorean theorem, for its sides we have $x^{2}+x^{2}=(4 \sqrt{2})^{2}$, from which we get $x=4$.

Let's denote the length of segment $AD$ by $y$. Triangle $ADB$ is a right triangle with vertex at $D$. According to the Pythagorean theorem, for its sides we have $x^{2}+y^{2}=5^{2}$, from which we get $y=3$.

Since triangle $ABC$ is isosceles with vertex at $A$, the height dropped from vertex $A$ to the base $BC$ (whose length we denote by $h$) bisects it. According to the Pythagorean theorem, we get

$$
h=\sqrt{5^{2}-(2 \sqrt{2})^{2}}=\sqrt{17}
$$

Let's denote the unknown height of the pyramid by $H$. The volume of the pyramid is

$$
\begin{aligned}
V & =\frac{1}{3} P_{ABC} H=\frac{1}{3} \cdot \frac{1}{2} \cdot 4 \sqrt{2} \cdot \sqrt{17} \cdot H \\
& =\frac{2}{3} \sqrt{34} \cdot H
\end{aligned}
$$

On the other hand, since the angles at vertex $D$ are right, the volume of the pyramid is

$$
\begin{aligned}
V & =\frac{1}{3} \cdot P_{BCD} \cdot y=\frac{1}{6} x \cdot x \cdot y \\
& =\frac{1}{6} \cdot 4 \cdot 4 \cdot 3=8
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_b87d47f24deda06cfa28g-03.jpg?height=599&width=613&top_left_y=609&top_left_x=847)

Since $V=\frac{2}{3} \cdot \sqrt{34} \cdot H$, we get $H=\frac{12}{\sqrt{34}}=\frac{6}{17} \sqrt{34}$.