[Theorem of Three Perpendiculars]

The height of a right triangle $ABC$, dropped to the hypotenuse, is 9.6. From the vertex $C$ of the right angle, a perpendicular $CM$ is erected to the plane of triangle $ABC$, with $CM=28$. Find the distance from point $M$ to the hypotenuse $AB$.
Apply the theorem of three perpendiculars.

## Solution

Let $C K$ be the height of the given right triangle. Then $M K$ is the oblique to the plane of triangle $A B C$, and $C K$ is the orthogonal projection of this oblique on the plane of triangle $A B C$. Since $C K \perp A B$, by the theorem of three perpendiculars, $M K \perp A B$. Therefore, the length of segment $M K$ is the distance from point $M$ to line $A B$. From the right triangle $M C K$, by the Pythagorean theorem, we find that

$$
\begin{gathered}
M K=\sqrt{C K^{2}+C M^{2}}=\sqrt{9.6^{2}+28^{2}}=\sqrt{(48 / 5)^{2}+28^{2}}= \\
=4 \sqrt{(12 / 5)^{2}+7^{2}}=4 \sqrt{144+49 \cdot 25} / 5= \\
=4 \sqrt{1369} / 5=4 \cdot 37 / 5=148 / 5=29.6
\end{gathered}
$$

## Answer

29.6.