The altitude from the right angle to the hypotenuse creates two smaller right triangles inside the original triangle. How are these smaller triangles related to the original triangle in terms of angles and similarity?

In a right triangle, the altitude from the right angle to the hypotenuse is the geometric mean of the two segments it creates on the hypotenuse: if the altitude is $h$ and the segments are $p$ and $q$, then $h^2 = p \cdot q$. Apply this with $p = 9$ and $q = 16$ to find $h$.

In Desmos, type the expression sqrt(9*16) on a new line. The numerical value that Desmos outputs for this expression is the length of the altitude $CD$.

You have right triangle $\triangle ABC$ with the right angle at $C$, so $\overline{AB}$ is the hypotenuse.

The altitude from $C$ to $\overline{AB}$ meets $\overline{AB}$ at $D$, splitting the hypotenuse into two segments:

• $AD = 9$
• $DB = 16$

We are asked to find the length of the altitude $\overline{CD}$.

The altitude from the right angle to the hypotenuse creates two smaller right triangles, $\triangle ACD$ and $\triangle CBD$, both similar to the original $\triangle ABC$ (all three share the same acute angles).

From this similarity, there is a key relationship:

• The altitude to the hypotenuse is the geometric mean of the two segments of the hypotenuse.

If the altitude is $h$ and the hypotenuse is split into segments of lengths $p$ and $q$, then:

• $h^2 = p \cdot q$.

In this problem, $h$ corresponds to $CD$, $p$ to $AD = 9$, and $q$ to $DB = 16$, so we get:

• $CD^2 = 9 \cdot 16$.

From the relationship we found:

• $CD^2 = 9 \cdot 16$
• $CD^2 = 144$

Take the square root of both sides:

• $CD = \sqrt{144}$
• $CD = 12$

So, the length of altitude $\overline{CD}$ is $12$.