The altitude from the right angle to the hypotenuse splits the original right triangle into two smaller right triangles. How are those smaller triangles related to the original one?

Use $\triangle ABC \sim \triangle ACD$. Match sides around angle $A$: in $\triangle ABC$, sides $AC$ and $AB$ surround angle $A$; in $\triangle ACD$, sides $AD$ and $AC$ surround angle $A$. Write a ratio using these pairs and solve for $AD$.

In Desmos, type sqrt(6^2 + 8^2) to find the value of $AB$.

In a new line, type (6^2) / sqrt(6^2 + 8^2) to evaluate $\dfrac{AC^2}{AB}$. The value Desmos displays for this expression is the length of $AD$.

You have a right triangle $\triangle ABC$ with the right angle at $C$, so $AC$ and $BC$ are the legs and $AB$ is the hypotenuse.

An altitude is drawn from $C$ to the hypotenuse $AB$, meeting $AB$ at $D$. That means $CD$ is perpendicular to $AB$, and $D$ lies between $A$ and $B$.

You are asked to find the length of segment $AD$ along the hypotenuse.

Since $\triangle ABC$ is a right triangle with legs $AC = 6$ and $BC = 8$, use the Pythagorean theorem to find $AB$:

So $AB = 10$.

The altitude from the right angle to the hypotenuse in a right triangle creates two smaller right triangles that are each similar to the original triangle.

Specifically, $\triangle ABC \sim \triangle ACD$ because:

• They both have angle $A$ in common.
• Each has a right angle (at $C$ in $\triangle ABC$ and at $D$ in $\triangle ACD$).

From the similarity $\triangle ABC \sim \triangle ACD$, match corresponding sides:

• Hypotenuse $AB$ in $\triangle ABC$ corresponds to hypotenuse $AC$ in $\triangle ACD$.
• Leg $AC$ in $\triangle ABC$ corresponds to leg $AD$ in $\triangle ACD$.

So the ratio

which leads to

From the equation $AC^2 = AB \cdot AD$, solve for $AD$:

Substitute $AC = 6$ and $AB = 10$:

So the length of $AD$ is $\dfrac{18}{5}$, which corresponds to choice C.