Question 8·Medium·Lines, Angles, and Triangles

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.