The problem is asking to find $AM$, where the following is given:

• $AB \parallel \alpha$, $AK \cap \alpha = M$,
• $\angle C = 90^\circ$,
• $AB = 15$,
• $AC = 9$.

Here are the steps to solve the problem:

Step 1: Analyze the Given Geometry

The triangle $\triangle ABC$ has:

• $\angle C = 90^\circ$, making $\triangle ABC$ a right triangle.
• $AB = 15$ is the hypotenuse.
• $AC = 9$ is one leg.

Step 2: Use the Pythagorean Theorem to Find $BC$

$AB^2 = AC^2 + BC^2$ $15^2 = 9^2 + BC^2$ $225 = 81 + BC^2$ $BC^2 = 144 \implies BC = 12$

Step 3: Find the Coordinates of $M$

Since $AB \parallel \alpha$ and $AK$ intersects $\alpha$ at $M$, $M$ must be the projection of $A$ onto the plane $\alpha$. This essentially forms a right triangle with $AM$ as the altitude.

Step 4: Use Triangle Similarity

From the geometry, $\triangle AMC \sim \triangle ABC$ (since they share $\angle C = 90^\circ$).

The ratio of corresponding sides is: $\frac{AM}{AC} = \frac{BC}{AB}$

Substitute the known values: $\frac{AM}{9} = \frac{12}{15}$

Simplify: $AM = 9 \cdot \frac{12}{15}$

$AM = 7.2$

Final Answer:

$AM = 7.2$

Would you like a more detailed explanation or clarifications? Here are 5 related questions you might explore:

1. How is the Pythagorean theorem applied to right triangles?
2. What are the properties of projections in geometry?
3. How does triangle similarity work in geometric problems?
4. Why is $AB \parallel \alpha$ significant in this problem?
5. Can this solution be visualized using a coordinate system?

Tip: Always identify key right triangles in geometry problems—they often simplify calculations!