The question asks to "Decompose the regular pentagon into only triangles."

To decompose a regular pentagon into triangles, follow these steps:

1. Draw diagonals from one vertex: Pick any vertex of the pentagon and draw lines (diagonals) to all non-adjacent vertices of the pentagon.
2. Count the triangles: This process divides the pentagon into exactly 3 triangles.

For example, if you choose a vertex and connect it to two opposite (non-adjacent) vertices, you will see that the pentagon breaks down into three triangular regions.

Would you like further details or clarification?

Here are five related questions to expand on this concept:

1. How many triangles can you decompose a regular hexagon into?
2. What is the sum of interior angles of a pentagon?
3. How does decomposing shapes into triangles help in calculating their area?
4. Can you decompose an irregular pentagon the same way as a regular one?
5. How many diagonals does a regular pentagon have?

Tip: The sum of the interior angles of any polygon can be calculated using the formula: $(n-2) \times 180^\circ$, where $n$ is the number of sides.