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To determine how many triangles can be formed from a decagon using diagonals from a common vertex, let's break down the problem step by step.

### Understanding the Decagon

A **decagon** is a polygon with **10 sides** and **10 vertices**. For the purpose of this problem, we'll focus on the diagonals drawn from a **specific common vertex**.

### Step 1: Identifying Diagonals from the Common Vertex

From any given vertex in a decagon:

- **Adjacent vertices**: These are the two vertices directly connected to the common vertex by sides of the decagon. Diagonals are **not** drawn to these vertices.
  
- **Non-adjacent vertices**: These are the remaining **7 vertices** (since 10 total vertices - 2 adjacent vertices - 1 common vertex = 7) to which diagonals can be drawn.

Therefore, from the common vertex, there are **7 possible diagonals**.

### Step 2: Forming Triangles with Diagonals

A triangle is formed by connecting the common vertex with two other vertices via diagonals. Essentially, each triangle is defined by choosing **2 distinct diagonals** from the common vertex.

- **Number of ways to choose 2 diagonals from 7**: This can be calculated using the combination formula:
  
  $$
  \text{Number of triangles} = \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21
  $$

### Conclusion

By selecting any two diagonals from the common vertex in a decagon, you can form **21 distinct triangles**.

**Answer:**  
21 triangles can be formed.
21 triangles can be formed from a decagon using diagonals from a common vertex.

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