The area and the perimeter of the triangle with sides $10,8,6$ are equal. Find all the triangles with integral sides whose area and perimeter are equal.
1. Let \(a, b, c\) be the sides of the triangle. The perimeter \(p\) is given by:
   \[
   p = a + b + c
   \]
   The semiperimeter \(s\) is:
   \[
   s = \frac{p}{2}
   \]
   The area \(A\) of the triangle can be found using Heron's formula:
   \[
   A = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   Given that the area and the perimeter are equal, we have:
   \[
   A = p
   \]

2. The inradius \(r\) of the triangle is given by:
   \[
   r = \frac{A}{s}
   \]
   Since \(A = p\), we have:
   \[
   r = \frac{p}{s} = \frac{2p}{p} = 2
   \]

3. Consider the figure below. Let \(x = r \tan \frac{\theta_1}{2}\), \(y = r \tan \frac{\theta_2}{2}\), and \(z = r \tan \left( \frac{360^\circ - \theta_1 - \theta_2}{2} \right)\). Using the tangent addition formula, we get:
   \[
   z = r \frac{\tan \frac{\theta_1}{2} + \tan \frac{\theta_2}{2}}{1 - \tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2}}
   \]
   Substituting \(x\) and \(y\), we have:
   \[
   z = \frac{x + y}{\frac{xy}{r^2} - 1} = \frac{r^2 (x + y)}{xy - r^2} = \frac{4(x + y)}{xy - 4}
   \]
   Since \(r = 2\), we have:
   \[
   z \in \mathbb{N} \implies xy - 4 \mid 4(x + y)
   \]
   This implies:
   \[
   xy - 4 \leq 4(x + y)
   \]
   Dividing both sides by \(xy\), we get:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \geq \frac{1}{4}
   \]
   Simplifying further, we get:
   \[
   \left(1 + \frac{1}{x}\right) \left(1 + \frac{1}{y}\right) \geq \frac{5}{4}
   \]

4. Let \(x \geq y\). Then:
   \[
   1 + \frac{1}{x} \leq 1 + \frac{1}{y}
   \]
   Therefore:
   \[
   \left(1 + \frac{1}{y}\right)^2 \geq \left(1 + \frac{1}{x}\right) \left(1 + \frac{1}{y}\right) \geq \frac{5}{4}
   \]
   Solving for \(y\), we get:
   \[
   y \leq 2(\sqrt{5} + 2) \approx 8
   \]

5. We now consider different cases for \(y\):

   - **Case I: \(y = 1\)**
     \[
     z = \frac{4 + 4x}{x - 4} = 4 + \frac{20}{x - 4}
     \]
     Solving for \(x\), we get:
     \[
     x - 4 \in \{1, 2, 4, 5, 10, 20\} \implies x \in \{5, 6, 8, 9, 14, 24\}
     \]
     Correspondingly:
     \[
     z \in \{24, 14, 9, 8, 5, 6\}
     \]
     This gives the following sets of sides:
     \[
     \{6, 25, 29\}, \{7, 15, 20\}, \{9, 10, 17\}
     \]

   - **Case II: \(y = 2\)**
     \[
     z = \frac{8 + 4x}{2x - 4}
     \]
     Solving similarly, we get the following pairs:
     \[
     \{5, 12, 13\}, \{6, 8, 10\}
     \]

   - **Case III: \(y = 3\)**
     \[
     \text{No solutions}
     \]

   - **Case IV: \(y = 4\)**
     \[
     \text{Same as } \{6, 8, 10\}
     \]

   - **Case V: \(y = 5\)**
     \[
     \text{No solutions (only solution has } y > x\text{)}
     \]

   - **Case VI: \(y = 6\)**
     \[
     \text{No solutions (all solutions have } y > x\text{)}
     \]

   - **Case VII: \(y = 7\)**
     \[
     \text{No solutions}
     \]

   - **Case VIII: \(y = 8\)**
     \[
     \text{Same as } \{9, 10, 17\}
     \]

6. Therefore, the only solutions are:
   \[
   \{6, 25, 29\}, \{7, 15, 20\}, \{9, 10, 17\}, \{5, 12, 13\}, \{6, 8, 10\}
   \]

The final answer is \( \boxed{ \{6, 25, 29\}, \{7, 15, 20\}, \{9, 10, 17\}, \{5, 12, 13\}, \{6, 8, 10\} } \)