An equilateral triangle's sides increase at the rate of 2cm/sec. If the area of its incircle increases at a rate of `kcm^(2)//sec` (when the length of the side is `(6)/(pi)cm`), then the value of k is

Answer

To solve the problem, we need to find the rate of increase of the area of the incircle of an equilateral triangle when the side length is \( \frac{6}{\pi} \) cm and the sides of the triangle are increasing at a rate of 2 cm/sec.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( x \) be the length of a side of the equilateral triangle.
- The radius \( r \) of the incircle can be expressed in terms of \( x \).

2. **Find the Radius of the Incircle**:
- For an equilateral triangle, the radius \( r \) of the incircle is given by:
\[ r = \frac{x \sqrt{3}}{6} \]

3. **Find the Area of the Incircle**:
- The area \( A \) of the incircle is given by:
\[ A = \pi r^2 = \pi \left(\frac{x \sqrt{3}}{6}\right)^2 = \pi \frac{3x^2}{36} = \frac{\pi x^2}{12} \]

4. **Differentiate the Area with Respect to Time**:
- To find the rate of change of the area with respect to time, differentiate \( A \):
\[ \frac{dA}{dt} = \frac{d}{dt}\left(\frac{\pi x^2}{12}\right) = \frac{\pi}{12} \cdot 2x \frac{dx}{dt} = \frac{\pi x}{6} \frac{dx}{dt} \]

5. **Substitute Known Values**:
- We know that \( \frac{dx}{dt} = 2 \) cm/sec and we need to evaluate this when \( x = \frac{6}{\pi} \):
\[ \frac{dA}{dt} = \frac{\pi \left(\frac{6}{\pi}\right)}{6} \cdot 2 = \frac{6}{6} \cdot 2 = 2 \text{ cm}^2/\text{sec} \]

6. **Conclusion**:
- Since we are given that the area of the incircle increases at a rate of \( k \) cm²/sec, we find that:
\[ k = 2 \]

### Final Answer:
The value of \( k \) is \( 2 \).