If the area bounded by `y+|x-pi|le pi and y ge (pi)/(2)` is `Kpi^(2)` sq. units, then the value of 8K is equal to

Answer

To solve the problem, we need to find the area bounded by the equations \( y + |x - \pi| \leq \pi \) and \( y \geq \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Understand the Equations**:
- The first inequality can be rewritten as:
\[ y \leq \pi - |x - \pi| \]
- The second inequality is:
\[ y \geq \frac{\pi}{2} \]

2. **Graph the First Inequality**:
- The equation \( y = \pi - |x - \pi| \) represents a V-shaped graph that opens downwards with its vertex at \( ( \pi, \pi ) \).
- The two lines that form the V are:
- \( y = \pi - (x - \pi) \) for \( x \geq \pi \) which simplifies to \( y = 2\pi - x \)
- \( y = \pi - (\pi - x) \) for \( x < \pi \) which simplifies to \( y = x \)

3. **Graph the Second Inequality**:
- The line \( y = \frac{\pi}{2} \) is a horizontal line that intersects the y-axis at \( \frac{\pi}{2} \).

4. **Find Intersection Points**:
- Set \( y = \frac{\pi}{2} \) equal to the two equations from the first inequality:
- For \( y = 2\pi - x \):
\[ \frac{\pi}{2} = 2\pi - x \implies x = \frac{3\pi}{2} \]
- For \( y = x \):
\[ \frac{\pi}{2} = x \implies x = \frac{\pi}{2} \]

5. **Determine the Area**:
- The area bounded by the lines \( y = \frac{\pi}{2} \), \( y = 2\pi - x \), and \( y = x \) forms a trapezium.
- The vertices of the trapezium are \( \left( \frac{\pi}{2}, \frac{\pi}{2} \right) \), \( \left( \frac{3\pi}{2}, \frac{\pi}{2} \right) \), \( \left( \frac{3\pi}{2}, \frac{\pi}{2} \right) \), and \( \left( \pi, \pi \right) \).

6. **Calculate the Area of the Trapezium**:
- The lengths of the parallel sides are:
- Top side: \( y = 2\pi - x \) at \( x = \frac{3\pi}{2} \) gives \( y = \frac{\pi}{2} \)
- Bottom side: \( y = \frac{\pi}{2} \)
- The height of the trapezium is the difference in y-values:
\[ \text{Height} = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]
- The area \( A \) of the trapezium is given by:
\[ A = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \]
- Substituting the values:
\[ A = \frac{1}{2} \times \left( \frac{3\pi}{2} - \frac{\pi}{2} \right) \times \frac{\pi}{2} = \frac{1}{2} \times \pi \times \frac{\pi}{2} = \frac{\pi^2}{4} \]

7. **Relate Area to K**:
- We know that the area is \( K\pi^2 \), so:
\[ K\pi^2 = \frac{\pi^2}{4} \implies K = \frac{1}{4} \]

8. **Find the Value of \( 8K \)**:
- Finally, calculate \( 8K \):
\[ 8K = 8 \times \frac{1}{4} = 2 \]

### Final Answer:
The value of \( 8K \) is \( 2 \).