If the area bounded by the curve `y+x^(2)=8x` and the line `y=12` is K sq. units, then the vlaue of `(3K)/(10)` is

Answer

To solve the problem, we need to find the area bounded by the curve \( y + x^2 = 8x \) and the line \( y = 12 \). Then we will compute the value of \( \frac{3K}{10} \), where \( K \) is the area we find.

### Step-by-step Solution:

1. **Rearranging the Curve Equation**:
The equation of the curve can be rearranged as:
\[ y = 8x - x^2 \]
This is a downward-opening parabola.

2. **Finding Points of Intersection**:
To find the area bounded by the curve and the line \( y = 12 \), we need to find the points where these two equations intersect. Set the equations equal to each other:
\[ 8x - x^2 = 12 \]
Rearranging gives:
\[ x^2 - 8x + 12 = 0 \]

3. **Solving the Quadratic Equation**:
We can factor the quadratic:
\[ (x - 2)(x - 6) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = 6 \]
The points of intersection are \( (2, 12) \) and \( (6, 12) \).

4. **Setting Up the Integral for Area**:
The area \( K \) between the curve and the line from \( x = 2 \) to \( x = 6 \) can be found using the integral:
\[ K = \int_{2}^{6} \left( (8x - x^2) - 12 \right) \, dx \]
Simplifying the integrand:
\[ K = \int_{2}^{6} (8x - x^2 - 12) \, dx = \int_{2}^{6} (-x^2 + 8x - 12) \, dx \]

5. **Calculating the Integral**:
Now we calculate the integral:
\[ K = \int_{2}^{6} (-x^2 + 8x - 12) \, dx \]
The antiderivative is:
\[ -\frac{x^3}{3} + 4x^2 - 12x \]
Evaluating from 2 to 6:
\[ K = \left[ -\frac{6^3}{3} + 4(6^2) - 12(6) \right] - \left[ -\frac{2^3}{3} + 4(2^2) - 12(2) \right] \]

6. **Calculating Each Term**:
For \( x = 6 \):
\[ -\frac{216}{3} + 4(36) - 72 = -72 + 144 - 72 = 0 \]
For \( x = 2 \):
\[ -\frac{8}{3} + 16 - 24 = -\frac{8}{3} - 8 = -\frac{32}{3} \]
Thus,
\[ K = 0 - \left(-\frac{32}{3}\right) = \frac{32}{3} \]

7. **Finding \( \frac{3K}{10} \)**:
Now we compute \( \frac{3K}{10} \):
\[ \frac{3K}{10} = \frac{3 \cdot \frac{32}{3}}{10} = \frac{32}{10} = 3.2 \]

### Final Answer:
The value of \( \frac{3K}{10} \) is \( \boxed{3.2} \).