Let ABCD be a square such that the side AB is on the line `2x-y=17` and two vertices C and D are on the parabola `y=x^(2)`. If length of side of ABCD is less than 10 and its area is A, then A is equal to

Answer

To solve the problem, we need to find the area \( A \) of the square \( ABCD \) given that two vertices \( C \) and \( D \) lie on the parabola \( y = x^2 \) and the side \( AB \) is on the line \( 2x - y = 17 \). We also know that the length of the side of the square is less than 10.

### Step-by-Step Solution:

1. **Identify the line equation**: The line equation is given as \( 2x - y = 17 \). We can rewrite it in slope-intercept form:
\[ y = 2x - 17 \]

2. **Determine the coordinates of points A and B**: Let's assume point \( A \) is at \( (x_1, y_1) \) and point \( B \) is at \( (x_2, y_2) \) on the line. Since \( AB \) is horizontal, we can set \( y_1 = y_2 \). Thus, we can take \( A = (x_1, 2x_1 - 17) \) and \( B = (x_1 + a, 2x_1 - 17) \), where \( a \) is the length of the side of the square.

3. **Vertices C and D on the parabola**: The vertices \( C \) and \( D \) will have coordinates \( C = (x_1, y_1 + a) \) and \( D = (x_1 + a, y_1 + a) \). Since \( C \) and \( D \) lie on the parabola \( y = x^2 \), we can express their coordinates as:
\[ C = (x_1, (x_1)^2) \quad \text{and} \quad D = (x_1 + a, (x_1 + a)^2) \]

4. **Set the y-coordinates equal**: Since \( C \) and \( D \) must have the same y-coordinate:
\[ (x_1 + a)^2 = (x_1)^2 + a \]

5. **Expand and simplify**:
\[ x_1^2 + 2ax_1 + a^2 = x_1^2 + a \]
\[ 2ax_1 + a^2 - a = 0 \]

6. **Factor the equation**:
\[ a(2x_1 + a - 1) = 0 \]
This gives us two cases: \( a = 0 \) or \( 2x_1 + a - 1 = 0 \).

7. **Solve for \( a \)**: From \( 2x_1 + a - 1 = 0 \):
\[ a = 1 - 2x_1 \]

8. **Find the area \( A \)**: The area of the square is given by:
\[ A = a^2 = (1 - 2x_1)^2 \]

9. **Maximize the area under the constraint**: Since the length of the side \( a < 10 \):
\[ 1 - 2x_1 < 10 \implies -2x_1 < 9 \implies x_1 > -\frac{9}{2} \]

10. **Evaluate the area**: To find the maximum area, we can substitute values for \( x_1 \) within the constraints. The maximum area occurs when \( a \) is maximized, which is when \( x_1 = 0 \):
\[ a = 1 \quad \text{and} \quad A = 1^2 = 1 \]

11. **Check the maximum area condition**: Since \( a < 10 \) is satisfied, we can conclude that the maximum area \( A \) is:
\[ A = 1 \]

### Final Answer:
The area \( A \) is equal to \( 1 \).