For a given point A(0,0), ABCD is a rhombus of side 10 units where slope of AB is `4/3`and slope of AD is `3/4`. The sum of abscissa and ordinate of point C (where C lies in first quadrant) is

Explanation

To solve the problem, we need to find the coordinates of point C in the rhombus ABCD, given the slopes of sides AB and AD, and the side length of the rhombus.

### Step-by-Step Solution:

1. **Identify the Coordinates of Points A and B**:
- Point A is given as (0, 0).
- Let the coordinates of point B be (x_B, y_B).
- The slope of AB is given as \( \frac{4}{3} \). Thus, we can write:
\[ \frac{y_B - 0}{x_B - 0} = \frac{4}{3} \implies y_B = \frac{4}{3} x_B \]

2. **Use the Length of AB**:
- The length of AB is given as 10 units. Therefore, we have:
\[ x_B^2 + y_B^2 = 10^2 \]
- Substituting \( y_B = \frac{4}{3} x_B \) into the equation:
\[ x_B^2 + \left(\frac{4}{3} x_B\right)^2 = 100 \]
\[ x_B^2 + \frac{16}{9} x_B^2 = 100 \]
\[ \frac{25}{9} x_B^2 = 100 \]
\[ x_B^2 = 100 \times \frac{9}{25} = 36 \implies x_B = 6 \quad (\text{since } x_B > 0) \]
- Now, substituting back to find \( y_B \):
\[ y_B = \frac{4}{3} \times 6 = 8 \]
- Thus, the coordinates of point B are \( (6, 8) \).

3. **Identify the Coordinates of Point D**:
- Let the coordinates of point D be \( (x_D, y_D) \).
- The slope of AD is given as \( \frac{3}{4} \). Thus:
\[ \frac{y_D - 0}{x_D - 0} = \frac{3}{4} \implies y_D = \frac{3}{4} x_D \]

4. **Use the Length of AD**:
- The length of AD is also 10 units. Therefore:
\[ x_D^2 + y_D^2 = 10^2 \]
- Substituting \( y_D = \frac{3}{4} x_D \):
\[ x_D^2 + \left(\frac{3}{4} x_D\right)^2 = 100 \]
\[ x_D^2 + \frac{9}{16} x_D^2 = 100 \]
\[ \frac{25}{16} x_D^2 = 100 \]
\[ x_D^2 = 100 \times \frac{16}{25} = 64 \implies x_D = 8 \quad (\text{since } x_D > 0) \]
- Now substituting back to find \( y_D \):
\[ y_D = \frac{3}{4} \times 8 = 6 \]
- Thus, the coordinates of point D are \( (8, 6) \).

5. **Find the Midpoint K of Diagonal AC**:
- The coordinates of point C can be denoted as \( (x_C, y_C) \).
- The midpoint K of diagonal AC is given by:
\[ K = \left( \frac{0 + x_C}{2}, \frac{0 + y_C}{2} \right) \]
- Since diagonals of a rhombus bisect each other, the midpoint K must also be the midpoint of BD:
\[ K = \left( \frac{6 + 8}{2}, \frac{8 + 6}{2} \right) = \left( 7, 7 \right) \]

6. **Set Up the Equations for C**:
- From the midpoint K:
\[ \frac{x_C}{2} = 7 \implies x_C = 14 \]
\[ \frac{y_C}{2} = 7 \implies y_C = 14 \]

7. **Calculate the Sum of Abscissa and Ordinate of Point C**:
- The sum of the coordinates of point C is:
\[ x_C + y_C = 14 + 14 = 28 \]

### Final Answer:
The sum of the abscissa and ordinate of point C is **28**.