The diagonal of square is changing at the rate of `0.5 cms^(-1)`. Then the rate of change of area, when the area is `400 cm^(2)`, is equal to

Options

A. `20 sqrt(2)cm^(2)s^(-1)`
B. `10sqrt(2)cm^(2)s^(-1)`
C. `1/(10sqrt(2))cm^(2)s^(-1)`
D. `10/(sqrt(2))cm^(2)s^(-1)`

Explanation

To solve the problem, we need to find the rate of change of the area of a square when the area is \(400 \, \text{cm}^2\) and the diagonal is changing at a rate of \(0.5 \, \text{cm/s}\).

### Step-by-Step Solution:

1. **Understand the relationship between the area and the diagonal of the square**:
- The area \(A\) of a square is given by \(A = a^2\), where \(a\) is the length of a side of the square.
- The diagonal \(d\) of the square can be expressed in terms of the side length as \(d = a\sqrt{2}\).

2. **Differentiate the area with respect to time**:
- We need to find \(\frac{dA}{dt}\). Using the chain rule:
\[ \frac{dA}{dt} = \frac{d(a^2)}{dt} = 2a \frac{da}{dt} \]

3. **Relate the change in diagonal to the change in side length**:
- Since the diagonal \(d\) is related to the side length \(a\) by \(d = a\sqrt{2}\), we can differentiate this with respect to time:
\[ \frac{dd}{dt} = \sqrt{2} \frac{da}{dt} \]
- Given that \(\frac{dd}{dt} = 0.5 \, \text{cm/s}\), we can solve for \(\frac{da}{dt}\):
\[ 0.5 = \sqrt{2} \frac{da}{dt} \implies \frac{da}{dt} = \frac{0.5}{\sqrt{2}} = \frac{0.5 \sqrt{2}}{2} = \frac{0.25\sqrt{2}}{1} \]

4. **Find the side length \(a\) when the area is \(400 \, \text{cm}^2\)**:
- Since the area \(A = 400 \, \text{cm}^2\), we have:
\[ a^2 = 400 \implies a = \sqrt{400} = 20 \, \text{cm} \]

5. **Substitute \(a\) and \(\frac{da}{dt}\) into the area change formula**:
- Now we substitute \(a = 20 \, \text{cm}\) and \(\frac{da}{dt} = \frac{0.5}{\sqrt{2}}\) into the formula for \(\frac{dA}{dt}\):
\[ \frac{dA}{dt} = 2a \frac{da}{dt} = 2 \cdot 20 \cdot \frac{0.5}{\sqrt{2}} = 40 \cdot \frac{0.5}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 20 \cdot \frac{\sqrt{2}}{2} = 10\sqrt{2} \, \text{cm}^2/\text{s} \]

### Final Answer:
The rate of change of the area when the area is \(400 \, \text{cm}^2\) is \(10\sqrt{2} \, \text{cm}^2/\text{s}\).