The actual area of a rectangle is `60cm^2`,but while measuring its length a student decreases it by `20%` and the breadth increases by `25%` . The percentage error in area, calculated by the student is :

Options

A. `5%`
B. `25%`
C. can't be determined
D. none of these

Answer

To solve the problem, we need to calculate the percentage error in the area of a rectangle when the length is decreased by 20% and the breadth is increased by 25%.

### Step-by-Step Solution:

1. **Understand the Actual Area**:
The actual area of the rectangle is given as \( 60 \, \text{cm}^2 \).

2. **Let the Original Length and Breadth be**:
Let the original length be \( L \) and the original breadth be \( B \). The area can be expressed as:
\[ \text{Area} = L \times B = 60 \, \text{cm}^2 \]

3. **Calculate the Decreased Length**:
The length is decreased by 20%. Therefore, the new length \( L' \) is:
\[ L' = L - 0.20L = 0.80L \]

4. **Calculate the Increased Breadth**:
The breadth is increased by 25%. Therefore, the new breadth \( B' \) is:
\[ B' = B + 0.25B = 1.25B \]

5. **Calculate the New Area**:
The new area \( A' \) calculated by the student using the new dimensions is:
\[ A' = L' \times B' = (0.80L) \times (1.25B) \]
Simplifying this, we get:
\[ A' = 0.80 \times 1.25 \times (L \times B) = 1.00 \times 60 = 60 \, \text{cm}^2 \]

6. **Calculate the Percentage Error**:
The percentage error in the area can be calculated using the formula:
\[ \text{Percentage Error} = \left( \frac{\text{Actual Area} - \text{New Area}}{\text{Actual Area}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Error} = \left( \frac{60 - 60}{60} \right) \times 100 = 0\% \]

### Final Answer:
The percentage error in the area calculated by the student is \( 0\% \).