The length and breadth of a square are increased by 60% and 40% respectively . The area of the resulting rectangle exceeds the area of the square by :

Options

A. 2.24
B. 0.24
C. 1.24
D. 1

Answer

To solve the problem, we need to follow these steps:

### Step 1: Define the side length of the square
Let the side length of the square be \( s \).

### Step 2: Calculate the area of the square
The area of the square \( A_{square} \) is given by the formula:
\[ A_{square} = s^2 \]

### Step 3: Calculate the new dimensions of the rectangle
The length of the rectangle is increased by 60%, so the new length \( L \) is:
\[ L = s + 0.6s = 1.6s \]
The breadth of the rectangle is increased by 40%, so the new breadth \( B \) is:
\[ B = s + 0.4s = 1.4s \]

### Step 4: Calculate the area of the rectangle
The area of the rectangle \( A_{rectangle} \) is given by:
\[ A_{rectangle} = L \times B = (1.6s) \times (1.4s) = 2.24s^2 \]

### Step 5: Find the difference in area between the rectangle and the square
To find how much the area of the rectangle exceeds the area of the square, we calculate:
\[ \text{Difference} = A_{rectangle} - A_{square} = 2.24s^2 - s^2 = 1.24s^2 \]

### Conclusion
The area of the resulting rectangle exceeds the area of the square by \( 1.24s^2 \).

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