The areas of a square and a rectangle are equal. The length of the rectangle is greater than the length of any side of the square by 5 cm and the breadth is less by 3 cm. Find the perimeter of the rectangle.

Options

A. 17 cm
B. 26 cm
C. 30 cm
D. 34 cm

Answer

To solve the problem step by step, we will define the variables and set up equations based on the information provided.

### Step 1: Define the variables
Let the side of the square be \( a \) cm.

### Step 2: Express the dimensions of the rectangle
According to the problem:
- The length of the rectangle \( l \) is greater than the side of the square by 5 cm, so:
\[ l = a + 5 \]
- The breadth of the rectangle \( b \) is less than the side of the square by 3 cm, so:
\[ b = a - 3 \]

### Step 3: Set up the equation for the areas
The area of the square is:
\[ \text{Area of square} = a^2 \]
The area of the rectangle is:
\[ \text{Area of rectangle} = l \times b = (a + 5)(a - 3) \]
Since the areas are equal, we can set up the equation:
\[ a^2 = (a + 5)(a - 3) \]

### Step 4: Expand the right-hand side
Expanding the right-hand side:
\[ (a + 5)(a - 3) = a^2 - 3a + 5a - 15 = a^2 + 2a - 15 \]
So, we have:
\[ a^2 = a^2 + 2a - 15 \]

### Step 5: Simplify the equation
Subtract \( a^2 \) from both sides:
\[ 0 = 2a - 15 \]

### Step 6: Solve for \( a \)
Rearranging gives:
\[ 2a = 15 \]
\[ a = \frac{15}{2} = 7.5 \text{ cm} \]

### Step 7: Find the dimensions of the rectangle
Now, substituting \( a \) back to find the length and breadth of the rectangle:
- Length \( l \):
\[ l = a + 5 = 7.5 + 5 = 12.5 \text{ cm} \]
- Breadth \( b \):
\[ b = a - 3 = 7.5 - 3 = 4.5 \text{ cm} \]

### Step 8: Calculate the perimeter of the rectangle
The perimeter \( P \) of the rectangle is given by:
\[ P = 2(l + b) = 2(12.5 + 4.5) = 2(17) = 34 \text{ cm} \]

### Final Answer
The perimeter of the rectangle is \( 34 \) cm.

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