There is a rectangular plot the area of a rectangular plot gets reduced by 9 square metres if its length is reduced by 5 metres and breadth is increased by 3 metres if there is an increase in length by 3 metres and breadth by 2 metres then area is increased by 67 square meter
what is approximate area of largest circle drawn in it ?

Options

A. 64 square metres
B. 40 squre metres
C. 45 square metres
D. 36 square metres

Answer

To solve the problem step-by-step, we need to set up equations based on the information given about the rectangular plot.

### Step 1: Define Variables
Let:
- \( L \) = Length of the rectangular plot (in meters)
- \( B \) = Breadth of the rectangular plot (in meters)

### Step 2: Set Up the First Equation
According to the problem, if the length is reduced by 5 meters and the breadth is increased by 3 meters, the area is reduced by 9 square meters. This can be expressed as:
\[ (L - 5)(B + 3) = LB - 9 \]
Expanding the left side:
\[ LB - 5B + 3L - 15 = LB - 9 \]
Cancelling \( LB \) from both sides gives:
\[ -5B + 3L - 15 = -9 \]
Simplifying this, we get:
\[ 3L - 5B = 6 \quad \text{(Equation 1)} \]

### Step 3: Set Up the Second Equation
The problem also states that if the length is increased by 3 meters and the breadth by 2 meters, the area increases by 67 square meters. This can be expressed as:
\[ (L + 3)(B + 2) = LB + 67 \]
Expanding the left side:
\[ LB + 2L + 3B + 6 = LB + 67 \]
Cancelling \( LB \) from both sides gives:
\[ 2L + 3B + 6 = 67 \]
Simplifying this, we get:
\[ 2L + 3B = 61 \quad \text{(Equation 2)} \]

### Step 4: Solve the System of Equations
Now, we have a system of two equations:
1. \( 3L - 5B = 6 \)
2. \( 2L + 3B = 61 \)

We can solve these equations simultaneously. Let's first multiply Equation 2 by 5 to eliminate \( B \):
\[ 10L + 15B = 305 \quad \text{(Equation 3)} \]
Now, multiply Equation 1 by 3:
\[ 9L - 15B = 18 \quad \text{(Equation 4)} \]

### Step 5: Add Equations 3 and 4
Now, add Equation 3 and Equation 4:
\[ 10L + 15B + 9L - 15B = 305 + 18 \]
This simplifies to:
\[ 19L = 323 \]
Thus, we find:
\[ L = \frac{323}{19} = 17 \]

### Step 6: Substitute to Find \( B \)
Now substitute \( L = 17 \) back into Equation 1:
\[ 3(17) - 5B = 6 \]
This simplifies to:
\[ 51 - 5B = 6 \]
Rearranging gives:
\[ 5B = 45 \quad \Rightarrow \quad B = 9 \]

### Step 7: Calculate the Area of the Circle
The largest circle that can be inscribed in the rectangle will have a diameter equal to the smaller dimension, which is the breadth \( B \). Therefore, the radius \( r \) of the circle is:
\[ r = \frac{B}{2} = \frac{9}{2} = 4.5 \]
The area \( A \) of the circle is given by:
\[ A = \pi r^2 = \pi (4.5)^2 = \pi \cdot 20.25 \approx 63.617 \]

### Final Answer
The approximate area of the largest circle drawn in the rectangular plot is:
\[ \boxed{63.62} \text{ square meters} \]