A rectangular room has an area of 60 `m^2` and perimeter of 34 m. The length of the diagonal of the rectangular room is same as the side of a square. The area of the square (in `m^2`) is:

Options

A. 169
B. 144
C. 300
D. 244

Answer

To solve the problem step by step, we will first determine the dimensions of the rectangular room using the given area and perimeter, then calculate the length of the diagonal, and finally find the area of the square.

### Step 1: Set up the equations
Let the length of the rectangular room be \( l \) meters and the width be \( b \) meters. We have two equations based on the problem statement:

1. Area:
\[ l \times b = 60 \quad \text{(1)} \]

2. Perimeter:
\[ 2(l + b) = 34 \quad \text{(2)} \]
Simplifying equation (2):
\[ l + b = 17 \quad \text{(3)} \]

### Step 2: Solve for dimensions
From equation (3), we can express \( b \) in terms of \( l \):
\[ b = 17 - l \quad \text{(4)} \]

Now, substitute equation (4) into equation (1):
\[ l \times (17 - l) = 60 \]
Expanding this gives:
\[ 17l - l^2 = 60 \]
Rearranging it into standard quadratic form:
\[ l^2 - 17l + 60 = 0 \quad \text{(5)} \]

### Step 3: Factor the quadratic equation
We need to factor equation (5):
\[ (l - 12)(l - 5) = 0 \]
Thus, the solutions for \( l \) are:
\[ l = 12 \quad \text{or} \quad l = 5 \]

### Step 4: Find the corresponding width
Using equation (4), if \( l = 12 \):
\[ b = 17 - 12 = 5 \]
If \( l = 5 \):
\[ b = 17 - 5 = 12 \]
So the dimensions of the room are \( 12 \, m \) and \( 5 \, m \).

### Step 5: Calculate the diagonal
The length of the diagonal \( d \) of the rectangle can be calculated using the Pythagorean theorem:
\[ d = \sqrt{l^2 + b^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \, m \]

### Step 6: Calculate the area of the square
Since the length of the diagonal of the rectangular room is the same as the side of the square, the side of the square \( s \) is:
\[ s = 13 \, m \]
The area \( A \) of the square is given by:
\[ A = s^2 = 13^2 = 169 \, m^2 \]

### Final Answer
The area of the square is \( \boxed{169} \, m^2 \).

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