Let the length of the dining room be \( l \) meters and the width be \( w \) meters. We know two things from the problem:
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The area of the rectangle is given by the formula: \[ l \times w = 49 \quad \text{(1)} \]
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The perimeter of the rectangle is given by the formula: \[ 2(l + w) = 28 \quad \text{(2)} \]
From equation (2), we can simplify to find \( l + w \): \[ l + w = 14 \quad \text{(3)} \]
Now, we can use equations (1) and (3) to find \( l \) and \( w \). From equation (3), we can express \( l \) in terms of \( w \): \[ l = 14 - w \]
Substituting this expression for \( l \) into equation (1): \[ (14 - w) w = 49 \]
Expanding and rearranging gives us a quadratic equation: \[ 14w - w^2 = 49 \] \[ w^2 - 14w + 49 = 0 \]
We can solve for \( w \) using the quadratic formula: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -14 \), and \( c = 49 \): \[ w = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 49}}{2 \cdot 1} \] \[ = \frac{14 \pm \sqrt{196 - 196}}{2} \] \[ = \frac{14 \pm \sqrt{0}}{2} \] \[ = \frac{14}{2} = 7 \]
Thus, \( w = 7 \) meters.
Now substituting \( w \) back into equation (3) to find \( l \): \[ l + 7 = 14 \] \[ l = 14 - 7 = 7 \]
Therefore, the dimensions of the dining room are: \[ \boxed{7 \text{ meters by } 7 \text{ meters}} \]
The dining room is a square with both dimensions equal to 7 meters.