Let's denote the length of the table as \( l \) and the width as \( w \).
We are given two equations based on the area and the perimeter of the table:
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From the area: \[ l \cdot w = 30 \]
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From the perimeter: \[ 2l + 2w = 26 \] Simplifying this, we get: \[ l + w = 13 \]
Now we can solve these two equations simultaneously.
From the second equation, we can express \( w \) in terms of \( l \): \[ w = 13 - l \]
Now, substituting this expression for \( w \) into the first equation: \[ l(13 - l) = 30 \] Expanding this, we have: \[ 13l - l^2 = 30 \] Rearranging it gives: \[ l^2 - 13l + 30 = 0 \]
Now, we can factor this quadratic equation: \[ (l - 3)(l - 10) = 0 \]
Setting each factor to zero gives us the possible values for \( l \): \[ l - 3 = 0 \quad \Rightarrow \quad l = 3 \] \[ l - 10 = 0 \quad \Rightarrow \quad l = 10 \]
Thus, \( l \) can be 3 or 10. Substituting back to find \( w \):
- If \( l = 3 \): \[ w = 13 - 3 = 10 \]
- If \( l = 10 \): \[ w = 13 - 10 = 3 \]
Thus, the dimensions of the table are: \[ \text{3 feet by 10 feet} \]
So, the answer is: 3 feet by 10 feet.