Let the length and width of the rectangular ink pad be \( l \) and \( w \), respectively. We are given two equations based on the area and perimeter of the rectangle:
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Area: \[ l \times w = 28 \]
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Perimeter: \[ 2l + 2w = 22 \]
We can simplify the perimeter equation: \[ l + w = 11 \]
Now we have the system of equations:
- \( l \times w = 28 \)
- \( l + w = 11 \)
From equation 2, we can express \( w \) in terms of \( l \): \[ w = 11 - l \]
Now we can substitute this expression for \( w \) into equation 1: \[ l \times (11 - l) = 28 \]
Expanding this gives: \[ 11l - l^2 = 28 \]
Rearranging gives us a standard quadratic equation: \[ l^2 - 11l + 28 = 0 \]
We can use the quadratic formula to solve for \( l \): \[ l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -11 \), and \( c = 28 \).
Calculating the discriminant: \[ b^2 - 4ac = (-11)^2 - 4 \times 1 \times 28 = 121 - 112 = 9 \]
Now substituting into the quadratic formula: \[ l = \frac{11 \pm \sqrt{9}}{2} = \frac{11 \pm 3}{2} \]
This gives us two possible solutions for \( l \): \[ l = \frac{14}{2} = 7 \quad \text{and} \quad l = \frac{8}{2} = 4 \]
So, the dimensions \( l \) and \( w \) are:
- If \( l = 7 \), then \( w = 11 - 7 = 4 \).
- If \( l = 4 \), then \( w = 11 - 4 = 7 \).
Thus, the dimensions of the ink pad are: \[ 7 \text{ cm by } 4 \text{ cm} \]
So the final answer is: \[ \boxed{7 \text{ centimeters by } 4 \text{ centimeters}} \]