Let the length of the ink pad be \( l \) cm and the width be \( w \) cm. We have two equations based on the information given:
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The area of the rectangle: \[ l \times w = 36 \]
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The perimeter of the rectangle: \[ 2l + 2w = 26 \] Simplifying the perimeter equation gives: \[ l + w = 13 \]
Now we can solve these two equations simultaneously. From the second equation, we can express \( w \) in terms of \( l \): \[ w = 13 - l \]
Now substitute this expression for \( w \) into the area equation: \[ l \times (13 - l) = 36 \]
Expanding and rearranging: \[ 13l - l^2 = 36 \] \[ l^2 - 13l + 36 = 0 \]
Now we can use the quadratic formula to solve for \( l \): \[ l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -13 \), and \( c = 36 \).
Calculating the discriminant: \[ b^2 - 4ac = (-13)^2 - 4 \times 1 \times 36 = 169 - 144 = 25 \]
Now solving for \( l \): \[ l = \frac{13 \pm \sqrt{25}}{2} = \frac{13 \pm 5}{2} \] This gives us two possible values for \( l \): \[ l = \frac{18}{2} = 9 \quad \text{and} \quad l = \frac{8}{2} = 4 \]
So, the two possible dimensions are:
- \( l = 9 \), \( w = 4 \)
- \( l = 4 \), \( w = 9 \)
Thus, the dimensions of the ink pad are: \[ \text{9 centimeters by 4 centimeters} \] or equivalently, \[ \text{4 centimeters by 9 centimeters} \]
Final answer: 9 centimeters by 4 centimeters.