Let the dimensions of the playing card be \( x \) centimeters (length) and \( y \) centimeters (width).
We know the following two equations based on the information given:
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The perimeter of the card is given by: \[ 2x + 2y = 34 \] This can be simplified to: \[ x + y = 17 \quad \text{(Equation 1)} \]
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The area of the card is given by: \[ x \cdot y = 70 \quad \text{(Equation 2)} \]
To solve for \( x \) and \( y \), we can use Equation 1 to express \( y \) in terms of \( x \): \[ y = 17 - x \]
Now, substitute this expression for \( y \) into Equation 2: \[ x(17 - x) = 70 \] This simplifies to: \[ 17x - x^2 = 70 \] Rearranging gives: \[ x^2 - 17x + 70 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our case, \( a = 1 \), \( b = -17 \), and \( c = 70 \): \[ x = \frac{17 \pm \sqrt{(-17)^2 - 4 \cdot 1 \cdot 70}}{2 \cdot 1} \] Calculating the discriminant: \[ x = \frac{17 \pm \sqrt{289 - 280}}{2} \] \[ x = \frac{17 \pm \sqrt{9}}{2} \] \[ x = \frac{17 \pm 3}{2} \]
This gives us two possible solutions for \( x \):
- \( x = \frac{20}{2} = 10 \)
- \( x = \frac{14}{2} = 7 \)
Now we find \( y \) for each case:
Case 1: If \( x = 10 \) \[ y = 17 - 10 = 7 \]
Case 2: If \( x = 7 \) \[ y = 17 - 7 = 10 \]
Thus, the dimensions of the playing card are \( 10 \) centimeters by \( 7 \) centimeters.
Therefore, the final answer is:
10 centimeters by 7 centimeters.