To find the width of the border that Hassan wants to add to his photo, we first need to calculate the area of the photo and then determine how the border affects the total area.
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The area of the original picture is calculated as: \[ \text{Area of picture} = \text{length} \times \text{width} = 10 , \text{inches} \times 6 , \text{inches} = 60 , \text{square inches} \]
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The total area of the framed picture is given as 96 square inches.
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Let \( x \) be the width of the border. When the border is added, the dimensions of the framed picture become:
- Length: \( 10 + 2x \)
- Width: \( 6 + 2x \)
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The area of the framed picture can thus be expressed as: \[ \text{Area of framed picture} = (10 + 2x)(6 + 2x) \]
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We set up the equation to find \( x \): \[ (10 + 2x)(6 + 2x) = 96 \]
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Expanding the left side: \[ 60 + 20x + 12x + 4x^2 = 96 \] Combining terms gives: \[ 4x^2 + 32x + 60 = 96 \]
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Now we subtract 96 from both sides: \[ 4x^2 + 32x + 60 - 96 = 0 \] Simplifying gives: \[ 4x^2 + 32x - 36 = 0 \]
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Dividing the entire equation by 4 to simplify: \[ x^2 + 8x - 9 = 0 \]
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We can factor this quadratic equation: \[ (x + 9)(x - 1) = 0 \]
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Setting each factor equal to zero gives us possible solutions for \( x \): \[ x + 9 = 0 \quad \Rightarrow \quad x = -9 \quad \text{(not physically meaningful)} \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
So the width of the border is 1 inch.
Thus, the answer is: • 1 inch
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