To find the width of the border, we first need to calculate the area of the original picture and then determine the dimensions of the entire framed picture.
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Calculate the area of the original picture: The area of the picture is given by the formula: \[ \text{Area} = \text{Length} \times \text{Width} \] Substituting in the values: \[ \text{Area} = 10 \text{ inches} \times 6 \text{ inches} = 60 \text{ square inches} \]
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Determine the area of the border: The area of the entire framed picture is given as 96 square inches. The area of the border alone can be calculated as follows: \[ \text{Area of Border} = \text{Area of Framed Picture} - \text{Area of Original Picture} \] \[ \text{Area of Border} = 96 \text{ square inches} - 60 \text{ square inches} = 36 \text{ square inches} \]
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Establish the relationship for dimensions with the border: Let \( x \) be the width of the border. The dimensions of the entire framed picture will then be: \[ \text{Length of framed picture} = 10 + 2x \] \[ \text{Width of framed picture} = 6 + 2x \]
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Set up the equation for the area of the framed picture: The area of the framed picture can be expressed as: \[ (10 + 2x)(6 + 2x) = 96 \]
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Expand and simplify the equation: \[ 60 + 20x + 12x + 4x^2 = 96 \] \[ 4x^2 + 32x + 60 = 96 \] \[ 4x^2 + 32x - 36 = 0 \]
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Divide the entire equation by 4: \[ x^2 + 8x - 9 = 0 \]
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Solve the quadratic equation using the quadratic formula: The quadratic formula is given by: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \] Here, \( a = 1 \), \( b = 8 \), and \( c = -9 \): \[ x = \frac{{-8 \pm \sqrt{{8^2 - 4(1)(-9)}}}}{2(1)} \] \[ x = \frac{{-8 \pm \sqrt{{64 + 36}}}}{2} \] \[ x = \frac{{-8 \pm \sqrt{100}}}{2} \] \[ x = \frac{{-8 \pm 10}}{2} \]
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Calculate the two potential solutions: \[ x = \frac{2}{2} = 1 \quad \text{or} \quad x = \frac{{-18}}{2} = -9 \] Since the width cannot be negative, we have: \[ x = 1 \]
The width of the border is 1 inch.
Final answer: 1 inch
1 answer