To find the width of the border that Hassan wants to add to his photo, we need to determine the area of the border plus the area of the photo itself and equate it to the final area of the framed picture.
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Area of the picture:
The area of the picture can be calculated using the formula for the area of a rectangle, A = length × width. \[ \text{Area of the picture} = 10 \text{ inches} \times 6 \text{ inches} = 60 \text{ square inches} \] -
Final area of the framed picture:
Given that the final area of the entire framed picture is 96 square inches. -
Area of the border:
The area of the border surrounding the picture can be determined by subtracting the area of the picture from the final area of the framed picture. \[ \text{Area of the border} = \text{Final area} - \text{Area of the picture} = 96 \text{ square inches} - 60 \text{ square inches} = 36 \text{ square inches} \] -
Dimensions of the framed picture:
If we let \( x \) be the width of the border, then the dimensions of the framed picture will be:- Length: \( 10 + 2x \)
- Width: \( 6 + 2x \)
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Calculating the total area using the dimensions: The area of the framed picture can also be expressed as: \[ (10 + 2x)(6 + 2x) = 96 \text{ square inches} \]
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Expanding the equation: \[ (10 + 2x)(6 + 2x) = 60 + 20x + 12x + 4x^2 = 60 + 32x + 4x^2 \]
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Setting the equation equal to the final area: \[ 60 + 32x + 4x^2 = 96 \]
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Rearranging the equation: \[ 4x^2 + 32x + 60 - 96 = 0 \] \[ 4x^2 + 32x - 36 = 0 \]
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Simplifying: Divide the whole equation by 4: \[ x^2 + 8x - 9 = 0 \]
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Factoring the quadratic: Now, we factor the quadratic: \[ (x + 9)(x - 1) = 0 \]
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Finding the solutions: \[ x + 9 = 0 \implies x = -9 \quad (\text{not valid as width can't be negative}) \] \[ x - 1 = 0 \implies x = 1 \]
Thus, the width of the border is 1 inch.
So the answer is 1 inch.