To find the width of the border that Hassan wants to add, we begin by calculating the area of the unframed picture, and then we can find the area of the framed picture and the border's width.
The area of the unframed picture is given by the formula for the area of a rectangle:
\[ \text{Area} = \text{length} \times \text{width} \]
The length of the picture is 10 inches and the width is 6 inches, so:
\[ \text{Area of the picture} = 10 , \text{inches} \times 6 , \text{inches} = 60 , \text{square inches} \]
According to the problem, the final area of the entire framed picture is 96 square inches. The area of the border can be found by subtracting the area of the picture from the area of the framed picture:
\[ \text{Area of the border} = \text{Area of framed picture} - \text{Area of picture} = 96 , \text{square inches} - 60 , \text{square inches} = 36 , \text{square inches} \]
Now, to find the dimensions of the framed picture, we can denote the width of the border as \(x\). The length of the framed picture will be the original length plus twice the border width (because there’s a border on both sides), and the same applies for the width. Therefore:
\[ \text{Length of framed picture} = 10 + 2x \] \[ \text{Width of framed picture} = 6 + 2x \]
The area of the framed picture can be expressed as:
\[ \text{Area of framed picture} = (10 + 2x)(6 + 2x) \]
Setting this equal to the area of the framed picture (96 square inches), we have:
\[ (10 + 2x)(6 + 2x) = 96 \]
Now, expand this equation:
\[ 60 + 20x + 12x + 4x^2 = 96 \] \[ 4x^2 + 32x + 60 = 96 \] \[ 4x^2 + 32x - 36 = 0 \]
Next, divide the entire equation by 4:
\[ x^2 + 8x - 9 = 0 \]
Now, we can factor this quadratic equation:
\[ (x + 9)(x - 1) = 0 \]
Setting each factor to zero gives us:
\[ x + 9 = 0 \quad \Rightarrow \quad x = -9 \quad \text{(not a valid width)} \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
Thus, the width of the border is:
\[ \boxed{1 \text{ inch}} \]
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