To find the width of the border, we will first determine the area of the picture itself and then use the total area of the framed picture to find the area added by the border.
-
Calculate the area of the picture: \[ \text{Area}_{\text{picture}} = \text{length} \times \text{width} = 10 , \text{inches} \times 6 , \text{inches} = 60 , \text{square inches} \]
-
Determine the area of the border and total area: \[ \text{Area}{\text{framed}} = \text{Area}{\text{border}} + \text{Area}{\text{picture}} \] Given that the total area is 96 square inches: \[ 96 = \text{Area}{\text{border}} + 60 \] \[ \text{Area}_{\text{border}} = 96 - 60 = 36 , \text{square inches} \]
-
Let \( x \) be the width of the border. The overall length and width including the border would then be: \[ \text{Final length} = 10 + 2x \] \[ \text{Final width} = 6 + 2x \]
-
The area of the whole framed picture can also be expressed as: \[ \text{Area}_{\text{framed}} = (10 + 2x)(6 + 2x) \]
-
Set this equal to the total area we found: \[ (10 + 2x)(6 + 2x) = 96 \]
-
Expanding this: \[ 60 + 20x + 12x + 4x^2 = 96 \] \[ 4x^2 + 32x + 60 = 96 \] \[ 4x^2 + 32x - 36 = 0 \]
-
Now, simplify the equation by dividing everything by 4: \[ x^2 + 8x - 9 = 0 \]
-
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 8, c = -9 \): \[ x = \frac{-8 \pm \sqrt{8^2 - 4 \times 1 \times (-9)}}{2 \times 1} \] \[ x = \frac{-8 \pm \sqrt{64 + 36}}{2} \] \[ x = \frac{-8 \pm \sqrt{100}}{2} \] \[ x = \frac{-8 \pm 10}{2} \]
Calculating the two potential solutions:
- \( x = \frac{-8 + 10}{2} = 1 \)
- \( x = \frac{-8 - 10}{2} = -9 \) (not a valid solution since width cannot be negative)
Therefore, the width of the border is 1 inch.
The answer is: 1 inch.