Question
Hassan wants to add a border to a photo he took before he frames it. The final area of the entire framed picture will be 96 square inches. The length of the picture is 10 inches, and the width is 6 inches. Which of the following is the width of the border?(1 point)
Responses
9 inches
9 inches
3 inches
3 inches
4 inches
4 inches
1 inch
1 inch
Responses
9 inches
9 inches
3 inches
3 inches
4 inches
4 inches
1 inch
1 inch
Answers
GPT-4o mini
To find the width of the border around the picture, we can start by determining the area of the picture itself and then figure out how much extra area the border adds.
1. Calculate the area of the picture:
\[
\text{Area of the picture} = \text{length} \times \text{width} = 10 \, \text{inches} \times 6 \, \text{inches} = 60 \, \text{square inches}
\]
2. The area of the entire framed picture, including the border, is given as 96 square inches.
3. To find the area of just the border, we subtract the area of the picture from the area of the entire framed picture:
\[
\text{Area of the border} = \text{Area of the framed picture} - \text{Area of the picture}
= 96 \, \text{square inches} - 60 \, \text{square inches}
= 36 \, \text{square inches}
\]
4. Let \( x \) be the width of the border. The dimensions of the entire framed picture will be:
\[
\text{Length} = 10 + 2x
\]
\[
\text{Width} = 6 + 2x
\]
5. The area of the entire framed picture can be expressed in terms of \( x \):
\[
\text{Area of the framed picture} = (10 + 2x)(6 + 2x)
\]
6. Set this expression equal to 96 square inches and solve for \( x \):
\[
(10 + 2x)(6 + 2x) = 96
\]
Expanding the left-hand side:
\[
60 + 20x + 12x + 4x^2 = 96
\]
Combining like terms:
\[
4x^2 + 32x + 60 = 96
\]
Simplifying:
\[
4x^2 + 32x - 36 = 0
\]
Dividing the entire equation by 4:
\[
x^2 + 8x - 9 = 0
\]
7. Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 1, \; b = 8, \; c = -9
\]
\[
x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1}
\]
\[
x = \frac{-8 \pm \sqrt{64 + 36}}{2}
\]
\[
x = \frac{-8 \pm \sqrt{100}}{2}
\]
\[
x = \frac{-8 \pm 10}{2}
\]
8. This gives us two potential solutions:
\[
x = \frac{2}{2} = 1 \quad \text{(only the positive width makes sense)}
\]
\[
x = \frac{-18}{2} = -9 \quad \text{(this is not a valid solution)}
\]
Thus, the width of the border is:
**1 inch**.
1. Calculate the area of the picture:
\[
\text{Area of the picture} = \text{length} \times \text{width} = 10 \, \text{inches} \times 6 \, \text{inches} = 60 \, \text{square inches}
\]
2. The area of the entire framed picture, including the border, is given as 96 square inches.
3. To find the area of just the border, we subtract the area of the picture from the area of the entire framed picture:
\[
\text{Area of the border} = \text{Area of the framed picture} - \text{Area of the picture}
= 96 \, \text{square inches} - 60 \, \text{square inches}
= 36 \, \text{square inches}
\]
4. Let \( x \) be the width of the border. The dimensions of the entire framed picture will be:
\[
\text{Length} = 10 + 2x
\]
\[
\text{Width} = 6 + 2x
\]
5. The area of the entire framed picture can be expressed in terms of \( x \):
\[
\text{Area of the framed picture} = (10 + 2x)(6 + 2x)
\]
6. Set this expression equal to 96 square inches and solve for \( x \):
\[
(10 + 2x)(6 + 2x) = 96
\]
Expanding the left-hand side:
\[
60 + 20x + 12x + 4x^2 = 96
\]
Combining like terms:
\[
4x^2 + 32x + 60 = 96
\]
Simplifying:
\[
4x^2 + 32x - 36 = 0
\]
Dividing the entire equation by 4:
\[
x^2 + 8x - 9 = 0
\]
7. Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 1, \; b = 8, \; c = -9
\]
\[
x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1}
\]
\[
x = \frac{-8 \pm \sqrt{64 + 36}}{2}
\]
\[
x = \frac{-8 \pm \sqrt{100}}{2}
\]
\[
x = \frac{-8 \pm 10}{2}
\]
8. This gives us two potential solutions:
\[
x = \frac{2}{2} = 1 \quad \text{(only the positive width makes sense)}
\]
\[
x = \frac{-18}{2} = -9 \quad \text{(this is not a valid solution)}
\]
Thus, the width of the border is:
**1 inch**.