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?
a 1 inch
b 3 inches
c 4 inches
d 9 inches
a 1 inch
b 3 inches
c 4 inches
d 9 inches
Answers
GPT-4o mini
To determine the width of the border that Hassan wants to add to the photo, we first need to understand the total area of the framed picture, which is 96 square inches.
1. **Calculate the area of the photo:**
The area of the photo can be calculated using the formula for the area of a rectangle:
\[
\text{Area} = \text{length} \times \text{width}
\]
For the photo:
\[
\text{Area} = 10 \, \text{inches} \times 6 \, \text{inches} = 60 \, \text{square inches}
\]
2. **Set up the equation for the framed picture:**
Let \( x \) be the width of the border. The dimensions of the framed picture will be:
- Length: \( 10 + 2x \)
- Width: \( 6 + 2x \)
Therefore, the area of the framed picture can be expressed as:
\[
(10 + 2x)(6 + 2x) = 96
\]
3. **Expand the equation:**
\[
(10 + 2x)(6 + 2x) = 60 + 20x + 12x + 4x^2 = 60 + 32x + 4x^2
\]
Set this equal to 96:
\[
60 + 32x + 4x^2 = 96
\]
4. **Rearrange the equation:**
\[
4x^2 + 32x + 60 - 96 = 0
\]
\[
4x^2 + 32x - 36 = 0
\]
5. **Simplify the equation:**
Divide the entire equation by 4:
\[
x^2 + 8x - 9 = 0
\]
6. **Use the quadratic formula to solve for \( x \):**
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 8 \), and \( c = -9 \):
\[
b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot (-9) = 64 + 36 = 100
\]
\[
x = \frac{-8 \pm \sqrt{100}}{2 \cdot 1} = \frac{-8 \pm 10}{2}
\]
This gives us two potential solutions:
\[
x = \frac{2}{2} = 1 \quad \text{and} \quad x = \frac{-18}{2} = -9
\]
Since width cannot be negative, we have:
\[
x = 1
\]
So, the width of the border is **1 inch**.
The answer is \( \text{a } 1 \text{ inch} \).
1. **Calculate the area of the photo:**
The area of the photo can be calculated using the formula for the area of a rectangle:
\[
\text{Area} = \text{length} \times \text{width}
\]
For the photo:
\[
\text{Area} = 10 \, \text{inches} \times 6 \, \text{inches} = 60 \, \text{square inches}
\]
2. **Set up the equation for the framed picture:**
Let \( x \) be the width of the border. The dimensions of the framed picture will be:
- Length: \( 10 + 2x \)
- Width: \( 6 + 2x \)
Therefore, the area of the framed picture can be expressed as:
\[
(10 + 2x)(6 + 2x) = 96
\]
3. **Expand the equation:**
\[
(10 + 2x)(6 + 2x) = 60 + 20x + 12x + 4x^2 = 60 + 32x + 4x^2
\]
Set this equal to 96:
\[
60 + 32x + 4x^2 = 96
\]
4. **Rearrange the equation:**
\[
4x^2 + 32x + 60 - 96 = 0
\]
\[
4x^2 + 32x - 36 = 0
\]
5. **Simplify the equation:**
Divide the entire equation by 4:
\[
x^2 + 8x - 9 = 0
\]
6. **Use the quadratic formula to solve for \( x \):**
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 8 \), and \( c = -9 \):
\[
b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot (-9) = 64 + 36 = 100
\]
\[
x = \frac{-8 \pm \sqrt{100}}{2 \cdot 1} = \frac{-8 \pm 10}{2}
\]
This gives us two potential solutions:
\[
x = \frac{2}{2} = 1 \quad \text{and} \quad x = \frac{-18}{2} = -9
\]
Since width cannot be negative, we have:
\[
x = 1
\]
So, the width of the border is **1 inch**.
The answer is \( \text{a } 1 \text{ inch} \).