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)
1 inch
3 inches
9 inches
4 inches
(1 point)
1 inch
3 inches
9 inches
4 inches
Answers
GPT-4o mini
To find the width of the border, we need to determine the total area of the picture including the border and then calculate the width based on the dimensions of the original picture.
1. **Calculate the area of the original picture**:
- Length = 10 inches
- Width = 6 inches
- Area of the original picture = Length × Width = 10 inches × 6 inches = 60 square inches.
2. **Given total area of the framed picture**:
- Total area = 96 square inches.
3. **Calculate the area of the border**:
- Area of the border = Total area - Area of the original picture
- Area of the border = 96 square inches - 60 square inches = 36 square inches.
4. **Let \( x \) be the width of the border**. The dimensions of the entire framed picture (including the border) will be:
- Length of framed picture = Length of original picture + 2x = \( 10 + 2x \)
- Width of framed picture = Width of original picture + 2x = \( 6 + 2x \)
5. **Set up the equation for the total area of the framed picture**:
\[
(10 + 2x)(6 + 2x) = 96
\]
6. **Expand the left side**:
\[
60 + 20x + 12x + 4x^2 = 96
\]
\[
4x^2 + 32x + 60 = 96
\]
7. **Rearranging the equation to set it to 0**:
\[
4x^2 + 32x + 60 - 96 = 0
\]
\[
4x^2 + 32x - 36 = 0
\]
8. **Divide the entire equation by 4 to simplify**:
\[
x^2 + 8x - 9 = 0
\]
9. **Now we can apply the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)** where \( a = 1, b = 8, c = -9 \):
\[
x = \frac{{-8 \pm \sqrt{{8^2 - 4 \cdot 1 \cdot (-9)}}}}{2 \cdot 1}
\]
\[
x = \frac{{-8 \pm \sqrt{{64 + 36}}}}{2}
\]
\[
x = \frac{{-8 \pm \sqrt{100}}}{2}
\]
\[
x = \frac{{-8 \pm 10}}{2}
\]
10. **Solving for \( x \)**:
- Possible solutions:
\[
x = \frac{2}{2} = 1 \quad \text{(positive solution)}
\]
\[
x = \frac{{-18}}{2} = -9 \quad \text{(not applicable)}
\]
Thus, the width of the border is **1 inch**.
1. **Calculate the area of the original picture**:
- Length = 10 inches
- Width = 6 inches
- Area of the original picture = Length × Width = 10 inches × 6 inches = 60 square inches.
2. **Given total area of the framed picture**:
- Total area = 96 square inches.
3. **Calculate the area of the border**:
- Area of the border = Total area - Area of the original picture
- Area of the border = 96 square inches - 60 square inches = 36 square inches.
4. **Let \( x \) be the width of the border**. The dimensions of the entire framed picture (including the border) will be:
- Length of framed picture = Length of original picture + 2x = \( 10 + 2x \)
- Width of framed picture = Width of original picture + 2x = \( 6 + 2x \)
5. **Set up the equation for the total area of the framed picture**:
\[
(10 + 2x)(6 + 2x) = 96
\]
6. **Expand the left side**:
\[
60 + 20x + 12x + 4x^2 = 96
\]
\[
4x^2 + 32x + 60 = 96
\]
7. **Rearranging the equation to set it to 0**:
\[
4x^2 + 32x + 60 - 96 = 0
\]
\[
4x^2 + 32x - 36 = 0
\]
8. **Divide the entire equation by 4 to simplify**:
\[
x^2 + 8x - 9 = 0
\]
9. **Now we can apply the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)** where \( a = 1, b = 8, c = -9 \):
\[
x = \frac{{-8 \pm \sqrt{{8^2 - 4 \cdot 1 \cdot (-9)}}}}{2 \cdot 1}
\]
\[
x = \frac{{-8 \pm \sqrt{{64 + 36}}}}{2}
\]
\[
x = \frac{{-8 \pm \sqrt{100}}}{2}
\]
\[
x = \frac{{-8 \pm 10}}{2}
\]
10. **Solving for \( x \)**:
- Possible solutions:
\[
x = \frac{2}{2} = 1 \quad \text{(positive solution)}
\]
\[
x = \frac{{-18}}{2} = -9 \quad \text{(not applicable)}
\]
Thus, the width of the border is **1 inch**.