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
4 inches
4 inches
3 inches
3 inches
9 inches
9 inches
1 inch
1 inch
Responses
4 inches
4 inches
3 inches
3 inches
9 inches
9 inches
1 inch
1 inch
Answers
GPT-4o mini
To find the width of the border Hassan wants to add to the photo, we first need to determine the area of the picture and then the dimensions of the entire framed picture with the border.
The area of the picture is given by:
\[
\text{Area}_{\text{picture}} = \text{length} \times \text{width} = 10 \text{ inches} \times 6 \text{ inches} = 60 \text{ square inches}
\]
The total area of the framed picture (picture plus border) is given as 96 square inches. Therefore, the area of the border can be calculated as:
\[
\text{Area}_{\text{border}} = \text{Area}_{\text{framed}} - \text{Area}_{\text{picture}} = 96 \text{ square inches} - 60 \text{ square inches} = 36 \text{ square inches}
\]
Next, let the width of the border be \( x \) inches. The dimensions of the entire framed picture will then be:
- Length: \( 10 + 2x \)
- Width: \( 6 + 2x \)
The area of the framed picture can also be expressed as:
\[
\text{Area}_{\text{framed}} = (10 + 2x)(6 + 2x)
\]
Setting this equal to the total area, we have:
\[
(10 + 2x)(6 + 2x) = 96
\]
Expanding the left side:
\[
60 + 20x + 12x + 4x^2 = 96
\]
This simplifies to:
\[
4x^2 + 32x + 60 = 96
\]
Subtracting 96 from both sides gives:
\[
4x^2 + 32x - 36 = 0
\]
We can simplify this by dividing the whole equation by 4:
\[
x^2 + 8x - 9 = 0
\]
Next, we will factor this quadratic equation:
\[
(x + 9)(x - 1) = 0
\]
Setting each factor to zero gives:
\[
x + 9 = 0 \implies x = -9 \quad (\text{not a valid solution})
\]
\[
x - 1 = 0 \implies x = 1
\]
Thus, the width of the border is \( \boxed{1} \) inch.
The area of the picture is given by:
\[
\text{Area}_{\text{picture}} = \text{length} \times \text{width} = 10 \text{ inches} \times 6 \text{ inches} = 60 \text{ square inches}
\]
The total area of the framed picture (picture plus border) is given as 96 square inches. Therefore, the area of the border can be calculated as:
\[
\text{Area}_{\text{border}} = \text{Area}_{\text{framed}} - \text{Area}_{\text{picture}} = 96 \text{ square inches} - 60 \text{ square inches} = 36 \text{ square inches}
\]
Next, let the width of the border be \( x \) inches. The dimensions of the entire framed picture will then be:
- Length: \( 10 + 2x \)
- Width: \( 6 + 2x \)
The area of the framed picture can also be expressed as:
\[
\text{Area}_{\text{framed}} = (10 + 2x)(6 + 2x)
\]
Setting this equal to the total area, we have:
\[
(10 + 2x)(6 + 2x) = 96
\]
Expanding the left side:
\[
60 + 20x + 12x + 4x^2 = 96
\]
This simplifies to:
\[
4x^2 + 32x + 60 = 96
\]
Subtracting 96 from both sides gives:
\[
4x^2 + 32x - 36 = 0
\]
We can simplify this by dividing the whole equation by 4:
\[
x^2 + 8x - 9 = 0
\]
Next, we will factor this quadratic equation:
\[
(x + 9)(x - 1) = 0
\]
Setting each factor to zero gives:
\[
x + 9 = 0 \implies x = -9 \quad (\text{not a valid solution})
\]
\[
x - 1 = 0 \implies x = 1
\]
Thus, the width of the border is \( \boxed{1} \) inch.