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

1 inch
1 inch

9 inches

Answer 1

To determine the width of the border, we need to first calculate the area of the original picture, then find the area of the entire framed picture, and finally find the dimensions of the framed picture to get the width of the border.

1. Calculate the area of the original picture:
\[
\text{Area of the original picture} = \text{length} \times \text{width} = 10 \text{ inches} \times 6 \text{ inches} = 60 \text{ square inches}
\]

2. The area of the entire framed picture is given as 96 square inches.

3. The area of the border can be calculated by subtracting the area of the original picture from the area of the framed picture:
\[
\text{Area of the border} = \text{Area of the framed picture} - \text{Area of the original picture} = 96 \text{ square inches} - 60 \text{ square inches} = 36 \text{ square inches}
\]

4. Let \( x \) be the width of the border. The dimensions of the framed picture will be:
\[
\text{Length of framed picture} = 10 + 2x
\]
\[
\text{Width of framed picture} = 6 + 2x
\]

5. The area of the framed picture can also be expressed as:
\[
\text{Area of the framed picture} = (10 + 2x)(6 + 2x)
\]
Setting this equal to 96 square inches:
\[
(10 + 2x)(6 + 2x) = 96
\]

6. Expand the left side:
\[
60 + 20x + 12x + 4x^2 = 96
\]
\[
4x^2 + 32x + 60 = 96
\]
\[
4x^2 + 32x - 36 = 0
\]

7. Simplifying the equation by dividing everything by 4:
\[
x^2 + 8x - 9 = 0
\]

8. Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 8 \), \( c = -9 \):
\[
x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1}
\]
\[
x = \frac{-8 \pm \sqrt{64 + 36}}{2}
\]
\[
x = \frac{-8 \pm \sqrt{100}}{2}
\]
\[
x = \frac{-8 \pm 10}{2}
\]

9. This gives two solutions:
\[
x = \frac{2}{2} = 1 \quad \text{and} \quad x = \frac{-18}{2} = -9
\]
We discard \( x = -9 \) since width cannot be negative.

Thus, the width of the border is \( \boxed{1} \) inch.