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 that Hassan wants to add to his photo, we need to find the dimensions of the framed picture and set up an equation based on the total area.

The dimensions of the unframed photo are:
- Length = 10 inches
- Width = 6 inches

The area of the photo is:
\[
\text{Area of the photo} = \text{Length} \times \text{Width} = 10 \times 6 = 60 \text{ square inches}
\]

Let \( x \) be the width of the border. When the border is added, the dimensions of the framed picture will be:
- Length = \( 10 + 2x \)
- Width = \( 6 + 2x \)

The area of the framed picture will be:
\[
\text{Area of the framed picture} = (10 + 2x)(6 + 2x)
\]

We know that the area of the entire framed picture is 96 square inches, so we can set up the equation:
\[
(10 + 2x)(6 + 2x) = 96
\]

Now, we expand the left side:
\[
10 \times 6 + 10 \times 2x + 6 \times 2x + 2x \times 2x = 60 + 20x + 12x + 4x^2 = 60 + 32x + 4x^2
\]

Setting this equal to 96:
\[
60 + 32x + 4x^2 = 96
\]

Rearranging this gives:
\[
4x^2 + 32x + 60 - 96 = 0 \\
4x^2 + 32x - 36 = 0 \\
x^2 + 8x - 9 = 0 \quad (\text{dividing the entire equation by 4})
\]

Now, we can factor the quadratic equation:
\[
(x + 9)(x - 1) = 0
\]

This gives us two potential solutions:
\[
x + 9 = 0 \quad \Rightarrow \quad x = -9 \quad \text{(not valid since width cannot be negative)} \\
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

Thus, the width of the border is:
\[
\boxed{1} \text{ inch}
\]