To find the dimensions of the original photo, we need to work with the enlarged dimensions given and the dilation factor. The problem states that the enlarged poster has dimensions of 24 inches in width and 32 inches in height, and it was enlarged using a dilation factor of 4.
### Step-by-step Solution
1. Understand the Relationship:
- The dimensions of the enlarged photo are a result of multiplying the original dimensions by the dilation factor.
2. Identify the Dilation Factor:
- The dilation factor provided is 4.
3. Set Up the Equations:
- Let the original width be \(W\) and the original height be \(H\).
- According to the dilation factor, the enlarged width can be given by the original width multiplied by 4:
[tex]\[
4W = 24
\][/tex]
- Similarly, the enlarged height can be given by the original height multiplied by 4:
[tex]\[
4H = 32
\][/tex]
4. Solve for the Original Width:
- To find the original width \(W\), divide both sides of the equation \(4W = 24\) by 4:
[tex]\[
W = \frac{24}{4} = 6
\][/tex]
5. Solve for the Original Height:
- To find the original height \(H\), divide both sides of the equation \(4H = 32\) by 4:
[tex]\[
H = \frac{32}{4} = 8
\][/tex]
6. Combine the Results:
- The original dimensions of the photo are 6 inches in width and 8 inches in height.
Therefore, the dimensions of the original photo are:
[tex]\[ \boxed{6 \times 8} \][/tex]
Among the given options:
- \(3 \times 8\)
- \(6 \times 8\)
- \(12 \times 16\)
- \(18 \times 24\)
The correct answer is [tex]\(6 \times 8\)[/tex] inches.
