To find the original dimensions of the photo before it was enlarged, we need to understand the concept of dilation and how it affects dimensions.
The poster was made using a dilation [tex]\( D_{Q, 4} \)[/tex], where 4 is the dilation factor. This means that each dimension of the original photo was multiplied by 4 to obtain the dimensions of the enlarged poster.
Given that the dimensions of the enlarged photo are [tex]\( 24 \)[/tex] inches in width and [tex]\( 32 \)[/tex] inches in height, we need to determine the original dimensions.
Let's denote the width and height of the original photo by [tex]\( \text{original\_width} \)[/tex] and [tex]\( \text{original\_height} \)[/tex].
Since the enlargement was done by a factor of 4, we have:
[tex]\[ \text{original\_width} \times 4 = 24 \][/tex]
[tex]\[ \text{original\_height} \times 4 = 32 \][/tex]
To find the original dimensions, we divide the dimensions of the enlarged photo by the dilation factor:
[tex]\[ \text{original\_width} = \frac{24}{4} = 6 \text{ inches} \][/tex]
[tex]\[ \text{original\_height} = \frac{32}{4} = 8 \text{ inches} \][/tex]
Thus, the dimensions of the original photo are [tex]\( 6 \)[/tex] inches by [tex]\( 8 \)[/tex] inches.
Given the options:
1. [tex]\( 3 \times 8 \)[/tex]
2. [tex]\( 6 \times 8 \)[/tex]
3. [tex]\( 12 \times 16 \)[/tex]
4. [tex]\( 18 \times 24 \)[/tex]
We can see that the correct option is [tex]\( 6 \times 8 \)[/tex]. Therefore, the dimensions of the original photo are [tex]\( 6 \)[/tex] inches by [tex]\( 8 \)[/tex] inches.
