10 of 1010 of 10 Items Question DeeDee draws a rectangle with sides measuring x inches and y inches. She dilates the rectangle using a scale factor of 1414. What is the area of the original rectangle in terms of x and y? What is the area of the dilated rectangle? How many times greater is the area of the original rectangle compared to the area of the dilated rectangle?(2 points) Responses The area of the original rectangle is xy The area of the dilated rectangle is xy16 The area of the original rectangle is 16 times the area of the dilated rectangle The area of the original rectangle is xy The area of the dilated rectangle is x y over 16 The area of the original rectangle is 16 times the area of the dilated rectangle The area of the original rectangle is xy The area of the dilated rectangle is 16xy The area of the original rectangle is 14 times the area of the dilated rectangle The area of the original rectangle is xy The area of the dilated rectangle is 16 over x y The area of the original rectangle is 14 times the area of the dilated rectangle The area of the original rectangle is x2 The area of the dilated rectangle is xy16 The area of the original rectangle is 36 times the area of the dilated rectangle The area of the original rectangle is x2 The area of the dilated rectangle is x y over 16 The area of the original rectangle is 36 times the area of the dilated rectangle Skip to navigation page 1 of 1v

Let's break down the question step by step:

1. **Calculate the area of the original rectangle:**
- The area \( A \) of a rectangle is given by the formula:
\[
A = \text{length} \times \text{width}
\]
- In terms of \( x \) and \( y \), the area of the original rectangle is:
\[
A = xy
\]

2. **Dilate the rectangle:**
- The problem states that DeeDee dilates the rectangle with a scale factor of \( \frac{1}{4} \) (note: the scale factor given is \( 1414 \) which appears to be written incorrectly, but we'll assume it should indicate a scale factor of \( \frac{1}{4} \)).
- After dilation, the new dimensions would be:
\[
\text{New length} = \frac{x}{4}, \quad \text{New width} = \frac{y}{4}
\]

3. **Calculate the area of the dilated rectangle:**
- The area of the dilated rectangle is:
\[
A_{\text{dilated}} = \left(\frac{x}{4}\right) \times \left(\frac{y}{4}\right) = \frac{xy}{16}
\]

4. **Compare the areas:**
- Now, we can compare the area of the original rectangle to the area of the dilated rectangle:
\[
\text{Ratio} = \frac{A_{\text{original}}}{A_{\text{dilated}}} = \frac{xy}{\frac{xy}{16}} = 16
\]
- Therefore, the area of the original rectangle is **16 times greater** than the area of the dilated rectangle.

**Summary:**
- The area of the original rectangle is \( xy \).
- The area of the dilated rectangle is \( \frac{xy}{16} \).
- The area of the original rectangle is 16 times greater than the area of the dilated rectangle.

### Final Answers:
- Area of original rectangle: \( xy \)
- Area of dilated rectangle: \( \frac{xy}{16} \)
- Area comparison: The area of the original rectangle is 16 times the area of the dilated rectangle.