DeeDee draws a rectangle with sides measuring x inches and y inches. She dilates the rectangle using a scale factor of 1/4 . 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? Explain your answers and show all your work.(4 points)

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The area of a rectangle is calculated using the formula \( \text{Area} = \text{length} \times \text{width} \). For DeeDee's original rectangle with sides measuring \( x \) inches and \( y \) inches, the area can be expressed as:

\[ \text{Area}_{\text{original}} = x \times y. \]

When she dilates the rectangle using a scale factor of \( \frac{1}{4} \), the new dimensions become:

\[ \text{New length} = \frac{x}{4} \quad \text{and} \quad \text{New width} = \frac{y}{4}. \]

The area of the dilated rectangle is then calculated as follows:

\[ \text{Area}_{\text{dilated}} = \left(\frac{x}{4}\right) \times \left(\frac{y}{4}\right) = \frac{xy}{16}. \]

To compare the areas of the original and dilated rectangles, we can calculate the ratio of the original area to the dilated area:

\[ \text{Ratio} = \frac{\text{Area}{\text{original}}}{\text{Area}{\text{dilated}}} = \frac{xy}{\frac{xy}{16}} = 16. \]

This means the area of the original rectangle is \( 16 \) times greater than the area of the dilated rectangle. Essentially, when a shape is dilated, both its dimensions and area are affected by the square of the scale factor. Since the scale factor is \( \frac{1}{4} \), the area is reduced by a factor of \( \left(\frac{1}{4}\right)^2 = \frac{1}{16} \). Thus, the area of the original rectangle is \( 16 \) times that of the dilated rectangle, confirming the result of our calculations.