To solve this problem, let's break it down step by step.
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Area of the Original Rectangle:
The area \( A \) of a rectangle can be calculated using the formula: \[ A = \text{length} \times \text{width} \] For the original rectangle with sides measuring \( x \) inches and \( y \) inches, the area is: \[ A_{\text{original}} = x \times y \] -
Area of the Dilated Rectangle:
When the rectangle is dilated by a scale factor of \( \frac{1}{4} \), both the length and width are multiplied by \( \frac{1}{4} \). Thus, the new dimensions of the dilated rectangle will be: \[ \text{new width} = \frac{1}{4}x \quad \text{and} \quad \text{new length} = \frac{1}{4}y \] The area of the dilated rectangle \( A_{\text{dilated}} \) can be calculated as: \[ A_{\text{dilated}} = \left(\frac{1}{4}x\right) \times \left(\frac{1}{4}y\right) = \frac{1}{16}xy \] -
Comparison of the Areas:
To find how many times greater the area of the original rectangle is compared to the dilated rectangle, we divide the area of the original rectangle by the area of the dilated rectangle: \[ \text{Comparison} = \frac{A_{\text{original}}}{A_{\text{dilated}}} = \frac{xy}{\frac{1}{16}xy} \] Simplifying this gives: \[ \text{Comparison} = \frac{xy}{\frac{1}{16}xy} = 16 \]
Therefore, the area of the original rectangle is 16 times greater than the area of the dilated rectangle.
In summary:
- Area of the original rectangle: \( xy \)
- Area of the dilated rectangle: \( \frac{1}{16}xy \)
- The area of the original rectangle is 16 times greater than the area of the dilated rectangle.
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